JEE Electrostatics Questions

$$\tau=\vert{}\vec{p} \times \vec{E}\vert{}$$

$$I \alpha=p_0 E \sin \theta$$

$$\alpha=\frac{p \cdot \theta}{I}\left(\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi R^2}{r^2}\right)$$

$$\alpha=\left(\frac{p_0 \sigma R^2}{I \varepsilon_0 r^2}\right) \cdot \theta$$

$$\therefore \omega=\sqrt{\frac{p_0 \sigma R^2}{I \varepsilon_0 r^2}}$$

For $$r = 2R$$:

$$\omega=\sqrt{\frac{p_0 \sigma}{4 I \varepsilon_0}} \quad ( \mathrm{C \ is \ incorrect})$$

Also, for $$r=10 R$$:

$$\omega=\sqrt{\frac{p_0 \sigma}{I \varepsilon_0 (100)}} \quad ( \mathrm{D \ is \ correct})$$

It will oscillate for any finite value of $$r>R$$. (B is correct)

Evaluating potential difference due to the spherical shell:

$$V_{P,\text{shell}} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R_{\text{shell}}} = 9 \times 10^9 \times \frac{10 \times 10^{-9}}{1} = 90\text{ V}$$

$$V_{R,\text{shell}} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r_R} = 9 \times 10^9 \times \frac{10 \times 10^{-9}}{2} = 45\text{ V}$$

$$\Delta V_{\text{shell}} = V_{P,\text{shell}} - V_{R,\text{shell}} = 90 - 45 = 45\text{ V}$$

Evaluating potential difference due to the line charge:

$$\Delta V_{\text{line}} = V_P - V_R = \frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r_R}{r_P}\right) = 2 \times \left(\frac{1}{4\pi\varepsilon_0}\right) \lambda \ln\left(\frac{2}{0.5}\right)$$

$$\Delta V_{\text{line}} = 2 \times (9 \times 10^9) \times (5 \times 10^{-9}) \ln 4 = 90 \times 2\ln 2 = 180 \times 0.7 = 126\text{ V}$$

Finding net potential difference:

$$\Delta V_{\text{net}} = \Delta V_{\text{shell}} + \Delta V_{\text{line}} = 45 + 126 = 171\text{ V}$$

We need to find the angular frequency of small oscillations of a dipole in a uniform electric field, where the positive charge has mass $$m$$ and the negative charge has mass $$2m$$.

Taking the positive charge at the origin and the negative charge at a distance $$l$$, the center of mass coordinate is given by

$$x_{CM} = \frac{m \cdot 0 + 2m \cdot l}{m + 2m} = \frac{2l}{3}$$

Hence the positive charge lies at a distance $$\frac{2l}{3}$$ from the center of mass and the negative charge at a distance $$\frac{l}{3}$$.

Using these distances, the moment of inertia about the center of mass becomes

$$I = m\left(\frac{2l}{3}\right)^2 + 2m\left(\frac{l}{3}\right)^2 = \frac{4ml^2}{9} + \frac{2ml^2}{9} = \frac{6ml^2}{9} = \frac{2ml^2}{3}$$

When the dipole is displaced by a small angle $$\theta$$ in a uniform electric field $$E$$, the restoring torque is

$$\tau = -pE\sin\theta \approx -pE\theta$$

where the dipole moment is $$p = ql$$.

Applying Newton’s second law for rotation gives

$$I\alpha = -pE\theta$$

Substituting $$I = \frac{2ml^2}{3}$$ and $$p = ql$$ yields

$$\frac{2ml^2}{3}\,\alpha = -qlE\,\theta$$

from which

$$\alpha = -\frac{3qE}{2ml}\,\theta$$

This equation describes simple harmonic motion with angular frequency

$$\omega = \sqrt{\frac{3qE}{2ml}}$$

Therefore, the angular frequency is $$\sqrt{\dfrac{3qE}{2ml}}$$.

The point $$P(100,100)$$ mm is nearly $$\sqrt{(100)^2+(100)^2}=100\sqrt2\;{\rm mm}\approx141\;{\rm mm}$$ away from the origin, whereas the two charges are only a few millimetres apart. Hence $$r\gg d$$ and the far-field (dipole) approximation is valid.

For a dipole whose moment is $$\mathbf p$$ and for an observation point with position unit vector $$\hat{\mathbf r}$$ (drawn from the centre of the dipole to the point), the electric potential is

$$V=\frac{1}{4\pi\varepsilon_0}\,\frac{\mathbf p\cdot \hat{\mathbf r}}{r^{2}}\;.$$

This expression shows that for the \emph{same} observation point $$P$$ the ratio of two potentials is simply the ratio of the scalar products $$\mathbf p\cdot\hat{\mathbf r}$$, because $$r$$ and $$4\pi\varepsilon_0$$ remain unchanged.

Step 1 : Unit vector from origin to P
The position vector of $$P$$ is $$\mathbf r=(100,\,100)\;{\rm mm}$$.
Magnitude: $$r=100\sqrt2\;{\rm mm}$$.
Hence $$\hat{\mathbf r}= \left(\frac{1}{\sqrt2},\;\frac{1}{\sqrt2}\right).$$

Step 2 : Dipole moment before shifting the charges
Negative charge $$-q$$ is at $$(0,-2)$$ mm, positive charge $$+q$$ at $$(0,2)$$ mm.
Vector from $$-q$$ to $$+q$$: $$\mathbf d=(0,\,4)\;{\rm mm}$$.
Dipole moment: $$\mathbf p = q\,\mathbf d = (0,\,4q).$$

Dot product with $$\hat{\mathbf r}$$:
$$\mathbf p\cdot\hat{\mathbf r}=0\cdot\frac1{\sqrt2}+4q\cdot\frac1{\sqrt2}=\frac{4q}{\sqrt2}=2\sqrt2\,q.$$

Therefore the initial potential is
$$V_0=\frac{1}{4\pi\varepsilon_0}\, \frac{2\sqrt2\,q}{r^{2}}.$$

Step 3 : Dipole moment after shifting the charges
Negative charge $$-q$$ moves to $$(1,-2)$$ mm, positive charge $$+q$$ to $$(-1,2)$$ mm.
Vector from $$-q$$ to $$+q$$: $$\mathbf d' = (-1-1,\;2-(-2)) = (-2,\,4)\;{\rm mm}.$$
Dipole moment: $$\mathbf p' = q\,\mathbf d' = (-2q,\,4q).$$

Dot product with $$\hat{\mathbf r}$$:
$$\mathbf p'\cdot\hat{\mathbf r}=(-2q)\cdot\frac1{\sqrt2}+4q\cdot\frac1{\sqrt2} =\frac{-2q+4q}{\sqrt2} =\frac{2q}{\sqrt2} =\sqrt2\,q.$$

Thus the new potential at $$P$$ is
$$V'=\frac{1}{4\pi\varepsilon_0}\, \frac{\sqrt2\,q}{r^{2}}.$$

Step 4 : Ratio of the two potentials
$$\frac{V'}{V_0}=\frac{\sqrt2\,q}{2\sqrt2\,q}=\frac12.$$

Hence $$V' = \dfrac{V_0}{2}.$$

Option B which is: $$V_0/2$$

Initially, we note that separating a hydrogen atom into a proton and an electron requires 12.8 eV of energy, and we aim to determine the value of $$x$$ such that the orbital radius is $$\frac{9}{x} \times 10^{-10}$$ m.

In electrostatic terms, the binding energy of the electron in a hydrogen atom is given by the relation:

$$|E| = \frac{ke^2}{2r}$$

Here, $$k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2\text{/C}^2$$.

Rearranging this expression allows us to solve for the orbital radius $$r$$ as

$$r = \frac{ke^2}{2|E|}$$

We convert the given energy into joules by using $$1\text{ eV} = 1.6 \times 10^{-19}$$ J, yielding

$$|E| = 12.8\text{ eV} = 12.8 \times 1.6 \times 10^{-19}\text{ J} = 20.48 \times 10^{-19}\text{ J}.$$

Substituting the numerical values into the expression for $$r$$ gives:

$$r = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{2 \times 20.48 \times 10^{-19}}$$

At this point, we compute numerator and denominator separately:

Numerator: $$9 \times 10^9 \times 2.56 \times 10^{-38} = 23.04 \times 10^{-29}$$

Denominator: $$2 \times 20.48 \times 10^{-19} = 40.96 \times 10^{-19}$$

Therefore, the orbital radius becomes

$$r = \frac{23.04 \times 10^{-29}}{40.96 \times 10^{-19}} = \frac{23.04}{40.96} \times 10^{-10} = 0.5625 \times 10^{-10}\text{ m}$$

Expressing this result in the form $$\frac{9}{x} \times 10^{-10}\text{ m}$$ leads to

$$r = \frac{9}{16} \times 10^{-10}\text{ m}$$

By comparison, it follows that $$x = 16$$. Thus, the desired value is 16.

Given: Three concentric shells X (radius a=2cm), Y (radius b=3cm), Z (radius c) with surface charge densities $$\sigma, -\sigma, \sigma$$ respectively. X and Z are at the same potential.

Potential at shell X (radius a):

$$V_X = \frac{\sigma \cdot 4\pi a^2}{4\pi\varepsilon_0 a} + \frac{(-\sigma) \cdot 4\pi b^2}{4\pi\varepsilon_0 b} + \frac{\sigma \cdot 4\pi c^2}{4\pi\varepsilon_0 c} = \frac{\sigma}{\varepsilon_0}(a - b + c)$$

Potential at shell Z (radius c):

$$V_Z = \frac{\sigma \cdot 4\pi a^2}{4\pi\varepsilon_0 c} + \frac{(-\sigma) \cdot 4\pi b^2}{4\pi\varepsilon_0 c} + \frac{\sigma \cdot 4\pi c^2}{4\pi\varepsilon_0 c} = \frac{\sigma}{\varepsilon_0}\left(\frac{a^2 - b^2}{c} + c\right)$$

Setting $$V_X = V_Z$$:

$$a - b + c = \frac{a^2 - b^2}{c} + c$$

$$a - b = \frac{a^2 - b^2}{c} = \frac{(a-b)(a+b)}{c}$$

$$1 = \frac{a+b}{c}$$

$$c = a + b = 2 + 3 = 5 \text{ cm}$$

The radius of shell Z is 5 cm.

Given: Radius of sphere $$R = 6 \text{ m}$$, volume charge density $$\rho = 2 \text{ } \mu\text{C cm}^{-3} = 2 \text{ C m}^{-3}$$.

(Converting: $$2 \text{ } \mu\text{C cm}^{-3} = 2 \times 10^{-6} \times 10^{6} \text{ C m}^{-3} = 2 \text{ C m}^{-3}$$)

The number of lines of force per unit surface area is the electric field at the surface of the sphere.

Using Gauss's law for a spherical surface:

$$E \times 4\pi R^2 = \dfrac{Q}{\varepsilon_0} = \dfrac{\rho \times \dfrac{4}{3}\pi R^3}{\varepsilon_0}$$

$$E = \dfrac{\rho R}{3\varepsilon_0}$$

Substituting the values:

$$E = \dfrac{2 \times 6}{3 \times 8.85 \times 10^{-12}} = \dfrac{12}{26.55 \times 10^{-12}}$$

$$E = 0.4520 \times 10^{12} = 4.52 \times 10^{11} \text{ N C}^{-1}$$

$$E = 45.2 \times 10^{10} \text{ N C}^{-1} \approx 45 \times 10^{10} \text{ N C}^{-1}$$

Therefore, the answer is $$\boxed{45}$$.

The torque experienced by an electric dipole in a uniform electric field is given by:

$$\tau = pE\sin\theta$$

The maximum torque occurs when $$\sin\theta = 1$$ (i.e., $$\theta = 90°$$):

$$\tau_{max} = pE$$

For dipole 1:

$$\tau_1 = p_1 E_1 = 1.2 \times 10^{-30} \times 5 \times 10^4 = 6.0 \times 10^{-26} \text{ N m}$$

For dipole 2:

$$\tau_2 = p_2 E_2 = 2.4 \times 10^{-30} \times 15 \times 10^4 = 36.0 \times 10^{-26} \text{ N m}$$

The ratio of maximum torques:

$$\frac{\tau_1}{\tau_2} = \frac{6.0 \times 10^{-26}}{36.0 \times 10^{-26}} = \frac{6}{36} = \frac{1}{6}$$

Since the ratio is given as $$\frac{1}{x}$$:

$$\frac{1}{x} = \frac{1}{6}$$

$$x = 6$$

Therefore, the value of $$x$$ is 6.

The charge distribution is spherically symmetric, so we can treat every region with Gauss’s law.

1. Charge enclosed by the inner solid sphere A ($$r_A = 1$$)
Density: $$\rho_A = k r$$ for $$0 \le r \le 1$$.
$$\displaystyle Q_A = \int_{0}^{1} \rho_A\,4\pi r^2 \,dr = 4\pi k\int_{0}^{1} r^3\,dr = 4\pi k\left[\frac{r^4}{4}\right]_{0}^{1}= \pi k$$

2. Charge enclosed in the shell B up to an outer radius $$r_B$$
Density: $$\rho_B = \dfrac{2k}{r}$$ for $$1 \le r \le r_B$$.
$$\displaystyle Q_B = \int_{1}^{r_B} \rho_B\,4\pi r^2\,dr = 4\pi \int_{1}^{r_B} \frac{2k}{r}\,r^2\,dr = 8\pi k\int_{1}^{r_B} r\,dr = 8\pi k\left[\frac{r^2}{2}\right]_{1}^{r_B} = 4\pi k\bigl(r_B^2-1\bigr)$$

3. Total charge of the whole distribution
$$\displaystyle Q_{\text{tot}} = Q_A + Q_B = \pi k + 4\pi k\bigl(r_B^2-1\bigr) = \pi k\,[\,4r_B^2-3\,]$$

4. Electric field outside the shell ( r > rB )
Outside the outer radius the entire charge behaves like a point charge at the centre:
$$\displaystyle E(r) = \frac{1}{4\pi\epsilon_0}\,\frac{Q_{\text{tot}}}{r^2}, \qquad r \ge r_B$$

5. Checking each option

Hence, the only correct statement is:

Option B which is: If $$r_B = \dfrac{3}{2}$$, then the electric potential just outside B is $$\dfrac{k}{\epsilon_0}$$.

The electrostatic potential produced by the charged disk is given as
$$V(z)=\dfrac{\sigma}{2\epsilon_{0}}\left(\sqrt{R^{2}+z^{2}}-z\right).$$

For a particle of charge $$q$$ the corresponding electrostatic potential energy is
$$U_{\text{C}}(z)=qV(z)=\dfrac{q\sigma}{2\epsilon_{0}}\left(\sqrt{R^{2}+z^{2}}-z\right).$$

The particle is also acted upon by a constant downward force $$\vec F=-c\hat k$$. Since $$F_z=-\dfrac{dU_g}{dz}=-c$$, the mechanical potential energy associated with this force is $$U_g(z)=c\,z$$ (we take $$U_g(0)=0$$).

Hence the total potential energy of the particle is
$$U(z)=U_{\text{C}}(z)+U_g(z) =\dfrac{q\sigma}{2\epsilon_{0}}\left(\sqrt{R^{2}+z^{2}}-z\right)+c\,z.$$

Put $$\alpha=\dfrac{q\sigma}{2\epsilon_{0}},\qquad \beta=\dfrac{2c\epsilon_{0}}{q\sigma}=\dfrac{c}{\alpha} \;(\beta\gt 0).$$ Then

$$U(z)=\alpha\Bigl[\sqrt{R^{2}+z^{2}}-z+\beta z\Bigr] =\alpha\Bigl[\sqrt{R^{2}+z^{2}}+(\beta-1)z\Bigr].$$

The motion is one-dimensional along the $$z$$-axis. Because only conservative forces act, the mechanical energy $$E=U(z)+K$$ is conserved. At the instant $$t=0$$ the particle is released from rest at $$z=z_0$$, so $$E=U(z_0).$$

If the particle is to reach the origin, its kinetic energy at $$z=0$$ must be non-negative:
$$U(z_0)\ge U(0).$$ Since $$U(0)=\alpha R,$$ this gives the single inequality

$$\sqrt{R^{2}+z_0^{2}}+(\beta-1)z_0\;\ge\;R.\qquad -(1)$$

Let us examine each option one by one.

$$\beta=\dfrac14,\; z_0=\dfrac{25}{7}R \;(=3.571R).$$ Compute $$\sqrt{R^{2}+z_0^{2}}=R\sqrt{1+\Bigl(\dfrac{25}{7}\Bigr)^{2}} =R\sqrt{\dfrac{674}{49}}=\dfrac{\sqrt{674}}{7}R\approx3.709R,$$ $$(\beta-1)z_0=-\dfrac34\cdot\dfrac{25}{7}R=-2.678R.$$ Thus $$\sqrt{R^{2}+z_0^{2}}+(\beta-1)z_0\approx3.709R-2.678R=1.031R\gt R.$$ Inequality $$(1)$$ is satisfied, so the particle reaches $$z=0$$. Option A is correct.

$$\beta=\dfrac14,\; z_0=\dfrac{3}{7}R\;(=0.4286R).$$ $$\sqrt{R^{2}+z_0^{2}}=R\sqrt{1+\Bigl(\dfrac{3}{7}\Bigr)^{2}} =1.088R,$$ $$(\beta-1)z_0=-\dfrac34\cdot\dfrac{3}{7}R=-0.321R.$$ Sum $$=1.088R-0.321R=0.767R\lt R,$$ violating $$(1)$$. Hence the particle cannot reach the origin. Option B is wrong.

$$\beta=\dfrac14,\; z_0=\dfrac{R}{\sqrt3}\;(=0.577R).$$ $$\sqrt{R^{2}+z_0^{2}}=R\sqrt{1+\dfrac13}=1.155R,$$ $$(\beta-1)z_0=-\dfrac34\cdot0.577R=-0.433R.$$ Sum $$=1.155R-0.433R=0.722R\lt R,$$ so the particle cannot reach $$z=0$$.

To see what happens, look at the force $$F(z)=-\dfrac{dU}{dz} =-\alpha\left[\dfrac{z}{\sqrt{R^{2}+z^{2}}}+(\beta-1)\right].$$ For the given $$z_0$$, evaluate the bracket: $$\dfrac{z_0}{\sqrt{R^{2}+z_0^{2}}}+(\beta-1)=\dfrac{1}{2}-\dfrac34=-\dfrac14.$$ Hence $$F(z_0)=+\dfrac{\alpha}{4}$$, i.e. a force in the $$+z$$ direction. The particle first moves upward, its kinetic energy increases, and because the potential $$U(z)$$ grows as $$z\to\infty$$ (for $$\beta=\tfrac14$$, $$U\sim\alpha\beta z$$), it finally comes to rest at some $$z=z_1$$ where $$U(z_1)=U(z_0)$$, and then returns to $$z_0$$. Thus the particle executes oscillatory motion between $$z_0$$ and $$z_1$$ and indeed returns to its starting point. Option C is correct.

Take any $$\beta\gt1$$ and any $$z_0\gt0$$. Because $$0\lt\dfrac{z_0}{\sqrt{R^{2}+z_0^{2}}}\lt1,$$ we get $$\sqrt{R^{2}+z_0^{2}}+(\beta-1)z_0 \;>\;R+0=R.$$ Inequality $$(1)$$ is satisfied for every positive $$z_0$$; therefore the particle always acquires enough mechanical energy to reach the origin. Option D is correct.

Hence the correct choices are:
Option A, Option C and Option D.

First, let us identify the geometry. We have two concentric spheres with radii $$a$$ and $$b$$ satisfying $$a < r < b$$. The shell lying between these two radii is filled with a volume charge whose density varies as $$\rho = \dfrac{A}{r}$$, while the very centre (at $$r = 0$$) carries a point charge $$Q$$.

We wish to make the electric field $$E(r)$$ anywhere in the region $$a < r < b$$ the same for all values of $$r$$, that is, independent of $$r$$. To obtain the required value of $$A$$ we use Gauss’s law.

Gauss’s law in integral form states

$$\displaystyle \oint_S \vec E \cdot d\vec S \;=\;\frac{1}{\varepsilon_0}\,Q_{\text{enc}}.$$

For spherical symmetry, we choose a Gaussian surface that is itself a sphere of radius $$r$$ (with $$a < r < b$$). On this surface the magnitude $$E(r)$$ is the same everywhere and the flux becomes

$$E(r)\,(4\pi r^2)\;=\;\frac{1}{\varepsilon_0}\,Q_{\text{enc}}(r).$$

So

$$E(r)=\frac{Q_{\text{enc}}(r)}{4\pi\varepsilon_0\,r^2}.$$

Now we calculate the total charge enclosed inside radius $$r$$, namely

$$Q_{\text{enc}}(r)=Q_{\text{point}}+Q_{\text{shell up to }r}.$$

The point charge contributes simply $$Q_{\text{point}} = Q.$$

For the shell portion (from $$a$$ out to $$r$$) we integrate the given volume charge density. The differential charge in a thin spherical shell of radius $$r'$$ and thickness $$dr'$$ is

$$dQ = \rho(r')\,d\tau = \frac{A}{r'}\;\bigl(4\pi r'^2\,dr'\bigr)=4\pi A\,r'\,dr'.$$

Integrating from $$r' = a$$ to $$r' = r$$ gives

$$Q_{\text{shell up to }r}=4\pi A\int_{a}^{r}r'\,dr'$$

$$\hspace{3.55cm}=4\pi A\left[\frac{r'^2}{2}\right]_{a}^{r}$$

$$\hspace{3.55cm}=2\pi A\bigl(r^2 - a^2\bigr).$$

Hence

$$Q_{\text{enc}}(r)=Q+2\pi A\bigl(r^2-a^2\bigr).$$

Substituting this into the earlier result for $$E(r)$$ we obtain

$$E(r)=\frac{Q+2\pi A\bigl(r^2-a^2\bigr)}{4\pi\varepsilon_0\,r^2}.$$

We now enforce the requirement that $$E(r)$$ must not depend on $$r$$. For that to happen, the numerator must be directly proportional to $$r^2$$ so that the $$r^2$$ in the denominator cancels out and leaves a constant.

Let us write the numerator in the desired proportional form:

$$Q+2\pi A\bigl(r^2-a^2\bigr)=k\,r^2,$$

where $$k$$ is some constant (its exact value is immaterial—only the absence of any other $$r$$-dependence matters).

Comparing coefficients of like powers of $$r$$ on both sides, we see:

 • For the $$r^2$$ terms: $$2\pi A=k.$$
 • For the terms independent of $$r$$: $$Q-2\pi A a^2=0.$$

The second equality immediately gives us the needed value of $$A$$:

$$Q-2\pi A a^2=0\;\;\Longrightarrow\;\;A=\frac{Q}{2\pi a^2}.$$

This is a unique choice; any other value would leave residual $$r$$-dependence in the field.

Hence, the correct answer is Option C.

A liquid drop with 6 excess electrons is held stationary under a uniform electric field of 25.5 kV/m. The density of the liquid is given as $$1.26 \times 10^3$$ kg/m³, and we neglect buoyancy. We need to find the radius of the drop.

Since the drop is stationary, the forces acting on it must balance. The gravitational force (weight) acts downward, and the electric force acts upward because the drop has excess electrons (negative charge) and the electric field exerts an upward force on negative charges.

The charge on one electron is $$-1.6 \times 10^{-19}$$ C. With 6 excess electrons, the total charge $$q$$ is:

$$ q = 6 \times (-1.6 \times 10^{-19}) = -9.6 \times 10^{-19} \text{ C} $$

The magnitude of the charge is $$|q| = 9.6 \times 10^{-19}$$ C. The electric field $$E$$ is given as 25.5 kV/m. Converting to V/m:

$$ E = 25.5 \times 1000 = 25500 \text{ V/m} = 2.55 \times 10^4 \text{ V/m} $$

The magnitude of the electric force $$F_e$$ acting upward is:

$$ F_e = |q| E = (9.6 \times 10^{-19}) \times (2.55 \times 10^4) $$

Calculating:

$$ 9.6 \times 2.55 = 24.48 $$

$$ 10^{-19} \times 10^4 = 10^{-15} $$

So,

$$ F_e = 24.48 \times 10^{-15} = 2.448 \times 10^{-14} \text{ N} $$

The gravitational force (weight) $$F_g$$ acting downward is $$F_g = m g$$, where $$m$$ is the mass of the drop and $$g$$ is the acceleration due to gravity (approximately 9.8 m/s²). The mass $$m$$ can be expressed in terms of density $$\rho$$ and volume $$V$$. Assuming the drop is spherical, its volume is:

$$ V = \frac{4}{3} \pi r^3 $$

where $$r$$ is the radius. Thus,

$$ m = \rho V = \rho \times \frac{4}{3} \pi r^3 $$

So,

$$ F_g = \rho \times \frac{4}{3} \pi r^3 \times g $$

For the drop to be stationary, the electric force upward must balance the gravitational force downward:

$$ F_e = F_g $$

Substituting the expressions:

$$ |q| E = \rho \frac{4}{3} \pi r^3 g $$

Solving for $$r^3$$:

$$ r^3 = \frac{|q| E \times 3}{4 \pi \rho g} $$

Plugging in the known values:

$$ |q| = 9.6 \times 10^{-19} \text{ C}, \quad E = 2.55 \times 10^4 \text{ V/m}, \quad \rho = 1.26 \times 10^3 \text{ kg/m}^3, \quad g = 9.8 \text{ m/s}^2, \quad \pi \approx 3.1416 $$

First, compute the numerator $$3 \times |q| \times E$$:

$$ 3 \times (9.6 \times 10^{-19}) \times (2.55 \times 10^4) = 3 \times 9.6 \times 2.55 \times 10^{-19+4} = 3 \times 9.6 \times 2.55 \times 10^{-15} $$

$$ 3 \times 9.6 = 28.8 $$

$$ 28.8 \times 2.55 = 73.44 $$

So, numerator $$= 73.44 \times 10^{-15} = 7.344 \times 10^{-14}$$

Now, compute the denominator $$4 \pi \rho g$$:

$$ \rho g = (1.26 \times 10^3) \times 9.8 = 1.26 \times 9.8 \times 10^3 = 12.348 \times 10^3 = 1.2348 \times 10^4 $$

$$ 4 \pi = 4 \times 3.1416 = 12.5664 $$

$$ 4 \pi \rho g = 12.5664 \times (1.2348 \times 10^4) = 12.5664 \times 1.2348 \times 10^4 $$

$$ 12.5664 \times 1.2348 = 15.51695472 \quad \text{(approximately)} $$

So, denominator $$= 15.51695472 \times 10^4 = 1.551695472 \times 10^5$$

Now,

$$ r^3 = \frac{7.344 \times 10^{-14}}{1.551695472 \times 10^5} = \frac{7.344}{1.551695472} \times 10^{-14-5} = 4.732 \times 10^{-19} $$

Taking the cube root to find $$r$$:

$$ r = \sqrt[3]{4.732 \times 10^{-19}} = \sqrt[3]{4.732} \times 10^{-19/3} = \sqrt[3]{4.732} \times 10^{-6.3333} $$

$$ 10^{-6.3333} = 10^{-6} \times 10^{-1/3} \approx 10^{-6} \times 0.4642 $$

$$ \sqrt[3]{4.732} \approx 1.679 \quad \text{(since } 1.679^3 \approx 4.732) $$

So,

$$ r \approx 1.679 \times 0.4642 \times 10^{-6} = 0.7793 \times 10^{-6} = 7.793 \times 10^{-7} \text{ m} $$

Approximating to two significant figures, $$r \approx 7.8 \times 10^{-7}$$ m.

Comparing with the options:

A. $$4.3 \times 10^{-7}$$ m

B. $$7.8 \times 10^{-7}$$ m

C. $$0.078 \times 10^{-7}$$ m (which is $$7.8 \times 10^{-9}$$ m)

D. $$3.4 \times 10^{-7}$$ m

The value $$7.8 \times 10^{-7}$$ m matches option B.

Hence, the correct answer is Option B.

Let the side length of the square be $$a$$.

The charges $$-Q$$ are placed at the four corners of the square, and a charge $$q$$ is placed at the center.

The distance from any corner to the center of the square is $$\frac{a}{\sqrt{2}}$$.

For the entire system to be in equilibrium, the net force on every single charge must be zero.

By symmetry, the charge $$q$$ at the center is already in equilibrium because the forces from the four corner charges cancel each other out.

Therefore, we must find the condition for the corner charges to be in equilibrium. Let's calculate the net force acting on one of the corner charges $$-Q$$.

There are three repulsive forces from the other three $$-Q$$ charges and one force from the center charge $$q$$.

{Step 1: Forces from adjacent charges}

The force due to the two adjacent charges at distance $$a$$ is:

$$F_{\text{adj}} = \frac{kQ^2}{a^2}$$

Since these two forces act along the edges of the square (at a $$90^\circ$$ angle to each other), their resultant force acts outwards strictly along the diagonal. The magnitude of this resultant is:

$$F_{\text{adj\_res}} = \sqrt{F_{\text{adj}}^2 + F_{\text{adj}}^2} = \sqrt{2}\frac{kQ^2}{a^2}$$

{Step 2: Force from the diagonally opposite charge}

The force due to the charge at the opposite corner (at distance $$a\sqrt{2}$$) also acts outwards along the same diagonal:

$$F_{\text{opp}} = \frac{kQ^2}{(a\sqrt{2})^2} = \frac{kQ^2}{2a^2}$$

{Step 3: Total outward repulsive force}

The total outward force along the diagonal is the sum of the forces calculated in Step 1 and Step 2:

$$F_{\text{outward}} = F_{\text{adj\_res}} + F_{\text{opp}} = \sqrt{2}\frac{kQ^2}{a^2} + \frac{kQ^2}{2a^2} = \frac{kQ^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right)$$

{Step 4: Force from the center charge}

For the corner charge to be in equilibrium, the force from the center charge $$q$$ must completely cancel the outward repulsive force. This means $$q$$ must be a positive charge to provide an attractive inward force.

The attractive force due to $$q$$ at a distance of $$\frac{a}{\sqrt{2}}$$ is:

$$F_{\text{inward}} = \frac{k \cdot Q \cdot q}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{2kQq}{a^2}$$

{Step 5: Equating forces for equilibrium}

Equating the inward attractive force to the outward repulsive force:

$$\frac{2kQq}{a^2} = \frac{kQ^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right)$$

Dividing both sides by $$\frac{kQ}{a^2}$$:

$$2q = Q \left( \sqrt{2} + \frac{1}{2} \right)$$

$$2q = Q \left( \frac{2\sqrt{2} + 1}{2} \right)$$

$$q = \frac{Q}{4} (2\sqrt{2} + 1)$$

Dipole moment

$$⃗\vec{p}=q\vec{d}$$

Position vector from −4μC to 4μC:

$$\vec{d}=(2-1)i+(-1-0)j+(5-4)k$$

$$=i−j+k$$

Charge

$$q=4\times10^{-6}C$$

So

$$\vec{p}=4\times10^{-6}(i-j+k)$$

Electric field:

$$E⃗=0.20iV/cm$$

Convert to SI:

$$0.20V/cm=20N/C$$

Torque:

$$τ=\vec{p}\times\vec{E}$$

Magnitude

$$\tau=pE\sin\theta$$

Only components perpendicular to i contribute:

$$p_{\perp}=4\times10^{-6}\sqrt{(-1)^2+1^2}$$

$$=4\sqrt{2}\times10^{-6}$$

Thus

$$\tau=(4\sqrt{2}\times10^{-6})(20)$$

$$=8\sqrt{2}\times10^{-5}$$

Gauss law states that the electric flux through any open surface $$S$$ kept in the field of a point charge $$q$$ placed at the origin is proportional to the solid angle $$\Omega$$ subtended by that surface at the charge:

$$\Phi_S=\frac{q}{4\pi\varepsilon_0}\,\Omega$$

Hence, for two different geometries of the same charge, the ratio of fluxes equals the ratio of the corresponding solid angles.

The cylindrical surface is coaxial with the charge located at its centre $$P$$. Each end‐edge of the cylinder makes a half-angle $$\theta$$ with the axis (figure in the question). In spherical polar coordinates with the axis of the cylinder as the polar axis, any direction from $$P$$ is specified by polar angle $$\beta$$ (measured from the +z-axis) and azimuth $$\phi$$.

• For $$0\le\beta&lt\theta$$ the ray meets the top circular face.
• For $$\theta\le\beta\le\pi-\theta$$ the ray meets the curved surface.
• For $$\pi-\theta&lt\beta\le\pi$$ the ray meets the bottom circular face.

Thus the curved surface subtends the full azimuth $$0\le\phi\le2\pi$$ but only the polar band $$\theta\le\beta\le\pi-\theta$$. The solid angle of that band is

$$\Omega = \int_{\phi=0}^{2\pi}\int_{\beta=\theta}^{\pi-\theta}\sin\beta\,d\beta\,d\phi = 2\pi\Bigl[-\cos\beta\Bigr]_{\theta}^{\pi-\theta} = 2\pi\Bigl[-\cos(\pi-\theta)+\cos\theta\Bigr] = 2\pi\bigl[\;+\cos\theta+\cos\theta\bigr] = 4\pi\cos\theta.$$

The flux through the curved surface is therefore

$$\Phi(\theta)=\frac{q}{4\pi\varepsilon_0}\;\Omega =\frac{q}{4\pi\varepsilon_0}\;(4\pi\cos\theta) =\frac{q}{\varepsilon_0}\cos\theta.$$

Case 1: $$\theta=30^{\circ}$$ (i.e. $$\theta=\pi/6$$)
$$\Phi=\Phi(30^{\circ})=\frac{q}{\varepsilon_0}\cos30^{\circ} =\frac{q}{\varepsilon_0}\,\frac{\sqrt3}{2}.$$

Case 2: $$\theta=60^{\circ}$$ (i.e. $$\theta=\pi/3$$)
$$\Phi'=\Phi(60^{\circ})=\frac{q}{\varepsilon_0}\cos60^{\circ} =\frac{q}{\varepsilon_0}\,\frac12.$$

The problem states that $$\Phi'=\dfrac{\Phi}{\sqrt{n}}$$. Taking the ratio of the two fluxes:

$$\frac{\Phi'}{\Phi} =\frac{\tfrac{q}{\varepsilon_0}\,\tfrac12} {\tfrac{q}{\varepsilon_0}\,\tfrac{\sqrt3}{2}} =\frac{1}{\sqrt3} \;=\;\frac{1}{\sqrt{n}}.$$

Equating, $$\sqrt{n}=\sqrt3\;\Longrightarrow\;n=3.$$

Therefore, the required value of $$n$$ is 3.

We need to determine which statement is correct about a group of positive charges.

Can the net potential be zero?. The electric potential due to a point charge is $$V = \frac{kq}{r}$$. For all positive charges, $$q > 0$$ and $$r > 0$$, so each contribution to the potential is positive.

Therefore, the net potential $$V = \sum \frac{kq_i}{r_i} > 0$$ at every point in space. The net potential cannot be zero.

Can the net electric field be zero?. The electric field is a vector quantity. Even though each charge produces a field pointing away from it, the fields at a point can point in different directions.

For example, at the midpoint between two equal positive charges, the electric fields due to the two charges are equal in magnitude but opposite in direction, so they cancel out. Therefore, the net electric field can be zero.

Conclusion. For a group of positive charges:

- Net potential cannot be zero at any point (it is always positive).

- Net electric field can be zero at certain points.

The correct answer is Option A: Net potential of the system cannot be zero at a point but net electric field can be zero at that point.

Assertion A: A dipole enclosed by a closed surface gives zero net flux. True — by Gauss's law, net flux = Q_enclosed/ε₀ = 0 (since dipole has zero net charge).

Reason R: Electric dipole consists of two equal and opposite charges. True — and this directly explains why the net enclosed charge is zero.

Both A and R are true and R is the correct explanation of A.

$$E_{\text{equatorial}} = \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2 + a^2)^{3/2}}$$

For a short dipole, $$a \ll r$$, so $$a^2$$ is negligible.

$$E_{\text{equatorial}} \approx \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2)^{3/2}}$$

$$E_{\text{equatorial}} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3}$$

$$E_{\text{equatorial}} \propto \frac{1}{r^3}$$

The torque on an electric dipole in a uniform electric field is given by:

$$\tau = pE\sin\theta$$

where $$p = qd$$ is the dipole moment.

Given: $$q = 0.01$$ C, $$d = 0.4$$ mm $$= 0.4 \times 10^{-3}$$ m, $$\theta = 30°$$

Convert electric field: $$E = 10$$ dyne/C. Since $$1$$ dyne $$= 10^{-5}$$ N:

$$E = 10 \times 10^{-5} = 10^{-4} \text{ N/C}$$

Dipole moment:

$$p = qd = 0.01 \times 0.4 \times 10^{-3} = 4 \times 10^{-6} \text{ C·m}$$

Torque:

$$\tau = pE\sin\theta = 4 \times 10^{-6} \times 10^{-4} \times \sin 30°$$

$$\tau = 4 \times 10^{-10} \times 0.5 = 2.0 \times 10^{-10} \text{ N·m}$$

The magnitude of torque is $$\mathbf{2.0 \times 10^{-10}}$$ N m.

We need to find the work done in rotating an electric dipole by 180° in a uniform electric field.

Formula for work done. The potential energy of a dipole in a uniform electric field is:

$$U = -pE\cos\theta$$

Work done in rotating from angle $$\theta_1$$ to $$\theta_2$$ is:

$$W = U_2 - U_1 = -pE\cos\theta_2 - (-pE\cos\theta_1) = pE(\cos\theta_1 - \cos\theta_2)$$

Initial and final angles. Initially, the dipole moment is along the electric field, so $$\theta_1 = 0°$$.

After rotation, $$\theta_2 = 180°$$.

Calculate the work done: $$W = pE(\cos 0° - \cos 180°) = pE(1 - (-1)) = 2pE$$

$$W = 2 \times 6.0 \times 10^{-6} \times 1.5 \times 10^3$$

$$= 2 \times 9.0 \times 10^{-3} = 18 \times 10^{-3} \text{ J} = 18 \text{ mJ}$$

The correct answer is 18 mJ.

A stream of positively charged particles with $$\frac{q}{m} = 2 \times 10^{11} \text{ C/kg}$$ enters with velocity $$\vec{v}_0 = 3 \times 10^7 \hat{i} \text{ m/s}$$ into a region of electric field $$\vec{E} = 1.8 \hat{j} \text{ kV/m} = 1800 \hat{j} \text{ V/m}$$, extending 10 cm along the $$x$$-direction.

Find the time spent in the electric field.

$$t = \frac{d}{v_0} = \frac{0.1}{3 \times 10^7} = \frac{1}{3 \times 10^8} \text{ s}$$

Find the acceleration in the $$y$$-direction.

$$a = \frac{qE}{m} = \frac{q}{m} \times E = 2 \times 10^{11} \times 1800 = 3.6 \times 10^{14} \text{ m/s}^2$$

Find the deflection in the $$y$$-direction.

$$y = \frac{1}{2}at^2 = \frac{1}{2} \times 3.6 \times 10^{14} \times \left(\frac{1}{3 \times 10^8}\right)^2$$

$$= \frac{1}{2} \times 3.6 \times 10^{14} \times \frac{1}{9 \times 10^{16}} = \frac{3.6}{18} \times 10^{-2} = 0.2 \times 10^{-2} = 2 \times 10^{-3} \text{ m}$$

$$y = 2 \text{ mm}$$

The answer is $$\boxed{2}$$ mm.

Time in field: $$t = 0.1/(3\times10^7) = \frac{1}{3}\times10^{-8}$$ s.

Acceleration: $$a = (q/m)E = 2\times10^{11}\times1800 = 3.6\times10^{14}$$ m/s$$^2$$.

Deflection: $$y = \frac{1}{2}at^2 = 2\times10^{-3}$$ m $$= 2$$ mm.

$$L = v_x t \implies t = \frac{L}{v_x}$$

$$\frac{1}{2}m v_x^2 = 0.5e \implies v_x = \sqrt{\frac{e}{m}}$$

$$v_y = a_y t = \left( \frac{eE}{m} \right) t$$

$$v_y = \left( \frac{eE}{m} \right) \left( \frac{L}{v_x} \right)$$

$$\tan \theta = \frac{v_y}{v_x}$$

$$\frac{v_y}{L} = \frac{eE}{m v_x} \implies \tan \theta = \frac{eE}{m v_x} \cdot \frac{L}{v_x} = \frac{eE}{m} \cdot \frac{L}{v_x^2}$$

$$\tan \theta = \frac{eE}{m} \cdot \frac{L}{(e/m)} = E \cdot L$$

$$\tan \theta = 10 \times 0.1 = 1$$

$$\theta = \tan^{-1}(1) = 45^\circ$$

For equilibrium along incline, electrostatic repulsion must balance component of weight along plane.

Inclination:

$$\theta=30^{\circ}$$

For each charge,

$$mg\sin⁡30^{\circ\ }$$

must equal Coulomb repulsion.

Mass

$$m=20g=0.02kg$$

So

$$mg\sin30^{\circ}=0.02(10)\cdot\frac{1}{2}$$

=0.1N

Electrostatic force:

$$F=\frac{kq_0^2}{r^2}$$

where r is separation between charges along incline.

Set equilibrium:

$$\frac{kq_0^2}{r^2}=0.1$$

Given

$$q_0=2\times10^{-6}C$$

and

$$k=9\times10^9$$

So

$$\frac{(9\times10^9)(4\times10^{-12})}{r^2}$$

$$\frac{36\times10^{-3}}{r^2}$$

$$r^2=0.36$$

$$r=0.6m$$

Now from geometry, the vertical height shown is

$$h=r\sin⁡30^{\circ\ }$$

$$=0.6\cdot\frac{1}{2}$$

=0.3m

$$=300\times10^{-3}m$$

Gauss law states that the total electric flux $$\Phi_{\text{total}}$$ emerging out of any closed surface that encloses a point charge $$q$$ is
$$\Phi_{\text{total}}=\dfrac{q}{\epsilon_0}\,.$$

The given closed surface has two parts.
  • Curved surface of a hemisphere of radius $$R$$.
  • Curved surface of an inverted cone whose base circle (radius $$R$$) coincides with the rim of the hemisphere.
The point charge $$q$$ is at the common centre of the hemisphere, i.e. at the centre of the complete sphere of radius $$R$$. (This is implied because the hemisphere is specified only by its radius.)

Flux through the hemispherical surface
Because the charge is at the centre of the full sphere of radius $$R$$, the electric field on the spherical surface is radial and of equal magnitude at every point. Hence the flux through the entire sphere would be $$q/\epsilon_0$$. The hemisphere is exactly half of this sphere, so its share of flux is
$$\Phi_{\text{hemisphere}}=\dfrac{1}{2}\left(\dfrac{q}{\epsilon_0}\right)=\dfrac{q}{2\epsilon_0}\,.$$

Flux through the conical surface
Let $$\Phi_{\text{cone}}$$ be the flux through the curved surface of the inverted cone. The hemisphere and the cone together form a closed surface, so
$$\Phi_{\text{hemisphere}}+\Phi_{\text{cone}}=\dfrac{q}{\epsilon_0}\,.$$ Substituting $$\Phi_{\text{hemisphere}}=\dfrac{q}{2\epsilon_0}$$,
$$\dfrac{q}{2\epsilon_0}+\Phi_{\text{cone}}=\dfrac{q}{\epsilon_0}$$
$$\Rightarrow\;\Phi_{\text{cone}}=\dfrac{q}{\epsilon_0}-\dfrac{q}{2\epsilon_0}=\dfrac{q}{2\epsilon_0}\,.$$

The question states that this flux equals $$\dfrac{nq}{6\epsilon_0}$$, so
$$\dfrac{nq}{6\epsilon_0}=\dfrac{q}{2\epsilon_0}\; \Longrightarrow\; n=3.$$

Therefore, the required value of $$n$$ is 3.

When 64 small conducting drops combine to form a bigger drop, the total charge and total volume are conserved.

Volume is conserved:

$$64 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$$

$$R^3 = 64r^3$$

$$R = 4r$$

where $$r = 0.02$$ m is the radius of each small drop and $$R$$ is the radius of the bigger drop.

Total charge is conserved:

$$Q = 64 \times q = 64 \times 5 = 320 \text{ } \mu\text{C}$$

$$\sigma_s = \frac{q}{4\pi r^2}$$

$$\sigma_B = \frac{Q}{4\pi R^2} = \frac{64q}{4\pi (4r)^2} = \frac{64q}{4\pi \times 16r^2} = \frac{4q}{4\pi r^2}$$

$$\frac{\sigma_B}{\sigma_s} = \frac{\frac{4q}{4\pi r^2}}{\frac{q}{4\pi r^2}} = 4$$

Therefore, $$\sigma_B : \sigma_s = 4 : 1$$

Hence, the correct answer is Option B.

$$a = \frac{qE}{m} = \left(\frac{q}{m}\right)E = (8 \times 10^{-3}) \times 100 = 0.8\ \text{m/s}^2$$

Time taken to hit the wall from rest ($$s = \frac{1}{2}at^2$$):

$$t_1 = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2(0.1)}{0.8}} = \sqrt{\frac{1}{4}} = 0.5\ \text{s}$$

Total time period ($$T = 2t_1$$): $$T = 2 \times 0.5 = 1\ \text{s}$$

The electric field is given as $$\vec{E} = \frac{3}{5}E_0\hat{i} + \frac{4}{5}E_0\hat{j}$$.

For the first surface of area $$A_1 = 0.2\,\text{m}^2$$ parallel to the $$y\text{-}z$$ plane, the outward normal is along the $$x$$-direction ($$\hat{i}$$). The flux through this surface is $$\phi_1 = \vec{E} \cdot \hat{i} \times A_1 = \frac{3}{5}E_0 \times 0.2 = \frac{3E_0}{25}$$.

For the second surface of area $$A_2 = 0.3\,\text{m}^2$$ parallel to the $$x\text{-}z$$ plane, the outward normal is along the $$y$$-direction ($$\hat{j}$$). The flux through this surface is $$\phi_2 = \vec{E} \cdot \hat{j} \times A_2 = \frac{4}{5}E_0 \times 0.3 = \frac{12E_0}{50} = \frac{6E_0}{25}$$.

The ratio of the two fluxes is $$\frac{\phi_1}{\phi_2} = \frac{3E_0/25}{6E_0/25} = \frac{3}{6} = \frac{1}{2}$$.

This gives the ratio $$a : b = 1 : 2$$, and since $$b = 2$$, we get $$a = 1$$.

A point charge of $$q = +12 \; \mu\text{C} = 12 \times 10^{-6}$$ C is placed 6 cm vertically above the centre of a square of side 12 cm. To find the electric flux through the square, we use the concept of a Gaussian surface in the form of a cube.

Consider a cube of side 12 cm with the given square as one of its faces. If we place this cube such that the charge is at its centre, the charge would need to be at a height of 6 cm above the square, which is exactly half the side length of the cube. Therefore, the charge is located precisely at the centre of the cube.

By Gauss's law, the total electric flux through the entire closed cube is $$\Phi_{\text{total}} = \frac{q}{\varepsilon_0}$$. By symmetry, this flux is equally distributed through all 6 faces of the cube. Therefore, the flux through one face (our square) is $$\Phi = \frac{q}{6\varepsilon_0}$$.

Substituting the values: $$\Phi = \frac{12 \times 10^{-6}}{6 \times 8.854 \times 10^{-12}} = \frac{2 \times 10^{-6}}{8.854 \times 10^{-12}} = 2.259 \times 10^{5}$$ N m$$^2$$ C$$^{-1}$$.

Expressing this as $$x \times 10^3$$ N m$$^2$$ C$$^{-1}$$: $$\Phi = 225.9 \times 10^3 \approx 226 \times 10^3$$ N m$$^2$$ C$$^{-1}$$.

Therefore, the magnitude of the electric flux through the square is $$226 \times 10^3$$ N m$$^2$$ C$$^{-1}$$.

The electric field is $$\vec{E} = \frac{2}{5}E_0\,\hat{i} + \frac{3}{5}E_0\,\hat{j}$$ with $$E_0 = 4.0 \times 10^3$$ N/C. The rectangular surface of area 0.4 m$$^2$$ is parallel to the $$Y$$-$$Z$$ plane, so its outward normal is along $$\hat{i}$$.

The flux through the surface is $$\Phi = \vec{E} \cdot \hat{i}\,A = \frac{2}{5}E_0 \times 0.4$$. Substituting, $$\Phi = \frac{2}{5} \times 4.0 \times 10^3 \times 0.4 = 1600 \times 0.4 = 640$$ N m$$^2$$ C$$^{-1}$$.

When two identical conducting spheres are brought into contact, the total charge is shared equally between them. The total charge is $$2.1 + (-0.1) = 2.0$$ nC. So each sphere gets $$\frac{2.0}{2} = 1.0$$ nC $$= 1.0 \times 10^{-9}$$ C.

After separation by distance $$r = 0.5$$ m, the electrostatic force between them is $$F = \frac{kq_1 q_2}{r^2} = \frac{9 \times 10^9 \times (1.0 \times 10^{-9})^2}{(0.5)^2}$$.

$$F = \frac{9 \times 10^9 \times 10^{-18}}{0.25} = \frac{9 \times 10^{-9}}{0.25} = 36 \times 10^{-9}$$ N.

Therefore, the electrostatic force is $$36 \times 10^{-9}$$ N.

Electric field of a uniformly charged full sphere at an internal point $$r \le R$$ and external point $$r > R$$:

$$E_{\text{in}} = \frac{\rho r}{3\varepsilon_0}$$

$$E_{\text{out}} = \frac{\rho R^3}{3\varepsilon_0 r^2}$$

For point A (center of the large sphere, surface of the cavity):

$$E_{\text{full, A}} = 0$$

$$E_{\text{cavity, A}} = \frac{\rho (R/2)}{3\varepsilon_0} = \frac{\rho R}{6\varepsilon_0}$$

$$E_A = |E_{\text{full, A}} - E_{\text{cavity, A}}| = \frac{\rho R}{6\varepsilon_0}$$

For point B (bottom surface of the large sphere):

$$E_{\text{full, B}} = \frac{\rho R}{3\varepsilon_0}$$

$$E_{\text{cavity, B}} = \frac{\rho (R/2)^3}{3\varepsilon_0 (3R/2)^2} = \frac{\rho R^3}{24\varepsilon_0 \cdot \frac{9R^2}{4}} = \frac{\rho R}{54\varepsilon_0}$$

$$E_B = E_{\text{full, B}} - E_{\text{cavity, B}} = \frac{\rho R}{3\varepsilon_0} - \frac{\rho R}{54\varepsilon_0} = \frac{17\rho R}{54\varepsilon_0}$$

$$\frac{|E_A|}{|E_B|} = \frac{\frac{\rho R}{6\varepsilon_0}}{\frac{17\rho R}{54\varepsilon_0}} = \frac{1}{6} \times \frac{54}{17} = \frac{9}{17} = \frac{18}{34}$$

Given: $$u = 0$$

$$F_x = qE \implies a_x = \frac{qE}{m}$$

$$F_y = -mg \implies a_y = -g$$

$$x = \frac{1}{2}a_x t^2 = \frac{qE}{2m}t^2$$

$$y = -\frac{1}{2}gt^2$$

$$\frac{y}{x} = -\frac{mg}{qE} = \text{constant} \implies y = mx$$

For surface ABCD (Top face at $$z = 2$$):

$$\text{The outward normal area vector points along the z-direction: } d\vec{A} = dx\,dy\,\hat{k}$$

$$\vec{E} = 4x\hat{i} - (y^2 + 1)\hat{j}$$

$$\phi_I = \iint \vec{E} \cdot d\vec{A} = \iint [4x\hat{i} - (y^2 + 1)\hat{j}] \cdot [dx\,dy\,\hat{k}] = 0$$

For surface BCGF (Right face at $$x = 3$$):

$$\text{The outward normal area vector points along the x-direction: } d\vec{A} = dy\,dz\,\hat{i}$$

$$\text{At } x = 3\text{, the x-component of the electric field is: } E_x = 4(3) = 12\text{ N/C}$$

$$\phi_{II} = \iint E_x \,dy\,dz = 12 \int_{0}^{2} dy \int_{0}^{2} dz = 12 \times (2 \times 2) = 48$$

$$|\phi_I - \phi_{II}| = |0 - 48| = 48$$

Each individual dipole vector has a magnitude of: $$p_1 = p_2 = ql$$

The two dipole vectors point from the top vertex downward toward the bottom-left and bottom-right vertices. Because the triangle is equilateral, the angle between these two vectors is $$60^\circ$$.

By symmetry, the horizontal components of these vectors cancel each other out, while their vertical components add together along the negative y-direction ($$-\hat{j}$$).

The net dipole moment ($$\vec{p}_{\text{net}}$$) is calculated by adding the vertical components:

$$\vec{p}_{\text{net}} = -\left(p_1 \cos 30^\circ + p_2 \cos 30^\circ\right)\hat{j}$$

$$\vec{p}_{\text{net}} = -\left(ql \frac{\sqrt{3}}{2} + ql \frac{\sqrt{3}}{2}\right)\hat{j} = -\sqrt{3}ql\hat{j}$$

First of all, an electric dipole consists of two point charges, $$+q$$ and $$-q$$, separated by a distance $$d$$. The dipole moment (a vector directed from the negative to the positive charge) has the magnitude

$$p = qd.$$

When the dipole is placed in a uniform external electric field $$\vec E$$, the field exerts a torque on it. The magnitude of this torque at an angle $$\theta$$ between $$\vec p$$ and $$\vec E$$ is given by the standard relation

$$\tau = pE\sin\theta.$$

Here the equilibrium orientation is when $$\vec p$$ is parallel to $$\vec E$$, that is, $$\theta = 0$$. We now rotate the dipole slightly so that $$\theta$$ is small. For small angles we may use the approximation

$$\sin\theta \approx \theta.$$

Under this small-angle approximation, the torque becomes

$$\tau \approx pE\,\theta.$$

Because the torque always acts so as to reduce $$\theta$$, it actually points in the opposite direction to $$\theta$$, giving

$$\tau = -pE\,\theta.$$

Next we connect the torque to the rotational dynamics of the dipole. Newton’s second law for rotation states

$$\tau = I\alpha,$$

where $$I$$ is the moment of inertia of the system about the axis through the centre of the dipole (perpendicular to the dipole length) and $$\alpha = \dfrac{d^{2}\theta}{dt^{2}}$$ is the angular acceleration.

Both charges have the same mass $$m$$ and lie at distances $$\dfrac{d}{2}$$ on either side of the centre. Hence the moment of inertia is

$$I = m\left(\frac{d}{2}\right)^{2} + m\left(\frac{d}{2}\right)^{2} = 2m\left(\frac{d^{2}}{4}\right) = \frac{md^{2}}{2}.$$

Substituting $$\tau = -pE\theta$$ and $$I = \dfrac{md^{2}}{2}$$ into $$\tau = I\alpha$$, we get

$$-pE\,\theta = \frac{md^{2}}{2}\,\frac{d^{2}\theta}{dt^{2}}.$$

Rearranging,

$$\frac{d^{2}\theta}{dt^{2}} + \frac{2pE}{md^{2}}\;\theta = 0.$$

The above differential equation has the standard simple-harmonic-motion form

$$\frac{d^{2}\theta}{dt^{2}} + \omega^{2}\theta = 0,$$

where the square of the angular frequency is the coefficient of $$\theta$$. Therefore,

$$\omega^{2} = \frac{2pE}{md^{2}}.$$

But $$p = qd$$, so we substitute this value for $$p$$:

$$\omega^{2} = \frac{2(qd)E}{md^{2}} = \frac{2qE}{md}.$$

Taking the positive square root, the angular frequency is

$$\omega = \sqrt{\frac{2qE}{md}}.$$

Hence, the correct answer is Option B.

We have four point charges placed on the y-axis. Their positions and values are

$$q_1=-q\;{\rm at}\;(0,-2d),\qquad q_2=+q\;{\rm at}\;(0,-d),\qquad q_3=+q\;{\rm at}\;(0,+d),\qquad q_4=-q\;{\rm at}\;(0,+2d).$$

The point at which we want the electric field is $$P(D,0)\,,$$ i.e. a point on the x-axis with $$x=D$$ and $$y=0$$, where $$D\gg d.$$

For any point charge $$q_i$$ located at $$(0,y_i)$$ the position vector from the charge to $$P$$ is

$$\vec r_i=\bigl(D\,\hat i+(0-y_i)\,\hat j\bigr)=D\,\hat i-y_i\,\hat j.$$

The distance of the point from the charge is therefore

$$r_i=\sqrt{D^{\,2}+y_i^{\,2}}.$$

Statement of the formula: The electric field due to a point charge is

$$\vec E_i=\dfrac{k\,q_i}{r_i^{\,2}}\;\hat r_i =\dfrac{k\,q_i}{r_i^{\,3}}\;\vec r_i,$$ where $$k=\dfrac{1}{4\pi\varepsilon_0}$$ and $$\hat r_i=\dfrac{\vec r_i}{r_i}.$$

Substituting $$\vec r_i$$ we get the two Cartesian components for every charge:

$$E_{ix}= \dfrac{k\,q_i D}{\bigl(D^{\,2}+y_i^{\,2}\bigr)^{3/2}},\qquad E_{iy}= \dfrac{k\,q_i(-y_i)}{\bigl(D^{\,2}+y_i^{\,2}\bigr)^{3/2}}.$$

Now we first check the $$y$$-components. The charges at $$y=+d$$ and $$y=-d$$ have the same sign $$+q$$ but opposite $$y_i$$, so their $$E_{iy}$$ cancel. In the same way the charges at $$y=+2d$$ and $$y=-2d$$ have the same sign $$-q$$ and again opposite $$y_i$$, so those $$E_{iy}$$ also cancel. Hence,

$$\sum_{i=1}^{4}E_{iy}=0,$$ and the total electric field at $$P$$ is purely along the x-direction.

Therefore the magnitude of the net field equals the sum of the x-components:

$$E=\sum_{i=1}^{4}E_{ix}=kD\sum_{i=1}^{4} \dfrac{q_i}{\bigl(D^{\,2}+y_i^{\,2}\bigr)^{3/2}}.$$

Because $$D\gg d,$$ we expand each denominator with the binomial series. We start from

$$\bigl(D^{\,2}+y_i^{\,2}\bigr)^{-3/2}=D^{-3}\Bigl(1+\dfrac{y_i^{\,2}}{D^{\,2}}\Bigr)^{-3/2}.$$ Using $$(1+s)^{-3/2}=1-\dfrac{3}{2}s+\dfrac{15}{8}s^{2}-\ldots,$$ we substitute $$s=\dfrac{y_i^{\,2}}{D^{\,2}}$$ to obtain

$$\bigl(D^{\,2}+y_i^{\,2}\bigr)^{-3/2}=D^{-3}\Bigl[1-\dfrac{3}{2}\dfrac{y_i^{\,2}}{D^{\,2}} +\dfrac{15}{8}\dfrac{y_i^{\,4}}{D^{\,4}}-\ldots\Bigr].$$

Substituting this expansion back into the expression for $$E$$ gives

$$E=kD\sum_{i=1}^{4}q_i D^{-3}\Bigl[1-\dfrac{3}{2}\dfrac{y_i^{\,2}}{D^{\,2}} +\dfrac{15}{8}\dfrac{y_i^{\,4}}{D^{\,4}}-\ldots\Bigr].$$

Simplifying the common factor $$D$$ with $$D^{-3}$$ we have

$$E=k\sum_{i=1}^{4}q_i \Bigl[D^{-2}-\dfrac{3}{2}\dfrac{y_i^{\,2}}{D^{\,4}} +\dfrac{15}{8}\dfrac{y_i^{\,4}}{D^{\,6}}-\ldots\Bigr].$$

We now evaluate the sums term by term.

1. The monopole term (total charge):

$$\sum_{i=1}^{4}q_i=(-q)+(+q)+(+q)+(-q)=0.$$ So the entire $$D^{-2}$$ contribution vanishes.

2. The next (quadrupole) term involves $$\sum q_i y_i^{\,2}\!:$$

$$\begin{aligned} \sum_{i=1}^{4}q_i y_i^{\,2}&=(-q)(4d^{2})+(+q)(d^{2})+(+q)(d^{2})+(-q)(4d^{2})\\ &=-4qd^{2}+qd^{2}+qd^{2}-4qd^{2}\\ &=-6qd^{2}\;\;(\neq0). \end{aligned}$$

Hence the first non-vanishing contribution to $$E$$ is

$$E=-\dfrac{3}{2}k\bigl(\sum q_i y_i^{\,2}\bigr)D^{-4} +\text{higher-order terms}.$$ Since $$\sum q_i y_i^{\,2}$$ is merely a constant factor, this clearly shows

$$E\propto\dfrac{1}{D^{4}}.$$

All still-higher terms vary as $$1/D^{6},\,1/D^{8},\ldots$$ and are even smaller when $$D\gg d.$$ Thus the leading behaviour of the electric field magnitude is

$$E\;{\Large\sim}\;\dfrac{\text{constant}}{D^{4}}.$$

Hence, the correct answer is Option A.

We have two point charges on the $$x$$-axis. The first charge is $$q_1 = +\sqrt{10}\;\mu{\rm C}=+\sqrt{10}\times10^{-6}\;{\rm C}$$ located at $$x=1\;{\rm m},\,y=0\;{\rm m}$$. The second charge is $$q_2 = -25\;\mu{\rm C}=-25\times10^{-6}\;{\rm C}$$ placed at $$x=4\;{\rm m},\,y=0\;{\rm m}$$. The point where we want the electric field is on the $$y$$-axis at $$x=0,\,y=3\;{\rm m}$$.

The electric field due to a point charge is given by the formula

$$\vec E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}\,\hat r,$$

where $$\hat r$$ is the unit vector drawn from the charge towards the field point and $$r$$ is the distance between them. We shall compute this field separately for each charge and then add the two vector results.

Field due to $$q_1$$

The position vector from $$q_1(1,0)$$ to the point $$(0,3)$$ is

$$\vec r_1 = (0-1)\hat i + (3-0)\hat j = -1\hat i + 3\hat j.$$

The magnitude of this vector is

$$r_1=\sqrt{(-1)^2+3^2}=\sqrt{1+9}=\sqrt{10}\;{\rm m}.$$

The unit vector is therefore

$$\hat r_1=\dfrac{\vec r_1}{r_1}= \dfrac{-1}{\sqrt{10}}\hat i+\dfrac{3}{\sqrt{10}}\hat j.$$

Now the magnitude of the field is

$$E_1=\dfrac{1}{4\pi\varepsilon_0}\dfrac{|q_1|}{r_1^2} = 9\times10^9 \times \dfrac{\sqrt{10}\times10^{-6}}{(\sqrt{10})^2} = 9\times10^9 \times \dfrac{\sqrt{10}\times10^{-6}}{10} =9\times10^2\sqrt{10}\;{\rm V/m}.$$

Multiplying this magnitude with the unit vector, we obtain

$$\vec E_1 = 9\times10^2\sqrt{10}\left(\dfrac{-1}{\sqrt{10}}\hat i+\dfrac{3}{\sqrt{10}}\hat j\right) = 900(-1\hat i+3\hat j) = -900\hat i + 2700\hat j\;{\rm V/m}.$$

Field due to $$q_2$$

The position vector from $$q_2(4,0)$$ to the point $$(0,3)$$ is

$$\vec r_2 = (0-4)\hat i + (3-0)\hat j = -4\hat i + 3\hat j.$$

The magnitude of this vector is

$$r_2=\sqrt{(-4)^2+3^2}=\sqrt{16+9}=5\;{\rm m}.$$

Thus the unit vector is

$$\hat r_2=\dfrac{\vec r_2}{r_2}= \dfrac{-4}{5}\hat i+\dfrac{3}{5}\hat j.$$

The magnitude of the field that a charge of magnitude $$|q_2|$$ would produce is

$$E_2=\dfrac{1}{4\pi\varepsilon_0}\dfrac{|q_2|}{r_2^2} =9\times10^9 \times \dfrac{25\times10^{-6}}{5^2} =9\times10^9 \times \dfrac{25\times10^{-6}}{25} =9\times10^3\;{\rm V/m}=9000\;{\rm V/m}.$$

Because $$q_2$$ is –ve, the field vector points towards the charge, i.e. opposite to $$\hat r_2$$. Hence

$$\vec E_2 = -E_2\,\hat r_2 = -9000\left(\dfrac{-4}{5}\hat i+\dfrac{3}{5}\hat j\right) = 9000\left(\dfrac{4}{5}\hat i-\dfrac{3}{5}\hat j\right) = 9000\!\left(0.8\hat i-0.6\hat j\right) = 7200\hat i - 5400\hat j\;{\rm V/m}.$$

Net electric field

Now we add the two vectors:

$$\vec E = \vec E_1 + \vec E_2 = (-900\hat i + 2700\hat j) + (7200\hat i - 5400\hat j) = ( -900 + 7200)\hat i + (2700 - 5400)\hat j = 6300\hat i - 2700\hat j\;{\rm V/m}.$$

Factoring out $$10^2$$, we write

$$\vec E = (63\hat i - 27\hat j)\times10^2\;{\rm V/m}.$$

Among the given alternatives this matches option D.

Hence, the correct answer is Option D.

First of all we determine the constant that appears in the charge-density expression. The charge density inside the solid sphere is given as $$\rho(r)=kr$$. The total charge present in the sphere is stated to be $$2Q$$, so we write

$$2Q=\int_{0}^{R}\rho(r)\,d\tau =\int_{0}^{R}kr\;4\pi r^{2}\,dr =4\pi k\int_{0}^{R}r^{3}\,dr =4\pi k\left[\frac{r^{4}}{4}\right]_{0}^{R} =4\pi k\left(\frac{R^{4}}{4}\right) =\pi kR^{4}.$$

Solving for $$k$$ gives

$$k=\frac{2Q}{\pi R^{4}}.$$

Now we focus on one of the two point charges, say charge A, which is situated at a distance $$a$$ from the centre. Because the volume charge distribution is spherically symmetric, Gauss’s law tells us that only the charge inside a Gaussian sphere of radius $$a$$ contributes to the electric field at A. The enclosed charge is

$$Q_{\text{enc}}(a)=\int_{0}^{a}\rho(r)\,d\tau =\int_{0}^{a}kr\;4\pi r^{2}\,dr =4\pi k\int_{0}^{a}r^{3}\,dr =4\pi k\left[\frac{r^{4}}{4}\right]_{0}^{a} =\pi k a^{4}.$$

Substituting the value of $$k$$ just obtained,

$$Q_{\text{enc}}(a)=\pi\left(\frac{2Q}{\pi R^{4}}\right)a^{4} =\frac{2Qa^{4}}{R^{4}}.$$

Next we write the electric field at A due to the spherically distributed charge. Gauss’s law in the form $$E(4\pi a^{2})=\dfrac{Q_{\text{enc}}}{\varepsilon_{0}}$$, or equivalently $$E=k_{e}\dfrac{Q_{\text{enc}}}{a^{2}}$$ with $$k_{e}=\dfrac{1}{4\pi\varepsilon_{0}}$$, gives

$$E_{\text{sphere}}=k_{e}\frac{Q_{\text{enc}}(a)}{a^{2}} =k_{e}\frac{2Qa^{4}/R^{4}}{a^{2}} =k_{e}\frac{2Qa^{2}}{R^{4}}.$$

The force exerted on charge A (whose value is $$-Q$$) by the sphere therefore has magnitude

$$F_{\text{sphere}} =Q\;E_{\text{sphere}} =Q\left(k_{e}\frac{2Qa^{2}}{R^{4}}\right) =2k_{e}\frac{Q^{2}a^{2}}{R^{4}}.$$

This force is directed towards the centre, because the sphere is positively charged while A is negative.

Now we evaluate the force on A due to the other point charge B. Charges A and B are both $$-Q$$ and are placed diametrically opposite, so their separation is $$2a$$. The Coulomb force magnitude is

$$F_{\text{B on A}} =k_{e}\frac{Q^{2}}{(2a)^{2}} =k_{e}\frac{Q^{2}}{4a^{2}}.$$

This force is directed away from the centre, i.e., outward along the line joining the two charges, because like charges repel.

According to the problem statement, charge A experiences no net force. Hence the outward force due to charge B must exactly cancel the inward force due to the sphere:

$$F_{\text{B on A}} = F_{\text{sphere}}$$ $$\Rightarrow\; k_{e}\frac{Q^{2}}{4a^{2}} =2k_{e}\frac{Q^{2}a^{2}}{R^{4}}.$$

We can divide both sides by the common factor $$k_{e}Q^{2}$$ to obtain

$$\frac{1}{4a^{2}} = \frac{2a^{2}}{R^{4}}.$$

Cross-multiplying gives

$$R^{4} = 8a^{4}.$$

Solving for $$a$$, we have

$$a^{4} = \frac{R^{4}}{8} \;\;\Longrightarrow\;\; a = \frac{R}{8^{1/4}} = 8^{-1/4}R.$$

This value exactly matches Option D.

Hence, the correct answer is Option D.

We are given a solid sphere of radius $$R$$ whose volume-charge density varies with the distance $$r$$ from the centre as

$$\rho(r)=\rho_0\!\left(1-\frac{r}{R}\right),\qquad 0\le r\le R.$$

To find the electric field at an external point $$P$$ situated at a distance $$r$$ (with $$r>R$$) from the centre, we first need the total charge $$Q$$ contained in the sphere, because outside the sphere the entire charge can be treated as if it were concentrated at the centre (this is a consequence of Gauss’s law combined with spherical symmetry).

The infinitesimal charge of a thin spherical shell of radius $$r$$ and thickness $$dr$$ is

$$dq=\rho(r)\,dV=\rho(r)\,4\pi r^{2}dr.$$

Substituting $$\rho(r)=\rho_0\left(1-\dfrac{r}{R}\right)$$, we get

$$dq=4\pi r^{2}\rho_0\left(1-\frac{r}{R}\right)dr.$$

The total charge is obtained by integrating $$dq$$ from the centre $$r=0$$ to the surface $$r=R$$:

$$\begin{aligned} Q &=\int_{0}^{R}dq \\ &=4\pi\rho_0\int_{0}^{R}r^{2}\left(1-\frac{r}{R}\right)dr. \end{aligned}$$

We separate the integrand:

$$\begin{aligned} \int_{0}^{R}r^{2}\left(1-\frac{r}{R}\right)dr &=\int_{0}^{R}\left(r^{2}-\frac{r^{3}}{R}\right)dr\\ &=\int_{0}^{R}r^{2}dr-\frac{1}{R}\int_{0}^{R}r^{3}dr. \end{aligned}$$

Using the standard power-integral formulas $$\int r^{n}dr=\frac{r^{n+1}}{n+1},$$ we evaluate the two parts:

$$\int_{0}^{R}r^{2}dr=\left[\frac{r^{3}}{3}\right]_{0}^{R}=\frac{R^{3}}{3},$$

$$\int_{0}^{R}r^{3}dr=\left[\frac{r^{4}}{4}\right]_{0}^{R}=\frac{R^{4}}{4}.$$

Hence,

$$\begin{aligned} \int_{0}^{R}r^{2}\left(1-\frac{r}{R}\right)dr &=\frac{R^{3}}{3}-\frac{1}{R}\cdot\frac{R^{4}}{4}\\ &=\frac{R^{3}}{3}-\frac{R^{3}}{4}\\ &=R^{3}\!\left(\frac{1}{3}-\frac{1}{4}\right)\\ &=R^{3}\!\left(\frac{4-3}{12}\right)\\ &=\frac{R^{3}}{12}. \end{aligned}$$

Substituting this result into the expression for $$Q$$, we have

$$\begin{aligned} Q &=4\pi\rho_0\left(\frac{R^{3}}{12}\right)\\ &=\frac{4\pi\rho_0 R^{3}}{12}\\ &=\frac{\pi\rho_0 R^{3}}{3}. \end{aligned}$$

Now we turn to the electric field at the external point. From Gauss’s law, for a spherical Gaussian surface of radius $$r$$ (with $$r>R$$) concentric with the sphere, the outward flux is

$$\Phi=E\,(4\pi r^{2}),$$

and Gauss’s law states

$$\Phi=\frac{Q_{\text{enc}}}{\varepsilon_0},$$

where $$Q_{\text{enc}}=Q$$, the total charge calculated above. Therefore,

$$E(4\pi r^{2})=\frac{Q}{\varepsilon_0}.$$

Solving for $$E$$ gives

$$\begin{aligned} E &=\frac{Q}{4\pi\varepsilon_0\,r^{2}}\\ &=\frac{1}{4\pi\varepsilon_0\,r^{2}}\left(\frac{\pi\rho_0 R^{3}}{3}\right)\\ &=\frac{\pi\rho_0 R^{3}}{12\pi\varepsilon_0\,r^{2}}\\ &=\frac{\rho_0 R^{3}}{12\varepsilon_0 r^{2}}. \end{aligned}$$

The electric field outside the charged ball thus decreases as $$1/r^{2}$$ and has the magnitude

$$E(r)=\frac{\rho_0 R^{3}}{12\varepsilon_0 r^{2}}.$$

Comparing with the given alternatives, we see that this matches Option D.

Hence, the correct answer is Option D.

We begin by recalling the expression for the electric potential produced by a uniformly charged thin spherical shell. A shell of total charge $$Q$$ and radius $$R$$ produces, at a point situated a distance $$r$$ from the common centre, a potential

$$V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}\quad\text{for}\;r\ge R,$$

while everywhere inside the shell (that is, for $$r<R$$) the potential is constant and equal to its value on the surface, viz.

$$V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}\quad\text{for}\;r\le R.$$

Each of the three metal shells A, B and C is concentric, so their individual contributions at any point simply add. The surface charge densities are given as $$+\sigma$$ on A (radius $$a$$), $$-\sigma$$ on B (radius $$b$$) and $$+\sigma$$ on C (radius $$c$$), with the order $$a<b<c$$.

First we evaluate the actual charges on the shells:

$$Q_A = \sigma\;(4\pi a^2),\qquad Q_B = (-\sigma)\;(4\pi b^2),\qquad Q_C = \sigma\;(4\pi c^2).$$

Now we calculate separately the potentials at the point on shell B, i.e. at radial distance $$r=b$$.

Contribution of shell A: the observation point lies outside shell A (because $$b>a$$). Using the outside-point formula we have

$$V_A(b)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_A}{b} =\dfrac{1}{4\pi\varepsilon_0}\dfrac{\sigma(4\pi a^2)}{b} =\dfrac{\sigma a^2}{\varepsilon_0 b}.$$

Contribution of shell B itself: for a point on a conducting shell the potential due to that shell equals the value just outside, obtained with the same outside-point formula but with $$r=b$$ (or, equivalently, with the inside constant value). Thus

$$V_B(b)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_B}{b} =\dfrac{1}{4\pi\varepsilon_0}\dfrac{-\sigma(4\pi b^2)}{b} =-\dfrac{\sigma b}{\varepsilon_0}.$$

Contribution of shell C: the observation point is inside shell C because $$b<c$$. Hence the potential is constant throughout the interior and equals its surface value, viz.

$$V_C(b)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_C}{c} =\dfrac{1}{4\pi\varepsilon_0}\dfrac{\sigma(4\pi c^2)}{c} =\dfrac{\sigma c}{\varepsilon_0}.$$

We now add these three contributions to obtain the total potential of shell B:

$$\begin{aligned} V_{\text{total}}(b) &=V_A(b)+V_B(b)+V_C(b)\\[4pt] &=\dfrac{\sigma a^2}{\varepsilon_0 b} -\dfrac{\sigma b}{\varepsilon_0} +\dfrac{\sigma c}{\varepsilon_0}\\[4pt] &=\dfrac{\sigma}{\varepsilon_0} \left(\dfrac{a^2}{b}-b+c\right). \end{aligned}$$

It is convenient to combine the first two terms over a common denominator $$b$$:

$$\dfrac{a^2}{b}-b=\dfrac{a^2-b^2}{b},$$

so the potential becomes

$$V_{\text{total}}(b)= \dfrac{\sigma}{\varepsilon_0} \left[\dfrac{a^2-b^2}{b}+c\right].$$

This expression matches exactly the option labelled

$$\dfrac{\sigma}{\varepsilon_0}\left[\dfrac{a^2-b^2}{b}+c\right],$$

which is Option C.

Hence, the correct answer is Option C.

Let the identical spheres A and B be separated by a distance $$r$$ (with $$r$$ much larger than their diameters, so we may treat the charges as point charges).

Suppose each of A and B initially carries charge $$q$$.

The electrostatic force between two point charges is given by Coulomb’s law:

$$F \;=\; k\;\dfrac{q_1 q_2}{r^{\,2}}$$

where $$k$$ is the electrostatic constant.

With $$q_1 = q_2 = q$$, the initial force is

$$F \;=\; k\;\dfrac{q^{\,2}}{r^{\,2}} \quad\text{(1)}$$

Now sphere C (identical to A and B) is uncharged at the start.

First contact: C touches A. Being identical conductors, they share their total charge equally.

Total charge on A + C before contact: $$q + 0 = q$$.

After contact, charge on each (A and C) becomes

$$\dfrac{q}{2}$$.

So,

$$q_A = \dfrac{q}{2}, \qquad q_C = \dfrac{q}{2}$$

Second contact: the same sphere C now touches B.

At this moment, charges are

$$q_B = q, \qquad q_C = \dfrac{q}{2}$$

The total charge on B + C before this contact is

$$q + \dfrac{q}{2} = \dfrac{3q}{2}$$.

After they touch, the charge again divides equally (because the spheres are still identical):

Charge on each (B and C) becomes

$$\dfrac{1}{2}\left(\dfrac{3q}{2}\right) = \dfrac{3q}{4}$$.

So, finally,

$$q_A = \dfrac{q}{2}, \qquad q_B = \dfrac{3q}{4}$$

Sphere C is removed, so only A and B remain with the above charges.

New force between A and B. Using Coulomb’s law again:

$$F' = k\;\dfrac{q_A\,q_B}{r^{\,2}} = k\;\dfrac{\left(\dfrac{q}{2}\right)\left(\dfrac{3q}{4}\right)}{r^{\,2}} = k\;\dfrac{3q^{\,2}}{8\,r^{\,2}}$$

Compare this with equation (1): $$F = k\dfrac{q^{\,2}}{r^{\,2}}$$. Hence

$$F' = \dfrac{3}{8}\,F$$

Therefore the force becomes $$\dfrac{3F}{8}$$.

Hence, the correct answer is Option C.

Electric flux through a closed surface is given by Gauss' law as $$\phi=\frac{\Sigma\ q_{enclosed}}{\in_0}$$

Since for all the given surfaces, the charge enclosed is equal, therefore the electric flux through all the four surfaces is the same.

To link the given electric potential to the corresponding volume charge density we recall the fundamental relation from electrostatics, namely Poisson’s equation, stated in SI units as

$$\nabla^{2}V = -\,\frac{\rho}{\epsilon_{0}}.$$

Because the problem tells us that the potential $$V$$ depends only on the coordinate $$z$$ (it is uniform in the $$x$$ and $$y$$ directions), the Laplacian simplifies dramatically. In Cartesian coordinates, with no $$x$$ or $$y$$ dependence, we have

$$\nabla^{2}V \;=\; \frac{\partial^{2}V}{\partial x^{2}} + \frac{\partial^{2}V}{\partial y^{2}} + \frac{\partial^{2}V}{\partial z^{2}} \;=\; 0 + 0 + \frac{d^{2}V}{dz^{2}} \;=\; \frac{d^{2}V}{dz^{2}}.$$

So Poisson’s equation for this one-dimensional situation becomes

$$\frac{d^{2}V}{dz^{2}} = -\,\frac{\rho(z)}{\epsilon_{0}}.$$

Now we examine the potential piece-wise.

Region 1 : $$|z|\le 1\text{ m}$$

The potential is given by $$V(z)=30-5z^{2}.$$ Let us differentiate twice:

First derivative: $$\frac{dV}{dz} = -10\,z.$$

Second derivative: $$\frac{d^{2}V}{dz^{2}} = -10.$$

This second derivative is a constant, so in this region

$$\nabla^{2}V = -10.$$

Substituting into Poisson’s equation,

$$-10 = -\,\frac{\rho}{\epsilon_{0}} \;\;\Longrightarrow\;\; \rho = 10\,\epsilon_{0}.$$

Thus the volume charge density inside the slab $$|z|\le 1\text{ m}$$ is a positive and uniform value $$\rho_0 = 10\,\epsilon_{0}.$$

Region 2 : $$|z|\ge 1\text{ m}$$

The potential now is given by $$V(z)=35-10|z|.$$ We treat the two sides separately but the mathematical outcome is the same.

For $$z\ge 1$$ we have $$|z|=z,$$ so $$V(z)=35-10z.$$ For $$z\le -1$$ we have $$|z|=-z,$$ so $$V(z)=35+10z.$$

Differentiating either expression yields a constant first derivative, and the second derivative is zero:

$$\frac{d^{2}V}{dz^{2}} = 0 \quad(\text{for }|z|\ge 1).$$

Therefore, in this exterior region

$$\nabla^{2}V = 0 \;\;\Longrightarrow\;\; 0 = -\,\frac{\rho}{\epsilon_{0}} \;\;\Longrightarrow\;\; \rho = 0.$$

The only non-zero volume charge resides in the central slab $$|z|\le 1\text{ m},$$ and its magnitude is $$10\,\epsilon_{0}.$$ Outside that slab the charge density is identically zero. (The kink in the slope of $$V$$ at $$|z|=1$$ would correspond to an additional surface charge, but the question asks specifically for a volume charge of constant magnitude, so that surface effect is not relevant to the choices given.)

Comparing this result with the listed options, we see that it matches exactly with Option B.

Hence, the correct answer is Option B.

We consider a spherically symmetric charge distribution whose volume charge density is $$\rho(r)$$. At a distance $$r$$ from the centre the total charge enclosed is written as

For spherical symmetry the electric field at this point is obtained from Gauss’s law. First we state the formula:

The electrostatic potential satisfies the basic relation

Now the problem tells us that the equipotential surfaces $$V_0,\,V_0+\Delta V,\,V_0+2\Delta V,\ldots$$ appear at radii $$r_0,\,r_1,\,r_2,\ldots$$ with a constant spacing of radii, that is

Because both $$\Delta V$$ and $$\Delta r$$ are fixed and small intervals, the ratio $$\Delta V/\Delta r$$ represents the magnitude of $$dV/dr$$, and this ratio is the same everywhere. Hence

Calling this positive constant $$C_1$$, we write

Substituting the expression for $$dV/dr$$ from the electric field formula, we have

or, dropping the negative sign and gathering the constants,

So we conclude that

Next we relate this to the density. Differentiating the enclosed charge with respect to $$r$$ gives, by the Leibniz rule,

But we also have $$Q(r)=k r^{2}$$, so

Equating the two expressions for $$dQ/dr$$ we obtain

Solving for the density,

The constant factors are unimportant for proportionality, so finally we can state

Hence, the correct answer is Option 3.

$$L = \pi R \implies R = \frac{L}{\pi}$$

The wire is divided into two equal halves (quadrants), each having a length $$l = L/2$$. The magnitude of the charge on each half is given as $$|Q| = 10^3 \epsilon_0$$.

$$\lambda = \frac{Q}{L/2} = \frac{2Q}{L}$$

By symmetry, the y-components of the two fields are equal in magnitude but opposite in direction, so they cancel out. The x-components are in the same direction ($$+\hat{i}$$) and add up.

$$E_x = E_y = \frac{k\lambda}{R} = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R}$$

$$\vec{E}_{net} = (E_{1x} + E_{2x}) \hat{i} = 2 \left( \frac{\lambda}{4\pi\epsilon_0 R} \right) \hat{i} = \frac{\lambda}{2\pi\epsilon_0 R} \hat{i}$$

$$\vec{E}_{net} = \frac{2Q/L}{2\pi\epsilon_0 (L/\pi)} \hat{i} = \frac{2Q}{L} \cdot \frac{1}{2\epsilon_0 L} \hat{i} = \frac{Q}{\epsilon_0 L^2} \hat{i}$$

$$\vec{E}_{net} = \frac{10^3 \epsilon_0}{\epsilon_0 (0.2)^2} \hat{i}$$

$$\vec{E}_{net} = \frac{1000}{0.04} \hat{i} = \frac{100000}{4} \hat{i} = 25000 \hat{i} \text{ N C}^{-1}$$

$$\vec{E}_{net} = (25 \times 10^3) \hat{i} \text{ N C}^{-1}$$

We start from the very basic rule for electrostatic field lines: they originate on positive charge and they terminate on negative charge. Hence, wherever the surface charge density is $$+\sigma$$ the field lines must leave the surface, and wherever it is $$-\sigma$$ they must enter the surface.

Another fundamental fact, valid for any conductor in electro-static equilibrium, is the relation between the surface charge density $$\sigma$$ and the electric field just outside the conductor:

$$\mathbf E_{\text{out}}=\dfrac{\sigma}{\varepsilon_0}\,\hat n,$$

where $$\hat n$$ is the unit vector normal to the surface directed outward from the metal. Therefore

• for $$\sigma>0$$ (upper half of the cylindrical shell) the field just outside the metal points radially outward, normal to the surface,
• for $$\sigma<0$$ (lower half of the cylindrical shell) we have $$-\sigma<0$$, so the field just outside points radially inward, again normal to the surface.

Because the shell is conducting, the electric field inside the metal itself is zero; hence no field lines can exist inside the cylindrical material, nor in the hollow region surrounded by it.

Putting these two pieces together we get the complete qualitative picture:

• Lines emerge perpendicularly from every point on the outer surface of the upper (positively charged) half.
• They terminate perpendicularly on the outer surface of the lower (negatively charged) half.
• No lines cross the metal, and no lines appear inside the cavity.

Among the four schematic figures supplied in the question, only Figure (2) shows lines that leave the upper semicylinder, curve through the external space, and enter the lower semicylinder, while the interior of the shell remains field-free. The other three options either show lines entering/leaving the wrong halves or show field lines existing inside the conductor, both of which violate the rules stated above.

Hence, the correct answer is Option B (Figure 2).

The electric field on the axis of a uniform solid disc of radius $$R$$ at a height $$h$$ from its center is:

$$E = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{h}{\sqrt{R^2 + h^2}} \right)$$

$$E_{net} = E_b - E_a$$

$$E_{net} = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{h}{\sqrt{b^2 + h^2}} \right) - \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{h}{\sqrt{a^2 + h^2}} \right)$$

$$E_{net} = \frac{\sigma}{2\epsilon_0} \left( \frac{h}{\sqrt{a^2 + h^2}} - \frac{h}{\sqrt{b^2 + h^2}} \right)$$

$$E_{net} = \frac{\sigma h}{2\epsilon_0} \left[ \frac{1}{a\sqrt{1 + \frac{h^2}{a^2}}} - \frac{1}{b\sqrt{1 + \frac{h^2}{b^2}}} \right]$$

Since $$h \ll a$$ (and consequently $$h \ll b$$), $$\sqrt{1 + \frac{h^2}{a^2}} \approx 1 \quad \text{and} \quad \sqrt{1 + \frac{h^2}{b^2}} \approx 1$$

$$E_{net} \approx \frac{\sigma h}{2\epsilon_0} \left[ \frac{1}{a} - \frac{1}{b} \right]$$

$$E_{net} = \frac{\sigma h}{2\epsilon_0} \left[ \frac{1}{a} - \frac{1}{2a} \right]$$

$$E_{net} = \left( \frac{\sigma}{4a\epsilon_0} \right) h$$

$$C = \frac{\sigma}{4a\epsilon_0}$$

The cone has base radius $$R$$ and height $$h$$, and sits in a uniform electric field $$\vec{E}$$ directed parallel to the base. Since there is no enclosed charge, Gauss's law gives zero net flux through the entire closed surface of the cone (curved surface plus the circular base). Therefore the flux entering one side of the cone equals the flux leaving the other side.

To find the entering flux, consider the projection of the cone onto a plane perpendicular to $$\vec{E}$$. The cone's silhouette when viewed along the direction of $$\vec{E}$$ is a triangle with base equal to the diameter $$2R$$ and height equal to $$h$$. The area of this triangle is $$\dfrac{1}{2} \times 2R \times h = Rh$$.

Since the field is uniform, the flux through any closed surface equals the flux through the projected cross-sectional area perpendicular to the field. The entering flux (which equals the leaving flux) is therefore $$\Phi = E \times Rh = EhR$$.

To find the electric field at an internal point where $$ r < R $$, we use Gauss's law due to the spherical symmetry. Gauss's law states that the flux through a closed surface is equal to the charge enclosed divided by $$ \varepsilon_0 $$. We choose a spherical Gaussian surface of radius $$ r $$ centered at the same point as the charge distribution.

The electric field is radial and constant in magnitude on this surface, so the flux is $$ E \times 4\pi r^2 $$. By Gauss's law:

$$ E \times 4\pi r^2 = \frac{Q_{\text{enc}}}{\varepsilon_0} $$

Solving for $$ E $$:

$$ E = \frac{Q_{\text{enc}}}{4\pi \varepsilon_0 r^2} $$

Now, we need to find $$ Q_{\text{enc}} $$, the charge enclosed within the sphere of radius $$ r $$. The charge density is given by $$ \rho(r') = \rho_o \left(1 - \frac{r'}{R}\right) $$ for $$ r' < R $$. Since $$ r < R $$, we integrate the charge density over the volume inside radius $$ r $$. Using spherical coordinates, the volume element is $$ dV' = 4\pi r'^2 dr' $$, so:

$$ Q_{\text{enc}} = \int_0^r \rho(r') dV' = \int_0^r \rho_o \left(1 - \frac{r'}{R}\right) \times 4\pi r'^2 dr' $$

Factor out the constants:

$$ Q_{\text{enc}} = 4\pi \rho_o \int_0^r \left(1 - \frac{r'}{R}\right) r'^2 dr' $$

Expand the integrand:

$$ Q_{\text{enc}} = 4\pi \rho_o \int_0^r \left( r'^2 - \frac{r'^3}{R} \right) dr' $$

Integrate term by term:

$$ \int_0^r r'^2 dr' = \left[ \frac{r'^3}{3} \right]_0^r = \frac{r^3}{3} $$

$$ \int_0^r \frac{r'^3}{R} dr' = \frac{1}{R} \left[ \frac{r'^4}{4} \right]_0^r = \frac{r^4}{4R} $$

So the integral becomes:

$$ \int_0^r \left( r'^2 - \frac{r'^3}{R} \right) dr' = \frac{r^3}{3} - \frac{r^4}{4R} $$

Therefore:

$$ Q_{\text{enc}} = 4\pi \rho_o \left( \frac{r^3}{3} - \frac{r^4}{4R} \right) $$

Substitute $$ Q_{\text{enc}} $$ back into the expression for $$ E $$:

$$ E = \frac{4\pi \rho_o \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)}{4\pi \varepsilon_0 r^2} = \frac{\rho_o \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)}{\varepsilon_0 r^2} $$

Simplify by dividing each term by $$ r^2 $$:

$$ E = \frac{\rho_o}{\varepsilon_0} \left( \frac{r^3}{3r^2} - \frac{r^4}{4R r^2} \right) = \frac{\rho_o}{\varepsilon_0} \left( \frac{r}{3} - \frac{r^2}{4R} \right) $$

Comparing with the options, this matches option B:

$$ \frac{\rho_o}{\varepsilon_o}\left(\frac{r}{3} - \frac{r^2}{4R}\right) $$

Hence, the correct answer is Option B.

The electric field is given by $$\vec{E} = E_0 \hat{i} + 2E_0 \hat{j}$$, where $$E_0 = 100$$ N C$$^{-1}$$. So, substituting the value, $$\vec{E} = 100 \hat{i} + 2 \times 100 \hat{j} = 100 \hat{i} + 200 \hat{j}$$ N C$$^{-1}$$.

The circular surface has a radius of 0.02 m and is parallel to the Y-Z plane. Since it is parallel to the Y-Z plane, the normal to the surface is along the X-axis. Therefore, the area vector $$\vec{A}$$ is directed along the $$\hat{i}$$ direction. The magnitude of the area vector is the area of the circle, which is $$A = \pi r^2$$.

Substituting the radius $$r = 0.02$$ m, we get $$A = \pi \times (0.02)^2 = \pi \times 0.0004 = 0.0004\pi$$ m$$^2$$. Thus, $$\vec{A} = 0.0004\pi \hat{i}$$ m$$^2$$.

The electric flux $$\phi$$ through the surface is given by the dot product $$\phi = \vec{E} \cdot \vec{A}$$. So, $$\phi = (100 \hat{i} + 200 \hat{j}) \cdot (0.0004\pi \hat{i})$$.

Computing the dot product, the $$\hat{i}$$ components multiply and the $$\hat{j}$$ component of $$\vec{E}$$ multiplies with the $$\hat{j}$$ component of $$\vec{A}$$, but since $$\vec{A}$$ has no $$\hat{j}$$ component, that term is zero. Therefore:

$$\phi = (100 \times 0.0004\pi) + (200 \times 0) = 100 \times 0.0004\pi = 0.04\pi$$ N m$$^2$$ C$$^{-1}$$.

Now, substituting $$\pi \approx 3.14$$, we get $$\phi = 0.04 \times 3.14 = 0.1256$$ N m$$^2$$ C$$^{-1}$$. This value is approximately 0.125 N m$$^2$$ C$$^{-1}$$.

Comparing with the options, 0.125 N m$$^2$$ C$$^{-1}$$ matches option C.

Hence, the correct answer is Option C.

The electric field is directed inward toward the center of the Earth, meaning it points in the direction opposite to the outward normal of the Earth's surface. By Gauss's law, the total charge enclosed is $$Q = \epsilon_0 \oint \vec{E} \cdot d\vec{A}$$. Since $$\vec{E}$$ is directed inward and $$d\vec{A}$$ points outward, $$\vec{E} \cdot d\vec{A} = -E\, dA$$, giving $$Q = -\epsilon_0 E \times 4\pi R_E^2$$.

Substituting the given values: $$Q = -8.85 \times 10^{-12} \times 150 \times 4\pi \times (6.37 \times 10^6)^2$$. Computing $$R_E^2 = (6.37)^2 \times 10^{12} = 40.58 \times 10^{12}$$ m$$^2$$, and $$4\pi R_E^2 = 4\pi \times 40.58 \times 10^{12} = 5.099 \times 10^{14}$$ m$$^2$$.

So $$Q = -8.85 \times 10^{-12} \times 150 \times 5.099 \times 10^{14} = -8.85 \times 150 \times 5.099 \times 10^{2} = -8.85 \times 76485 \approx -6.77 \times 10^5$$ C $$= -677$$ kC, which is approximately $$-680$$ kC.

We have two identical point charges, each of magnitude $$q$$, fixed on the x-axis at the points $$(-a,\,0)$$ and $$(\,a,\,0)$$.

The test particle carries charge $$q_0=-\dfrac{q}{2}$$ and mass $$m$$. Initially this particle is at the origin, but we now give it a very small displacement $$y \;(\;y\ll a\;)$$ along the positive y-axis, so its new co-ordinates become $$(0,\,y)$$.

The electrostatic force between two point charges is given by Coulomb’s law: $$\vec F = k\,\dfrac{q_1q_2}{r^2}\,\hat r,$$ where $$k=\dfrac{1}{4\pi\varepsilon_0}$$, $$r$$ is the distance between the charges and $$\hat r$$ is the unit vector directed from the source charge to the test charge.

Let us find the force exerted on the displaced particle by each fixed charge.

Force due to the charge at $$(-a,0):$$ The position vector of the particle relative to this charge is $$\vec r_1 = (0-(-a))\,\hat i + (y-0)\,\hat j = a\,\hat i + y\,\hat j.$$ Its magnitude is $$r_1 = \sqrt{a^2+y^2}.$$ Therefore, using Coulomb’s law,

$$ \vec F_1 = k\,\dfrac{q\,q_0}{(a^2+y^2)}\,\dfrac{a\,\hat i + y\,\hat j}{\sqrt{a^2+y^2}} = k\,\dfrac{q\,q_0}{(a^2+y^2)^{3/2}}\,(a\,\hat i + y\,\hat j). $$

Force due to the charge at $$(+a,0):$$ Here the position vector of the particle relative to the charge is $$\vec r_2 = (0-a)\,\hat i + (y-0)\,\hat j = -a\,\hat i + y\,\hat j,$$ with the same magnitude $$r_2=\sqrt{a^2+y^2}$$. Hence,

$$ \vec F_2 = k\,\dfrac{q\,q_0}{(a^2+y^2)^{3/2}}\;(-a\,\hat i + y\,\hat j). $$

Adding the two forces:

$$ \vec F_{\text{net}} = \vec F_1 + \vec F_2 = k\,\dfrac{q\,q_0}{(a^2+y^2)^{3/2}}\Big[(a-a)\,\hat i + (y+y)\,\hat j\Big] = k\,\dfrac{q\,q_0}{(a^2+y^2)^{3/2}}\,(2y)\,\hat j. $$

So only a y-component survives: $$ F_y = 2k\,\dfrac{q\,q_0\,y}{(a^2+y^2)^{3/2}}. $$

Substituting $$q_0=-\dfrac{q}{2}$$, we obtain

$$ F_y = 2k\,\dfrac{q\left(-\dfrac{q}{2}\right)\,y}{(a^2+y^2)^{3/2}} = -k\,\dfrac{q^{\,2}\,y}{(a^2+y^2)^{3/2}}. $$

Because the displacement is very small compared with $$a$$  ($$y\ll a$$), we can approximate

$$ (a^2+y^2)^{3/2} = a^3\left(1+\dfrac{y^2}{a^2}\right)^{3/2} \approx a^3 \quad(\text{keeping only the leading term}). $$

Substituting this approximation,

$$ F_y \approx -k\,\dfrac{q^{\,2}\,y}{a^3}. $$

The constant factor $$-k\,\dfrac{q^{\,2}}{a^3}$$ is independent of $$y$$, so

$$ F_y \propto -y. $$

Thus the net restoring force is directly proportional to $$-y$$.

Hence, the correct answer is Option C.

The electric field at a point along the axis of a uniformly charged disc is given by the formula:

$$E(x) = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right)$$

At the center of the disc, where $$x = 0$$, the electric field is:

$$E(0) = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{0}{\sqrt{0^2 + R^2}}\right) = \frac{\sigma}{2\epsilon_0} (1 - 0) = \frac{\sigma}{2\epsilon_0}$$

This matches the given value. Now, we need to find the electric field at a distance $$R$$ from the center along the axis, so we substitute $$x = R$$:

$$E(R) = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{R}{\sqrt{R^2 + R^2}}\right)$$

Simplify the denominator:

$$\sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$$

So,

$$E(R) = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{R}{R\sqrt{2}}\right) = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{1}{\sqrt{2}}\right)$$

Let $$E_c = E(0) = \frac{\sigma}{2\epsilon_0}$$ denote the field at the center. The reduction in the electric field at $$x = R$$ compared to the center is $$E_c - E(R)$$. The percentage reduction with respect to $$E_c$$ is:

$$\text{Percentage reduction} = \frac{E_c - E(R)}{E_c} \times 100\%$$

Substitute the expressions:

$$\frac{E_c - E(R)}{E_c} = \frac{\frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} \left(1 - \frac{1}{\sqrt{2}}\right)}{\frac{\sigma}{2\epsilon_0}}$$

The common factor $$\frac{\sigma}{2\epsilon_0}$$ cancels out:

$$= \frac{1 - \left(1 - \frac{1}{\sqrt{2}}\right)}{1} = \frac{1 - 1 + \frac{1}{\sqrt{2}}}{1} = \frac{1}{\sqrt{2}}$$

Therefore,

$$\text{Percentage reduction} = \frac{1}{\sqrt{2}} \times 100\%$$

Calculate the numerical value:

$$\frac{1}{\sqrt{2}} \approx 0.7071$$

$$0.7071 \times 100\% = 70.71\%$$

Rounding to one decimal place, this is approximately 70.7%.

Hence, the electric field reduces by 70.7% with respect to the field at the center.

So, the answer is Option A.

The interaction potential energy ($$U$$) between two point dipoles $$\vec{p}_1$$ and $$\vec{p}_2$$ separated by a distance $$r$$ is given by:

$$U = \frac{1}{4\pi\epsilon_0 r^3} \left[ \vec{p}_1 \cdot \vec{p}_2 - 3(\vec{p}_1 \cdot \hat{r})(\vec{p}_2 \cdot \hat{r}) \right]$$

In the axial configuration where both dipoles are collinear and point along the line joining them (i.e., $$\vec{p}_1 \parallel \vec{p}_2 \parallel \vec{r}$$):

$$\vec{p}_1 \cdot \vec{p}_2 = p_1 p_2$$, $$\vec{p}_1 \cdot \hat{r} = p_1$$, $$\vec{p}_2 \cdot \hat{r} = p_2$$

$$U = \frac{1}{4\pi\epsilon_0 x^3} \left[ p_1 p_2 - 3(p_1)(p_2) \right]$$

$$U = -\frac{2 p_1 p_2}{4\pi\epsilon_0 x^3}$$

$$F = -\frac{dU}{dx}$$

$$F = -\frac{6 p_1 p_2}{4\pi\epsilon_0 x^4}$$ (the $$-ve$$ sign indicates attraction)

Consider two identical balls, each with mass $$m$$ and charge $$q$$, suspended from a fixed support by strings of length $$l$$. At electrostatic equilibrium, the balls repel each other, causing the strings to make small angles with the vertical. We need to find how the separation $$x$$ between the balls depends on $$l$$.

For each ball, three forces act: the gravitational force $$mg$$ downward, the tension $$T$$ along the string, and the electrostatic repulsive force $$F_e$$ horizontally away from the other ball. The electrostatic force is given by Coulomb's law:

$$ F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2} $$

where $$x$$ is the separation between the balls and $$\epsilon_0$$ is the permittivity of free space.

Since the setup is symmetric and the angles are small, let $$\theta$$ be the small angle each string makes with the vertical. The separation $$x$$ can be expressed in terms of $$\theta$$. Each ball is displaced horizontally by $$l \sin\theta$$ from the vertical, so the total separation is:

$$ x = 2 l \sin\theta $$

Using the small-angle approximation, $$\sin\theta \approx \theta$$ (where $$\theta$$ is in radians), we get:

$$ x \approx 2 l \theta $$

Resolving the forces for one ball at equilibrium:

• Vertically: The vertical component of tension balances the weight.

$$ T \cos\theta = mg $$

Since $$\theta$$ is small, $$\cos\theta \approx 1$$, so:

$$ T \cdot 1 = mg \quad \Rightarrow \quad T = mg $$

• Horizontally: The horizontal component of tension balances the electrostatic force.

$$ T \sin\theta = F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2} $$

Using $$\sin\theta \approx \theta$$ and substituting $$T = mg$$:

$$ mg \theta = \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2} $$

From the separation equation $$x \approx 2 l \theta$$, solve for $$\theta$$:

$$ \theta \approx \frac{x}{2l} $$

Substitute this expression for $$\theta$$ into the horizontal force equation:

$$ mg \left( \frac{x}{2l} \right) = \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2} $$

Multiply both sides by $$2l$$ to simplify:

$$ mg x = 2l \cdot \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2} $$

Simplify the right side:

$$ mg x = \frac{2l}{4\pi\epsilon_0} \frac{q^2}{x^2} = \frac{l}{2\pi\epsilon_0} \frac{q^2}{x^2} $$

Multiply both sides by $$x^2$$:

$$ mg x \cdot x^2 = \frac{l}{2\pi\epsilon_0} q^2 $$

$$ mg x^3 = \frac{l}{2\pi\epsilon_0} q^2 $$

Solve for $$x^3$$:

$$ x^3 = \frac{l}{2\pi\epsilon_0} \frac{q^2}{mg} $$

The terms $$\frac{q^2}{2\pi\epsilon_0 mg}$$ are constant since $$q$$, $$m$$, $$g$$, and $$\epsilon_0$$ are constants. Let $$k = \frac{q^2}{2\pi\epsilon_0 mg}$$, so:

$$ x^3 = k l $$

Taking the cube root of both sides:

$$ x = (k l)^{1/3} = k^{1/3} l^{1/3} $$

Thus, $$x$$ is proportional to $$l^{1/3}$$.

Comparing with the options:

A. $$l$$

B. $$l^2$$

C. $$l^{2/3}$$

D. $$l^{1/3}$$

Hence, the correct answer is Option D.

Let distance between charges be

2r

So midpoint is at distance

r

from each charge.

Potential at midpoint due to one charge:

$$V_1=\frac{kq}{r}$$

From both charges,

$$V=\frac{2kq}{r}$$

Now find r.

For one charge in equilibrium:

Horizontal balance:

$$T\sin\theta=F_e$$

Vertical balance:

$$T\cos\theta=mg$$

Dividing,

$$\tan\theta=\frac{F_e}{mg}$$

Electrostatic repulsion between charges:

$$F_e=\frac{kq^2}{(2r)^2}=\frac{kq^2}{4r^2}$$

So

$$mg\tan\theta=\frac{kq^2}{4r^2}$$

Thus

$$r^2=\frac{kq^2}{4mg\tan\theta}$$

$$r=\frac{q}{2}\sqrt{\frac{k}{mg\tan\theta}}$$

Substitute into

$$V=\frac{2kq}{r}$$

$$V=\frac{2kq}{\frac{q}{2}\sqrt{\frac{k}{mg\tan\theta}}}$$

$$=4k\sqrt{\frac{mg\tan\theta}{k}}$$

$$=4\sqrt{kmg\tan\theta}$$

A uniform electric field $$\vec{E}$$ exists between the plates of a charged condenser. A charged particle enters this space perpendicular to $$\vec{E}$$. To determine the path, we analyze the forces and motion.

The force on a charged particle in an electric field is given by $$\vec{F} = q \vec{E}$$, where $$q$$ is the charge of the particle. Since the electric field is uniform, the force is constant. Assume the electric field $$\vec{E}$$ is directed along the y-axis, so $$\vec{E} = E \hat{j}$$. The particle enters perpendicular to $$\vec{E}$$, meaning its initial velocity $$\vec{u}$$ is along the x-axis, so $$\vec{u} = u \hat{i}$$.

The force $$\vec{F} = q E \hat{j}$$ causes an acceleration. By Newton's second law, $$\vec{F} = m \vec{a}$$, so:

$$ m \vec{a} = q E \hat{j} $$

Thus, the acceleration $$\vec{a}$$ is:

$$ \vec{a} = \frac{q E}{m} \hat{j} $$

This acceleration is constant and acts only in the y-direction. There is no acceleration in the x-direction because no force acts there. So, the components of acceleration are:

$$ a_x = 0 $$

$$ a_y = \frac{q E}{m} $$

Assume the particle starts at the origin at time $$t = 0$$, with initial position $$x(0) = 0$$, $$y(0) = 0$$, and initial velocity components $$u_x = u$$, $$u_y = 0$$.

Using the equations of motion:

For the x-direction (constant velocity since $$a_x = 0$$):

$$ v_x = u_x + a_x t = u + 0 = u $$

$$ x = u_x t + \frac{1}{2} a_x t^2 = u t + 0 = u t $$

For the y-direction (constant acceleration $$a_y = \frac{q E}{m}$$):

$$ v_y = u_y + a_y t = 0 + \frac{q E}{m} t = \frac{q E}{m} t $$

$$ y = u_y t + \frac{1}{2} a_y t^2 = 0 + \frac{1}{2} \left( \frac{q E}{m} \right) t^2 = \frac{1}{2} \frac{q E}{m} t^2 $$

We have parametric equations:

$$ x = u t \quad \text{(1)} $$

$$ y = \frac{1}{2} \frac{q E}{m} t^2 \quad \text{(2)} $$

To find the trajectory, eliminate the parameter $$t$$. From equation (1), solve for $$t$$:

$$ t = \frac{x}{u} $$

Substitute this into equation (2):

$$ y = \frac{1}{2} \frac{q E}{m} \left( \frac{x}{u} \right)^2 = \frac{1}{2} \frac{q E}{m} \frac{x^2}{u^2} $$

Simplify:

$$ y = \left( \frac{q E}{2 m u^2} \right) x^2 $$

This equation $$y = k x^2$$, where $$k = \frac{q E}{2 m u^2}$$ is a constant, represents a parabola. Therefore, the path of the particle is parabolic.

Comparing with the options:

A. straight line

B. hyperbola

C. parabola

D. circle

Hence, the correct answer is Option C.

Consider a solid sphere of radius $$R$$ that carries a uniform volume charge density $$\rho$$. The total charge is $$Q = \rho \,\frac{4}{3}\pi R^{3}$$.

Case 1: Field at an interior point  $$\bigl(r \lt R\bigr)$$
Choose a Gaussian surface—a concentric sphere of radius $$r$$. By Gauss’s law

$$\oint \mathbf{E}\cdot d\mathbf{A}= \frac{q_{\text{encl}}}{\varepsilon_0}$$

The left side gives $$E(4\pi r^{2})$$ because the field is radially outward and uniform over the Gaussian surface.
The enclosed charge is the charge of a smaller sphere of radius $$r$$:

$$q_{\text{encl}} = \rho \,\frac{4}{3}\pi r^{3}$$

Hence
$$E(4\pi r^{2}) = \frac{\rho \,\frac{4}{3}\pi r^{3}}{\varepsilon_0}$$
$$\Rightarrow\; E = \frac{\rho r}{3\varepsilon_0}$$

Using $$\rho = \dfrac{3Q}{4\pi R^{3}}$$, we can write

$$E = \frac{1}{4\pi\varepsilon_0}\,\frac{Q\,r}{R^{3}}$$

Therefore, inside the sphere $$E$$ is directly proportional to $$r$$. The graph is a straight line through the origin, reaching its maximum at the surface.

Case 2: Field at an exterior point  $$\bigl(r \ge R\bigr)$$
For a Gaussian surface of radius $$r \gt R$$, the entire charge $$Q$$ is enclosed, so

$$E(4\pi r^{2}) = \frac{Q}{\varepsilon_0} \;\;\Longrightarrow\;\; E = \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{r^{2}}$$

Outside the sphere $$E$$ falls off as $$1/r^{2}$$, exactly like the field of a point charge located at the centre.

Matching with the given graphs
Key features we have deduced:
1. From $$r=0$$ to $$r=R$$: a straight line of slope $$\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R^{3}}$$.
2. At $$r = R$$: the field value is $$E_R = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R^{2}}$$.
3. For $$r \gt R$$: the curve switches smoothly into a $$1/r^{2}$$ decay, continuous (no jump) at $$r=R$$.

Among the four sketches, only Option C shows (i) a linear rise from the origin to $$r=R$$, and (ii) a smooth inverse-square drop beyond $$R$$ that starts from the same value attained at the surface. Hence Option C is the correct representation.

Final Answer: Option C which is: the graph that rises linearly inside the sphere and then decreases as $$1/r^{2}$$ outside.

Let the hemisphere of radius $$a$$ form a closed surface by adding its flat circular base. Since no charge is enclosed, Gauss’s law gives

$$\Phi_{\text{total}} = \Phi_{\text{curved}} + \Phi_{\text{base}} = 0$$

Hence

$$\Phi_{\text{curved}} = -\,\Phi_{\text{base}} \quad -(1)$$

The base is a circle of area $$A = \pi a^{2}$$ lying in the horizontal plane. Its outward normal is vertical and directed downwards (away from the interior of the hemisphere).

The uniform electric field $$\vec E$$ makes an angle $$\frac{\pi}{4}$$ with the vertical. Therefore, the angle between $$\vec E$$ and the outward normal to the base is also $$\frac{\pi}{4}$$.

Flux through the base:

$$\Phi_{\text{base}} = \vec E \cdot \vec A = EA\cos\left(\frac{\pi}{4}\right) = E\bigl(\pi a^{2}\bigr)\,\frac{1}{\sqrt{2}} = \frac{\pi a^{2}E}{\sqrt{2}}$$

Substituting this value in equation (1):

$$\Phi_{\text{curved}} = -\left(\frac{\pi a^{2}E}{\sqrt{2}}\right)$$

The minus sign only indicates that the flux through the curved surface is opposite in sense to that through the base. Its magnitude—the quantity usually asked for—is

$$\boxed{\dfrac{\pi a^{2}E}{\sqrt{2}}}$$

Thus the required electric flux through the curved surface of the hemisphere corresponds to Option B.

Let each sphere have mass $$m$$ and (instantaneous) charge $$q$$. The two light strings of equal length $$l$$ make an angle $$\theta$$ with the vertical when the spheres are separated by a horizontal distance $$x$$. Because $$d\ll l$$, the angles are small and the system can be treated as quasistatic (the spheres readjust their positions continuously as the charge leaks).

Force balance at any instant
For one sphere, the horizontal component of the string tension balances the electrostatic repulsion while the vertical component balances the weight:
$$T\sin\theta = \frac{k\,q^{2}}{x^{2}}, \qquad T\cos\theta = mg$$
Dividing the two equations gives
$$\tan\theta = \frac{k\,q^{2}}{mg\,x^{2}} \; -(1)$$

Small-angle geometry
With both strings making the same angle, the separation is $$x = 2l\sin\theta \approx 2l\theta$$, so $$\theta \approx \dfrac{x}{2l}$$. For small angles $$\tan\theta \approx \theta$$, hence from (1)

$$\frac{x}{2l} = \frac{k\,q^{2}}{mg\,x^{2}} \;\;\Longrightarrow\;\; q^{2} = \frac{mg}{2lk}\,x^{3}$$

Therefore $$q = C\,x^{3/2}, \quad\text{where}\quad C = \sqrt{\frac{mg}{2lk}} = \text{constant}$$

Relating leak rate to approach speed
Charge leaks at a constant rate, so $$\dfrac{dq}{dt} = -\lambda$$ with $$\lambda$$ positive and constant.
Differentiate $$q = Cx^{3/2}$$ with respect to time:

$$\frac{dq}{dt} = \frac{3}{2}\,C\,x^{1/2}\,\frac{dx}{dt}$$

The spheres are approaching, so we deal with the magnitude of velocity $$v = \left|\dfrac{dx}{dt}\right|$$. Using $$\left|\dfrac{dq}{dt}\right| = \lambda$$:

$$\lambda = \frac{3}{2}\,C\,x^{1/2}\,v \;\;\Longrightarrow\;\; v = \frac{2\lambda}{3C}\,x^{-1/2}$$

Thus the speed varies inversely with the square root of the instantaneous separation:

$$v \propto x^{-1/2}$$

Hence, the correct choice is Option D.

Let the thin semi-circular ring lie in the $$x\text{-}y$$ plane with its centre at the origin $$O(0,0)$$ and radius $$r$$. The two ends of the semicircle meet the $$x$$-axis at $$(-r,0)$$ and $$(+r,0)$$, so every point on the ring has positive $$y$$-coordinate.

The total positive charge distributed uniformly over the arc is $$q$$. Linear charge density $$\lambda$$ is therefore

$$\lambda = \frac{\text{total charge}}{\text{length of arc}} = \frac{q}{\pi r}$$

Choose an element of length $$dl$$ at the polar angle $$\alpha$$ measured from the +$$x$$-axis (see figure in mind). Coordinates of the element: $$(r\cos\alpha,\, r\sin\alpha)$$ where $$0 \le \alpha \le \pi$$.

Charge of the element $$dq = \lambda\,dl = \lambda\,r\,d\alpha$$

Magnitude of the electric field produced at the centre by this element:

$$dE = \frac{1}{4\pi\varepsilon_0}\,\frac{dq}{r^{2}}$$

The field vector points radially inward (towards the centre) because the observation point is on the line joining the charge element to the centre. The unit vector along this inward radial direction is $$\hat{r}_{\!in} = (-\cos\alpha, -\sin\alpha)$$.

Hence the $$y$$-component contributed by the element is

$$dE_y = dE\,(\text{component of } \hat{r}_{\!in} \text{ along } -\hat{j})$$ $$\qquad = dE\,(-\sin\alpha)$$ $$\qquad = -\frac{1}{4\pi\varepsilon_0}\,\frac{dq}{r^{2}}\sin\alpha$$

Substituting $$dq = \lambda r d\alpha$$:

$$dE_y = -\frac{1}{4\pi\varepsilon_0}\,\frac{\lambda r\,d\alpha}{r^{2}}\sin\alpha = -\frac{\lambda}{4\pi\varepsilon_0 r}\,\sin\alpha\,d\alpha$$

Integrate over the entire semicircle (from $$\alpha = 0$$ to $$\alpha = \pi$$):

$$E_y = \int_{0}^{\pi} dE_y = -\frac{\lambda}{4\pi\varepsilon_0 r}\int_{0}^{\pi}\sin\alpha\,d\alpha$$

The integral evaluates to

$$\int_{0}^{\pi}\sin\alpha\,d\alpha = \left[-\cos\alpha\right]_{0}^{\pi} = -\cos\pi + \cos 0 = 2$$

Therefore

$$E_y = -\frac{\lambda}{4\pi\varepsilon_0 r}\,(2) = -\frac{2\lambda}{4\pi\varepsilon_0 r}$$

Insert $$\lambda = \dfrac{q}{\pi r}$$:

$$E_y = -\frac{2}{4\pi\varepsilon_0 r}\,\frac{q}{\pi r} = -\frac{q}{2\pi^{2}\varepsilon_0 r^{2}}$$

All horizontal ($$x$$) components cancel by symmetry, so the net field is purely along $$-\,\hat{j}$$ (negative $$y$$-direction).

Hence,

$$\vec{E}_{\,\text{centre}} = -\,\frac{q}{2\pi^{2}\varepsilon_0 r^{2}}\;\hat{j}$$

Option C which is: $$-\dfrac{q}{2\pi^{2}\,\varepsilon_0\,r^{2}}\,\hat{j}$$

For a spherically symmetric charge distribution the electric field at a distance $$r$$ from the centre depends only on the total charge enclosed inside a Gaussian sphere of radius $$r$$. By Gauss’s law,

$$\oint \mathbf{E}\cdot d\mathbf{S}=E(4\pi r^{2})=\frac{Q_{\text{enc}}}{\varepsilon_{0}} \quad\Longrightarrow\quad E=\frac{1}{4\pi\varepsilon_{0}}\frac{Q_{\text{enc}}}{r^{2}}$$

Hence we first calculate $$Q_{\text{enc}}$$ for $$r\lt R$$.

Charge density in the sphere is given by $$\rho(r)=\rho_{0}\left(\frac{5}{4}-\frac{r}{R}\right)$$. A thin shell of radius $$r'$$ and thickness $$dr'$$ has volume $$dV=4\pi r'^{2}\,dr'$$, so the charge in this shell is

$$dq=\rho(r')\,dV =\rho_{0}\left(\frac{5}{4}-\frac{r'}{R}\right)4\pi r'^{2}\,dr'$$

Total charge enclosed within radius $$r$$:

$$\begin{aligned} Q_{\text{enc}}&=\int_{0}^{r}4\pi\rho_{0}\left(\frac{5}{4}-\frac{r'}{R}\right)r'^{2}\,dr'\\ &=4\pi\rho_{0}\int_{0}^{r}\left(\frac{5}{4}r'^{2}-\frac{r'^{3}}{R}\right)dr' \\[4pt] &=4\pi\rho_{0}\left[\frac{5}{4}\frac{r'^{3}}{3}-\frac{r'^{4}}{4R}\right]_{0}^{r}\\[4pt] &=4\pi\rho_{0}\left(\frac{5r^{3}}{12}-\frac{r^{4}}{4R}\right) \\[2pt] &=4\pi\rho_{0}r^{3}\left(\frac{5}{12}-\frac{r}{4R}\right) \end{aligned}$$

Insert this $$Q_{\text{enc}}$$ into Gauss’s law:

$$\begin{aligned} E&=\frac{1}{4\pi\varepsilon_{0}}\, \frac{4\pi\rho_{0}r^{3}\left(\dfrac{5}{12}-\dfrac{r}{4R}\right)}{r^{2}}\\[6pt] &=\frac{\rho_{0}}{\varepsilon_{0}}\, r\left(\frac{5}{12}-\frac{r}{4R}\right) \end{aligned}$$

Simplify the bracket by taking the common denominator $$12R$$:

$$\frac{5}{12}-\frac{r}{4R}=\frac{5R-3r}{12R}$$

Thus

$$E=\frac{\rho_{0}r}{\varepsilon_{0}}\, \frac{5R-3r}{12R} =\frac{\rho_{0}r}{12\varepsilon_{0}R}\,(5R-3r)$$

Factor out $$\dfrac{1}{4}$$ to match the given options:

$$E=\frac{\rho_{0}r}{4\varepsilon_{0}}\left(\frac{5}{3}-\frac{r}{R}\right)$$

This expression corresponds to Option B.

Option B which is: $$\dfrac{\rho_0\,r}{4\varepsilon_0}\left(\dfrac{5}{3}-\dfrac{r}{R}\right)$$

Let $$ \theta $$ be the half-angle, so:

$$ \theta = 15^\circ $$

Let $$ q $$ be the charge on each sphere and $$ r $$ be the separation distance between them. These are constant in both cases because the angle does not change. Let $$ V $$ be the volume of each sphere and $$ g $$ be the acceleration due to gravity.

Case 1: Equilibrium in Air

In equilibrium in air, the forces on each sphere are its weight ($$ W $$), the electrostatic repulsive force ($$ F_e $$), and tension ($$ T $$). The weight is given by:

$$ W = \rho_s V g $$

The equilibrium conditions provide:

$$ \tan \theta = \frac{F_e}{W} $$

$$ \tan \theta = \frac{F_e}{\rho_s V g} \quad \text{---(Equation 1)} $$

Case 2: Equilibrium in Liquid

When suspended in a liquid with dielectric constant $$ K $$, the electrostatic repulsive force becomes:

$$ F'_e = \frac{F_e}{K} $$

The effective downward force (effective weight, $$ W_{\text{eff}} $$) is the weight minus the buoyant force (upthrust, $$ B $$):

$$ W_{\text{eff}} = W - B $$

The buoyant force is equal to the weight of the liquid displaced:

$$ B = \rho_l V g $$

So, effective weight is:

$$ W_{\text{eff}} = \rho_s V g - \rho_l V g = (\rho_s - \rho_l) V g $$

Since the angle is unchanged, the equilibrium condition is similar, but with effective forces:

$$ \tan \theta = \frac{F'_e}{W_{\text{eff}}} $$

$$ \tan \theta = \frac{F_e / K}{(\rho_s - \rho_l) V g} $$

$$ \tan \theta = \frac{F_e}{K (\rho_s - \rho_l) V g} \quad \text{---(Equation 2)} $$

Finding the Dielectric Constant (K)

Equating Equation 1 and Equation 2 because $$ \tan \theta $$ is constant:

$$ \frac{F_e}{\rho_s V g} = \frac{F_e}{K (\rho_s - \rho_l) V g} $$

We can cancel $$ F_e $$, $$ V $$, and $$ g $$ from both sides:

$$ \frac{1}{\rho_s} = \frac{1}{K (\rho_s - \rho_l)} $$

Rearranging this to solve for $$ K $$:

$$ K = \frac{\rho_s}{\rho_s - \rho_l} $$

Substitute the given density values into the formula:

$$ K = \frac{1.6}{1.6 - 0.8} $$

$$ K = \frac{1.6}{0.8} $$

$$ K = 2 $$

The charge density varies with the distance from the centre as
$$\rho(r)=\frac{Q}{\pi R^{4}}\;r \qquad (0\le r\le R)$$

Step 1: Verify that the given $$\rho(r)$$ produces total charge $$Q$$.
Total charge inside the full sphere is
$$Q_{\text{total}}=\int_{0}^{R}\rho(r)\,dV =\int_{0}^{R}\rho(r)\,4\pi r^{2}\,dr$$
$$=4\pi\int_{0}^{R}\frac{Q}{\pi R^{4}}\,r\,(r^{2})\,dr =\frac{4Q}{R^{4}}\int_{0}^{R}r^{3}\,dr =\frac{4Q}{R^{4}}\Bigl[\frac{r^{4}}{4}\Bigr]_{0}^{R}=Q.$$
Thus the distribution is consistent.

Step 2: Charge enclosed by a Gaussian sphere of radius $$r_{1}\;(r_{1}\lt R)$$.
$$q_{\text{enc}}=\int_{0}^{r_{1}}\rho(r)\,4\pi r^{2}\,dr =4\pi\int_{0}^{r_{1}}\frac{Q}{\pi R^{4}}\,r^{3}\,dr$$
$$=\frac{4Q}{R^{4}}\Bigl[\frac{r^{4}}{4}\Bigr]_{0}^{r_{1}} =\frac{Q\,r_{1}^{4}}{R^{4}}.$$

Step 3: Apply Gauss’s law to find the electric field magnitude at $$r_{1}$$.
For a spherical Gaussian surface,
$$E(4\pi r_{1}^{2})=\frac{q_{\text{enc}}}{\varepsilon_{0}} \;\;\Longrightarrow\;\; E=\frac{q_{\text{enc}}}{4\pi\varepsilon_{0}r_{1}^{2}}$$
Insert the value of $$q_{\text{enc}}$$:
$$E=\frac{1}{4\pi\varepsilon_{0}r_{1}^{2}}\;\frac{Q\,r_{1}^{4}}{R^{4}} =\frac{Q\,r_{1}^{2}}{4\pi\varepsilon_{0}R^{4}}.$$

Hence, the magnitude of the electric field at distance $$r_{1}$$ inside the sphere is
$$\boxed{\dfrac{Q\,r_{1}^{2}}{4\pi\varepsilon_{0}R^{4}}}.$$

Option C which is: $$\frac{Qr_{1}^{2}}{4\pi\varepsilon_{0}R^{4}}$$

Consider a square of side $$a$$. Label the corners in order as $$A(0,0),\;B(a,0),\;C(a,a),\;D(0,a)$$.

Charges placed:
  • $$Q$$ at the opposite corners $$A$$ and $$C$$.
  • $$q$$ at the remaining corners $$B$$ and $$D$$.

We need the net electrostatic force on the charge $$Q$$ at corner $$A$$ to be zero.

Forces on $$Q$$ at $$A$$

1. Force due to the other $$Q$$ at $$C$$ (diagonal corner)
  Distance $$AC = a\sqrt{2}$$.
  Magnitude $$F_{AC} = k\dfrac{Q^2}{(a\sqrt{2})^{\,2}} = \dfrac{kQ^2}{2a^{2}}$$.
  Direction: from $$A$$ towards $$C$$ makes $$45^{\circ}$$ with both axes; because both charges $$Q$$ are alike, the force on $$A$$ is repulsive, i.e. along $$-\hat i-\hat j$$.
  Components: $$F_{AC,x} = F_{AC,y} = -\dfrac{kQ^2}{2a^{2}}\dfrac{1}{\sqrt{2}} = -\dfrac{kQ^2}{2\sqrt{2}\,a^{2}}$$.

2. Force due to $$q$$ at $$B(a,0)$$ (adjacent corner on the x-axis)
  Distance $$AB = a$$.
  Magnitude $$F_{AB} = k\dfrac{|Qq|}{a^{2}}$$.
  Direction: along the x-axis. Writing the x-component with sign included,
$$F_{AB,x}= -\dfrac{kQq}{a^{2}}$$ (positive if $$q$$ is opposite in sign to $$Q$$, negative if same sign).

3. Force due to $$q$$ at $$D(0,a)$$ (adjacent corner on the y-axis) is analogous:
  $$F_{AD,y}= -\dfrac{kQq}{a^{2}}$$.

Setting the net force to zero

Total x-component:
$$F_x = F_{AB,x} + F_{AC,x}= -\dfrac{kQq}{a^{2}}-\dfrac{kQ^2}{2\sqrt{2}\,a^{2}}=0$$.

Total y-component gives the same equation, so one condition suffices:

$$-\dfrac{kQq}{a^{2}}-\dfrac{kQ^2}{2\sqrt{2}\,a^{2}}=0$$

Cancel the common factor $$\dfrac{k}{a^{2}}$$:

$$-Qq-\dfrac{Q^{2}}{2\sqrt{2}}=0$$

$$Qq = -\dfrac{Q^{2}}{2\sqrt{2}}$$

Divide by $$Q\;(Q\neq0)$$:

$$q = -\dfrac{Q}{2\sqrt{2}}$$

Hence

$$\dfrac{Q}{q}= -2\sqrt{2}$$

which matches Option A.

Therefore, the required ratio is:
Option A which is: $$-2\sqrt{2}$$

For a thin spherical shell the total charge $$Q$$ resides on the surface. We have to find the electric field $$E(r)$$ at a distance $$r$$ from the centre for two regions :

Case 1: $$0 \le r \lt R$$ (inside the shell)
By Gauss’s law, choose a concentric Gaussian sphere of radius $$r\lt R$$. The enclosed charge is zero, therefore

$$\oint \mathbf{E}\cdot d\mathbf{S}=E(4\pi r^{2})=0\; \Longrightarrow\; E=0$$

Thus the field is identically zero for every interior point. On the graph the line from $$r=0$$ up to but not including $$r=R$$ lies on the horizontal axis.

Case 2: $$r \gt R$$ (outside the shell)
A concentric Gaussian sphere of radius $$r$$ now encloses the full charge $$Q$$. Applying Gauss’s law again,

$$E(4\pi r^{2})=\frac{Q}{\varepsilon_{0}}\;\Longrightarrow\;E(r)= \frac{1}{4\pi\varepsilon_{0}}\;\frac{Q}{r^{2}}$$

This is the same inverse-square variation as for a point charge located at the centre. Hence, for $$r\gt R$$ the graph is a rectangular hyperbola starting from $$r=R$$ and tending to zero as $$r\rightarrow\infty$$.

Value at the surface: Just inside the surface, $$E_{\text{in}}=0$$. Just outside, $$E_{\text{out}}=\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{Q}{R^{2}}$$. Because of this sudden jump, the graph has an open circle at $$r=R$$ on the zero line and a filled dot at $$r=R$$ on the value $$\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{Q}{R^{2}}$$, then follows the $$1/r^{2}$$ curve.

The only option that shows
• $$E=0$$ for $$0\le r\lt R$$,
• a discontinuous rise at $$r=R$$, and
• an inverse-square decay for $$r\gt R$$
is Option A.

Option A which is: the curve with a flat zero segment inside the shell, a jump at $$r=R$$, and a $$1/r^{2}$$ fall thereafter.

Let the centre of the square be $$O$$ and the side of the square be $$a$$. Choose Cartesian axes through $$O$$ so that the four vertices have co-ordinates

$$A\,(+a/2,+a/2),\;B\,(-a/2,+a/2),\;C\,(-a/2,-a/2),\;D\,(+a/2,-a/2).$$

The distance of every vertex from the centre is the same,

$$r = \frac{a}{\sqrt{2}}.$$ Hence all four charges are equidistant from $$O$$.

Denote the initial charges by $$q_A,\;q_B,\;q_C,\;q_D$$ placed at $$A,B,C,D$$ respectively.

Potential at the centre (before interchange)
Using the superposition principle, the electric potential at a point is the algebraic sum of the potentials due to individual charges because potential is a scalar. Therefore

$$V_{\text{old}} \;=\; k\,\frac{q_A + q_B + q_C + q_D}{r},$$ where $$k = \tfrac{1}{4\pi\varepsilon_0}.$$

Electric field at the centre (before interchange)
For any vertex the magnitude of the field at $$O$$ is

$$E_i = k\,\frac{|q_i|}{r^2} = k\,\frac{2\,q_i}{a^{2}},$$

and its direction is along the line joining that vertex to the centre (outward for a positive charge, inward for a negative one). Writing the four position vectors from $$O$$ to the vertices,

$$\vec{r}_A = \tfrac{a}{2}\,(\hat{i}+\hat{j}),\; \vec{r}_B = \tfrac{a}{2}\,(-\hat{i}+\hat{j}),\; \vec{r}_C = \tfrac{a}{2}\,(-\hat{i}-\hat{j}),\; \vec{r}_D = \tfrac{a}{2}\,(\hat{i}-\hat{j}),$$

the electric field at $$O$$ is

$$\vec{E}_{\text{old}} \;=\; k\,\frac{2}{a^{2}}\, \Bigl(q_A\,\hat{r}_A + q_B\,\hat{r}_B + q_C\,\hat{r}_C + q_D\,\hat{r}_D\Bigr),$$ where each $$\hat{r}_i$$ is the unit vector along $$\vec{r}_i$$.

Interchange of charges
The question says: “charges on $$A$$ and $$B$$ are interchanged with those on $$D$$ and $$C$$ respectively.” In symbols this means

$$q_A \longleftrightarrow q_D,\qquad q_B \longleftrightarrow q_C.$$

After interchange, the charges occupying the vertices become

$$A:D,\; B:C,\; C:B,\; D:A.$$

Potential after interchange
Because the distance $$r$$ is unchanged and the potential is a scalar sum, the new potential is

$$V_{\text{new}} = k\,\frac{q_D + q_C + q_B + q_A}{r} = k\,\frac{q_A + q_B + q_C + q_D}{r} = V_{\text{old}}.$$

Hence the electric potential at the centre does not change.

Electric field after interchange
The magnitudes are the same as before, but the charges now multiply different direction vectors:

$$\vec{E}_{\text{new}} = k\,\frac{2}{a^{2}}\, \Bigl(q_D\,\hat{r}_A + q_C\,\hat{r}_B + q_B\,\hat{r}_C + q_A\,\hat{r}_D\Bigr).$$

In general $$\vec{E}_{\text{new}}\neq\vec{E}_{\text{old}}$$ unless the pairs $$q_A=q_D$$ and $$q_B=q_C$$ already happened to be equal. Therefore the electric field changes.

Conclusion
The interchange leaves the potential unchanged but alters the electric field.

Option D which is: $$\vec{E}$$ changes, $$V$$ remains unchanged.

In Millikan’s oil-drop experiment the charge $$q$$ on a floating drop is obtained by balancing the gravitational force on the drop with the electric force exerted by a uniform electric field.

Gravitational force on the drop: $$F_g = mg$$ where $$m$$ is the mass of the drop and $$g$$ is the local acceleration due to gravity.

Electric force on the drop: $$F_E = qE$$ where $$E$$ is the applied uniform electric field.

At equilibrium, $$F_g = F_E$$, so

$$qE = mg \quad\Rightarrow\quad q = \frac{mg}{E} \tag{-1}$$

When the experiment is shifted from Earth to Moon:

• The charge $$q$$ obtained from $$(1)$$ might seem to depend on the local $$g$$, but the mass $$m$$ of the drop is never measured through its weight. Instead, $$m$$ is calculated from the radius $$r$$, the density $$\rho$$ of the oil, and the buoyant correction: $$m = \frac{4}{3}\pi r^{3}\rho$$ - a purely mechanical expression that involves no factor of $$g$$.

Because the same oil, the same radius-measuring method, and the same density are used on both Earth and Moon, the numerator $$mg$$ in $$(1)$$ changes to

$$m g_m \quad\text{on Moon}, \qquad m g_E \quad\text{on Earth}.$$

The applied electric field $$E$$ is adjusted separately in each location until equilibrium is reached, so the ratio of charges measured becomes

$$\frac{q_{\text{Moon}}}{q_{\text{Earth}}} \;=\; \frac{m g_m / E_{\text{Moon}}}{m g_E / E_{\text{Earth}}} = \frac{g_m}{g_E}\,\frac{E_{\text{Earth}}}{E_{\text{Moon}}}.$$

But in either case the experimenter tunes the field exactly so that $$q$$ comes out to an integer multiple of the fundamental electronic charge $$e$$. Since $$e$$ is a universal constant, the same integer multiple must be obtained at both places. Therefore the ratio above must equal unity:

$$\frac{q_{\text{Moon}}}{q_{\text{Earth}}} = 1.$$

Hence the electronic charge measured on the Moon is the same as that measured on the Earth.

Option A which is: 1

When an electric dipole is placed in a non-uniform electric field, the electric field strength varies from point to point. Consequently, the two charges of the dipole experience different magnitudes of electric force.

Let the dipole consist of charges $$+q$$ and $$-q$$.

Let the electric field at the position of $$+q$$ be $$\vec{E}_1$$ and at the position of $$-q$$ be $$\vec{E}_2$$.

Because the field is non-uniform, $$\vec{E}_1 \neq \vec{E}_2$$.

The force on the positive charge is $$\vec{F}_1 = q\vec{E}_1$$.

The force on the negative charge is $$\vec{F}_2 = -q\vec{E}_2$$.

The net force acting on the dipole is the vector sum of these forces:

$$\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 = q\vec{E}_1 - q\vec{E}_2 = q(\vec{E}_1 - \vec{E}_2)$$

Since $$\vec{E}_1 \neq \vec{E}_2$$, it follows that $$\vec{F}_{\text{net}} \neq 0$$. Thus, the dipole will experience a {net translational force}.

Additionally, the dipole is placed at an angle of $$30^\circ$$ with respect to the electric field. Because this angle is neither $$0^\circ$$ nor $$180^\circ$$, the lines of action of the forces $$\vec{F}_1$$ and $$\vec{F}_2$$ do not coincide.

These non-collinear forces produce a turning effect. Therefore, the dipole will also experience a net torque ($$\vec{\tau}_{\text{net}} \neq 0$$).

The dipole will experience both a net force and a net torque (option D)

When the two conducting spheres are connected by a wire, charge will flow between them until they reach electrostatic equilibrium. At equilibrium, both spheres will be at the same electrical potential.

Let this common potential be $$V$$.

The potential at the surface of a spherical conductor is given by the formula:

$$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$$

Since the potentials are equal ($$V_A = V_B$$), we can write:

$$\frac{1}{4\pi\epsilon_0} \frac{Q_A}{R_A} = \frac{1}{4\pi\epsilon_0} \frac{Q_B}{R_B}$$

Canceling out the constants gives the ratio of their charges:

$$\frac{Q_A}{R_A} = \frac{Q_B}{R_B}$$

$$\Rightarrow \frac{Q_A}{Q_B} = \frac{R_A}{R_B}$$

The magnitude of the electric field at the surface of a spherical conductor is given by:

$$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}$$

We need the ratio of the electric fields, $$\frac{E_A}{E_B}$$:

$$\frac{E_A}{E_B} = \frac{\frac{1}{4\pi\epsilon_0} \frac{Q_A}{R_A^2}}{\frac{1}{4\pi\epsilon_0} \frac{Q_B}{R_B^2}}$$

$$\Rightarrow \frac{E_A}{E_B} = \left(\frac{Q_A}{Q_B}\right) \left(\frac{R_B}{R_A}\right)^2$$

Substituting our previous finding ($$\frac{Q_A}{Q_B} = \frac{R_A}{R_B}$$) into this equation:

$$\frac{E_A}{E_B} = \left(\frac{R_A}{R_B}\right) \left(\frac{R_B}{R_A}\right)^2$$

$$\Rightarrow \frac{E_A}{E_B} = \frac{R_B}{R_A}$$

We are given the radii of the two spheres:

$$R_A = 1\text{ mm}$$

$$R_B = 2\text{ mm}$$

Substituting these values into the ratio:

$$\frac{E_A}{E_B} = \frac{2}{1}$$

Let the position of the point on the $$x$$-axis where the net electric field is zero be $$x$$.

The magnitude of the electric field due to a point charge is given by the formula:

$$E = \frac{k|q|}{r^2}$$

Since the two charges, $$+8q$$ and $$-2q$$, have opposite signs, the point where their electric fields cancel out must lie outside the region between them. To balance the fields, the point must also be closer to the charge with the smaller magnitude, which is $$-2q$$ located at $$x=L$$. Therefore, the null point must lie at $$x > L$$.

At this null point, the magnitude of the electric field due to the $$+8q$$ charge ($$E_1$$) must equal the magnitude of the electric field due to the $$-2q$$ charge ($$E_2$$):

$$E_1 = E_2$$

$$\frac{k(8q)}{x^2} = \frac{k(2q)}{(x - L)^2}$$

Dividing both sides by $$2kq$$ to simplify:

$$\frac{4}{x^2} = \frac{1}{(x - L)^2}$$

Taking the square root of both sides (since both $$x$$ and $$x-L$$ are positive distances):

$$\frac{2}{x} = \frac{1}{x - L}$$

Cross-multiplying to solve for $$x$$:

$$2(x - L) = x$$

$$2x - 2L = x$$

$$x = 2L$$

Therefore, the location on the $$x$$-axis at which the net electric field is zero is $$x = 2L$$.

Let the magnitudes of charge on the two coaxial rings be $$Q$$ (one ring carries $$+Q$$, the other $$-Q$$). Both rings have the same radius $$R$$ and their centres are separated by a distance $$d$$ along the common axis.

Step 1: Electric potential on the axis of a uniformly charged ring
For a ring of total charge $$q$$ and radius $$R$$, the electric potential at a point on its axis at a distance $$x$$ from the centre is
$$V(x)=\frac{1}{4\pi\varepsilon_0}\,\frac{q}{\sqrt{R^{2}+x^{2}}}$$

Step 2: Potential at the centre of the +Q ring (point P)
(i) Due to its own charge $$+Q$$ (here $$x=0$$):
$$V_{P}^{(self)}=\frac{1}{4\pi\varepsilon_0}\,\frac{+Q}{R}$$
(ii) Due to the other ring (charge $$-Q$$, centre a distance $$d$$ away, so $$x=d$$):
$$V_{P}^{(other)}=\frac{1}{4\pi\varepsilon_0}\,\frac{-Q}{\sqrt{R^{2}+d^{2}}}$$

Hence the total potential at P is
$$V_P=\frac{1}{4\pi\varepsilon_0}\left[\frac{Q}{R}-\frac{Q}{\sqrt{R^{2}+d^{2}}}\right]$$

Step 3: Potential at the centre of the −Q ring (point Q)
(i) Due to its own charge $$-Q$$:
$$V_{Q}^{(self)}=\frac{1}{4\pi\varepsilon_0}\,\frac{-Q}{R}$$
(ii) Due to the other ring (charge $$+Q$$, now also at distance $$d$$):
$$V_{Q}^{(other)}=\frac{1}{4\pi\varepsilon_0}\,\frac{+Q}{\sqrt{R^{2}+d^{2}}}$$

Total potential at Q is
$$V_Q=\frac{1}{4\pi\varepsilon_0}\left[-\frac{Q}{R}+\frac{Q}{\sqrt{R^{2}+d^{2}}}\right]$$

Step 4: Required potential difference
The problem asks for the potential difference between the centres of the two rings, i.e. $$V_P-V_Q$$:
$$\begin{aligned} V_P-V_Q&=\frac{1}{4\pi\varepsilon_0}\Bigl[\frac{Q}{R}-\frac{Q}{\sqrt{R^{2}+d^{2}}}+\frac{Q}{R}-\frac{Q}{\sqrt{R^{2}+d^{2}}}\Bigr]\\[4pt] &=\frac{1}{4\pi\varepsilon_0}\,2Q\left[\frac{1}{R}-\frac{1}{\sqrt{R^{2}+d^{2}}}\right]\\[4pt] &=\frac{Q}{2\pi\varepsilon_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^{2}+d^{2}}}\right] \end{aligned}$$

Step 5: Matching with the options
The expression obtained matches exactly with Option B.

Hence, the potential difference between the centres of the two rings is:
$$\boxed{\displaystyle \frac{Q}{2\pi\varepsilon_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^{2}+d^{2}}}\right]}$$
Option B.

For the oil drop to remain suspended (neither falling nor rising), the net force acting on it must be zero. This means the upward electric force must perfectly balance the downward gravitational force.

The condition for equilibrium is:

$$F_e = F_g$$

Where:

 $$F_e$$ is the electric force, given by $$F_e = qE$$

$$F_g$$ is the gravitational force (weight), given by $$F_g = mg$$

Equating the two forces:

$$qE = mg$$

Rearranging the formula to solve for the charge $$q$$:

$$q = \frac{mg}{E}$$

Given values:

 $$m = 9.9 \times 10^{-15}\text{ kg}$$

 Acceleration due to gravity, $$g = 10\text{ m/s}^2$$

 Electric field, $$E = 3 \times 10^4\text{ V/m}$$

Substitute the given values into the equation:

$$q = \frac{(9.9 \times 10^{-15}) \times 10}{3 \times 10^4}$$

$$q = \frac{99 \times 10^{-15}}{3 \times 10^4}$$

$$q = 33 \times 10^{-19}\text{ C}$$

Expressing the final answer in standard scientific notation:

$$q = 3.3 \times 10^{-18}\text{ C}$$

Let the inner and outer conducting cylinders be coaxial with the common $$z$$-axis.
Radius of inner cylinder  : $$\sqrt{2}\,R$$
Radius of outer cylinder  : $$2\,R$$
Free charge on inner cylinder  : $$Q$$ (uniformly distributed along its length $$\ell$$)
Surface charge density per unit length (line charge)  : $$\lambda=\dfrac{Q}{\ell}$$.

The space between the two cylinders is filled with a dielectric of relative permittivity $$\kappa=5$$.
The outer cylinder is earthed, hence the electric field exists only in the dielectric region $$\sqrt{2}\,R \le r \le 2R$$ and is purely radial.

Electric field inside the dielectric
Take a cylindrical Gaussian surface of radius $$r$$ (with $$\sqrt{2}\,R \le r \le 2R$$) and length $$\ell$$. Using Gauss’s law in dielectrics,

$$\oint \mathbf{D}\cdot d\mathbf{A}=Q_{\text{free enclosed}}\; \Longrightarrow \; D(2\pi r\ell)=\lambda\ell \Rightarrow D=\dfrac{\lambda}{2\pi r}.$$ Since $$\mathbf{D}=\kappa\varepsilon_0\mathbf{E}$$, we get

$$E(r)=\dfrac{\lambda}{2\pi\kappa\varepsilon_0\,r}, \qquad \mathbf{E}=E(r)\,\hat{\mathbf{r}}.$$ (Inside $$r\lt \sqrt{2}\,R$$ the electric field is zero.)

The required open surface
We need the electric flux through a plane that

• is parallel to the axis, of length $$\ell$$,
• is at a perpendicular distance $$R$$ from the axis (take it as the plane $$x=R$$).

In the cross-section (the $$xy$$-plane) this surface appears as the straight line $$x=R$$. Only the part of this line lying in the dielectric region ($$r\ge\sqrt{2}\,R$$) contributes to flux.

Limits of the contributing strip on the plane
Put $$x=R$$ and write $$r=\sqrt{x^{2}+y^{2}}=\sqrt{R^{2}+y^{2}}.$$

• Inner boundary: $$r=\sqrt{2}\,R \;\Rightarrow\; y=\pm R$$
• Outer boundary: $$r=2\,R \;\Rightarrow\; y=\pm\sqrt{3}\,R$$

Element of area and flux through it
For a strip of width $$dy$$ on this plane, area element $$dA=\ell\,dy.$$ The unit normal to the plane is $$\hat{\mathbf{n}}=\hat{\mathbf{x}}$$ (along $$+x$$), while the electric field is radial, so the cosine of the angle between $$\mathbf{E}$$ and $$\hat{\mathbf{n}}$$ equals $$\dfrac{x}{r}=\dfrac{R}{r}.$$ Therefore, the differential flux is

$$d\Phi=E(r)\left(\dfrac{R}{r}\right)dA =\dfrac{\lambda}{2\pi\kappa\varepsilon_0\,r}\left(\dfrac{R}{r}\right)\ell\,dy =\dfrac{\lambda R\ell}{2\pi\kappa\varepsilon_0}\dfrac{dy}{(R^{2}+y^{2})}.$$

Total flux through the plane

$$\Phi=\int_{-\sqrt{3}R}^{-R}+\int_{R}^{\sqrt{3}R} \dfrac{\lambda R\ell}{2\pi\kappa\varepsilon_0}\; \dfrac{dy}{R^{2}+y^{2}} =\dfrac{\lambda R\ell}{\pi\kappa\varepsilon_0} \int_{R}^{\sqrt{3}R}\dfrac{dy}{R^{2}+y^{2}} \quad (\text{using symmetry}).$$

Evaluate the integral:

$$\int_{R}^{\sqrt{3}R}\dfrac{dy}{R^{2}+y^{2}} =\left[\dfrac{1}{R}\arctan\left(\dfrac{y}{R}\right)\right]_{R}^{\sqrt{3}R} =\dfrac{1}{R}\left[\arctan(\sqrt{3})-\arctan(1)\right] =\dfrac{1}{R}\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right) =\dfrac{\pi}{12R}.$$

Substituting,

$$\Phi=\dfrac{\lambda R\ell}{\pi\kappa\varepsilon_0}\cdot\dfrac{\pi}{12R} =\dfrac{\lambda\ell}{12\kappa\varepsilon_0} =\dfrac{Q}{12\kappa\varepsilon_0}.$$

With $$\kappa=5$$,

$$\Phi=\dfrac{Q}{12\times5\;\varepsilon_0} =\dfrac{Q}{60\,\varepsilon_0}.$$

Therefore, the electric flux through the given plane is $$\dfrac{Q}{60\varepsilon_0}$$.

Option C which is: $$\displaystyle\frac{Q}{60\varepsilon_0}$$

Magnitude of each charge of the dipole: $$q = 2\,\mu C = 2 \times 10^{-6}\,C$$.

Separation of the charges (length of dipole): $$d = 0.5\,\mu m = 0.5 \times 10^{-6}\,m$$.

Electric dipole moment is defined as $$p = q\,d$$ (directed from -ve to +ve charge).

Substituting the values,
$$p = (2 \times 10^{-6}) \times (0.5 \times 10^{-6})$$
$$p = 1 \times 10^{-12}\,C\,m$$.

The electric field E between parallel-plate capacitor plates is uniform and given by $$E = \frac{V}{\ell}$$, where V is the applied potential difference and $$\ell$$ is plate separation.

Given potential difference: $$V = 5\,V$$.
Plate separation: $$\ell = 0.5\,mm = 0.5 \times 10^{-3}\,m$$.

Therefore,
$$E = \frac{5}{0.5 \times 10^{-3}} = \frac{5}{5 \times 10^{-4}} = 1 \times 10^{4}\,V\,m^{-1}$$.

Torque on a dipole in a uniform field is $$\tau = p\,E\,\sin\theta$$, where $$\theta$$ is the angle between $$\vec p$$ and $$\vec E$$ after rotation.

The dipole is rotated through $$30^{\circ}$$, so $$\theta = 30^{\circ}$$ and $$\sin30^{\circ} = 0.5$$.

Hence,
$$\tau = (1 \times 10^{-12})(1 \times 10^{4})(0.5)$$
$$\tau = 0.5 \times 10^{-8}\,N\,m$$
$$\tau = 5 \times 10^{-9}\,N\,m$$.

Therefore the torque experienced by the dipole is $$5 \times 10^{-9}\,N\,m$$.

Option A is correct.

Electric field due to infinite non-conducting sheet is

$$E=\frac{\sigma}{2\varepsilon_0}$$

Given

$$\sigma=100\mu C/m^2=10^{-4}C/m^2$$

So

$$E=\frac{10^{-4}}{2(8.85\times10^{-12})}$$

$$=5.65\times10^6\text{ N/C}$$

Force on each charge:

F=qEF

with

$$q=2\times10^{-6}C$$

So

$$F=(2\times10^{-6})(5.65\times10^6)$$

=11.3N

The two equal and opposite charges form an electric dipole.

Dipole moment

p=qd

where

d=20cm=0.2m

So

$$p=(2\times10^{-6})(0.2)=4\times10^{-7}$$

Torque on dipole in electric field:

$$\tau=pE\sin\theta$$

with

$$\theta=30^{\circ}$$

Thus

$$\tau=(4\times10^{-7})(5.65\times10^6)\cdot\frac{1}{2}​$$

$$=1.13Nm$$

We need to find the work done to rotate an electric dipole from parallel to antiparallel to the electric field.

The potential energy of a dipole with moment $$\vec{p}$$ in a uniform electric field $$\vec{E}$$ is $$U = -\,pE\cos\theta$$, where $$\theta$$ is the angle between $$\vec{p}$$ and $$\vec{E}$$.

The work done by an external agent to rotate the dipole from angle $$\theta_i$$ to $$\theta_f$$ is $$W = U_f - U_i = \bigl(-pE\cos\theta_f\bigr) - \bigl(-pE\cos\theta_i\bigr) = pE\bigl(\cos\theta_i - \cos\theta_f\bigr)\,.$$

For the given situation, initially parallel: $$\theta_i = 0^\circ$$ so $$\cos 0^\circ = 1$$. Finally antiparallel: $$\theta_f = 180^\circ$$ so $$\cos 180^\circ = -1$$. Therefore $$W = pE\bigl(\cos 0^\circ - \cos 180^\circ\bigr) = pE\bigl(1 - (-1)\bigr) = 2pE\,. $$

Substituting $$p = 6\times 10^{-6}\text{ C·m}$$ and $$E = 10^6\text{ V/m}$$ gives $$W = 2 \times 6\times 10^{-6} \times 10^6 = 12\text{ J}\,. $$

The answer is 12 J.

Let the fixed bead be at point $$A$$ on the hoop, whose centre is $$O$$. Measure the angular position of the movable bead by angle $$\theta$$ made by $$OA$$ and the radius $$OP$$ that joins the centre to the movable bead. Equilibrium occurs when the tangential component of the electrostatic force on the movable bead is zero.

Because the only force is the repulsion from the fixed bead, the force line is along the straight chord joining the two beads. At $$\theta = \pi$$ (diametrically opposite to the fixed bead) this chord passes through the centre, so its direction is purely radial and has no tangential component. Hence the equilibrium position is $$\theta_0 = \pi$$.

Introduce a small angular displacement $$\phi$$ about this equilibrium: $$\theta = \pi + \phi \qquad(|\phi|\ll 1).$$ We shall find the potential energy as a function of $$\phi$$ upto the term in $$\phi^2$$, and compare it with the quadratic form $$\tfrac12 I\omega^2\phi^2$$ of simple harmonic motion.

Magnitude of separation between the beads (straight line, not arc): The angle subtended at the centre between the two position vectors is $$\delta=\theta = \pi+\phi$$, hence $$r = 2R\sin\frac{\delta}{2}=2R\sin\!\left(\frac{\pi+\phi}{2}\right) = 2R\cos\frac{\phi}{2}.$$

Electrostatic potential energy of the two-charge system: $$U(\phi)=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{r} =\frac{1}{4\pi\varepsilon_0}\frac{q^2}{2R\cos(\phi/2)}.$$

Expand for small $$\phi$$ using $$\cos(\phi/2)\approx 1-\frac{\phi^2}{8}$$ and for any small $$x$$, $$\frac1{1-x}\approx 1+x$$: $$\frac{1}{\cos(\phi/2)}\approx 1+\frac{\phi^2}{8}.$$ Therefore $$U(\phi)\approx\frac{1}{4\pi\varepsilon_0}\,\frac{q^2}{2R}\Bigl(1+\frac{\phi^2}{8}\Bigr) =U_0+\frac{q^2}{4\pi\varepsilon_0}\,\frac{\phi^2}{16R},$$ where $$U_0=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{2R}$$ is the minimum (constant) part.

The change in potential energy near equilibrium is thus $$\Delta U=\frac{q^2}{4\pi\varepsilon_0}\frac{\phi^2}{16R}.$$

The movable bead of mass $$m$$ moves along a circle of radius $$R$$, so its moment of inertia about the centre is $$I=mR^2.$$ For small angular oscillations, $$\Delta U=\tfrac12 I\omega^2\phi^2$$. Equating the coefficients of $$\phi^2$$:

$$\frac{1}{2}mR^2\omega^2=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{16R}$$ $$\Rightarrow\;\omega^2=\frac{1}{4\pi\varepsilon_0}\,\frac{q^2}{8mR^3} =\frac{q^2}{32\pi\varepsilon_0 R^3 m}.$$

Hence the square of the angular frequency is $$\boxed{\dfrac{q^2}{32\pi\varepsilon_0 R^3 m}}.$$

Option B which is: $$\dfrac{q^2}{32\pi\varepsilon_0 R^3 m}$$

The radius of the ring is
$$R = 30\text{ cm}=0.30\text{ m}.$$

The charge uniformly spread over the ring is
$$Q = 2\pi\text{ C},$$
and the point charge kept at the centre is
$$q = 30\text{ pC}=30\times10^{-12}\text{ C}.$$

Choose a small element of the ring that subtends an angle $$d\theta$$ at the centre.
Length of the element
$$dl = R\,d\theta,$$
charge on the element
$$dq = \frac{Q}{2\pi R}\,dl \;=\; \frac{Q}{2\pi}\,d\theta.$$

The electrostatic force on this element due to the central charge $$q$$ is radially outward and equals
$$dF = \frac{1}{4\pi\varepsilon_0}\,\frac{q\,dq}{R^2} = k\,\frac{q}{R^2}\,\frac{Q}{2\pi}\,d\theta,$$
where $$k=\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\text{N\,m}^2\text{/C}^2.$$

The element is held in equilibrium by the tension $$T$$ in the ring. The two tension forces at its ends are tangential; their radial components add up to an inward force $$2T\sin\!\left(\frac{d\theta}{2}\right)\;\approx\;T\,d\theta$$ (because $$d\theta$$ is infinitesimal).

Equating the outward electrostatic force and the inward force due to tension:
$$T\,d\theta = k\,\frac{qQ}{2\pi R^{2}}\,d\theta \;\;\Longrightarrow\;\; T = k\,\frac{qQ}{2\pi R^{2}}.$$

In the given data $$Q = 2\pi\,$C, so the factors $$2$$\pi$$$$ cancel, giving a very simple result:
$$T = k\,$$\frac{q}{R^{2}$$}.$$

Substituting the numerical values:
$$T = 9$$\times$$10^{9}\,$$\frac{30\times10^{-12}$$}{(0.30)^{2}} = 9$$\times$$10^{9}$$\times$$30$$\times$$10^{-12}$$\times$$$$\frac{1}{0.09}$$$$ $$\phantom{T} = 9$$\times$$30$$\times$$10^{-3}$$\times$$$$\frac{1}{0.09}$$ = 270$$\times$$10^{-3}$$\times$$$$\frac{1}{0.09}$$ = 0.27$$\times$$$$\frac{1}{0.09}$$ = 3\;$$\text{N}$$.$$

Hence the tension in the metallic ring is $$\mathbf{3\;N}.$$

At point (0,0,2):

Electric field due to infinite sheet (on xy-plane with +σ_s): $$E_{sheet} = \frac{\sigma_s}{2\varepsilon_0}$$ (directed in +z direction at z=2)

Electric field due to line charge at z=4 parallel to y-axis: distance from line to point = $$\sqrt{0^2 + (4-2)^2} = 2$$ m

$$E_{line} = \frac{\lambda_e}{2\pi\varepsilon_0 \times 2} = \frac{\lambda_e}{4\pi\varepsilon_0}$$

Ratio: $$\frac{E_{sheet}}{E_{line}} = \frac{\sigma_s/(2\varepsilon_0)}{\lambda_e/(4\pi\varepsilon_0)} = \frac{2\pi\sigma_s}{\lambda_e}$$

Given $$|\sigma_s| = 2|\lambda_e|$$: $$\frac{E_{sheet}}{E_{line}} = \frac{2\pi \times 2\lambda_e}{\lambda_e} = 4\pi$$

We need $$4\pi = \pi\sqrt{n}$$, so $$\sqrt{n} = 4$$, $$n = 16$$.

The answer is 16.

Two identical charged spheres suspended by equal-length strings make angle $$\theta$$ in air. When suspended in water, the angle remains the same. Density of sphere = $$1.5$$ g/cc, density of water = $$1$$ g/cc.

In air, each sphere is in equilibrium under the action of electrostatic repulsion, gravity, and string tension; the balance of horizontal and vertical forces yields $$\tan(\theta/2) = \frac{F_e}{mg}$$, where $$F_e$$ is the electrostatic force and $$mg$$ is the weight of the sphere.

When the spheres are suspended in water, the electrostatic force is reduced by the dielectric constant $$K$$ so that $$F_e' = F_e/K$$, and the effective weight is reduced by the buoyant force: $$W' = mg - \rho_w V g = Vg(\rho_s - \rho_w)$$, where $$\rho_s$$ and $$\rho_w$$ are the densities of the sphere and water respectively and $$V$$ is the volume of the sphere. The equilibrium in water therefore gives $$\tan(\theta/2) = \frac{F_e/K}{Vg(\rho_s - \rho_w)}$$.

Since the angle remains the same in both media, we equate the two expressions: $$\frac{F_e}{mg} = \frac{F_e/K}{Vg(\rho_s - \rho_w)}$$. Noting that $$m = \rho_s V$$, this reduces to $$\frac{1}{\rho_s} = \frac{1}{K(\rho_s - \rho_w)}$$, which leads to $$K = \frac{\rho_s}{\rho_s - \rho_w} = \frac{1.5}{1.5 - 1.0} = \frac{1.5}{0.5} = 3$$.

The dielectric constant of water is 3.

Solution :

Work done by electric field is given by :

$$W = q\vec{E}\cdot\vec{d}$$

Given :

$$q = 2 \times 10^{-2}\text{ C}$$

Electric field is along positive x-axis :

$$\vec{E} = 30\hat{i}\ \text{N C}^{-1}$$

Coordinates :

$$P(1,2,0)$$

$$S(0,0,0)$$

Displacement from P to S :

$$\vec{d} = (0-1)\hat{i} + (0-2)\hat{j}$$

$$= -\hat{i} -2\hat{j}$$

Now,

$$\vec{E}\cdot\vec{d} = 30\hat{i}\cdot(-\hat{i}-2\hat{j})$$

$$= -30$$

Therefore,

$$W = (2 \times 10^{-2})(-30)$$

$$= -0.6\text{ J}$$

Final Answer :

$$-0.6\text{ J}$$

In a medium of dielectric constant $$k$$, the electrostatic force between charges $$q_1$$ and $$q_2$$ separated by distance $$d$$ is given by

$$F = \frac{q_1 q_2}{4\pi\epsilon_0 k d^2}$$

Similarly, if the same force acts in air at an equivalent distance $$d'$$, it can be expressed as

$$F = \frac{q_1 q_2}{4\pi\epsilon_0 d'^2}$$

Equating these two expressions yields $$\frac{1}{k d^2} = \frac{1}{d'^2}$$, which implies $$d'^2 = k d^2$$ and therefore $$d' = d\sqrt{k}$$, showing that the equivalent distance in air is $$d\sqrt{k}$$.

We need to find the electric flux through the surface opposite to $$S$$ in a cuboid of dimensions $$2L \times 2L \times L$$, where charge $$q$$ is placed at the centre of surface $$S$$ (the $$2L \times 2L$$ face).

We use the mirror image method. The charge $$q$$ sits at the centre of face $$S$$, and the cuboid extends a distance $$L$$ perpendicular to this face. By placing an identical mirror cuboid on the other side of $$S$$, we form a cube of dimensions $$2L \times 2L \times 2L$$ with $$q$$ at its centre.

By Gauss's Law, the total flux through this cube is $$\frac{q}{\epsilon_0}$$. Since the cube has 6 identical faces and $$q$$ is at the centre, the flux through each face is $$\frac{q}{6\epsilon_0}$$.

Now, the surface opposite to $$S$$ in the original cuboid is exactly one face of the combined cube (the face at distance $$L$$ from the charge along the perpendicular direction). So the flux through it is $$\frac{q}{6\epsilon_0}$$.

Hence, the correct answer is Option D.

Let the unit vector $$\hat{n}$$ point from left to right (the positive direction). Both sheets have a positive surface charge density $$+\sigma$$.

For Region I (left of both sheets):

The electric field from the first sheet points to the left: $$\vec{E}_1 = -\frac{\sigma}{2\epsilon_0}\hat{n}$$

The electric field from the second sheet points to the left: $$\vec{E}_2 = -\frac{\sigma}{2\epsilon_0}\hat{n}$$

Total field: $$\vec{E}_I = \vec{E}_1 + \vec{E}_2 = -\frac{\sigma}{2\epsilon_0}\hat{n} - \frac{\sigma}{2\epsilon_0}\hat{n} = -\frac{\sigma}{\epsilon_0}\hat{n}$$

For Region II (between the sheets):

The electric field from the first (left) sheet points to the right: $$\vec{E}_1 = \frac{\sigma}{2\epsilon_0}\hat{n}$$

The electric field from the second (right) sheet points to the left: $$\vec{E}_2 = -\frac{\sigma}{2\epsilon_0}\hat{n}$$

Total field: $$\vec{E}_{II} = \vec{E}_1 + \vec{E}_2 = \frac{\sigma}{2\epsilon_0}\hat{n} - \frac{\sigma}{2\epsilon_0}\hat{n} = 0$$

For Region III (right of both sheets):

The electric field from the first sheet points to the right: $$\vec{E}_1 = \frac{\sigma}{2\epsilon_0}\hat{n}$$

The electric field from the second sheet points to the right: $$\vec{E}_2 = \frac{\sigma}{2\epsilon_0}\hat{n}$$

Total field: $$\vec{E}_{III} = \vec{E}_1 + \vec{E}_2 = \frac{\sigma}{2\epsilon_0}\hat{n} + \frac{\sigma}{2\epsilon_0}\hat{n} = \frac{\sigma}{\epsilon_0}\hat{n}$$

At point $$x = 2$$ cm, the net electric field must be zero.

Charge at origin: $$q_1 = 10$$ μC (positive). At $$x = 2$$ cm, the field due to $$q_1$$ points in the +x direction.

For the net field to be zero, the 40 μC charge must be placed to the right of $$x = 2$$ (so its field at x=2 points in the -x direction).

Let the 40 μC charge be at position $$x = d$$ cm where $$d > 2$$.

At $$x = 2$$:

$$\frac{kq_1}{(2)^2} = \frac{kq_2}{(d-2)^2}$$

$$\frac{10}{4} = \frac{40}{(d-2)^2}$$

$$(d-2)^2 = \frac{40 \times 4}{10} = 16$$

$$d - 2 = 4$$

$$d = 6$$ cm

The correct answer is Option 1: $$x = 6$$ cm.

Given: An infinite sheet charge ($$+\sigma$$) and an infinite line charge ($$+\lambda$$) are parallel, separated by 5 m. Points P and Q are at distances $$\frac{3}{\pi}$$ m and $$\frac{4}{\pi}$$ m from the line charge, respectively, toward the sheet charge. Given $$2|\sigma| = |\lambda|$$.

Electric fields:

Sheet charge: $$E_{sheet} = \frac{\sigma}{2\varepsilon_0}$$ (constant, directed away from the sheet)

Line charge at distance $$r$$: $$E_{line} = \frac{\lambda}{2\pi\varepsilon_0 r}$$ (directed away from the line)

Both P and Q lie between the line and the sheet, so $$E_{line}$$ points toward the sheet and $$E_{sheet}$$ points toward the line (they oppose each other).

At point P (distance $$\frac{3}{\pi}$$ from line):

$$E_{line,P} = \frac{\lambda}{2\pi\varepsilon_0 \cdot \frac{3}{\pi}} = \frac{\lambda}{6\varepsilon_0}$$

$$E_P = \left|\frac{\lambda}{6\varepsilon_0} - \frac{\sigma}{2\varepsilon_0}\right|$$

At point Q (distance $$\frac{4}{\pi}$$ from line):

$$E_{line,Q} = \frac{\lambda}{2\pi\varepsilon_0 \cdot \frac{4}{\pi}} = \frac{\lambda}{8\varepsilon_0}$$

$$E_Q = \left|\frac{\lambda}{8\varepsilon_0} - \frac{\sigma}{2\varepsilon_0}\right|$$

Substituting $$\lambda = 2\sigma$$:

$$E_P = \left|\frac{2\sigma}{6\varepsilon_0} - \frac{\sigma}{2\varepsilon_0}\right| = \left|\frac{\sigma}{3\varepsilon_0} - \frac{\sigma}{2\varepsilon_0}\right| = \frac{\sigma}{6\varepsilon_0}$$

$$E_Q = \left|\frac{2\sigma}{8\varepsilon_0} - \frac{\sigma}{2\varepsilon_0}\right| = \left|\frac{\sigma}{4\varepsilon_0} - \frac{\sigma}{2\varepsilon_0}\right| = \frac{\sigma}{4\varepsilon_0}$$

The ratio:

$$\frac{E_P}{E_Q} = \frac{\sigma/(6\varepsilon_0)}{\sigma/(4\varepsilon_0)} = \frac{4}{6} = \frac{2}{3} = \frac{4}{6}$$

Comparing with $$\frac{E_P}{E_Q} = \frac{4}{a}$$:

$$a = 6$$

Therefore, the answer is $$\mathbf{6}$$.

Solution :

Given :

$$q_1 = 4q_0$$ at origin

$$q_2 = -q_0$$ at $$x = 12\text{ cm}$$

Let the proton be placed at a distance $$x$$ from the origin where net electrostatic force is zero.

Since the charges are opposite in sign, the zero force point cannot lie between them. It lies on the right side of $$q_2$$.

Let proton be at $$x > 12$$ cm.

Force due to $$q_1$$ on proton :

$$F_1 = \frac{k(4q_0)(q_0)}{x^2}$$

Force due to $$q_2$$ on proton :

$$F_2 = \frac{k(q_0)(q_0)}{(x-12)^2}$$

For net force to be zero :

$$F_1 = F_2$$

$$\frac{4}{x^2} = \frac{1}{(x-12)^2}$$

Taking square root :

$$\frac{2}{x} = \frac{1}{x-12}$$

$$2(x-12)=x$$

$$2x-24=x$$

$$x=24\text{ cm}$$

Final Answer :

$$24$$

Electric field at distance r from infinite line charge: $$E = \frac{\lambda}{2\pi\varepsilon_0 r}$$

For circular motion: $$eE = \frac{mv^2}{r}$$, so $$\frac{e\lambda}{2\pi\varepsilon_0 r} = \frac{mv^2}{r}$$

$$v^2 = \frac{e\lambda}{2\pi\varepsilon_0 m} = \frac{1.6 \times 10^{-19} \times 2 \times 10^{-8}}{2\pi \times 8.85 \times 10^{-12} \times 9 \times 10^{-31}}$$

$$= \frac{3.2 \times 10^{-27}}{5.0 \times 10^{-41}} = 6.4 \times 10^{13}$$

$$v = 8 \times 10^6$$ m/s. The answer is 8.

Use Gauss’s law in differential form:

$$\nabla\cdot\vec{E}=\frac{\rho}{\varepsilon_0}$$

and total charge enclosed

$$Q=ε_0∭(∇⋅E⃗)dV$$

Given

$$E = 2x^2\hat{i} - 4y\hat{j} + 6\hat{k}$$ N C$$^{-1}$$.

Divergence is

$$\nabla\cdot\vec{E}=\frac{\partial(2x^2)}{\partial x}+\frac{\partial(-4y)}{\partial y}+\frac{\partial(6)}{\partial z}$$

$$=4x−4+0$$

$$=4x−4$$

Now cuboid dimensions from figure:

$$0\le x\le1$$

$$0\le y\le2$$

$$0\le z\le3$$

So

$$Q=ε_0\int\int\int(4x−4)dzdydx$$

First integrate over z:

$$=ε_0\int\int3(4x−4)dydx$$

Integrate over y:

$$=ε_0\int6(4x−4)dx$$

$$=6\varepsilon_0\int_0^1(4x-4)dx$$

$$=6\varepsilon_0\left(2x^2-4x\right)_0^1$$

$$=6\varepsilon_0(2-4)$$

$$=-12\varepsilon_0$$

Magnitude of charge:

$$|Q|=12\varepsilon_0$$

Given charges:

• +q at x=0
• $$−2q\ at\ x=\frac{3R}{4}​\ (this\ is\ the\ ch\arg e\ on\ which\ force\ is\ required)\ $$

• $$+2q\ at\ x=R$$

Given:

$$q=2\times10^{-6}C$$

$$R=2cm=0.02m$$

Force on -2q due to charge q

Distance between them:

$$r_1=\frac{3R}{4}$$

Coulomb force:

$$F_1=\frac{k(q)(2q)}{(3R/4)^2}$$

$$\frac{2kq^2}{9R^2/16}$$

$$\frac{32kq^2}{9R^2}$$

Using

$$k=9\times10^9$$

and

$$q^2=(2\times10^{-6})^2=4\times10^{-12}$$

$$R^2=(0.02)^2=4\times10^{-4}$$

$$F_1=\frac{32(9\times10^9)(4\times10^{-12})}{9(4\times10^{-4})}=320N$$

This force is toward left (attraction).

Force on -2q due to charge 2q

Distance:

$$r_2=R-\frac{3R}{4}=\frac{R}{4}$$

Force:

$$F_2=\frac{k(2q)(2q)}{(R/4)^2}$$

$$=\frac{4kq^2}{R^2/16}$$

$$=\frac{64kq^2}{R^2}$$

Substituting:

$$F_2=\frac{64(9\times10^9)(4\times10^{-12})}{4\times10^{-4}}=5760N$$

This is toward right (attraction).

Net force on -2q is

Opposite directions, so subtract:

$$F=5760-320=5440N$$

Let the two positive charges $$+q$$ be placed at $$x = -a$$ and $$x = +a$$.

Let the test charge $$q_0$$ be placed at a distance $$y$$ from the center along the perpendicular bisector (the y-axis).

The distance from either charge to $$q_0$$ is $$r = \sqrt{a^2 + y^2}$$.

The magnitude of the force exerted by one charge is $$F' = \frac{k q q_0}{r^2} = \frac{k q q_0}{a^2 + y^2}$$

By symmetry, the horizontal components of the two forces cancel each other. The vertical components add together.

$$F_{\text{net}} = 2F' \cos\theta$$

$$\cos\theta = \frac{y}{r} = \frac{y}{\sqrt{a^2 + y^2}}$$

$$F_{\text{net}} = 2 \left( \frac{k q q_0}{a^2 + y^2} \right) \left( \frac{y}{\sqrt{a^2 + y^2}} \right)$$

$$F_{\text{net}} = \frac{2 k q q_0 y}{(a^2 + y^2)^{3/2}}$$

To find the distance $$y$$ where the force is maximum, we differentiate $$F_{\text{net}}$$ with respect to $$y$$ and set it to zero ($$\frac{dF_{\text{net}}}{dy} = 0$$):

Since $$2 k q q_0$$ is a constant, we only need to maximize the term $$y(a^2 + y^2)^{-3/2}$$:

$$\frac{d}{dy} \left[ y(a^2 + y^2)^{-3/2} \right] = 0$$

$$1 \cdot (a^2 + y^2)^{-3/2} + y \left( -\frac{3}{2} \right)(a^2 + y^2)^{-5/2}(2y) = 0$$

$$(a^2 + y^2)^{-3/2} - 3y^2(a^2 + y^2)^{-5/2} = 0$$

$$(a^2 + y^2)^1 - 3y^2 = 0$$

$$a^2 - 2y^2 = 0$$

$$2y^2 = a^2 \implies y = \frac{a}{\sqrt{2}}$$

The electric field is  $$\vec{E}=E_0x\,\hat{i}$$

Using Gauss's law,  $$\Phi=\frac{Q}{\epsilon_0}$$

The field is along the $$x$$-direction. Hence, there is no flux through the faces at $$x=0$$, $$y=0$$, $$y=a$$, $$z=0$$ and $$z=a$$.

Only the face at $$x=a$$ contributes to the flux.

At $$x=a$$,  $$E=E_0a$$

Area of the face is  $$A=a^2$$

Therefore, the total electric flux is  $$\Phi=E_0a\times a^2=E_0a^3$$

By Gauss's law,  $$Q=\epsilon_0E_0a^3$$

Given,  $$\epsilon_0=9\times10^{-12}\,C^2N^{-1}m^{-2}$$

$$E_0=4\times10^4\,NC^{-1}m^{-1}$$

$$a=2\,cm=2\times10^{-2}\,m$$

Therefore,  $$Q=(9\times10^{-12})(4\times10^4)(2\times10^{-2})^3$$

$$Q=(9\times10^{-12})(4\times10^4)(8\times10^{-6})$$

$$Q=288\times10^{-14}\,C$$

Hence,  $${Q=288}$$

Therefore, the correct answer is $$288$$.

The electric field is uniform in direction (along $$+\hat{i}$$) but varies with the $$x$$-coordinate:
$$\vec E = 4000\,x^{2}\,\hat{i}\;{\text{V m}}^{-1}$$.

Place the cube so that its faces are perpendicular to the coordinate axes and the two faces perpendicular to the $$x$$-axis lie at $$x = 0$$ and $$x = a$$, where the edge length of the cube is
$$a = 20 \text{ cm} = 0.20 \text{ m}$$.

Electric flux through a closed surface is
$$\Phi = \oint_S \vec E \cdot d\vec A$$.

The field has no $$y$$ or $$z$$ components, so the four faces whose outward normals are $$\pm\hat{j}$$ or $$\pm\hat{k}$$ give zero contribution because $$\vec E \cdot d\vec A = 0$$ there.

Only the two faces perpendicular to the $$x$$-axis contribute.

Outward normal $$\hat n = +\hat{i}$$,
$$d\vec A = \hat{i}\,dA$$,
$$\vec E \cdot d\vec A = E_x\,dA = (4000\,a^{2})\,dA$$.

Total flux through this face:
$$\Phi_{x=a} = E_x(a)\,A = 4000\,a^{2}\,A$$, where $$A = a^{2}$$ is the area of one face.

Outward normal $$\hat n = -\hat{i}$$,
$$d\vec A = -\hat{i}\,dA$$,
$$\vec E \cdot d\vec A = -E_x(0)\,dA = 0$$ because $$E_x(0)=0$$.

Hence $$\Phi_{x=0} = 0$$.

Net outward flux through the entire cube:
$$\Phi = \Phi_{x=a} - \Phi_{x=0} = 4000\,a^{2}\,A$$.

Substitute $$a = 0.20 \text{ m}$$:
$$A = a^{2} = (0.20)^{2} = 0.040 \text{ m}^{2}$$,
$$E_x(a) = 4000\,(0.20)^{2} = 4000 \times 0.040 = 160 \text{ V m}^{-1}$$.

Therefore
$$\Phi = 160 \text{ V m}^{-1} \times 0.040 \text{ m}^{2} = 6.4 \text{ V m}.$$

The question asks for the answer in $$\text{V cm}$$. Since $$1 \text{ m} = 100 \text{ cm}$$,
$$6.4 \text{ V m} = 6.4 \times 100 \text{ V cm} = 640 \text{ V cm}.$$

Electric flux through the cube = 640 V cm.

We need to evaluate both statements about electric charges and fields.

Checking Statement I:

When a positive point charge is brought near a point in an existing electric field, the electric field at that point is the vector sum of the original field and the field due to the point charge.

If the point charge is positive and the direction of its field at the given point is along (or has a component along) the original electric field, the net field at that point will increase.

So yes, the value of electric field at a point near the charge may increase if the charge is positive. Statement I is TRUE.

Checking Statement II:

An electric dipole placed in a uniform electric field experiences:

- A torque: $$\vec{\tau} = \vec{p} \times \vec{E}$$

- Net force = $$\vec{0}$$ (since the field is uniform, the forces on the positive and negative charges are equal and opposite)

The net force on the dipole in a uniform field is zero. Statement II is FALSE.

Statement I is true but Statement II is false. The correct answer is Option C.

We have two identical metallic spheres A and B, each carrying the same charge $$q$$ (since they repel with force $$F$$, both have like charges). The initial force between them at distance $$d$$ is $$F = \frac{kq^2}{d^2}$$.

An uncharged sphere C is first placed in contact with A. Since C is identical to A, the charge is shared equally, so after contact A has $$\frac{q}{2}$$ and C has $$\frac{q}{2}$$.

Now C (with charge $$\frac{q}{2}$$) is placed in contact with B (which still has charge $$q$$). The total charge is $$\frac{q}{2} + q = \frac{3q}{2}$$, which is shared equally, giving B a charge of $$\frac{3q}{4}$$ and C a charge of $$\frac{3q}{4}$$.

Sphere C is now placed at the midpoint between A and B, at a distance $$\frac{d}{2}$$ from each.

The force on C due to A is $$F_{CA} = \frac{k \cdot \frac{q}{2} \cdot \frac{3q}{4}}{(d/2)^2} = \frac{k \cdot \frac{3q^2}{8}}{d^2/4} = \frac{3kq^2}{2d^2} = \frac{3F}{2}$$, directed away from A (repulsive, towards B).

The force on C due to B is $$F_{CB} = \frac{k \cdot \frac{3q}{4} \cdot \frac{3q}{4}}{(d/2)^2} = \frac{k \cdot \frac{9q^2}{16}}{d^2/4} = \frac{9kq^2}{4d^2} = \frac{9F}{4}$$, directed away from B (repulsive, towards A).

These two forces act in opposite directions along the line joining A and B. The net force on C is $$F_{CB} - F_{CA} = \frac{9F}{4} - \frac{3F}{2} = \frac{9F - 6F}{4} = \frac{3F}{4}$$, directed from B towards A.

Hence, the correct answer is Option 2.

We are given: electric field $$E = 4.9 \times 10^5$$ N/C (vertical, upward), mass of water droplet $$m = 0.1$$ g $$= 0.1 \times 10^{-3}$$ kg $$= 10^{-4}$$ kg, $$g = 9.8$$ m/s$$^2$$. The electric force must balance the gravitational force to prevent the droplet from falling:

$$ qE = mg $$

Solving for the charge $$q$$, we get:

$$ q = \frac{mg}{E} = \frac{10^{-4} \times 9.8}{4.9 \times 10^5} $$

$$ q = \frac{9.8 \times 10^{-4}}{4.9 \times 10^5} $$

$$ q = 2 \times 10^{-9} \text{ C} $$

Therefore, the correct answer is Option B.

A positive charge particle of mass 100 mg is thrown opposite to a uniform electric field of $$1 \times 10^5$$ N C$$^{-1}$$. Charge = 40 $$\mu$$C, initial velocity = 200 m s$$^{-1}$$.

First, the particle is thrown opposite to the electric field. Since the particle is positively charged, the electric force acts along the field direction, opposing the motion.

$$F = qE = 40 \times 10^{-6} \times 1 \times 10^5 = 4 \text{ N}$$

$$a = \frac{F}{m} = \frac{4}{100 \times 10^{-6}} = \frac{4}{10^{-4}} = 4 \times 10^4 \text{ m s}^{-2}$$

Next, using $$v^2 = u^2 - 2as$$ (deceleration):

$$0 = (200)^2 - 2(4 \times 10^4)s$$

$$s = \frac{40000}{80000} = 0.5 \text{ m}$$

The correct answer is Option A: 0.5 m.

We need to evaluate the Assertion and Reason about non-polar materials and dipole moments.

The assertion that non-polar materials do not have any permanent dipole moment is correct. In non-polar materials (like $$O_2$$, $$N_2$$, $$CH_4$$), the centers of positive and negative charge distributions coincide in the absence of an external electric field, resulting in zero permanent dipole moment.

The reason states that when a non-polar material is placed in an electric field, the centre of positive charge distribution coincides with the centre of negative charge distribution; however, this is incorrect. When a non-polar material is placed in an external electric field, the centres of positive and negative charge distributions get separated (they no longer coincide). This separation creates an induced dipole moment. In the absence of an external field, the centres coincide, but the reason specifically refers to the situation under an external field, making it wrong.

Since the assertion is correct but the reason is not, the correct answer is Option C.

Given three charges at the vertices of a right-angled triangle with the right angle at A, where $$q_A = 5 \, \mu C$$, $$q_B = 0.16 \, \mu C$$, and $$q_C = 0.3 \, \mu C$$. The side lengths are $$AB = 3 \text{ cm} = 0.03 \text{ m}$$, $$CA = 3 \text{ cm} = 0.03 \text{ m}$$, and $$BC = 3\sqrt{2} \text{ cm}$$.

Using Coulomb's law, the magnitude of the force on charge A due to charge B is given by $$F_{AB} = \frac{kq_Aq_B}{AB^2} = \frac{9 \times 10^9 \times 5 \times 10^{-6} \times 0.16 \times 10^{-6}}{(0.03)^2} = \frac{9 \times 10^9 \times 8 \times 10^{-13}}{9 \times 10^{-4}} = \frac{7.2 \times 10^{-3}}{9 \times 10^{-4}} = 8 \text{ N}$$.

Similarly, the force on charge A due to charge C is $$F_{AC} = \frac{kq_Aq_C}{CA^2} = \frac{9 \times 10^9 \times 5 \times 10^{-6} \times 0.3 \times 10^{-6}}{(0.03)^2} = \frac{9 \times 10^9 \times 1.5 \times 10^{-12}}{9 \times 10^{-4}} = \frac{1.35 \times 10^{-2}}{9 \times 10^{-4}} = 15 \text{ N}$$.

Since the angle at A is $$90°$$, the forces $$F_{AB}$$ and $$F_{AC}$$ are perpendicular to each other. Therefore, the net force on A is $$F_{net} = \sqrt{F_{AB}^2 + F_{AC}^2} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \text{ N}$$.

Hence, the electrostatic force on the charge at A is $$\textbf{17}$$ N.

Let the six charges be placed at the six vertices of a regular hexagon of side length $$a$$. For a regular hexagon the distance of every vertex from the centre $$O$$ (the circumscribed radius) is also equal to $$a$$. Therefore, for every charge the distance to the point $$O$$ is

$$r = a$$

Denote $$k = \dfrac{1}{4\pi \epsilon_0}$$ for brevity.

Electric field at $$O$$
For a point charge the magnitude of the electric field at a distance $$r$$ is $$E = k \dfrac{q}{r^{2}}$$ and it is directed along the line joining the charge to the point of observation.

Label the six vertices in order as $$1,2,3,4,5,6$$ with vertex 1 on the positive $$x$$-axis. Five of the charges are $$q$$ and the sixth (placed at vertex 1) is $$x$$.

The six radial directions are separated by $$60^\circ$$. If all six charges were equal to $$q$$ the vector sum of the six fields would be zero because the six vectors form a closed regular hexagon. Hence

$$\sum_{i=1}^{6} \mathbf{E}_i(q) = \mathbf{0}$$

When the charge at vertex 1 is changed from $$q$$ to $$x$$, every field vector except the one coming from vertex 1 remains the same. Thus the resultant field becomes

$$\mathbf{E}_{\text{net}} = \mathbf{E}_1(x) + \sum_{i=2}^{6} \mathbf{E}_i(q) = \mathbf{E}_1(x) - \mathbf{E}_1(q)$$

The two vectors on the right-hand side have the same line of action (the radius through vertex 1), so the net magnitude is

$$E_{\text{net}} = k\,\dfrac{|x-q|}{a^{2}}$$

Special cases:
• When $$x = q$$, $$E_{\text{net}} = 0$$.
• When $$x = -q$$, $$E_{\text{net}} = k\,\dfrac{2q}{a^{2}} = \dfrac{q}{2\pi\epsilon_0 a^{2}}$$.

Option B claims the value $$\dfrac{q}{6\pi\epsilon_0 a^{2}}$$, which differs by a factor of $$\dfrac13$$ and is therefore incorrect.

Electric potential at $$O$$
The potential produced by a point charge at distance $$r$$ is $$V = k \dfrac{q}{r}$$ (scalar quantity).

With five charges equal to $$q$$ and one charge equal to $$x$$ the total potential is

$$V_{\text{net}} = k\,\dfrac{5q + x}{a}$$

Special cases:
• For $$x = 2q$$, $$V_{\text{net}} = k\,\dfrac{7q}{a} = \dfrac{7q}{4\pi\epsilon_0 a}$$.
Option C contains the extra factor $$\sqrt{3}$$ in the denominator, so it is incorrect.
• For $$x = -3q$$, $$V_{\text{net}} = k\,\dfrac{2q}{a} = \dfrac{q}{2\pi\epsilon_0 a}$$.
Option D again has the spurious factor $$\sqrt{3}$$ and a different numerical multiplier, hence is incorrect.

Conclusion: only the statement in Option A is correct.

Correct choice: Option A which is: When $$x = q$$, the magnitude of the electric field at $$O$$ is zero.

We need to determine the total electric charge enclosed inside a cube of side length $$a = 0.5\text{ m}$$ placed in a position-dependent electric field.

1. Analyze the Electric Field and Cube Orientation

From the diagram , the electric field vector is given as:

$$\vec{E} = 150 y^2 \hat{j}$$

This tells us two very important things:

• The electric field points exclusively along the $$y$$-axis ($$\hat{j}$$ component). Because there are no components along the $$\hat{i}$$ or $$\hat{k}$$ directions, the net electric flux through the four side faces parallel to the $$y$$-axis (left, right, front, and back faces) is exactly zero.
• The strength of the electric field depends entirely on the height coordinate $$y$$. Therefore, flux will only pass through the bottom face and the top face of the cube.

2. Calculate the Net Electric Flux ($$\Phi_{\text{net}}$$)

Let's evaluate the electric flux through the two active faces:

1. Bottom Face (located at $$y = 0$$):
   Substitute $$y = 0$$ into the electric field equation:

   $$\vec{E}_{\text{bottom}} = 150(0)^2 \hat{j} = 0$$

   Since the electric field strength is zero at the base plane, the flux passing through the bottom face is:

   $$\Phi_{\text{bottom}} = 0$$

2. Top Face (located at $$y = a = 0.5\text{ m}$$):
   The area of this square face is $$A = a^2 = (0.5)^2 = 0.25\text{ m}^2$$. The outward normal vector for the top surface points upwards ($$+\hat{j}$$).
   Substitute $$y = 0.5\text{ m}$$ into the electric field equation:

   $$\vec{E}_{\text{top}} = 150(0.5)^2 \hat{j} = 150 \times 0.25 \hat{j} = 37.5 \hat{j}\text{ N C}^{-1}$$

   The flux passing out of the top face is:

   $$\Phi_{\text{top}} = \vec{E}_{\text{top}} \cdot \vec{A} = 37.5 \times 0.25 = 9.375\text{ N m}^2\text{ C}^{-1}$$

The total net electric flux ($$\Phi_{\text{net}}$$) emerging from the entire cube is:

$$\Phi_{\text{net}} = \Phi_{\text{top}} + \Phi_{\text{bottom}} = 9.375 + 0 = 9.375\text{ N m}^2\text{ C}^{-1}$$

3. Find the Enclosed Charge ($q_{\text{in}}$) Using Gauss's Law

According to Gauss's Law, the net electric flux through any closed surface is proportional to the net charge enclosed inside it:

$$\Phi_{\text{net}} = \frac{q_{\text{in}}}{\varepsilon_0} \implies q_{\text{in}} = \Phi_{\text{net}} \cdot \varepsilon_0$$

Using the standard permittivity constant value ($$\varepsilon_0 = 8.854 \times 10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}$$):

$$q_{\text{in}} = 9.375 \times 8.854 \times 10^{-12}$$

$$q_{\text{in}} \approx 83.006 \times 10^{-12}\text{ C} = 8.3 \times 10^{-11}\text{ C}$$

Conclusion

The total net charge inside the cube is $$8.3 \times 10^{-11}\text{ C}$$, which corresponds exactly to Option A.

Statement I: An electric dipole is placed at the centre of a hollow sphere. A dipole consists of two equal and opposite charges, so the total charge enclosed by the sphere is zero. By Gauss's law, the total electric flux through the sphere is $$\Phi = \frac{q_{enc}}{\epsilon_0} = 0$$. However, the electric field due to the dipole is not zero at points inside the sphere (the dipole creates a non-uniform field). Therefore, Statement I is true.

Statement II: For a solid metallic sphere of radius $$R$$ carrying charge $$Q$$, all the charge resides on the outer surface of the conductor. For a Gaussian surface of radius $$r < R$$ (inside the conductor), the electric field is zero everywhere on this surface (as expected inside a conductor in electrostatic equilibrium). By Gauss's law, the flux through this surface is $$\Phi = \frac{q_{enc}}{\epsilon_0} = 0$$, since no charge is enclosed within radius $$r$$. Statement II claims the flux is not zero, which is incorrect. Therefore, Statement II is false.

Hence, Statement I is true but Statement II is false.

The point charges of $$1\,\mu C$$ each are placed at $$y = 1, 2, 4, 8, \ldots$$ m along the y-axis. A charge of $$1$$ C is placed at the origin. The total force on the 1 C charge is the sum of the forces due to each point charge, all directed along the y-axis.

The force due to the charge at distance $$r_n$$ is $$F_n = \frac{kQq}{r_n^2}$$. Here $$r_n = 2^{n-1}$$ m for $$n = 1, 2, 3, \ldots$$, so $$r_n^2 = 4^{n-1}$$.

The total force is $$F = kQq \sum_{n=1}^{\infty}\frac{1}{4^{n-1}} = kQq \cdot \frac{1}{1 - 1/4} = \frac{4}{3}kQq$$.

Substituting $$k = 9 \times 10^9$$, $$Q = 1$$ C, and $$q = 1 \times 10^{-6}$$ C: $$F = \frac{4}{3} \times 9 \times 10^9 \times 1 \times 10^{-6} = \frac{4}{3} \times 9 \times 10^3 = 12 \times 10^3$$ N.

Since $$F = x \times 10^3$$ N, the value of $$x$$ is $$12$$.

The electric field inside conducting materials is zero in electrostatic equilibrium, and outside a spherically symmetric charge distribution, it follows Gauss's Law: $$E = \frac{k q}{r^2}$$.

For $$0 \le r < R$$ (inside solid metal sphere):   $$E = 0$$

For $$R \le r < a$$ (space between sphere and shell):     $$E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$$

For $$a \le r \le b$$ (inside the conducting spherical shell):    $$E = 0$$

For $$r > b$$ (outside the entire system):   $$E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$$

Corrected Solution & Explanation

1. Total Flux through the Closed Symmetric Structure

By applying Gauss's Law, a charge $$q$$ enclosed fully within a symmetric boundary emits a total flux of $$\Phi = \frac{q}{\varepsilon_0}$$. To completely enclose a charge residing precisely on a corner vertex, we must place $$8$$ identical cubes together. Thus, the total flux passing through our single designated cube is:

$$\Phi_{\text{cube}} = \frac{q}{8\varepsilon_0}$$

2. Flux Distribution Across the Faces

A single cube consists of $$6$$ individual faces:

• Three Adjoining Faces: The 3 faces that intersect to form the corner vertex holding the charge lie completely parallel to the outward electric field lines. Since the field vectors run along the surfaces, the angle with the face area vectors is $$\theta = 90^\circ$$, meaning the net flux through each of these faces is exactly zero ($$\Phi = 0$$).
• Three Opposite Faces: The entire flux entering the cube must exit symmetrically through the 3 remaining outer faces opposite to the charge. Therefore, the total flux passing through one entire opposite square face is:

  $$\Phi_{\text{face}} = \frac{1}{3} \times \Phi_{\text{cube}} = \frac{1}{3} \times \frac{q}{8\varepsilon_0} = \frac{q}{24\varepsilon_0}$$

3. Analyzing the Shaded Geometric Region

Let's look closely at how the shaded area is defined in this classic configuration:

• The shaded area is a triangle formed by connecting the diagonals of the three opposite faces.
• It does not lie flat on a single outer square face; rather, it forms an inclined planar surface that bounds the inner corner.
• By projecting the flux lines passing through the three open outer faces, this specific triangular cross-section intercepts the entire regular flux flow directed through those three outer surfaces combined.

Because any field line passing through the inner shaded triangle must inevitably pass through the three outer symmetric faces to exit the cube, the total flux through the shaded area is exactly equal to the shared contribution of one full face path:

$$\Phi_{\text{shaded}} = \Phi_{\text{face}} = \frac{q}{24\varepsilon_0}$$

Correct Option: B ($$\frac{q}{24\varepsilon_0}$$)

$$E_+ = \frac{|\sigma_+|}{2\varepsilon_0}$$

$$E_- = \frac{|\sigma_-|}{2\varepsilon_0}$$

$$|\sigma_+| > |\sigma_-| \implies E_+ > E_-$$

$$\tan\alpha = \frac{E_-}{E_+}$$

$$\tan\alpha < 1 \implies \alpha < 45^\circ$$

The electric field lines must be straight lines within each quadrant, directed away from the positively charged plane ($$\sigma_+$$) and pointing towards the negatively charged plane ($$\sigma_-$$), making a sharper angle with the $$\sigma_+$$ axis than the $$\sigma_-$$ axis.

$$\text{The electric field in the region above it points straight up along the positive y-axis:}$$

$$\vec{E}_1 = \frac{\sigma}{2\varepsilon_0}\hat{y}$$

For the inclined sheet (inclined at $$30^\circ$$ to the horizontal with charge density $$+\sigma$$):

$$\text{The normal to this sheet points into the region between them, directed downwards and leftwards.}$$

$$\text{The angle of this electric field vector with the negative y-axis is } 30^\circ$$ making a $$60^\circ$$ angle with the negative $$x$$-axis:

$$\vec{E}_2 = E \cos(60^\circ)(-\hat{x}) + E \sin(60^\circ)(-\hat{y})$$

$$\vec{E}_2 = \frac{\sigma}{2\varepsilon_0}\left(-\frac{1}{2}\hat{x} - \frac{\sqrt{3}}{2}\hat{y}\right)$$

For the net electric field $$\vec{E}_{\text{net}}$$ in the region between them:

$$\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2$$

$$\vec{E}_{\text{net}} = \frac{\sigma}{2\varepsilon_0}\hat{y} + \frac{\sigma}{2\varepsilon_0}\left(-\frac{1}{2}\hat{x} - \frac{\sqrt{3}}{2}\hat{y}\right)$$

$$\vec{E}_{\text{net}} = \frac{\sigma}{2\varepsilon_0}\left[\left(1 - \frac{\sqrt{3}}{2}\right)\hat{y} - \frac{1}{2}\hat{x}\right]$$

We consider a ring of uniform charge $$Q$$ and radius $$R$$. The point of observation lies on the axis of the ring at a distance $$x$$ (later we shall rename it $$h$$) from the centre. For such a geometry, the standard result for the magnitude of the electric field on the axis is first recalled.

Formula stated: For a uniformly charged ring, the axial electric field magnitude is

$$E(x)=\frac{1}{4\pi\varepsilon_0}\,\frac{Q\,x}{\left(R^{2}+x^{2}\right)^{3/2}}.$$

Here $$\frac{1}{4\pi\varepsilon_0}$$ is the Coulomb constant, $$Q$$ is the total charge, $$R$$ is the ring’s radius and $$x$$ is the axial distance from the centre. We now wish to find the value of $$x$$ for which this function $$E(x)$$ is maximum in magnitude.

To locate an extremum, we differentiate $$E(x)$$ with respect to $$x$$ and set the derivative equal to zero.

First, write $$E(x)$$ in a clearer algebraic form for differentiation:

$$E(x)=kQ\,x\,(R^{2}+x^{2})^{-3/2}, \qquad \text{where }k=\frac{1}{4\pi\varepsilon_0}.$$

We have to find $$\dfrac{dE}{dx}$$. Using the product rule,

$$\frac{dE}{dx}=kQ\Bigg[\frac{d}{dx}\big(x\big)\cdot (R^{2}+x^{2})^{-3/2}+x\cdot\frac{d}{dx}\big((R^{2}+x^{2})^{-3/2}\big)\Bigg].$$

The derivative of the first factor $$x$$ is simply $$1$$. To differentiate the second factor we use the chain rule:

$$\frac{d}{dx}(R^{2}+x^{2})^{-3/2}=-\frac{3}{2}(R^{2}+x^{2})^{-5/2}\cdot 2x =-3x\,(R^{2}+x^{2})^{-5/2}.$$

Substituting this back, we obtain

$$\frac{dE}{dx}=kQ\Big[(R^{2}+x^{2})^{-3/2}+x\big(-3x\,(R^{2}+x^{2})^{-5/2}\big)\Big].$$

Simplify the expression inside the brackets by taking the common factor $$(R^{2}+x^{2})^{-5/2}$$ outside:

$$\frac{dE}{dx}=kQ\,(R^{2}+x^{2})^{-5/2}\Big[(R^{2}+x^{2})-3x^{2}\Big].$$

Combine like terms in the numerator:

$$\frac{dE}{dx}=kQ\,(R^{2}+x^{2})^{-5/2}\big(R^{2}-2x^{2}\big).$$

For an extremum we set $$\dfrac{dE}{dx}=0$$. The constant factors $$kQ$$ and $$(R^{2}+x^{2})^{-5/2}$$ are never zero, so the only factor that can vanish is

$$R^{2}-2x^{2}=0.$$

Solving this simple algebraic equation,

$$R^{2}=2x^{2}\quad\Rightarrow\quad x^{2}=\frac{R^{2}}{2}\quad\Rightarrow\quad x=\frac{R}{\sqrt{2}}.$$

Thus the electric field on the axis is largest when the observation point is at a distance $$h=\dfrac{R}{\sqrt{2}}$$ from the centre of the ring.

Finally, one may check that the second derivative is negative at this value of $$x$$, confirming that it is indeed a maximum, but this step is not required for the answer.

Hence, the correct answer is Option A.

We recall the vector relation for the torque experienced by an electric dipole in a uniform electric field. The formula is stated first:

$$\vec{\tau} \;=\; \vec{p}\;\times\;\vec{E}$$

Here $$\vec{p}$$ is the dipole moment and $$\vec{E}$$ is the electric field. The dipole moment is fixed in the xy-plane and makes an angle $$\theta$$ with the positive $$x$$-axis, so we may write it component-wise as

$$\vec{p}=p\bigl(\cos\theta\,\hat{i}+\sin\theta\,\hat{j}\bigr).$$

Now the first electric field is purely along the $$x$$-axis:

$$\vec{E}_{1}=E\,\hat{i}.$$

Using the cross-product formula, the torque produced by $$\vec{E}_{1}$$ is

$$\vec{\tau}_{1}= \vec{p}\times\vec{E}_{1} = p(\cos\theta\,\hat{i}+\sin\theta\,\hat{j})\times (E\,\hat{i}).$$

Because $$\hat{i}\times\hat{i}=0$$ and $$\hat{j}\times\hat{i}=-\hat{k},$$ only the second term survives:

$$\vec{\tau}_{1}=pE\sin\theta\;(-\hat{k}) = -\,pE\sin\theta\,\hat{k}.$$

The question tells us that for this same field the torque vector is $$\vec{T}_{1}=+\tau\,\hat{k}.$$ Thus the magnitude is clearly

$$\tau_{1}=pE\sin\theta.$$

Next, consider the second electric field, which points along the positive $$y$$-axis and is $$\sqrt{3}$$ times stronger than $$\vec{E}_{1}$$ in magnitude:

$$\vec{E}_{2}=\sqrt{3}\,E\,\hat{j}.$$

The torque produced by this field is obtained in the same way:

$$\vec{\tau}_{2}= \vec{p}\times\vec{E}_{2} = p(\cos\theta\,\hat{i}+\sin\theta\,\hat{j})\times(\sqrt{3}\,E\,\hat{j}).$$

Here $$\hat{j}\times\hat{j}=0$$ and $$\hat{i}\times\hat{j}=+\hat{k},$$ giving

$$\vec{\tau}_{2}=p\sqrt{3}E\cos\theta\,\hat{k}.$$

Therefore the magnitude of the second torque is

$$\tau_{2}=p\sqrt{3}E\cos\theta.$$

The statement in the problem is that the second torque is the negative of the first:

$$\vec{T}_{2}=-\,\vec{T}_{1}.$$

This implies that their magnitudes are equal while their directions are opposite, so we set

$$\tau_{2}=\tau_{1}.$$

Substituting the expressions just obtained, we have

$$p\sqrt{3}E\cos\theta \;=\; pE\sin\theta.$$

The factors $$p$$ and $$E$$ cancel immediately, leaving

$$\sqrt{3}\cos\theta=\sin\theta.$$

Dividing both sides by $$\cos\theta$$ (which is non-zero for the angles listed) gives the tangent:

$$\tan\theta=\sqrt{3}.$$

The standard value that satisfies this equation in the interval $$0^{\circ}\le\theta\le 90^{\circ}$$ is

$$\theta=60^{\circ}.$$

Hence, the correct answer is Option D.

1. Inner surface:

According to Gauss’s Law, the electric field inside the material of a conducting shell must be zero ($$\vec{E} = 0$$). If we draw a Gaussian surface entirely within the metal of the shell, the total enclosed charge must be zero. $$Q_{enclosed} = Q_{net, cavity} + Q_1 = 0$$

Net Charge ($$Q_1$$): Since the charges inside the cavity are $$+Q$$ and $$-Q$$, their sum is zero. Therefore, $$0 + Q_1 = 0 \implies \mathbf{Q_1 = 0}$$.

Surface Density ($$\sigma_1$$): Although the net charge is zero, the point charges inside are not at the same location. The $$+Q$$ charge will attract negative charges to the nearby inner surface, and the $$-Q$$ charge will attract positive charges. This electrostatic induction causes a redistribution of charge.

Because the distribution is not uniform, the surface charge density $$\sigma_1 \neq 0$$ (it is positive in some areas and negative in others).

2. Outer surface:

The shielding effect of a conductor means that the distribution of charges on the outer surface depends only on the net charge enclosed by the shell and any external electric fields.

Net Charge ($$Q_2$$): Since the shell is neutral and the net charge induced on the inner surface ($$Q_1$$) is zero, the net charge on the outer surface must also be $$Q_2 = 0$$ to maintain charge conservation.

Surface Density ($$\sigma_2$$): Because the net charge enclosed by the inner surface is zero, there is no net electric flux leaking out to induce a charge on the exterior. In the absence of an external field, the charge remains perfectly balanced.

Therefore, the outer surface charge density is $$\sigma_2 = 0$$.

$$\sigma_1 \neq 0, Q_1 = 0, \sigma_2 = 0, Q_2 = 0$$

Place a sample unit charge on the plane of these charges and see how it behaves under the electric field lines from the 3 charges placed on the surface ,electric field lines originated from charge of same sign never meet each other and will always tend to never meet and flow out from each other.
Here only option C is having all these requirements

Let the mass of the charged ball $$B$$ be $$m$$ and its charge be $$q$$.

The electric field $$E$$ produced by a large charged conducting sheet $$P$$ with surface charge density $$\sigma$$ is given by:

$$E = \frac{\sigma}{\epsilon_0}$$

The forces acting on the charged ball are:

1. The weight of the ball, $$mg$$, acting vertically downwards.

2. The electric force, $$F_e = qE$$, acting horizontally away from the sheet.

3. The tension $$T$$ in the silk thread $$S$$.

In equilibrium, the thread makes an angle $$\theta$$ with the vertical sheet. We can resolve the tension $$T$$ into horizontal and vertical components:

The vertical component balances the weight:

$$T \cos\theta = mg$$

The horizontal component balances the electric force:

$$T \sin\theta = qE$$

Dividing the horizontal equilibrium equation by the vertical equilibrium equation, we get:

$$\frac{T \sin\theta}{T \cos\theta} = \frac{qE}{mg}$$

$$\Rightarrow \quad \tan\theta = \frac{qE}{mg}$$

Now, substitute the value of the electric field $$E$$ into the equation:

$$\tan\theta = \frac{q \sigma}{\epsilon_0 mg}$$

Rearranging the equation to solve for $$\sigma$$, we find:

$$\sigma = \left( \frac{\epsilon_0 mg}{q} \right) \tan\theta$$

Since $$\epsilon_0$$, $$m$$, $$g$$, and $$q$$ are all constants for this system, we can conclude that the surface charge density $$\sigma$$ is directly proportional to $$\tan\theta$$:

$$\sigma \propto \tan\theta$$

Let the initial charges on the identical spherical conductors $$B$$ and $$C$$ be $$Q$$.

Let the distance between their centers be $$r$$.

The initial force of repulsion $$F$$ between them is given by Coulomb's Law:

$$F = k \frac{Q \cdot Q}{r^2} = k \frac{Q^2}{r^2}$$

Let the third identical, uncharged sphere be $$A$$. Its initial charge is $$0$$.

{Step 1: Sphere $$A$$ is brought in contact with $$B$$}

When two identical conductors are brought into contact, they share the total charge equally due to symmetry.

The new charge on $$B$$ ($$Q_B'$$) and the new charge on $$A$$ ($$Q_A'$$) will be the average of their initial charges:

$$Q_B' = Q_A' = \frac{Q + 0}{2} = \frac{Q}{2}$$

{Step 2: Sphere $$A$$ is brought in contact with $$C$$}

Sphere $$A$$ now carries a charge of $$\frac{Q}{2}$$. When it touches $$C$$ (which still has its original charge of $$Q$$), they will again share the total charge equally.

The new charge on $$C$$ ($$Q_C'$$) and the final charge on $$A$$ will be:

$$Q_C' = \frac{Q_A' + Q_C}{2} = \frac{\frac{Q}{2} + Q}{2} = \frac{\frac{3Q}{2}}{2} = \frac{3Q}{4}$$

{Step 3: Calculating the new force}

Sphere $$A$$ is now removed. The final charges on the remaining spheres are:

 Final charge on $$B$$:  $$Q_B' = \frac{Q}{2}$$

 Final charge on $$C$$:  $$Q_C' = \frac{3Q}{4}$$

The distance $$r$$ remains unchanged. The new force of repulsion $$F'$$ between $$B$$ and $$C$$ is:

$$F' = k \frac{Q_B' \cdot Q_C'}{r^2}$$

Substitute the new charge values:

$$F' = k \frac{\left(\frac{Q}{2}\right) \cdot \left(\frac{3Q}{4}\right)}{r^2}$$

$$F' = k \frac{3Q^2}{8r^2}$$

Factor out the constant fraction:

$$F' = \frac{3}{8} \left( k \frac{Q^2}{r^2} \right)$$

Since the term in the parentheses is exactly our initial force $$F$$, we substitute it back in:

$$F' = \frac{3}{8}F$$

Given: point charge $$q = +10^{-8}\,\text{C}$$ at a distance $$d = 20\,\text{cm} = 0.20\,\text{m}$$ from the centre $$O$$ of a conducting sphere of radius $$R = 10\,\text{cm} = 0.10\,\text{m}$$.
Throughout we use $$k = \dfrac{1}{4\pi\varepsilon_0}=9\times10^9\;\text{N m}^2\!\text{/C}^2$$.

Step 1 : Image-charge data for a charge outside a sphere
For a point charge $$q$$ kept at distance $$d\,(d\gt R)$$ from the centre of a conducting sphere of radius $$R$$:

• An image charge $$q' = -\dfrac{R}{d}\,q$$ is located on the same line at a distance $$b = \dfrac{R^{2}}{d}$$ from the centre.
• The combination $$(q,q')$$ alone makes the potential of the sphere zero (grounded condition).
• If the sphere is isolated (not grounded) and must keep a specified net charge $$Q_{\text{sphere}}$$, an additional charge $$q_c$$ is placed at the centre. $$q_c$$ is fixed by the condition $$q_c + q' = Q_{\text{sphere}}$$. The potential of the sphere then becomes a constant $$V_{\text{sphere}} = k\dfrac{q_c}{R}$$.

Part (A) : Potential before grounding
Initially the sphere is neutral $$(Q_{\text{sphere}}=0)$$ and isolated.

Image charge: $$q' = -\dfrac{R}{d}\,q = -\dfrac{0.10}{0.20}\,10^{-8} = -0.5\times10^{-8} = -5\times10^{-9}\,\text{C}$$.
To keep the sphere neutral, add a central charge $$q_c = -q' = +5\times10^{-9}\,\text{C}$$.

The potential of the sphere is therefore $$V_{\text{before}} = k\dfrac{q_c}{R} = 9\times10^9 \times\dfrac{5\times10^{-9}}{0.10} = 9\times10^9 \times 5\times10^{-8} = 45\times10^{1} = 450\;\text{V}.$$

Hence Option A is correct.

Part (B) : Charge that flows when the sphere is grounded
On connecting the sphere to earth, its potential must become zero. That is achieved by removing the central charge $$q_c$$ (ground provides/absorbs charge) so that only $$q'$$ remains on the sphere.

Charge leaving the sphere to the ground $$\;\Delta Q = q_c = +5\times10^{-9}\,\text{C}.$$

Thus $$5\times10^{-9}\,\text{C}$$ of positive charge flows from the sphere to the earth. Option B is correct.

Part (C) : Charge on the isolated sphere after the grounding wire is removed
With the earth connection opened, the induced charge that remains on the sphere equals $$q' = -5\times10^{-9}\,\text{C}$$.

Option C is correct.

Part (D) : Potential after the external charge is shifted to 30 cm
The grounding has already been removed, so the total charge on the sphere stays fixed at $$Q_{\text{sphere}} = -5\times10^{-9}\,\text{C}.$$ Now the point charge is shifted to $$d_2 = 30\,\text{cm} = 0.30\,\text{m}$$.

New image charge for the new position:
$$q'_2 = -\dfrac{R}{d_2}\,q = -\dfrac{0.10}{0.30}\,10^{-8} = -\dfrac{1}{3}\times10^{-8} = -3.33\times10^{-9}\,\text{C}.$$

To keep the sphere’s total charge fixed, the central charge becomes $$q_{c2} = Q_{\text{sphere}} - q'_2 = \bigl(-5\times10^{-9}\bigr) - \bigl(-3.33\times10^{-9}\bigr) = -1.67\times10^{-9}\,\text{C}.$$

The potential of the sphere is therefore $$V_{\text{final}} = k\dfrac{q_{c2}}{R} = 9\times10^9 \times\dfrac{-1.67\times10^{-9}}{0.10} \approx -150\;\text{V}.$$

This is neither 300 V nor positive. Hence Option D is incorrect.

Result
Correct statements: Option A, Option B, Option C.

Since the wire passes through two corners of the cube that are maximally separated, it passes along the body diagonal of the cube.

The side of the cube is  $$a=\sqrt{3}\,cm$$

Therefore, the length of the wire inside the cube is the body diagonal:  $$l=a\sqrt{3}$$

$$l=\sqrt{3}\times\sqrt{3}\,cm=3\,cm$$

$$l=3\times10^{-2}\,m$$

Hence, the charge enclosed by the cube is  $$q=\lambda l$$

Given,  $$\lambda=2\times10^{-9}\,C\,m^{-1}$$

Therefore,  $$q=(2\times10^{-9})(3\times10^{-2})$$

$$q=6\times10^{-11}\,C$$

By Gauss's law,  $$\Phi=\frac{q}{\epsilon_0}$$

Using  $$\frac{1}{4\pi\epsilon_0}=9\times10^9$$

we get  $$\frac{1}{\epsilon_0}=4\pi(9\times10^9)$$

Therefore,  $$\Phi=6\times10^{-11}\times4\pi\times9\times10^9$$

$$\Phi=2.16\pi\,Nm^2C^{-1}$$

Hence,  $${x=2.16\pi}$$

Hence, the correct option is D.

Field due to one infinite positively charged sheet is

$$E=\frac{\sigma}{2\varepsilon_0}$$

directed away from the sheet.

There are two identical positive sheets.

At points A and B, which lie between the sheets:

• Left sheet produces field to the right.
• Right sheet produces field to the left.

Magnitudes are equal, so they cancel.

So between the sheets only the charged sphere contributes.

Outside a uniformly charged sphere,

$$E\propto\frac{1}{r^2}$$

Point A is closer to the sphere than BBB, hence

$$E_A>E_B$$

Now point C:

It lies outside both sheets.

Both sheet fields point left and add:

$$E_{\text{sheets}}=\frac{\sigma}{\varepsilon_0}$$

Sphere field at C also points left, so it adds.

Thus

At point D:

Again sheet contribution is

$$\frac{\sigma}{\varepsilon_0}$$

Sphere field also adds.

But D is closer to the sphere than C, so

$$E_{\text{sphere at D}}>E_{\text{sphere at C}}$$

Therefore

$$E_D>E_C$$

By Gauss law,

For inner cube $$C_1$$​, only charge 2Q is enclosed.

$$\Phi_1=\frac{2Q}{\epsilon_0}$$

For outer cube $$C_2$$​, both charges 2Q and 3Q are enclosed.

Therefore,

$$\frac{\Phi_1}{\Phi_2}=\frac{2}{5}$$

So ratio is

$$2:5$$

At point P, net electric field along y-axis is zero.

So y-components of fields due to $$q_2\ and\ q_3$$​ must cancel.

Distance of P from charges:

For $$q_2$$​,

$$r_2=\sqrt{2^2+4^2}$$

$$=\sqrt{20}$$

For $$q_3$$​,

$$r_3=\sqrt{3^2+4^2}$$

=5

Electric field magnitude due to a point charge:

$$E=\frac{kq}{r^2}$$

Y-component is

$$E_y=E\sin⁡θ$$

For $$q_2$$​,

$$\sin\theta_2=\frac{4}{\sqrt{20}}$$

So

$$E_{2y}=\frac{kq_2}{20}\cdot\frac{4}{\sqrt{20}}$$

For $$q_3$$​,

$$\sin\theta_3=\frac{4}{5}$$

So

$$E_{3y}=\frac{kq_3}{25}\cdot\frac{4}{5}$$

For cancellation:

$$|E_{2y}|=|E_{3y}|$$

$$\frac{|q_2|}{20}\cdot\frac{4}{\sqrt{20}}=\frac{|q_3|}{25}\cdot\frac{4}{5}$$

Cancel 4:

$$\frac{|q_2|}{20\sqrt{20}}=\frac{|q_3|}{125}​$$

Thus

$$\frac{|q_2|}{|q_3|}=\frac{20\sqrt{20}}{125}$$

$$x=\frac{8}{5\sqrt{\ 5}}$$

so x=5

Given

$$E⃗=(2x)x$$

Cube side:

a=2m

From figure, cube extends from

x=2

to

x=4

Only two faces perpendicular to x-axis contribute flux, because field is along x-direction.

At left face:

x=2

$$E_1=2(2)=4$$

Area of face

$$A=2\times2=4$$

Flux through left face (inward, negative):

$$\Phi_1=-E_1A$$

$$=-4(4)=-16$$

At right face:

x=4

$$E_2=2(4)=8$$

Flux through right face:

$$\Phi_2=E_2$$

$$=8(4)=32$$

Net flux

$$\Phi=\Phi_1+\Phi_2$$

$$=32-16$$