JEE Electrostatics Questions

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: $$\displaystyle\frac{Q}{60\varepsilon_0}$$

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.

Using field contribution per sheet: $$E_i = \frac{\sigma_i}{2\varepsilon_0}$$

In Configuration I, summing fields from left to right:

$$E_{\text{region 4}} = \frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0} + \frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0} - \left(\frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0}\right) = 0$$

Finding fields in all regions of Configuration I:

$$E_1 = \frac{\sigma_0}{\varepsilon_0},\ E_2 = 0,\ E_3 = \frac{\sigma_0}{\varepsilon_0},\ E_4 = 0,\ E_5 = \frac{\sigma_0}{\varepsilon_0}$$

Evaluating potential difference for Configuration I with $$d = 1\ \mu\text{m}$$:

$$\Delta V_{\text{I}} = E_1 d + E_2 d + E_3 d + E_4 d + E_5 d = \frac{\sigma_0}{\varepsilon_0}d + 0 + \frac{\sigma_0}{\varepsilon_0}d + 0 + \frac{\sigma_0}{\varepsilon_0}d = \frac{3\sigma_0 d}{\varepsilon_0}$$

$$\Delta V_{\text{I}} = \frac{3 \times (9 \times 10^{-6}) \times 10^{-6}}{9 \times 10^{-12}} = 3\text{ V}$$

In Configuration II, summing fields from left to right in region 3:

$$E_{\text{region 3}} = \frac{\sigma_0/2}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0} + \frac{\sigma_0}{2\varepsilon_0} - \left(-\frac{\sigma_0}{2\varepsilon_0} + \frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0/2}{2\varepsilon_0}\right) = \frac{\sigma_0}{2\varepsilon_0}$$

Finding fields in all regions of Configuration II:

$$E_1 = \frac{\sigma_0}{2\varepsilon_0},\ E_2 = -\frac{\sigma_0}{2\varepsilon_0},\ E_3 = \frac{\sigma_0}{2\varepsilon_0},\ E_4 = -\frac{\sigma_0}{2\varepsilon_0},\ E_5 = \frac{\sigma_0}{2\varepsilon_0}$$

Evaluating potential difference for Configuration II:

$$\Delta V_{\text{II}} = \left(\frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0} + \frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0} + \frac{\sigma_0}{2\varepsilon_0}\right)d = \frac{\sigma_0 d}{2\varepsilon_0} \neq 0$$

A polar dielectric molecule (for example $$H_2O$$) possesses a permanent dipole moment $$\vec p_{0}$$ even when no external field is present.

However, in a macroscopic sample of such a dielectric, every molecule can point in any direction with equal probability when there is no external electric field. Hence the molecular dipoles are randomly oriented.

When vectors of equal magnitude are randomly oriented, their vector (vectorial) sum averages to zero: $$\vec P_{\text{net}} = \sum_i \vec p_{0i} = \vec 0$$.

Therefore the bulk or net dipole moment of the whole specimen is zero in the absence of an external electric field.

Now examine the statements:
Assertion (A): Net dipole moment is not zero without the field — this contradicts the conclusion above, so (A) is incorrect.
Reason (R): Permanent dipoles are oriented randomly when there is no field — this is exactly what happens, so (R) is correct.

Because (A) is false while (R) is true, (R) cannot be the correct explanation of (A). Thus the correct choice is Option D: $$(A)\ \text{is not correct but}\ (R)\ \text{is correct}$$.

An air-craft is essentially a hollow metallic body. When lightning (or any external electrostatic discharge) strikes it, charges get redistributed only on the outer metallic surface because of electrostatic induction.

For a conductor in electrostatic equilibrium, two well-known facts hold:
  • The electric field $$\mathbf{E}$$ anywhere inside the conducting material is zero.
  • Consequently, the electric field in any cavity completely surrounded by the conductor is also zero (principle of electrostatic shielding).

Hence, passengers seated inside the cabin experience no electric field; the metal fuselage acts as a Faraday cage that protects them from lightning. This confirms that Assertion (A) is a correct statement.

Reason (R) restates the principle used: “The electric field inside the cavity enclosed by a conductor is zero.” This statement is true and is exactly the physical explanation for the protection mentioned in (A).

Therefore, both Assertion (A) and Reason (R) are correct, and (R) is the correct explanation of (A).

The appropriate choice is Option A.

Electrostatic force is a conservative force, so the work $$W$$ done by (or against) the electric field in taking a test charge $$q_0$$ from point $$A$$ to point $$B$$ is related to the potential difference by
$$W = q_0\,(V_B - V_A)\,\,\,\,\,\,\,\, -(1)$$

For a uniformly charged thin spherical shell of radius $$R$$, the magnitude of the electric field $$E$$ at any point whose distance from the centre is $$r$$ satisfies
$$E = 0 \quad \text{for} \; r \lt R$$
$$E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^{2}} \quad \text{for} \; r \gt R$$

Since $$E = 0$$ everywhere inside the shell, the line integral of $$\mathbf{E}\cdot d\mathbf{l}$$ between any two interior points is zero. Therefore the electrostatic potential throughout the interior is the same constant value, equal to the potential on the surface:
$$V_{\text{inside}} = V_{\text{surface}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}\,\,\,\,\,\,\,\, -(2)$$

Using $$(2)$$ in $$(1)$$, for any two interior points $$A$$ and $$B$$ we get
$$V_B = V_A \;\; \Longrightarrow \;\; W = q_0\,(V_B - V_A) = 0$$

This result does not depend on the path followed because the electric force is conservative. Hence:

Case 1 : Assertion A
"Work done in moving a test charge between two points inside a uniformly charged spherical shell is zero, no matter which path is chosen."
We have just proven $$W = 0$$, so Assertion A is true.

Case 2 : Reason R
"Electrostatic potential inside a uniformly charged spherical shell is constant and is same as that on the surface of the shell."
Equation $$(2)$$ confirms this, so Reason R is also true.

Because the constancy of potential (Reason R) directly leads to zero potential difference, which in turn makes the work done zero (Assertion A), R is the correct explanation of A.

Therefore the correct choice is Option B: Both A and R are true and R is the correct explanation of A.

The surface charge density at any point on a conductor is inversely proportional to its local radius of curvature ($$R$$): $$\sigma \propto \frac{1}{R}$$

$$\sigma_1$$ is located at the sharpest point (pointed tip) at the top. This region has the absolute smallest radius of curvature ($$R_1$$).

$$\sigma_3$$ is located at the curved base at the bottom, which has a distinct curvature but is less sharp than the tip ($$R_3 > R_1$$).

$$\sigma_2$$ and $$\sigma_4$$ are situated on the symmetrical side flanks, which are relatively flat and identical in curvature ($$R_2 = R_4 > R_3 > R_1$$).

$$\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$$

Dipole has charge $$q = 4\mu C = 4 \times 10^{-6}$$ C and separation $$d = 18$$ cm $$= 0.18$$ m in a uniform electric field of magnitude $$E = 10^4$$ N/C.

The dipole moment is given by $$p = qd = 4 \times 10^{-6} \times 0.18 = 7.2 \times 10^{-7}$$ C m.

The work done in rotating the dipole from angle $$\theta_1$$ to angle $$\theta_2$$ in a uniform electric field is given by $$W = pE(\cos\theta_1 - \cos\theta_2)$$.

For rotation from $$0°$$ to $$180°$$, we have $$\cos 0° = 1$$ and $$\cos 180° = -1$$, so

$$W = pE(1-(-1)) = 2pE = 2 \times 7.2 \times 10^{-7} \times 10^4 = 14.4 \times 10^{-3}$$ J = 14.4 mJ.

The correct answer is Option 2: 14.4 mJ.

Field due to an infinite sheet of charge is

$$E=\frac{\sigma}{2\varepsilon_0}$$

For the left plate (+σ):

Field points away from positive sheet, so at midpoint it is toward right.

$$E_1=\frac{\sigma}{2\varepsilon_0}$$

For the right plate (−2σ):

Field points toward negative sheet, so at midpoint it is also toward right.

Magnitude

$$E_2=\frac{2\sigma}{2\varepsilon_0}=\frac{\sigma}{\varepsilon_0}$$

Net field

$$E=E_1+E_2$$

$$=\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{\varepsilon_0}$$

$$=\frac{3\sigma}{2\varepsilon_0}$$

Force on charge +q:

F=qE

so 

$$F=\frac{3q\sigma}{2\varepsilon_0}$$

To identify the physical quantity represented by $$\epsilon_0\dfrac{d\Phi_E}{dt}$$ we must compare its dimensions with those of the given options.

Step 1: Dimension of electric flux $$\Phi_E$$
Electric flux is defined as $$\Phi_E=\displaystyle\oint \mathbf E\cdot d\mathbf A$$.
• Dimension of electric field $$\mathbf E$$:
  $$\mathbf E=\dfrac{\text{Force}}{\text{Charge}}=\dfrac{MLT^{-2}}{IT}=MLT^{-3}I^{-1}$$.
• Area has dimension $$L^2$$.
Hence,
$$[\Phi_E]=MLT^{-3}I^{-1}\times L^2=ML^{3}T^{-3}I^{-1}$$.

Step 2: Dimension of $$\dfrac{d\Phi_E}{dt}$$
Time derivative multiplies the time dimension by $$T^{-1}$$:
$$\Bigl[\dfrac{d\Phi_E}{dt}\Bigr]=ML^{3}T^{-3}I^{-1}\times T^{-1}=ML^{3}T^{-4}I^{-1}.$$

Step 3: Dimension of $$\epsilon_0$$
From Coulomb’s law $$F=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r^2}$$ we get
$$\epsilon_0=\dfrac{q^2}{Fr^{2}}.$$
• $$[q]=IT,$$ so $$[q^2]=I^{2}T^{2}$$.
• $$[F]=MLT^{-2},\quad [r^2]=L^{2}.$$
Therefore,
$$[\epsilon_0]=\dfrac{I^{2}T^{2}}{MLT^{-2}\times L^{2}}=M^{-1}L^{-3}T^{4}I^{2}.$$

Step 4: Dimension of $$\epsilon_0\dfrac{d\Phi_E}{dt}$$
Multiply the results of Steps 2 and 3:
$$[\epsilon_0]\,[d\Phi_E/dt]=(M^{-1}L^{-3}T^{4}I^{2})(ML^{3}T^{-4}I^{-1})=I.$$

The final dimension is that of electric current. Thus $$\epsilon_0\dfrac{d\Phi_E}{dt}$$ represents a quantity having the dimensions of electric current.

Hence the correct choice is Option D: Electric current.

$$\text{Distance from each dipole to midpoint } X = r$$

$$\vec{E}_{\text{axial}} = \frac{2p}{4\pi\varepsilon_0 r^3}\hat{u}_{\text{axial}}, \quad \vec{E}_{\text{equatorial}} = -\frac{p}{4\pi\varepsilon_0 r^3}\hat{u}_{\text{equatorial}}$$

Evaluating entry (P):

$$\vec{p}_1 = p\hat{j} \quad (\text{equatorial at } X) \implies \vec{E}_1 = -\frac{p}{4\pi\varepsilon_0 r^3}\hat{j}$$

$$\vec{p}_2 = p\hat{j} \quad (\text{equatorial at } X) \implies \vec{E}_2 = -\frac{p}{4\pi\varepsilon_0 r^3}\hat{j}$$

$$\vec{E}_X = \vec{E}_1 + \vec{E}_2 = -\frac{p}{2\pi\varepsilon_0 r^3}\hat{j} \implies \text{(P)} \rightarrow \text{(2)}$$

Evaluating entry (Q):

$$\vec{p}_1 = p\hat{j} \quad (\text{equatorial at } X) \implies \vec{E}_1 = -\frac{p}{4\pi\varepsilon_0 r^3}\hat{j}$$

$$\vec{p}_2 = -p\hat{j} \quad (\text{equatorial at } X) \implies \vec{E}_2 = -\frac{-p}{4\pi\varepsilon_0 r^3}\hat{j} = \frac{p}{4\pi\varepsilon_0 r^3}\hat{j}$$

$$\vec{E}_X = \vec{E}_1 + \vec{E}_2 = 0 \implies \text{(Q)} \rightarrow \text{(1)}$$

Evaluating entry (R):

$$\vec{p}_1 = p\hat{j} \quad (\text{equatorial at } X) \implies \vec{E}_1 = -\frac{p}{4\pi\varepsilon_0 r^3}\hat{j}$$

$$\vec{p}_2 = p\hat{i} \quad (\text{axial at } X) \implies \vec{E}_2 = \frac{2p}{4\pi\varepsilon_0 r^3}\hat{i}$$

$$\vec{E}_X = \vec{E}_1 + \vec{E}_2 = \frac{p}{4\pi\varepsilon_0 r^3}(2\hat{i} - \hat{j}) \implies \text{(R)} \rightarrow \text{(4)}$$

Evaluating entry (S):

$$\vec{p}_1 = p\hat{i} \quad (\text{axial at } X) \implies \vec{E}_1 = \frac{2p}{4\pi\varepsilon_0 r^3}\hat{i}$$

$$\vec{p}_2 = p\hat{i} \quad (\text{axial at } X) \implies \vec{E}_2 = \frac{2p}{4\pi\varepsilon_0 r^3}\hat{i}$$

$$\vec{E}_X = \vec{E}_1 + \vec{E}_2 = \frac{p}{\pi\varepsilon_0 r^3}\hat{i} \implies \text{(S)} \rightarrow \text{(5)}$$

Arc AB is a quarter circle subtending

$$90^{\circ}$$

and its electric field at center is given as

E

Now consider arc ABC.

It consists of two quarter arcs:

• arc AB
• arc BC

Each subtends $$90^{\circ}$$, so each produces field of magnitude

E

at center.

For uniformly charged arc, field at center lies along bisector of the arc.

• For arc AB, field is along bisector of first quadrant diagonal toward southwest.
• For arc BC, field is along bisector of fourth quadrant diagonal toward northwest.

These two field vectors are perpendicular because the bisectors differ by

$$90^{\circ}$$

So resultant field due to arc ABC is

$$\sqrt{E^2+E^2}$$

$$=\sqrt{2}E$$

Let the first charge $$+q$$ be at the origin $$O(0,0,0)$$ and the second charge $$+9q$$ be at $$P(d,0,0)$$ on the x-axis.

Because both charges are positive, the electric field produced by each charge at any point on the line joining them points away from the respective charge.
Hence, at a point in between the two charges the field contributions will be in opposite directions and can cancel out.

Assume the required point $$R$$ is at $$x$$ (with $$0 \lt x \lt d$$).
Position of $$R$$: $$(x,0,0)$$.

Magnitude of field at $$R$$ due to $$+q$$ at the origin:
$$E_1 = \frac{kq}{x^{2}}$$ directed along $$+\hat{i}$$.

Magnitude of field at $$R$$ due to $$+9q$$ at $$P(d,0,0)$$:
$$E_2 = \frac{k(9q)}{(d-x)^{2}}$$ directed along $$-\hat{i}$$ (to the left).

For the net field to vanish, the magnitudes must be equal:
$$E_1 = E_2 \Longrightarrow \frac{kq}{x^{2}} = \frac{k(9q)}{(d-x)^{2}}$$.

Cancelling the common factors $$kq$$ gives
$$\frac{1}{x^{2}} = \frac{9}{(d-x)^{2}}$$.

Taking square roots:
$$\frac{1}{x} = \frac{3}{d-x} \Longrightarrow d - x = 3x$$.

Solving for $$x$$:
$$d = 4x \Longrightarrow x = \frac{d}{4}$$.

Therefore the electric field is zero at the point $$(d/4, 0, 0)$$.

Hence, the correct option is Option B.

A long straight wire carries uniform linear charge density $$\lambda = 2\,\text{nC m}^{-1}=2\times10^{-9}\,\text{C m}^{-1}$$.
We have to find the electric-flux through a Gaussian cube of side length $$a=\sqrt{3}\,\text{cm}=0.01\sqrt{3}\,\text{m}$$ when the wire passes through the two diagonally opposite corners of the cube.

Step 1 : Length of the wire lying inside the cube
For a cube of side $$a$$, the space-diagonal length is $$a\sqrt{3}$$.
Because the straight wire enters at one corner and exits at the diametrically opposite corner, the portion of the wire enclosed by the cube equals the space-diagonal.
Hence
$$ \ell_{\text{in}} = a\sqrt{3} $$

Substituting $$a=\sqrt{3}\,\text{cm}$$,
$$ \ell_{\text{in}} = (\sqrt{3}\,\text{cm})\sqrt{3}=3\,\text{cm}=0.03\,\text{m} $$

Step 2 : Charge enclosed by the Gaussian surface
Using $$q_{\text{enc}} = \lambda \ell_{\text{in}}$$,
$$ q_{\text{enc}} = (2\times10^{-9})\times(0.03)=6\times10^{-11}\,\text{C} $$

Step 3 : Electric-flux through the cube
Gauss’s law states $$\Phi = \dfrac{q_{\text{enc}}}{\varepsilon_0}$$ $$-(1)$$
With $$\varepsilon_0 = 8.854\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}$$, substitute in $$(1)$$:
$$ \Phi = \dfrac{6\times10^{-11}}{8.854\times10^{-12}} = 6.784\times10^{0}\;\text{N m}^2\text{C}^{-1} $$

Step 4 : Expressing the result in terms of $$\pi$$
Calculate $$2.16\pi$$:
$$ 2.16\pi = 2.16 \times 3.1416 \approx 6.79 $$ which matches the obtained value (6.784). Hence
$$ \Phi = 2.16\pi\; \text{N m}^2\text{C}^{-1} $$

Therefore, $$x = 2.16\pi$$ and the correct option is Option D.

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.

We need to find the distance between two identical spheres after charge sharing.

When identical conducting spheres are brought into contact, total charge distributes equally:

Each sphere gets $$\frac{4 \times 10^{-8} + 0}{2} = 2 \times 10^{-8}$$ C.

$$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$$

$$9 \times 10^{-3} = 9 \times 10^9 \times \frac{(2 \times 10^{-8})^2}{r^2}$$

$$9 \times 10^{-3} = 9 \times 10^9 \times \frac{4 \times 10^{-16}}{r^2}$$

$$r^2 = \frac{9 \times 10^9 \times 4 \times 10^{-16}}{9 \times 10^{-3}} = 4 \times 10^{-4}$$

$$r = 2 \times 10^{-2} \text{ m} = 2 \text{ cm}$$

The correct answer is Option 2: 2 cm.

Let the radii of the two conducting spheres be $$R$$ (smaller) and $$3R$$ (bigger).
Both spheres initially have the same surface charge density $$\sigma$$.

Initial charges
Charge on the smaller sphere: $$Q_{s}= \sigma \times 4\pi R^{2}$$
Charge on the bigger sphere: $$Q_{b}= \sigma \times 4\pi (3R)^{2}= \sigma \times 4\pi \times 9R^{2}= 9\sigma \,4\pi R^{2}$$

Total initial charge on the system: $$Q_{\text{total}} = Q_{s}+Q_{b} = \sigma \,4\pi R^{2}(1+9)=10\sigma \,4\pi R^{2}$$

When the spheres are connected by a conducting wire, they come to the same potential. For an isolated charged sphere of radius $$r$$, the electrostatic potential is

$$V = \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{r}$$

After contact, let the final charges be $$Q'_s$$ (smaller) and $$Q'_b$$ (bigger).

Equality of potentials gives
$$\frac{Q'_s}{R} = \frac{Q'_b}{3R} \;\;\Longrightarrow\;\; Q'_s = \frac{Q'_b}{3} \quad -(1)$$

Charge conservation gives
$$Q'_s + Q'_b = Q_{\text{total}} = 10\sigma \,4\pi R^{2} \quad -(2)$$

Substituting $$Q'_s$$ from $$(1)$$ into $$(2)$$:

$$\frac{Q'_b}{3} + Q'_b = 10\sigma \,4\pi R^{2}$$
$$\frac{4}{3}Q'_b = 10\sigma \,4\pi R^{2}$$
$$Q'_b = \frac{30}{4}\,\sigma \,4\pi R^{2}= \frac{15}{2}\,\sigma \,4\pi R^{2}$$

Using $$(1)$$,
$$Q'_s = \frac{Q'_b}{3}= \frac{15}{2}\times \frac{1}{3}\,\sigma \,4\pi R^{2}= \frac{5}{2}\,\sigma \,4\pi R^{2}$$

Final surface charge densities

Smaller sphere:
$$\sigma_1 = \frac{Q'_s}{4\pi R^{2}} = \frac{5}{2}\sigma$$

Bigger sphere:
$$\sigma_2 = \frac{Q'_b}{4\pi (3R)^{2}} = \frac{\frac{15}{2}\,\sigma \,4\pi R^{2}}{4\pi \times 9R^{2}} = \frac{15}{2}\times\frac{1}{9}\,\sigma = \frac{5}{6}\sigma$$

Required ratio

$$\frac{\sigma_1}{\sigma_2} = \frac{\frac{5}{2}\sigma}{\frac{5}{6}\sigma} = \frac{1}{2}\times\frac{6}{1} = 3$$

The ratio $$\dfrac{\sigma_1}{\sigma_2}$$ equals 3, which matches Option D.

Use Gauss law.

Length of line charge placed on edge BC:

$$\frac{a}{2}$$

Charge enclosed in cube:

$$q=\lambda\cdot\frac{a}{2}$$

Now, because charge lies along an edge, it is shared by 4 identical cubes if we imagine 4 cubes arranged around that edge.

So this cube encloses only one-fourth of that charge effectively for flux through this cube.

Effective enclosed charge:

$$q_{\text{eff}}=\frac{1}{4}\left(\lambda\frac{a}{2}\right)=\frac{\lambda a}{8}$$

By Gauss law,

$$\Phi=\frac{q_{\text{eff}}}{\epsilon_0}$$

therefore it is

$$\frac{\lambda}{8\epsilon_o} a$$

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

$$E_D>E_C$$

For an infinite non-conducting sheet, electric field is

$$E=\frac{\sigma}{2\varepsilon_0}$$

Force on charged bob due to field:

$$F_e=qE$$

$$=q\frac{\sigma}{2\varepsilon_0}$$

At equilibrium forces balance.

Weight downward:

mg

Electric force horizontal.

Tension along string.

Resolving tension:

$$T\cos45^{\circ}=mg$$

$$T\sin45^{\circ}=F_e$$

Dividing,

$$\tan45^{\circ}=\frac{F_e}{mg}$$

Since

$$\tan45^{\circ}=1$$

$$F_e=mg$$

So

$$q\frac{\sigma}{2\varepsilon_0}=mg$$

Thus

$$\sigma=\frac{2\varepsilon_0mg}{q}$$

Now

Mass:

$$100mg=100\times10^{-6}kg=10^{-4}kg$$

Charge:

$$q=10\mu C=10^{-5}C$$

Substitute:

σ

$$=\frac{2(8.85\times10^{-12})(10^{-4})(10)}{10^{-5}}$$

$$=17.7\times10^{-11}$$

$$=1.77\times10^{-10}\ \text{C/m}^2$$

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

$$\tau=(4\times10^{-7})(5.65\times10^6)\cdot\frac{1}{2}​$$

$$=1.13Nm$$

The ring lies in the $$xy$$-plane with its centre at the origin and has radius $$R = a\sqrt{2}$$. The linear charge density is uniform; therefore the axial electric field depends only on the distance $$z$$ from the centre along the $$+z$$-axis.

Electric field on the axis of a uniformly charged ring (standard result):
$$E(z)=\frac{1}{4\pi\varepsilon_0}\,\frac{Q\,z}{\left(z^{2}+R^{2}\right)^{3/2}}$$ $$-(1)$$
where $$Q$$ is the total charge on the ring.

To locate the position where the magnitude $$E(z)$$ is maximum, differentiate $$E(z)$$ with respect to $$z$$ and set the derivative to zero.

Because the constant factor $$\frac{1}{4\pi\varepsilon_0}Q$$ is positive, maximising $$E(z)$$ is equivalent to maximising the function
$$f(z)=\frac{z}{\left(z^{2}+R^{2}\right)^{3/2}}$$ $$-(2)$$

Differentiate $$f(z)$$ using the quotient rule:

$$\frac{df}{dz}= \frac{(z^{2}+R^{2})^{3/2}-z\cdot\frac{3}{2}(z^{2}+R^{2})^{1/2}\,(2z)}{(z^{2}+R^{2})^{3}}$$

Simplify the numerator step by step:

$$\begin{aligned} \text{Numerator} &= (z^{2}+R^{2})^{1/2}\Big[(z^{2}+R^{2})-3z^{2}\Big] \\ &= (z^{2}+R^{2})^{1/2}\Big[-2z^{2}+R^{2}\Big] \end{aligned}$$

Setting $$\frac{df}{dz}=0$$ gives
$$-2z^{2}+R^{2}=0 \quad\Longrightarrow\quad z^{2}=\frac{R^{2}}{2}$$

Since we need the point on the positive $$z$$-axis,
$$z=\frac{R}{\sqrt{2}}$$ $$-(3)$$

Insert the actual radius $$R=a\sqrt{2}$$:

$$z=\frac{a\sqrt{2}}{\sqrt{2}}=a$$

Therefore the electric field produced by the uniformly charged ring is maximum at a distance $$a$$ from the centre along the positive $$z$$-axis.

Final answer: Option C, $$a$$.

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.

Solution :

Given electric field :

$$\vec{E} = (2\hat{i}+4\hat{j}+6\hat{k}) \times 10^3\text{ N/C}$$

Surface is parallel to x-z plane.

Therefore, area vector is along y-axis.

Hence, only y-component of electric field contributes to flux.

So,

$$E_y = 4 \times 10^3\text{ N/C}$$

Electric flux is given by :

$$\phi = \vec{E}\cdot\vec{A}$$

$$= E_y A$$

Given,

$$\phi = 6.0\text{ Nm}^2\text{C}^{-1}$$

Therefore,

$$6 = (4 \times 10^3)A$$

$$A = \frac{6}{4 \times 10^3}$$

$$= 1.5 \times 10^{-3}\text{ m}^2$$

Converting into cm\(^2\) :

$$A = 1.5 \times 10^{-3} \times 10^4$$

= 15cm$$^2$$.

Final Answer :

$$15$$

Electric field due to infinite line charge at distance $$r$$: $$E = \frac{2k\lambda}{r}$$.

For circular motion: $$qE = \frac{mv^2}{r}$$.

$$q \cdot \frac{2k\lambda}{r} = \frac{mv^2}{r} \Rightarrow v^2 = \frac{2k\lambda q}{m}$$

$$v = \sqrt{\frac{2k\lambda q}{m}}$$

Time period: $$T = \frac{2\pi r}{v} = 2\pi r\sqrt{\frac{m}{2k\lambda q}}$$

The answer is Option (2): $$\boxed{T = 2\pi r\sqrt{\frac{m}{2k\lambda q}}}$$.

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

$$2:5$$

We need to find the electric flux through a sphere of radius $$4a$$ centered at origin with charges $$5Q$$ at $$(3a, 0)$$ and $$-2Q$$ at $$(-5a, 0)$$.

The sphere has radius $$4a$$ and center at origin.

Charge $$5Q$$ at $$(3a, 0)$$: distance from origin = $$3a < 4a$$. Inside the sphere.

Charge $$-2Q$$ at $$(-5a, 0)$$: distance from origin = $$5a > 4a$$. Outside the sphere.

$$\Phi = \frac{Q_{enclosed}}{\varepsilon_0} = \frac{5Q}{\varepsilon_0}$$

The electric flux is $$\frac{5Q}{\varepsilon_0}$$, which matches Option B.

Therefore, the answer is Option B.

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: $$\dfrac{q^2}{32\pi\varepsilon_0 R^3 m}$$

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

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

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

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

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

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)

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.

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

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.

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

$$\Phi=\Phi_1+\Phi_2$$

$$=32-16$$

$$=16$$

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

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.

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:

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

Inside surface SSS enclosed charges are:

$$+5q,\quad+q,\quad-2q$$

Charges +3q and −4q are outside, so contribute zero net flux through closed surface.

Net enclosed charge:

$$Q_{enc}=5q+q-2q=4q$$

By Gauss law,

$$\Phi=\frac{Q_{enc}}{\epsilon_0}$$

So,

$$\frac{4q}{\epsilon_0}$$

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 :

$$-0.6\text{ J}$$

For a point charge Q placed at the centre of a conducting spherical shell:

Regions are:

• Region I:r<a(inside cavity)
• Region II: a<r<b (inside conducting material)
• Region III: r>b (outside shell)

We examine electric field in each region.

For Region I (r<a)

Using Gauss law, take a spherical Gaussian surface of radius r<ar

Only charge enclosed is Q.

$$\oint\vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_0}$$

Since symmetry makes E constant over the surface,

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

$$E_I=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\ne0$$

For Region II (a<r<b):

This is the conducting material.

In electrostatic equilibrium, electric field inside a conductor is zero.

$$E_{II}=0$$

This happens because charge Q induces

• charge −Q on inner surface
• charge +Q on outer surface

so field inside the conductor cancels.

For Region III (r>b):

Take a Gaussian sphere of radius r>b.

Total enclosed charge is

$$Q+(-Q)+(+Q)=Q$$

Using Gauss law,

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

$$E_{III}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\ne0$$

So field outside is not zero; the whole system behaves like a point charge Q at the centre.

Therefore,

$$EI\ne0,E_I\ne0,\quad\quad E_{III}\ne0$$

Correct option:

B

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.

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 two parts be $$q$$ and $$(Q - q)$$ where $$Q = 10 \mu$$C.

The force between them: $$F = k\frac{q(Q-q)}{r^2}$$

For maximum force, we maximize $$q(Q-q)$$. Taking derivative and setting to zero:

$$\frac{d}{dq}[q(Q-q)] = Q - 2q = 0$$

$$q = \frac{Q}{2} = 5 \mu\text{C}$$

The two parts are $$5\mu$$C and $$5\mu$$C.

This matches option 3.

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 $$\sqrt{\dfrac{3qE}{2ml}}$$.

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.

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: $$\vec{E}_{III} = \vec{E}_1 + \vec{E}_2 = \frac{\sigma}{2\epsilon_0}\hat{n} + \frac{\sigma}{2\epsilon_0}\hat{n} = \frac{\sigma}{\epsilon_0}\hat{n}$$

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

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

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

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

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

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.

For a uniformly charged insulating solid sphere of radius RRR, the electric field varies with distance rrr from the center as follows

$$Inside\ the\ sphere\ (r\le R)$$

Using Gauss law:

Charge enclosed within radius r,

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

where

$$\rho=\frac{Q}{\frac{4}{3}\pi R^3}$$

Using Gauss law,

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

Substituting,

$$E=\frac{Qr}{4\pi\varepsilon_0R^3}$$

Thus,

$$E\propto r$$

So electric field increases linearly from zero at the center to a maximum at the surface.

At

$$r=R$$

$$E_{\max}=\frac{Q}{4\pi\varepsilon_0R^2}$$
$$Outside\ the\ sphere(r>R)$$

Entire charge acts as if concentrated at the center:

$$E=\frac{Q}{4\pi\varepsilon_0r^2}$$

Thus,

$$E\propto\frac{1}{r^2}$$

So electric field decreases inversely as square of distance.

Graph

• From $$r=0\ to\ r=R$$: straight line increasing linearly
• At r=R : maximum field
• For r>R : decreasing inverse-square curve

So the graph starts from origin, rises linearly, then falls as a hyperbola-like curve.

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.

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: $$V_0/2$$

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 :

$$24$$

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.

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:

$$|Q|=12\varepsilon_0$$

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 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 $$y(a^2 + y^2)^{-3/2}$$:

$$\frac{d}{dy} \left[ y(a^2 + y^2)^{-3/2} \right] = 0$$

$$1 \cdot (a^2 + y^2)^{-3/2} + y \left( -\frac{3}{2} \right)(a^2 + y^2)^{-5/2}(2y) = 0$$

$$(a^2 + y^2)^{-3/2} - 3y^2(a^2 + y^2)^{-5/2} = 0$$

$$(a^2 + y^2)^1 - 3y^2 = 0$$

$$a^2 - 2y^2 = 0$$

$$2y^2 = a^2 \implies y = \frac{a}{\sqrt{2}}$$

A cubical volume is bounded by $$x = 0, x = a, y = 0, y = a, z = 0, z = a$$ and it is placed in an electric field $$\vec{E} = E_0 x \hat{i}$$ with $$E_0 = 4 \times 10^4$$ NC$$^{-1}$$m$$^{-1}$$. Here $$a = 2$$ cm. We need to find the charge $$Q$$ enclosed in units of $$10^{-14}$$ C.

Gauss's law states that the enclosed charge equals $$\epsilon_0 \oint \vec{E} \cdot d\vec{A}$$. Since $$\vec{E} = E_0 x \hat{i}$$, only the faces perpendicular to $$\hat{i}$$ give nonzero flux. At $$x = 0$$ the field is zero, so its contribution is zero. At $$x = a$$ the field is $$\vec{E} = E_0 a \hat{i}$$ and the outward normal is $$\hat{i}$$, with area $$a^2$$.

The resulting flux through the face at $$x = a$$ is $$\Phi = E_0 a \cdot a^2 = E_0 a^3$$.

The enclosed charge becomes $$Q = \epsilon_0 E_0 a^3$$. Substituting $$\epsilon_0 = 9 \times 10^{-12}$$, $$E_0 = 4 \times 10^4$$, and $$a = 2 \times 10^{-2}$$ gives
$$Q = (9 \times 10^{-12}) \times (4 \times 10^4) \times (2 \times 10^{-2})^3$$
$$= 9 \times 10^{-12} \times 4 \times 10^4 \times 8 \times 10^{-6}$$
$$= (9 \times 4 \times 8) \times 10^{-14}$$
$$= 288 \times 10^{-14} \text{ C}$$.

Therefore, $$Q = 288$$.

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

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

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.

Total charge available: $$Q = 4\,\mu C$$. This charge is split into two parts: one part is $$q$$ and the other is $$Q-q = 4 - q$$, with $$0 \le q \le 4$$.

The two charges are kept at a fixed separation $$r$$. Coulomb’s law gives the magnitude of the electrostatic force between them as

$$F = k\,\frac{q\,(4-q)}{r^{2}}$$

where $$k = 9 \times 10^{9}\,{\rm N\,m^{2}\,C^{-2}}$$ is Coulomb’s constant.
Since $$k$$ and $$r$$ are constants, maximising $$F$$ is the same as maximising the product

$$f(q) = q\,(4-q)$$

Expand $$f(q)$$:

$$f(q) = 4q - q^{2}$$

This is a concave-down parabola in $$q$$. The maximum of a quadratic $$-q^{2}+4q$$ occurs at its vertex.
For a quadratic $$ax^{2}+bx+c$$ with $$a \lt 0$$, the vertex is at $$x = -\frac{b}{2a}$$.
Here, $$a = -1$$ and $$b = 4$$, so

$$q_{\text{max}} = -\frac{4}{2(-1)} = 2$$

Thus the first charge should be $$q = 2\,\mu C$$ and the second charge is

$$(4 - q)\,\mu C = (4 - 2)\,\mu C = 2\,\mu C$$

Therefore, to obtain the maximum possible electrostatic force, the charge should be divided equally:

Case of maximum force: $$2\,\mu C$$ and $$2\,\mu C$$

Option B is correct.

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 need to find the electric field at a point $$P$$ located at a distance $$r$$ ($$r < R$$) from the center of a spherically symmetric charge distribution.

The charge density is given as:

$$\rho(r) = \rho_0 \left( \frac{3}{4} - \frac{r}{R} \right) \quad \text{for } r \le R$$

Step 1: Applying Gauss's Law

According to Gauss's Law, the electric flux through a spherical Gaussian surface of radius $$r$$ is equal to the enclosed charge ($$Q_{\text{encl}}$$) divided by the permittivity of free space ($$\varepsilon_0$$):

$$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{encl}}}{\varepsilon_0}$$

Due to spherical symmetry, the electric field $$E$$ is radial and constant in magnitude at a distance $$r$$. The surface area of the sphere is $$4\pi r^2$$, so the left side of the equation simplifies to:

$$E \cdot (4\pi r^2) = \frac{Q_{\text{encl}}}{\varepsilon_0} \implies E = \frac{Q_{\text{encl}}}{4\pi \varepsilon_0 r^2}$$

Step 2: Finding the Enclosed Charge ($$Q_{\text{encl}}$$)

The total charge enclosed within a sphere of radius $$r$$ is found by integrating the charge density over the volume of thin spherical shells of thickness $$dx$$ from $$0$$ to $$r$$:

$$Q_{\text{encl}} = \int_0^r \rho(x) \cdot (4\pi x^2 \, dx)$$

Substituting the given charge density formula into the integral expression:

$$Q_{\text{encl}} = 4\pi \int_0^r \rho_0 \left( \frac{3}{4} - \frac{x}{R} \right) x^2 \, dx$$

$$Q_{\text{encl}} = 4\pi \rho_0 \int_0^r \left( \frac{3}{4}x^2 - \frac{x^3}{R} \right) dx$$

Performing the integration term by term:

$$Q_{\text{encl}} = 4\pi \rho_0 \left[ \frac{3}{4} \cdot \frac{x^3}{3} - \frac{x^4}{4R} \right]_0^r$$

$$Q_{\text{encl}} = 4\pi \rho_0 \left( \frac{r^3}{4} - \frac{r^4}{4R} \right)$$

Factoring out $$\frac{r^3}{4}$$ from the expression:

$$Q_{\text{encl}} = 4\pi \rho_0 \cdot \frac{r^3}{4} \left( 1 - \frac{r}{R} \right) = \pi \rho_0 r^3 \left( 1 - \frac{r}{R} \right)$$

Step 3: Calculating the Electric Field ($$E$$)

Now, substitute the enclosed charge value back into the Gauss's Law expression:

$$E = \frac{\pi \rho_0 r^3 \left( 1 - \frac{r}{R} \right)}{4\pi \varepsilon_0 r^2}$$

Canceling out $$\pi$$ and $$r^2$$ from the numerator and denominator simplifies the equation to:

$$E = \frac{\rho_0 r}{4\varepsilon_0} \left( 1 - \frac{r}{R} \right)$$

Therefore, the correct choice is matching the marked green option:

$$\frac{\rho_0 r}{4\varepsilon_0} \left( 1 - \frac{r}{R} \right)$$

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.

Charges are:

At A:

$$\frac{q}{2}$$

At D:

q

At C:

$$\frac{q}{2}$$

Side of square:

a

Find net electric field at B.

Field Charge due to A

Distance

AB=a

So

$$E_A=\frac{k(q/2)}{a^2}=\frac{kq}{2a^2}$$

Direction: along +x (to right).

Field due to Charge at C

Distance

BC=a

So

$$E_C=\frac{kq}{2a^2}$$

Direction downward (−y).

Field due to charge at D

Diagonal distance

DB= 2a

So

$$E_D=\frac{kq}{(\sqrt{2}a)^2}$$

Direction along diagonal DB, making $$45^{\circ}$$.

Components:

$$E_{Dx}=\frac{kq}{2\sqrt{\ 2}a^2}$$

$$E_{Dy}=\frac{kq}{2\sqrt{\ 2}a^2}$$
Net components

X-component:

$$E_x=\frac{kq}{2a^2}+\frac{kq}{2\sqrt{2}a^2}$$

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

Y-component:

$$E_y=-\frac{kq}{2a^2}-\frac{kq}{2\sqrt{2}a^2}$$

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

So

$$|E_x|=|E_y|$$

Hence magnitude

$$E=\sqrt{\ 2}E_x$$

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

Simplify:

$$E=\frac{kq}{2a^2}(1+\sqrt{2})$$

$$\frac{q}{4\pi\varepsilon_0 a^2}\left(\frac{1}{\sqrt{2}} + \frac{1}{2}\right)$$

Let the two identical charges $$Q$$ be positioned along the horizontal axis, separated by a distance $$d$$. The midpoint acts as the origin, placing the charges at $$(-d/2, 0)$$ and $$(d/2, 0)$$.

The third charge $$q$$ is placed on the perpendicular bisector at a distance $$x$$ from the origin, at coordinates $$(0, x)$$.

The distance $$r$$ from either charge $$Q$$ to the charge $$q$$ is:    $$r = \sqrt{x^2 + \left(\frac{d}{2}\right)^2}$$

The magnitude of the Coulomb force exerted by one charge $$Q$$ on $$q$$ is:   $$F_1 = \frac{kQq}{r^2} = \frac{kQq}{x^2 + \left(\frac{d}{2}\right)^2}$$

By symmetry, the horizontal components of the forces cancel each other out, while the vertical components along the bisector add up:  $$F_{\text{net}} = 2F_1\cos\theta$$

Where $$\cos\theta = \frac{x}{r} = \frac{x}{\sqrt{x^2 + \left(\frac{d}{2}\right)^2}}$$. Substituting this back gives:   $$F_{\text{net}} = \frac{2kQqx}{\left[x^2 + \left(\frac{d}{2}\right)^2\right]^{3/2}}$$

To find the value of $$x$$ that yields the maximum force, we set the first derivative with respect to $$x$$ equal to zero:

$$\frac{d}{dx}\left[\frac{x}{\left(x^2 + \left(\frac{d}{2}\right)^2\right)^{3/2}}\right] = 0$$ $$\implies $$ $$\frac{\left(x^2 + \left(\frac{d}{2}\right)^2\right)^{3/2}(1) - x \cdot \frac{3}{2}\left(x^2 + \left(\frac{d}{2}\right)^2\right)^{1/2}(2x)}{\left(x^2 + \left(\frac{d}{2}\right)^2\right)^3} = 0$$

$$\left(x^2 + \left(\frac{d}{2}\right)^2\right) - 3x^2 = 0$$ $$\implies$$ $$\left(\frac{d}{2}\right)^2 = 2x^2$$ $$\implies$$ $$x^2 = \frac{d^2}{8}$$

$$\implies x = \frac{d}{2\sqrt{2}}$$

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.

A charge $$q$$ is placed at the centre of a closed hemispherical non-conducting surface. We need to find the total flux through the flat surface.

Apply Gauss's Law to the closed hemispherical surface.

The closed surface consists of two parts: the curved hemispherical surface and the flat circular base. Since the charge $$q$$ is enclosed within this closed surface, by Gauss's law:

$$\Phi_{\text{total}} = \Phi_{\text{curved}} + \Phi_{\text{flat}} = \frac{q}{\varepsilon_0}$$

Calculate the flux through the curved surface using symmetry.

Imagine a complete sphere centred at the charge $$q$$. The total flux through the full sphere is $$\frac{q}{\varepsilon_0}$$. Since the charge is at the centre, the flux is uniformly distributed over the sphere. The curved hemisphere is exactly half of this full sphere, so:

$$\Phi_{\text{curved}} = \frac{1}{2} \times \frac{q}{\varepsilon_0} = \frac{q}{2\varepsilon_0}$$

Find the flux through the flat surface.

$$\Phi_{\text{flat}} = \Phi_{\text{total}} - \Phi_{\text{curved}} = \frac{q}{\varepsilon_0} - \frac{q}{2\varepsilon_0} = \frac{q}{2\varepsilon_0}$$

Therefore, the total flux passing through the flat surface is $$\dfrac{q}{2\varepsilon_0}$$.

The correct answer is Option B.

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.

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.

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.

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.

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.

The electrostatic force between the two charges $$q$$ and $$(Q - q)$$ separated by distance $$r$$ is $$F = k\frac{q(Q-q)}{r^2}$$.

To maximize $$F$$, we treat $$r$$ and $$Q$$ as constants and differentiate with respect to $$q$$: $$\frac{dF}{dq} = \frac{k}{r^2}(Q - 2q) = 0$$, giving $$q = \frac{Q}{2}$$, i.e., $$Q = 2q$$.

This can also be verified by the second derivative: $$\frac{d^2F}{dq^2} = -\frac{2k}{r^2} < 0$$, confirming a maximum. The repulsion is maximised when the charge $$Q$$ is divided into two equal halves, so $$Q = 2q$$.

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.

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

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 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 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.

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.

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 $$x = 3$$):

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

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

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

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

$$\text{The electric field in the region above it points straight up along the positive y-axis:}$$

$$\vec{E}_1 = \frac{\sigma}{2\varepsilon_0}\hat{y}$$

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:

$$\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2$$

$$\vec{E}_{\text{net}} = \frac{\sigma}{2\varepsilon_0}\hat{y} + \frac{\sigma}{2\varepsilon_0}\left(-\frac{1}{2}\hat{x} - \frac{\sqrt{3}}{2}\hat{y}\right)$$

$$\vec{E}_{\text{net}} = \frac{\sigma}{2\varepsilon_0}\left[\left(1 - \frac{\sqrt{3}}{2}\right)\hat{y} - \frac{1}{2}\hat{x}\right]$$

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

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

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

Under this small-angle approximation, the torque becomes

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

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

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

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

Rearranging,

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

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

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

Taking the positive square root, the angular frequency is

Hence, the correct answer is Option B.

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 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.