JEE Electric Charges & Fields 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: $$\displaystyle\frac{Q}{60\varepsilon_0}$$

AtPP, force on charge Q is zero if net electric field is zero.

Field due to dipole 1 (axial point):

$$E_1=\frac{1}{4\pi\epsilon_0}\frac{2pr}{(r^2-a^2)^2}$$

where

p=2aq

Direction is leftward (toward −q).

Field due to dipole 2 (equatorial point):

$$E_2=\frac{1}{4\pi\epsilon_0}\frac{p}{(r^2+a^2)^{3/2}}$$

Direction is rightward (opposite to dipole moment).

For zero net force,

$$E_1=E_2$$

So,

$$\frac{2r}{(r^2-a^2)^2}=\frac{1}{(r^2+a^2)^{3/2}}$$

Let

$$x=\frac{a}{r}$$

Then

$$r^2-a^2=r^2(1-x^2)$$

$$r^2+a^2=r^2(1+x^2)$$

Substitute:

$$\frac{2}{(1-x^2)^2}=\frac{1}{(1+x^2)^{3/2}}$$

Solving,

$$x=\sqrt{5}-2$$

moment of inertia about center:

$$I=2m\left(\frac{l}{2}\right)^2=\frac{ml^2}{2}$$

For small oscillations,

$$I\omega^2=−pE_0θ$$

with

p=ql

So

$$\omega=\sqrt{\frac{pE_0}{I}}=\sqrt{\frac{qlE_0}{ml^2/2}}=\sqrt{\frac{2qE_0}{ml}}$$

Hence

$$T=\frac{2\pi}{\omega}$$

$$T=2\pi\sqrt{\frac{ml}{2qE_0}}$$

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.

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.

Electric field acts along +x, so force on charge is

F=qE

Initially particle is at angle

$$60^{\circ}$$

with x-axis and string length l.

Initial coordinates:

$$x_i=l\cos60^{\circ}=\frac{l}{2}$$

When it crosses x-axis, particle is at

$$x_f=l$$

(since still constrained on circle radius lll).

Displacement along field:

$$\Delta x=l-\frac{l}{2}=\frac{l}{2}$$

Work done by electric field:

$$W=qE\Delta x$$

$$=\frac{qEl}{2}$$

Tension does no work.

Using work-energy theorem:

$$\frac{1}{2}mv^2=\frac{qEl}{2}$$

So

$$mv^2=qEl$$

so

$$\sqrt{\frac{qEl}{m}}$$

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 

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

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.

The electric flux is given as $$\phi = \alpha \sigma + \beta \lambda$$, where $$\sigma$$ is the surface charge density and $$\lambda$$ is the linear charge density. We need to find what $$\frac{\alpha}{\beta}$$ represents.

Electric flux has dimensions derived from Gauss's law. The dimensional formula for electric flux is $$[\phi] = [\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2} \cdot \text{C}^{-1}]$$.

The surface charge density $$\sigma = \frac{\text{charge}}{\text{area}}$$, so its dimensional formula is $$[\sigma] = [\text{C} \cdot \text{m}^{-2}]$$.

The linear charge density $$\lambda = \frac{\text{charge}}{\text{length}}$$, so its dimensional formula is $$[\lambda] = [\text{C} \cdot \text{m}^{-1}]$$.

In the expression $$\phi = \alpha \sigma + \beta \lambda$$, both terms must have the same dimensions as flux.

For the term $$\alpha \sigma$$:

$$[\alpha \sigma] = [\alpha] \cdot [\sigma] = [\alpha] \cdot [\text{C} \cdot \text{m}^{-2}] = [\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2} \cdot \text{C}^{-1}]$$

Solving for $$[\alpha]$$:

$$[\alpha] = \frac{[\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2} \cdot \text{C}^{-1}]}{[\text{C} \cdot \text{m}^{-2}]} = [\text{kg} \cdot \text{m}^5 \cdot \text{s}^{-2} \cdot \text{C}^{-2}]$$

For the term $$\beta \lambda$$:

$$[\beta \lambda] = [\beta] \cdot [\lambda] = [\beta] \cdot [\text{C} \cdot \text{m}^{-1}] = [\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2} \cdot \text{C}^{-1}]$$

Solving for $$[\beta]$$:

$$[\beta] = \frac{[\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2} \cdot \text{C}^{-1}]}{[\text{C} \cdot \text{m}^{-1}]} = [\text{kg} \cdot \text{m}^4 \cdot \text{s}^{-2} \cdot \text{C}^{-2}]$$

Now, the ratio $$\frac{\alpha}{\beta}$$ has dimensions:

$$\left[\frac{\alpha}{\beta}\right] = \frac{[\alpha]}{[\beta]} = \frac{[\text{kg} \cdot \text{m}^5 \cdot \text{s}^{-2} \cdot \text{C}^{-2}]}{[\text{kg} \cdot \text{m}^4 \cdot \text{s}^{-2} \cdot \text{C}^{-2}]} = [\text{m}]$$

The dimension $$[\text{m}]$$ corresponds to length. Comparing with the options:

• A. Electric field has dimensions $$[\text{kg} \cdot \text{m} \cdot \text{s}^{-2} \cdot \text{C}^{-1}]$$
• B. Area has dimensions $$[\text{m}^2]$$
• C. Charge has dimensions $$[\text{C}]$$
• D. Displacement has dimensions $$[\text{m}]$$

Displacement has the same dimension as length, so $$\frac{\alpha}{\beta}$$ represents displacement.

Therefore, the correct answer is displacement (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

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

Configuration (1): charges at four corners.

Side pairs (4):

$$4\cdot\frac{kq_0^2}{a}$$

Diagonal pairs (2):

$$2\cdot\frac{kq_0^2}{a\sqrt{2}}$$

So,

$$U_1=\frac{kq_0^2}{a}(4+\sqrt{2})$$

Configuration (2): charges at midpoints of four sides.

Distance between adjacent midpoints:

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

4 such pairs:

$$4\cdot\frac{kq_0^2}{a/\sqrt{2}}=\frac{4\sqrt{2}kq_0^2}{a}$$

Distance between opposite midpoints:

a

2 such pairs:

$$2\cdot\frac{kq_0^2}{a}$$

Thus,

$$U_2=\frac{kq_0^2}{a}(4\sqrt{2}+2)$$

Work required:

$$W=U_2-U_1$$

$$=\frac{kq_0^2}{a}(3\sqrt{2}-2)$$

Field due to one infinite positively charged sheet is

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

directed away from the sheet.

There are two identical positive sheets.

At points A and B, which lie between the sheets:

• Left sheet produces field to the right.
• Right sheet produces field to the left.

Magnitudes are equal, so they cancel.

So between the sheets only the charged sphere contributes.

Outside a uniformly charged sphere,

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

Point A is closer to the sphere than BBB, hence

$$E_A>E_B$$

Now point C:

It lies outside both sheets.

Both sheet fields point left and add:

$$E_{\text{sheets}}=\frac{\sigma}{\varepsilon_0}$$

Sphere field at C also points left, so it adds.

Thus

At point D:

Again sheet contribution is

$$\frac{\sigma}{\varepsilon_0}$$

Sphere field also adds.

But D is closer to the sphere than C, so

$$E_{\text{sphere at D}}>E_{\text{sphere at C}}$$

Therefore

$$E_D>E_C$$

By Gauss law,

$$\Phi=\frac{q}{\epsilon_0}$$

So

$$q=ϵ_0Φ$$

Given

$$\Phi=-2\times10^4$$

Thus

$$q=(8.85\times10^{-12})(-2\times10^4)$$

$$=-17.7\times10^{-8}$$

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

Torque on dipole in electric field:

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

with

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

Thus

$$=1.13Nm$$

Field due to infinite positive sheet is uniform and directed away from sheet.

Dipole moment is from −q to +q, which here is also away from sheet, so

$$⃗\vec{p}\parallel\vec{E}$$

Torque on dipole:

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

Here

$$\theta=0$$

so

$$τ=0$$

Potential energy:

U=−p​⋅E=−pEcosθ

$$U=-pE$$

which is minimum.

Also forces on +q and −q are equal and opposite, so net force is zero.

Hence correct statement:

Potential energy is minimum and torque is zero.

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, $$a$$.

Use solid angle idea.

Flux through any surface due to point charge is

$$\Phi=\frac{q}{4\pi\varepsilon_0}\Omega$$

where Ω\OmegaΩ is solid angle subtended by the surface at the charge.

The charge is located on the normal through the center of the square, and from the geometry shown, the whole square subtends a solid angle corresponding to a face of a cube with charge at cube center.

Flux through entire square is therefore

$$\Phi_{\text{square}}=\frac{q}{6\varepsilon_0}$$

since total flux

$$\frac{q}{\varepsilon_0}$$

gets equally shared among 6 cube faces.

Given

q=1C

so

$$\Phi_{\text{square}}=\frac{1}{6\varepsilon_0}$$

Now look at shaded region.

its area is

$$\frac{5a^2}{8}$$

So shaded fraction of the square is

$$\frac{5}{8}$$

Flux through the whole square (one face of the imaginary cube) is

$$\frac{q}{6\varepsilon_0}$$

Hence flux through shaded region is

$$\Phi=\frac{5}{8}\cdot\frac{q}{6\varepsilon_0}$$

With

q=1

$$\Phi=\frac{5}{48\varepsilon_0}$$

Comparing with

$$\frac{5}{p}\cdot\frac{1}{\varepsilon_0}$$

we get

$$p=48$$

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.

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 force between charges in a dielectric medium at a reduced distance.

In vacuum at distance $$r$$ the force is given by $$F = \frac{kq_1q_2}{r^2}$$. Since the dielectric constant is $$K = 5$$ and the separation is reduced to $$r/5$$, Coulomb's law in the medium yields:

$$F' = \frac{kq_1q_2}{K \cdot (r/5)^2} = \frac{kq_1q_2}{5 \cdot r^2/25} = \frac{25kq_1q_2}{5r^2} = \frac{5kq_1q_2}{r^2} = 5F$$

This shows that although the dielectric reduces the force by a factor of $$K = 5$$, the shorter separation increases it by a factor of 25, resulting in a net factor of $$\frac{25}{5} = 5$$ times the original force. Therefore, the correct answer is Option 2: $$5F$$.

The electrostatic potential due to an electric dipole at a point at distance $$r$$ from the center of the dipole is given by:

$$V = \frac{1}{4\pi\epsilon_0} \cdot \frac{p\cos\theta}{r^2}$$

where $$p$$ is the dipole moment and $$\theta$$ is the angle between the dipole axis and the line joining the center of the dipole to the point.

From this expression, we can see that:

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

Note: This is different from a point charge, where $$V \propto \frac{1}{r}$$, and from the electric field of a dipole, where $$E \propto \frac{1}{r^3}$$.

The correct answer is $$\frac{1}{r^2}$$.

This question asks why vehicles carrying inflammable (flammable) fluids have metallic chains that touch the ground.

When a vehicle moves along a road, friction between the tyres and the road surface, as well as friction between the vehicle body and the air, causes a build-up of static electric charge on the vehicle's metallic body. This is similar to how rubbing a balloon on cloth creates static charge.

For ordinary vehicles, this small accumulation of charge is not a major concern. However, for vehicles carrying inflammable fluids (such as petroleum tankers), even a tiny spark can ignite the vapours and cause a catastrophic explosion.

A spark occurs when the accumulated charge reaches a high enough potential difference to ionise the surrounding air and discharge suddenly. To prevent this, metallic chains are attached to the vehicle so that they drag along the ground. Since metals are good conductors of electricity, the chain provides a continuous conducting path from the vehicle body to the ground (earth). This allows the excess charge to flow harmlessly to the earth as it builds up, keeping the vehicle at ground potential and preventing any dangerous accumulation of charge.

This is an application of the principle of earthing or grounding, which is widely used in electrical safety.

Therefore, the correct answer is Option (4): To conduct excess charge due to air friction to ground and prevent sparking.

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.

Initial: P and S each have Q. Force = $$kQ^2/d^2 = 16$$ N.

R (uncharged) touches P: charge on each = Q/2. Then R (Q/2) touches S (Q): charge on each = 3Q/4.

New charges: P = Q/2, S = 3Q/4.

Force = $$k(Q/2)(3Q/4)/d^2 = \frac{3}{8}kQ^2/d^2 = \frac{3}{8}×16 = 6$$ N.

The correct answer is Option (2): 6 N.

Charge q is at the center of one face of the cube.

Imagine attaching an identical cube to this face so the charge lies on the common face center, which becomes inside a rectangular box made of two cubes.

Now the charge is at the center of this combined box.

By Gauss law, total flux through the combined closed surface is

$$\Phi_{\text{total}}=\frac{q}{\varepsilon_0}$$

The two cubes are identical and symmetric, so flux splits equally between them.

Hence flux linked with one cube is

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

Electric flux is

$$⃗\phi=\vec{E}\cdot\vec{A}$$

Given

$$E=\left(6i+5j^​+3k\right)$$

Surface lies in YZ-plane, so area vector is along x-axis:

$$A⃗=30i$$

Therefore,

Dot product would be $$=\ 6\times\ 30$$

=180

We need to find the electric field at any point on the surface of a thin spherical shell with uniform surface charge density $$\sigma$$ and radius $$R$$.

The total charge on the shell is $$Q = \sigma \cdot 4\pi R^2$$.

By Gauss's law, the electric field just outside the shell (at distance $$R$$ from center) is:

$$E_{\text{outside}} = \frac{Q}{4\pi\epsilon_0 R^2} = \frac{\sigma \cdot 4\pi R^2}{4\pi\epsilon_0 R^2} = \frac{\sigma}{\epsilon_0}$$

The electric field just inside the shell is $$E_{\text{inside}} = 0$$ (no enclosed charge for a Gaussian surface inside).

For a conducting shell or a charged surface, the electric field AT the surface is taken as the value just outside the surface, which is $$\frac{\sigma}{\epsilon_0}$$. This is the standard convention used in JEE for the field "on the surface" of a spherical shell.

The correct answer is Option B: $$\sigma/\epsilon_0$$.

Two conducting spheres of radii $$a$$ and $$b$$ are connected by a wire. We need the ratio of their charges. When two conductors are connected by a wire, charge flows until they reach the same electric potential.

The potential of a conducting sphere with charge $$Q$$ and radius $$r$$ is:

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

Equating the potentials of the two spheres gives $$V_1 = V_2 \implies \frac{Q_1}{4\pi\varepsilon_0 a} = \frac{Q_2}{4\pi\varepsilon_0 b},$$ which simplifies to $$\frac{Q_1}{a} = \frac{Q_2}{b} \implies \frac{Q_1}{Q_2} = \frac{a}{b}.$$

The correct answer is Option (1): $$\frac{a}{b}$$.

Electric field due to an infinite line charge:

$$E=\frac{\lambda}{2\pi\epsilon_0r}$$

Force on electron provides centripetal force:

$$eE=\frac{mv^2}{r}$$

Substitute E:

$$e\cdot\frac{\lambda}{2\pi\epsilon_0r}=\frac{mv^2}{r}$$

r cancels:

$$\frac{e\lambda}{2\pi\epsilon_0}=mv^2$$

So

$$v^2=\frac{e\lambda}{2\pi\epsilon_0m}$$

which is constant.

Kinetic energy:

$$K=\frac{1}{2}mv^2$$

Also constant, independent of r.

so its constant

Two charges $$q$$ and $$3q$$ are separated by a distance $$r$$. We need to find the point between them where the resultant electric field is zero. Let the charge $$q$$ be at position O and the charge $$3q$$ at position $$r$$; let the point where the electric field vanishes be at a distance $$x$$ from $$q$$ (and hence at distance $$(r - x)$$ from $$3q$$). Since both charges are positive, the zero-field point lies between them where their fields are oppositely directed.

At distance $$x$$ from $$q$$, its field points away from $$q$$ (toward $$3q$$), while the field due to $$3q$$ at distance $$(r-x)$$ points away from $$3q$$ (toward $$q$$). Setting their magnitudes equal gives:

$$\frac{kq}{x^2} = \frac{k(3q)}{(r-x)^2}$$

Canceling $$kq$$ yields $$\frac{1}{x^2} = \frac{3}{(r-x)^2}$$. Cross-multiplying gives $$(r-x)^2 = 3x^2$$, and taking the positive root (since $$x>0$$ and $$r-x>0$$) leads to $$r - x = \sqrt{3}\,x$$, hence $$r = x(1 + \sqrt{3})$$ and

$$x = \frac{r}{1 + \sqrt{3}}$$

Numerically, with $$\sqrt{3}\approx1.732$$, this yields $$x\approx\frac{r}{2.732}\approx0.366r$$, confirming the point is between the charges and closer to the smaller charge $$q$$, as expected.

The correct answer is Option (3): $$\frac{r}{1 + \sqrt{3}}$$

Let the fixed bead be at point $$A$$ on the hoop, whose centre is $$O$$. Measure the angular position of the movable bead by angle $$\theta$$ made by $$OA$$ and the radius $$OP$$ that joins the centre to the movable bead. Equilibrium occurs when the tangential component of the electrostatic force on the movable bead is zero.

Because the only force is the repulsion from the fixed bead, the force line is along the straight chord joining the two beads. At $$\theta = \pi$$ (diametrically opposite to the fixed bead) this chord passes through the centre, so its direction is purely radial and has no tangential component. Hence the equilibrium position is $$\theta_0 = \pi$$.

Introduce a small angular displacement $$\phi$$ about this equilibrium: $$\theta = \pi + \phi \qquad(|\phi|\ll 1).$$ We shall find the potential energy as a function of $$\phi$$ upto the term in $$\phi^2$$, and compare it with the quadratic form $$\tfrac12 I\omega^2\phi^2$$ of simple harmonic motion.

Magnitude of separation between the beads (straight line, not arc): The angle subtended at the centre between the two position vectors is $$\delta=\theta = \pi+\phi$$, hence $$r = 2R\sin\frac{\delta}{2}=2R\sin\!\left(\frac{\pi+\phi}{2}\right) = 2R\cos\frac{\phi}{2}.$$

Electrostatic potential energy of the two-charge system: $$U(\phi)=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{r} =\frac{1}{4\pi\varepsilon_0}\frac{q^2}{2R\cos(\phi/2)}.$$

Expand for small $$\phi$$ using $$\cos(\phi/2)\approx 1-\frac{\phi^2}{8}$$ and for any small $$x$$, $$\frac1{1-x}\approx 1+x$$: $$\frac{1}{\cos(\phi/2)}\approx 1+\frac{\phi^2}{8}.$$ Therefore $$U(\phi)\approx\frac{1}{4\pi\varepsilon_0}\,\frac{q^2}{2R}\Bigl(1+\frac{\phi^2}{8}\Bigr) =U_0+\frac{q^2}{4\pi\varepsilon_0}\,\frac{\phi^2}{16R},$$ where $$U_0=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{2R}$$ is the minimum (constant) part.

The change in potential energy near equilibrium is thus $$\Delta U=\frac{q^2}{4\pi\varepsilon_0}\frac{\phi^2}{16R}.$$

The movable bead of mass $$m$$ moves along a circle of radius $$R$$, so its moment of inertia about the centre is $$I=mR^2.$$ For small angular oscillations, $$\Delta U=\tfrac12 I\omega^2\phi^2$$. Equating the coefficients of $$\phi^2$$:

$$\frac{1}{2}mR^2\omega^2=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{16R}$$ $$\Rightarrow\;\omega^2=\frac{1}{4\pi\varepsilon_0}\,\frac{q^2}{8mR^3} =\frac{q^2}{32\pi\varepsilon_0 R^3 m}.$$

Hence the square of the angular frequency is $$\boxed{\dfrac{q^2}{32\pi\varepsilon_0 R^3 m}}.$$

Option B which is: $$\dfrac{q^2}{32\pi\varepsilon_0 R^3 m}$$

The radius of the ring is
$$R = 30\text{ cm}=0.30\text{ m}.$$

The charge uniformly spread over the ring is
$$Q = 2\pi\text{ C},$$
and the point charge kept at the centre is
$$q = 30\text{ pC}=30\times10^{-12}\text{ C}.$$

Choose a small element of the ring that subtends an angle $$d\theta$$ at the centre.
Length of the element
$$dl = R\,d\theta,$$
charge on the element
$$dq = \frac{Q}{2\pi R}\,dl \;=\; \frac{Q}{2\pi}\,d\theta.$$

The electrostatic force on this element due to the central charge $$q$$ is radially outward and equals
$$dF = \frac{1}{4\pi\varepsilon_0}\,\frac{q\,dq}{R^2} = k\,\frac{q}{R^2}\,\frac{Q}{2\pi}\,d\theta,$$
where $$k=\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\text{N\,m}^2\text{/C}^2.$$

The element is held in equilibrium by the tension $$T$$ in the ring. The two tension forces at its ends are tangential; their radial components add up to an inward force $$2T\sin\!\left(\frac{d\theta}{2}\right)\;\approx\;T\,d\theta$$ (because $$d\theta$$ is infinitesimal).

Equating the outward electrostatic force and the inward force due to tension:
$$T\,d\theta = k\,\frac{qQ}{2\pi R^{2}}\,d\theta \;\;\Longrightarrow\;\; T = k\,\frac{qQ}{2\pi R^{2}}.$$

In the given data $$Q = 2\pi\,$C, so the factors $$2$$\pi$$$$ cancel, giving a very simple result:
$$T = k\,$$\frac{q}{R^{2}$$}.$$

Substituting the numerical values:
$$T = 9$$\times$$10^{9}\,$$\frac{30\times10^{-12}$$}{(0.30)^{2}} = 9$$\times$$10^{9}$$\times$$30$$\times$$10^{-12}$$\times$$$$\frac{1}{0.09}$$$$ $$\phantom{T} = 9$$\times$$30$$\times$$10^{-3}$$\times$$$$\frac{1}{0.09}$$ = 270$$\times$$10^{-3}$$\times$$$$\frac{1}{0.09}$$ = 0.27$$\times$$$$\frac{1}{0.09}$$ = 3\;$$\text{N}$$.$$

Hence the tension in the metallic ring is $$\mathbf{3\;N}.$$

Take away from sheet as positive.

Then electron starts moving away:

u=1

Displacement to reach sheet:

s=−1

Acceleration is toward sheet, hence negative:

$$a=-\frac{e\sigma}{2m\epsilon_0}$$

Using

$$s=ut+\frac{1}{2}at^2$$

with t=1,

$$-1=1+\frac{1}{2}a$$

$$\frac{a}{2}=-2$$

a=−4

Now

$$-\frac{e\sigma}{2m\epsilon_0}=-4$$

$$\frac{e\sigma}{2m\epsilon_0}=4$$

$$\sigma=8\frac{m\epsilon_0}{e}$$

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

Given

$$E=2\times10^4N/C$$

$$m=2g=2\times10^{-3}kg$$

l=20cm

Particle stays 10 cm from wall, so horizontal displacement is

x=10cm

From geometry,

$$\sin\theta=\frac{10}{20}=\frac{1}{2}$$

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

For equilibrium,

$$T\sin\theta=qE$$

$$T\cos\theta=mg$$

Dividing,

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

So

$$q=\frac{mg\tan\theta}{E}$$

Substitute,

$$q=\frac{(2\times10^{-3})(10)\cdot\frac{1}{\sqrt{3}}}{2\times10^4}$$

$$=\frac{10^{-6}}{\sqrt{3}}C$$

$$=\frac{1}{\sqrt{3}}\mu C$$

Dipole moment magnitude:

$$p=q(2r)=2qr$$

At point P on axial line, distance from center is

$$OP=r$$

Axial field of dipole:

$$E_P=\frac{2kp}{r^3}$$

Given this is

E

so

$$E=\frac{2kp}{r^3}$$

Now point R lies on equatorial line at distance

OR=2r

Equatorial field of dipole:

Then

$$E_R=\frac{kp}{(2r)^3}$$

$$=\frac{kp}{8r^3}$$

Using

$$E=\frac{2kp}{r^3}$$

we get

$$E_R=\frac{1}{16}E$$

So

$$\frac{E}{x}=\frac{E}{16}$$

Two identical charged spheres suspended by equal-length strings make angle $$\theta$$ in air. When suspended in water, the angle remains the same. Density of sphere = $$1.5$$ g/cc, density of water = $$1$$ g/cc.

In air, each sphere is in equilibrium under the action of electrostatic repulsion, gravity, and string tension; the balance of horizontal and vertical forces yields $$\tan(\theta/2) = \frac{F_e}{mg}$$, where $$F_e$$ is the electrostatic force and $$mg$$ is the weight of the sphere.

When the spheres are suspended in water, the electrostatic force is reduced by the dielectric constant $$K$$ so that $$F_e' = F_e/K$$, and the effective weight is reduced by the buoyant force: $$W' = mg - \rho_w V g = Vg(\rho_s - \rho_w)$$, where $$\rho_s$$ and $$\rho_w$$ are the densities of the sphere and water respectively and $$V$$ is the volume of the sphere. The equilibrium in water therefore gives $$\tan(\theta/2) = \frac{F_e/K}{Vg(\rho_s - \rho_w)}$$.

Since the angle remains the same in both media, we equate the two expressions: $$\frac{F_e}{mg} = \frac{F_e/K}{Vg(\rho_s - \rho_w)}$$. Noting that $$m = \rho_s V$$, this reduces to $$\frac{1}{\rho_s} = \frac{1}{K(\rho_s - \rho_w)}$$, which leads to $$K = \frac{\rho_s}{\rho_s - \rho_w} = \frac{1.5}{1.5 - 1.0} = \frac{1.5}{0.5} = 3$$.

The dielectric constant of water is 3.

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

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

$$=32-16$$

$$=16$$

The relation between current $$I$$ and charge $$Q$$ is given by the formula $$I = \frac{dQ}{dt}$$ Integrating both sides with respect to time, we get $$Q = \int I\,dt \quad\quad -(1)$$

Here, the current is given as $$I = 3t^2 + 4t^3 \quad\quad -(2)$$

Substitute equation $$(2)$$ into equation $$(1)$$: $$Q = \int (3t^2 + 4t^3)\,dt$$

We split the integral into two parts: $$Q = \int 3t^2\,dt \;+\; \int 4t^3\,dt$$

Now we integrate each term separately. Using the power rule $$\int t^n\,dt = \frac{t^{n+1}}{n+1}$$, we have:

$$\int 3t^2\,dt = 3 \cdot \frac{t^{3}}{3} = t^3$$ $$\int 4t^3\,dt = 4 \cdot \frac{t^{4}}{4} = t^4$$

Therefore, the antiderivative is $$Q(t) = t^3 + t^4 + C_1$$ where $$C_1$$ is the constant of integration. Since we are calculating the definite charge flow between two times, the constant cancels out.

The amount of charge flowing from $$t = 1\text{ s}$$ to $$t = 2\text{ s}$$ is given by the definite integral: $$Q_{1 \to 2} = \Bigl[t^3 + t^4\Bigr]_{1}^{2} \quad\quad -(3)$$

Evaluate at the upper limit $$t = 2$$: $$2^3 + 2^4 = 8 + 16 = 24$$

Evaluate at the lower limit $$t = 1$$: $$1^3 + 1^4 = 1 + 1 = 2$$

Subtract to find the net charge: $$Q_{1 \to 2} = 24 - 2 = 22\text{ C}$$

Final Answer: The amount of electric charge that flows during $$t=1\text{ s}$$ to $$t=2\text{ s}$$ is $$22\text{ C}$$.

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

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

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

In air: $$\tan(\theta/2) = \frac{F_e}{mg}$$ where $$F_e$$ is the electrostatic force.

In liquid: The angle remains the same, so $$\tan(\theta/2)$$ is unchanged. This means:

$$\frac{F_e'}{m'g} = \frac{F_e}{mg}$$

where $$F_e' = \frac{F_e}{K}$$ (dielectric constant $$K$$) and $$m'g = mg - \text{buoyancy} = mg(1 - \frac{\rho_l}{\rho_s})$$.

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

$$\frac{1}{K} = 1 - \frac{\rho_l}{\rho_s} = 1 - \frac{0.7}{1.4} = 1 - 0.5 = 0.5$$

$$K = 2$$.

Therefore, the answer is $$\boxed{2}$$.

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.

Coulomb force:

$$F_e=\frac{1}{4\pi\epsilon_0}\frac{e^2}{r^2}$$

Gravitational force:

$$F_g=\frac{Gm_pm_e}{r^2}$$

Ratio,

Using approximate values,

$$\frac{1}{4\pi\epsilon_0}=9\times10^9$$

$$G=6.67\times10^{-11}$$

$$m_p=1.67\times10^{-27}$$

Substituting gives order

$$\frac{F_e}{F_g}\sim10^{39}$$

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:

By Gauss law,

So,

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

The electrostatic force between two charges in a medium with dielectric constant $$k$$ is given by Coulomb's law:

$$F_{\text{medium}} = \frac{1}{4\pi\epsilon_0 k} \frac{q_1 q_2}{d^2}$$

In air (which is approximately vacuum), the force is:

$$F_{\text{air}} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$$

where $$r$$ is the equivalent distance in air for the same force.

Since the force is the same in both cases, set $$F_{\text{air}} = F_{\text{medium}}$$:

$$\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} = \frac{1}{4\pi\epsilon_0 k} \frac{q_1 q_2}{d^2}$$

Cancel the common terms $$\frac{1}{4\pi\epsilon_0}$$ and $$q_1 q_2$$:

$$\frac{1}{r^2} = \frac{1}{k d^2}$$

Take the reciprocal of both sides:

$$r^2 = k d^2$$

Take the square root of both sides:

$$r = d \sqrt{k}$$

Thus, the equivalent distance in air is $$d\sqrt{k}$$.

Comparing with the options, the correct answer is A. $$d\sqrt{k}$$.

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.

By Gauss's law, the electric flux through any closed surface equals the total charge enclosed divided by $$\varepsilon_0$$:

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

The positive plate of the capacitor carries charge $$Q = CV$$.

Since the closed surface encloses only the positive plate:

$$\Phi = \frac{CV}{\varepsilon_0}$$

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.

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 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: $$V_0/2$$

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.

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.

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.

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.

Two identical positive charges $$Q$$ are fixed at a distance $$2a$$ apart, and a negative charge $$q_0$$ (mass $$m$$) is placed at the midpoint. We wish to find the time period of small oscillations of $$q_0$$ along the line joining the charges.

We place the two fixed charges at $$x=-a$$ and $$x=+a$$, with the negative charge $$q_0$$ initially at the origin. Suppose it is displaced by a small distance $$x$$ (where $$x\ll a$$) toward the charge at $$+a$$.

Since $$q_0$$ is negative and both fixed charges are positive, it experiences attractive forces from each. The force due to the charge at $$+a$$ (pulling toward $$+a$$) is $$F_{\text{right}}=\frac{1}{4\pi\epsilon_0}\frac{q_0Q}{(a-x)^2},$$ while the force due to the charge at $$-a$$ (pulling toward $$-a$$) is $$F_{\text{left}}=\frac{1}{4\pi\epsilon_0}\frac{q_0Q}{(a+x)^2}.$$

Taking the $$+x$$ direction as positive, the net force on $$q_0$$ is $$F_{\text{net}}=\frac{q_0Q}{4\pi\epsilon_0}\Bigl[\frac{1}{(a-x)^2}-\frac{1}{(a+x)^2}\Bigr].$$ Because the charge is closer to $$+a$$ when displaced, the pull from that side is stronger, while the farther charge at $$-a$$ pulls back toward the origin. For negative $$q_0$$, this difference produces a restoring force toward $$x=0$$.

For small $$x$$ ($$x\ll a$$), we expand by the binomial theorem: $$\frac{1}{(a-x)^2}=\frac{1}{a^2}\Bigl(1-\frac{x}{a}\Bigr)^{-2}\approx\frac{1}{a^2}\Bigl(1+\frac{2x}{a}\Bigr),$$ $$\frac{1}{(a+x)^2}=\frac{1}{a^2}\Bigl(1+\frac{x}{a}\Bigr)^{-2}\approx\frac{1}{a^2}\Bigl(1-\frac{2x}{a}\Bigr).$$

Subtracting these approximations gives $$\frac{1}{(a-x)^2}-\frac{1}{(a+x)^2}\approx\frac{1}{a^2}\Bigl(\frac{4x}{a}\Bigr)=\frac{4x}{a^3}.$$

Substituting back into the expression for the force yields the magnitude of the restoring force $$|F|=\frac{q_0Q}{4\pi\epsilon_0}\cdot\frac{4x}{a^3}=\frac{q_0Q}{\pi\epsilon_0a^3}\,x.$$ This is of the form $$F=-kx$$, where $$k=\frac{q_0Q}{\pi\epsilon_0a^3}.$$

From the above, the angular frequency satisfies $$\omega^2=\frac{k}{m}=\frac{q_0Q}{\pi\epsilon_0ma^3},$$ and the time period is $$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{\pi\epsilon_0ma^3}{q_0Q}}=\sqrt{\frac{4\pi^3\epsilon_0ma^3}{q_0Q}}\,. $$

The correct answer is Option A: $$\sqrt{\dfrac{4\pi^3 \epsilon_0 m a^3}{q_0 Q}}$$.

Two conducting spheres $$A$$ (radius $$r_A = 5 \text{ mm}$$) and $$B$$ (radius $$r_B = 10 \text{ mm}$$) are connected by a conducting wire, so they reach the same potential in equilibrium. Thus $$V_A = V_B$$ gives $$\dfrac{kQ_A}{r_A} = \dfrac{kQ_B}{r_B}$$. From this, $$\dfrac{Q_A}{Q_B} = \dfrac{r_A}{r_B} = \dfrac{5}{10} = \dfrac{1}{2}$$.

The electric field at the surface of a conducting sphere is $$E = \dfrac{kQ}{r^2} = \dfrac{V}{r}$$. Since both spheres are at the same potential $$V$$, $$\dfrac{E_A}{E_B} = \dfrac{V/r_A}{V/r_B} = \dfrac{r_B}{r_A} = \dfrac{10}{5} = \dfrac{2}{1}$$. Therefore, $$E_A : E_B = 2 : 1$$.

The correct answer is Option B: $$2:1$$.

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:

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

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

The mass of each particle is $$m = 10$$ g $$= 0.01$$ kg, the charge is $$q = 2.0 \times 10^{-7}$$ C, the coefficient of friction is $$\mu = 0.25$$, and $$g = 10$$ m/s$$^2$$.

The two identical charged particles repel each other with the Coulomb force, and they remain in equilibrium because friction balances this repulsive force. At limiting equilibrium the Coulomb repulsive force equals the maximum static friction force:

$$F_{\text{Coulomb}} = F_{\text{friction}}$$ $$\frac{kq^2}{L^2} = \mu mg$$

Solving for $$L^2$$ gives $$L^2 = \frac{kq^2}{\mu mg}$$. Substituting numerical values: $$L^2 = \frac{9 \times 10^9 \times (2.0 \times 10^{-7})^2}{0.25 \times 0.01 \times 10}$$. Calculating the numerator: $$(2.0 \times 10^{-7})^2 = 4.0 \times 10^{-14}$$ and $$9 \times 10^9 \times 4.0 \times 10^{-14} = 36 \times 10^{-5} = 3.6 \times 10^{-4}$$. The denominator is $$0.25 \times 0.01 \times 10 = 0.025$$. Therefore, $$L^2 = \frac{3.6 \times 10^{-4}}{0.025} = \frac{3.6 \times 10^{-4}}{2.5 \times 10^{-2}} = 1.44 \times 10^{-2} \text{ m}^2$$.

Thus, $$L = \sqrt{1.44 \times 10^{-2}} = 0.12 \text{ m} = 12 \text{ cm}$$, so the separation $$L = 12$$ cm.

The correct answer is Option A.

Two point charges $$A = +8 \times 10^{-6}$$ C and $$B = -8 \times 10^{-6}$$ C are placed at a distance $$d$$ apart. The electric field at the midpoint $$O$$ is $$6.4 \times 10^4$$ N C$$^{-1}$$. We need to find the distance $$d$$.

Understand the configuration and direction of electric fields at the midpoint.

The midpoint $$O$$ is at a distance of $$d/2$$ from each charge.

The electric field due to the positive charge $$A$$ at point $$O$$ points away from $$A$$ (i.e., from $$A$$ toward $$B$$).

The electric field due to the negative charge $$B$$ at point $$O$$ points toward $$B$$ (i.e., also from $$A$$ toward $$B$$).

Since both fields point in the same direction, they add up.

Calculate the electric field due to each charge at the midpoint.

The magnitude of the electric field due to charge $$A$$ at the midpoint:

$$E_A = \frac{kq}{(d/2)^2} = \frac{4kq}{d^2}$$

Similarly, the magnitude of the electric field due to charge $$B$$ at the midpoint:

$$E_B = \frac{kq}{(d/2)^2} = \frac{4kq}{d^2}$$

where $$q = 8 \times 10^{-6}$$ C and $$k = 9 \times 10^9$$ N m$$^2$$ C$$^{-2}$$.

Write the total electric field at the midpoint.

Since both fields are in the same direction:

$$E_{total} = E_A + E_B = \frac{4kq}{d^2} + \frac{4kq}{d^2} = \frac{8kq}{d^2}$$

Substitute the known values and solve for $$d$$.

$$6.4 \times 10^4 = \frac{8 \times 9 \times 10^9 \times 8 \times 10^{-6}}{d^2}$$

Calculating the numerator:

$$8 \times 9 \times 10^9 \times 8 \times 10^{-6} = 8 \times 72 \times 10^{3} = 576 \times 10^{3} = 5.76 \times 10^5$$

So:

$$6.4 \times 10^4 = \frac{5.76 \times 10^5}{d^2}$$

Solve for $$d$$.

$$d^2 = \frac{5.76 \times 10^5}{6.4 \times 10^4} = \frac{576}{64} = 9$$

$$d = 3 \text{ m}$$

The correct answer is Option B.

Between the two plates of a parallel plate capacitor, the electric field is due to both plates. Each plate produces a uniform electric field of $$\frac{\sigma}{2\epsilon_0}$$.

Between the plates, the fields from both plates add up:

$$E_{between} = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$$

The force on the charged particle between the plates is:

$$F = qE_{between} = \frac{q\sigma}{\epsilon_0} = 10 \text{ N}$$

When one plate is removed, only one plate remains. The electric field due to a single infinite plate is:

$$E_{single} = \frac{\sigma}{2\epsilon_0}$$

The force on the charged particle now becomes:

$$F' = qE_{single} = \frac{q\sigma}{2\epsilon_0} = \frac{F}{2} = \frac{10}{2} = 5 \text{ N}$$

The correct answer is Option A.

A long cylindrical volume of radius $$R$$ contains a uniformly distributed charge of density $$\rho$$. A charge particle $$q$$ revolves around the cylinder in a circular path.

Find the electric field using Gauss's Law: Consider a cylindrical Gaussian surface of radius $$r$$ (where $$r > R$$) and length $$L$$, coaxial with the charged cylinder.

The charge enclosed by the Gaussian surface:

$$Q_{\text{enc}} = \rho \cdot \pi R^2 \cdot L$$

By Gauss's Law:

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

$$E \cdot (2\pi r L) = \frac{\rho \pi R^2 L}{\varepsilon_0}$$

Solving for $$E$$:

$$E = \frac{\rho R^2}{2\varepsilon_0 r}$$

Apply the condition for circular motion: For the charged particle $$q$$ revolving in a circular orbit of radius $$r$$, the electrostatic force provides the centripetal force:

$$qE = \frac{mv^2}{r}$$

Find the kinetic energy: The kinetic energy is:

$$KE = \frac{1}{2}mv^2 = \frac{1}{2}qEr$$

Substituting the expression for $$E$$:

$$KE = \frac{1}{2} \cdot q \cdot \frac{\rho R^2}{2\varepsilon_0 r} \cdot r$$

$$KE = \frac{1}{2} \cdot \frac{q\rho R^2}{2\varepsilon_0}$$

$$KE = \frac{q\rho R^2}{4\varepsilon_0}$$

The kinetic energy of the particle is $$\dfrac{q\rho R^2}{4\varepsilon_0}$$.

The correct answer is Option A.

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.

Here $$B_1,\ B_2,and\ B_3$$​ form an equilateral triangle.

Let the force between any two balls be F.

Consider ball $$B_1$$. It experiences two repulsive forces:

• Force F due to $$B_2$$​
• Force F due to $$B_3$$

These two forces act along $$B_1B_2\ and\ B_1B_3$$

Since $$B_1B_2B_3$$​ is an equilateral triangle, the angle between these two forces is

$$60^{\circ}$$

Hence the resultant force on $$B_1$$ is

$$R=\sqrt{F^2+F^2+2F^2\cos60^{\circ}}$$

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

$$R=\sqrt{3}F$$

Therefore,

Net force on one ballForce between any two balls=

$$\frac{\text{Net force on one ball}}{\text{Force between any two balls}}=\frac{\sqrt{\ 3}F}{F}$$

$$=\sqrt{3}:1$$

Hence the required ratio is

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.

We have a long cylindrical volume with uniform charge density $$\rho \text{ C m}^{-3}$$. We need to find the electric field at a distance $$x = \dfrac{2\epsilon_0}{\rho} \text{ m}$$ from the axis.

Consider a cylindrical Gaussian surface of radius $$x$$ and length $$l$$, coaxial with the charged cylinder.

By Gauss's law:

$$\oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0}$$

The enclosed charge is:

$$q_{enc} = \rho \times \pi x^2 l$$

The flux through the curved surface of the Gaussian cylinder is:

$$E \times 2\pi x l = \frac{\rho \pi x^2 l}{\epsilon_0}$$

$$E = \frac{\rho x}{2\epsilon_0}$$

$$E = \frac{\rho}{2\epsilon_0} \times \frac{2\epsilon_0}{\rho} = 1 \text{ V m}^{-1}$$

The electric field at the given distance is $$1 \text{ V m}^{-1}$$.

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:

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

For dipole 1:

For dipole 2:

The ratio of maximum torques:

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

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.

For a large plane sheet of positive charge with a uniform surface charge density $\sigma$, the electric field intensity at nearby points is determined using Gauss's Law.

1. Electric Field Expression

The electric field E generated by an infinite non-conducting plane sheet is given by the formula:

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

Where:

• $$\sigma$$ = Surface charge density
• $$\varepsilon_0$$ = Permittivity of free space

2. Independence of Distance

A key characteristic of a very large (infinite) plane sheet is that the electric field it produces is uniform. This means the magnitude of the field does not depend on the distance r from the sheet, provided the distance is small compared to the dimensions of the sheet.

3. Comparison of Points $$P_1$$ and $$P_2$$

Since points $$P_1$$ and $$P_2$$ are at different distances but are both near the large sheet:

• The field at $$P_1$$ is $$E_1 = \frac{\sigma}{2\varepsilon_0}$$
• The field at $$P_2$$ is $$E_2 = \frac{\sigma}{2\varepsilon_0}$$

Final Conclusion:

$$E_1 = E_2 = \frac{\sigma}{2\varepsilon_0}$$

Correct Option: (C)

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 have a thin, uniformly charged disc lying in the $$xy$$-plane with its centre at the origin. Its radius is $$R$$ and the surface charge density is $$\sigma$$. We wish to find the electric field intensity at a point located on the positive $$z$$-axis, a distance $$Z$$ from the origin. Because of circular symmetry, the field at this point must point purely along the $$z$$-direction; the transverse components cancel pair-wise.

To calculate the field, we break the disc into many thin concentric rings. Consider an elemental ring of radius $$r$$ and width $$dr$$. The area of this ring is

$$dA = 2\pi r\,dr.$$

Since the surface charge density is uniform, the charge on this ring is

$$dq = \sigma\,dA = \sigma\,(2\pi r\,dr).$$

Next we recall the standard formula for the magnitude of the electric field on the axis of a ring of radius $$r$$ carrying a total charge $$dq$$. For a point a distance $$Z$$ from the plane of the ring, the magnitude of the field contributed by that ring element is

$$dE = \frac{1}{4\pi\varepsilon_0}\;\frac{dq\,Z}{\left(Z^{2}+r^{2}\right)^{3/2}}.$$

Only the component along the axis survives; the horizontal components cancel by symmetry. Substituting the expression for $$dq$$ we obtain

$$dE = \frac{1}{4\pi\varepsilon_0}\;\frac{\sigma(2\pi r\,dr)\,Z}{\left(Z^{2}+r^{2}\right)^{3/2}} = \frac{\sigma Z}{2\varepsilon_0}\;\frac{r\,dr}{\left(Z^{2}+r^{2}\right)^{3/2}}.$$

To find the total field, we integrate $$r$$ from $$0$$ to $$R$$:

$$E = \int_{0}^{R} dE = \frac{\sigma Z}{2\varepsilon_0}\int_{0}^{R}\frac{r\,dr}{\left(Z^{2}+r^{2}\right)^{3/2}}.$$

We now evaluate the integral. Let

$$u = Z^{2}+r^{2}\;\;\Longrightarrow\;\;du = 2r\,dr\;\;\Longrightarrow\;\;r\,dr = \frac{du}{2}.$$

Substituting into the integral, we get

$$\int_{0}^{R}\frac{r\,dr}{\left(Z^{2}+r^{2}\right)^{3/2}} = \frac{1}{2}\int_{u=Z^{2}}^{u=Z^{2}+R^{2}} u^{-3/2}\;du.$$

Using the power-rule integral $$\int u^{n}\,du = \frac{u^{n+1}}{n+1}$$ (valid for $$n\neq-1$$), we first note that $$n=-\frac{3}{2}$$, so $$n+1=-\frac{1}{2}$$, and hence

$$\int u^{-3/2}\,du = \frac{u^{-1/2}}{-1/2} = -2\,u^{-1/2}.$$

Therefore,

$$\frac{1}{2}\int u^{-3/2}\,du = \frac{1}{2}\left[-2\,u^{-1/2}\right] = -u^{-1/2}.$$

Putting back the limits:

$$-u^{-1/2}\Big|_{Z^{2}}^{Z^{2}+R^{2}} = -\left[(Z^{2}+R^{2})^{-1/2} - (Z^{2})^{-1/2}\right] = \frac{1}{Z} - \frac{1}{\sqrt{Z^{2}+R^{2}}}.$$

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

$$E = \frac{\sigma Z}{2\varepsilon_0}\left(\frac{1}{Z} - \frac{1}{\sqrt{Z^{2}+R^{2}}}\right).$$

Simplifying, the factor $$Z$$ in the numerator of the first term cancels with the $$Z$$ in the denominator:

$$E = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{Z}{\sqrt{Z^{2}+R^{2}}}\right).$$

This expression matches the option labelled C:

$$E = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{Z}{(Z^{2}+R^{2})^{1/2}}\right).$$

Hence, the correct answer is Option C.

Two electrons are fixed at positions $$(-d, 0)$$ and $$(d, 0)$$. A proton is placed at the midpoint (origin) and displaced slightly by a distance $$x$$ perpendicular to the line joining the electrons, where $$x \ll d$$.

The distance from the displaced proton to each electron is $$r = \sqrt{d^2 + x^2}$$. The attractive Coulomb force on the proton due to each electron has magnitude $$F = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{d^2 + x^2}$$, where $$q$$ is the elementary charge.

By symmetry, the components of force along the line joining the electrons cancel out. The net restoring force is along the perpendicular direction. The perpendicular component from each electron is $$F\sin\theta = F\frac{x}{\sqrt{d^2 + x^2}}$$. The total restoring force is $$F_{\text{net}} = 2 \times \frac{1}{4\pi\varepsilon_0}\frac{q^2}{(d^2 + x^2)} \times \frac{x}{\sqrt{d^2 + x^2}} = \frac{q^2 x}{2\pi\varepsilon_0(d^2 + x^2)^{3/2}}$$.

Since $$x \ll d$$, we approximate $$(d^2 + x^2)^{3/2} \approx d^3$$. Therefore, $$F_{\text{net}} \approx \frac{q^2}{2\pi\varepsilon_0 d^3}x$$.

This is a restoring force proportional to $$x$$, confirming SHM. Comparing with $$F = m\omega^2 x$$, we get $$\omega^2 = \frac{q^2}{2\pi\varepsilon_0 m d^3}$$.

Therefore, the angular frequency is $$\omega = \left(\frac{q^2}{2\pi\varepsilon_0 m d^3}\right)^{\frac{1}{2}}$$.

The correct answer is $$\left(\frac{q^2}{2\pi\varepsilon_0 md^3}\right)^{\frac{1}{2}}$$.

An electron enters between the parallel plates of a capacitor at angle $$\alpha$$ with the plates and leaves at angle $$\beta$$. The electric field between the plates is perpendicular to the plates, so only the component of velocity perpendicular to the plates changes. The component of velocity parallel to the plates remains constant.

Let the speed of the electron on entering be $$v_1$$ and on leaving be $$v_2$$. The component of velocity parallel to the plates on entry is $$v_1\cos\alpha$$ and on exit is $$v_2\cos\beta$$.

Since the electric field is perpendicular to the plates, there is no force along the plates. Therefore, the parallel component of velocity is conserved: $$v_1\cos\alpha = v_2\cos\beta$$.

The kinetic energies are $$K_1 = \frac{1}{2}m v_1^2$$ and $$K_2 = \frac{1}{2}m v_2^2$$. Taking the ratio:

$$\frac{K_1}{K_2} = \frac{v_1^2}{v_2^2} = \left(\frac{v_1}{v_2}\right)^2 = \left(\frac{\cos\beta}{\cos\alpha}\right)^2 = \frac{\cos^2\beta}{\cos^2\alpha}$$

Therefore, $$K_1 : K_2 = \frac{\cos^2\beta}{\cos^2\alpha}$$.

Let us consider one of the two identical tennis balls. The ball has mass $$m$$, charge $$q$$ and is attached to a thread of length $$l$$ which makes a small angle $$\theta$$ with the vertical. Because both balls carry the same charge they repel each other and settle at a horizontal separation $$x$$. Geometrically the two equal threads form an isosceles triangle, so the horizontal distance between the balls is

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

We now analyse the forces acting on a single ball. Three forces are present:

(i) its weight $$\;mg$$ acting vertically downward,

(ii) the tension $$T$$ in the thread acting along the thread, and

(iii) the electrostatic repulsive force $$F$$ exerted horizontally by the other ball.

According to Coulomb’s law, the magnitude of the electrostatic force between two point charges is

$$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r^2},$$

where $$r$$ is the distance between the charges. In the present situation

$$r = 2l\sin\theta,$$

so

$$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{(2l\sin\theta)^2} = \dfrac{q^2}{16\pi\varepsilon_0 l^2\sin^2\theta}.$$

For equilibrium we resolve the tension into components. The vertical component balances the weight and the horizontal component balances the electrostatic repulsion:

$$T\cos\theta = mg,$$

$$T\sin\theta = F.$$

Dividing the second equation by the first gives

$$\tan\theta = \dfrac{F}{mg}.$$

Because the angle is small we may use the small-angle approximations

$$\sin\theta \approx \theta,\qquad \tan\theta \approx \theta,\qquad \cos\theta \approx 1.$$

Thus the previous relation becomes

$$\theta = \dfrac{F}{mg} = \dfrac{1}{mg}\left(\dfrac{q^2}{16\pi\varepsilon_0 l^2\theta^2}\right).$$

Rearranging, we collect all powers of $$\theta$$ on one side:

$$\theta^3 = \dfrac{q^2}{16\pi\varepsilon_0 l^2 mg}.$$

Taking the cube root,

$$\theta = \left(\dfrac{q^2}{16\pi\varepsilon_0 l^2 mg}\right)^{1/3}.$$

The required separation $$x$$ is related to $$\theta$$ by $$x = 2l\sin\theta \approx 2l\theta,$$ so

$$x = 2l \left(\dfrac{q^2}{16\pi\varepsilon_0 l^2 mg}\right)^{1/3}.$$

To simplify, we cube both sides temporarily:

$$x^3 = 8l^3\left(\dfrac{q^2}{16\pi\varepsilon_0 l^2 mg}\right) = \dfrac{8}{16}\,\dfrac{q^2 l}{\pi\varepsilon_0 mg} = \dfrac{q^2 l}{2\pi\varepsilon_0 mg}.$$

Finally taking the cube root once more we obtain

$$x = \left(\dfrac{q^2 l}{2\pi\varepsilon_0 mg}\right)^{1/3}.$$

This expression matches Option 2 (Option B).

Hence, the correct answer is Option 2.

Using x, y, z components only:

Take the corner with charge −Q as origin

(0,0,0)

Center of cube is at

First imagine all 8 corners have charge +Q.
By symmetry, electric field at cube center is zero.

Now replacing the +Q at origin by −Q is equivalent to adding an extra charge

−2Q

at origin.

So net field at center is simply field due to charge −2Q at origin.

Distance of center from origin:

$$r=\frac{\sqrt{3}a}{2}$$

Magnitude of field:

Direction is from center toward origin, i.e. along

(−1,−1,−1)

Unit vector in this direction:

So components are

$$E_y=-\frac{8kQ}{3\sqrt{3}a^2}$$

$$E_z=-\frac{8kQ}{3\sqrt{3}a^2}$$

Hence field is

We begin with the basic definition of electric flux. By definition, the differential electric flux $$d\Phi$$ through an infinitesimal area vector $$d\vec A$$ is

$$d\Phi \;=\; \vec E \cdot d\vec A \;=\; E\,A\cos\theta,$$

where $$\theta$$ is the angle between the electric field $$\vec E$$ and the outward-drawn area vector $$d\vec A$$. Using this, we shall test each statement.

Statement (a): “The electric lines of force entering into a Gaussian surface provide negative flux.” When the field lines enter the surface, $$\theta$$ is greater than $$90^{\circ}$$, so $$\cos\theta<0$$ and $$\vec E\cdot d\vec A<0$$. Hence $$d\Phi$$ is negative. So the statement is correct.

Statement (b): “A charge $$q$$ is placed at the centre of a cube. The flux through all the faces will be the same.” By Gauss’s law, the total flux leaving the cube is

$$\Phi_{\text{total}}=\dfrac{q}{\varepsilon_{0}}.$$

Because the charge is exactly at the centre, the cube is perfectly symmetric; therefore the flux is distributed equally among all six faces. Hence the flux through one face is

$$\Phi_{\text{one face}}=\dfrac{1}{6}\left(\dfrac{q}{\varepsilon_{0}}\right).$$

This shows that each face indeed receives the same flux, so statement (b) is correct.

Statement (c): “In a uniform electric field, net flux through a closed Gaussian surface containing no net charge is zero.” Gauss’s law states

$$\oint\vec E\cdot d\vec A=\dfrac{q_{\text{enc}}}{\varepsilon_{0}}.$$

If the surface encloses no charge, then $$q_{\text{enc}}=0$$, and therefore

$$\oint\vec E\cdot d\vec A = 0.$$

This is true regardless of whether the external field is uniform or not, so the statement is certainly correct.

Statement (d): “When an electric field is parallel to a Gaussian surface, it provides a finite non-zero flux.” If the field is exactly parallel to the surface, the angle $$\theta$$ between $$\vec E$$ and the outward normal $$d\vec A$$ is $$90^{\circ}$$. Hence

$$\cos\theta=\cos90^{\circ}=0,$$

so each elemental flux $$d\Phi=E\,A\cos\theta=0.$$ Integrating over the whole surface, the total flux also remains zero. Therefore the flux is not finite and non-zero; it is strictly zero. Thus statement (d) is incorrect.

We have found that statements (a), (b) and (c) are correct, while statement (d) is incorrect. Among the given options, only Option D states “(d) Only”.

Hence, the correct answer is Option D.

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.

Let us first mark the fixed, identical charges. They are kept 2 m apart, so each is 1 m away from the mid-point. The free particle of charge $$q$$ and mass $$m$$ (here $$m = 1\text{ mg} = 1\times10^{-6}\,\text{kg}$$) is initially at this mid-point, where the forces cancel out.

Suppose the free particle is displaced very slightly ($$x\ll1\text{ m}$$) along the line joining the fixed charges. Choose the positive x-axis from the left charge towards the right charge. After a displacement $$x$$ to the right, its coordinates and distances become:

$$ \begin{aligned} \text{Position of particle}&:& +x,\\[4pt] \text{Distance to right charge}&:& r_R = (1 - x),\\[4pt] \text{Distance to left charge}&:& r_L = (1 + x). \end{aligned} $$

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

$$ F = k\frac{q^2}{r^2},\qquad k = \frac1{4\pi\varepsilon_0}=9\times10^{9}\,\text{N\,m}^2\text{C}^{-2}. $$

Because the charges are like, each fixed charge repels the free charge. Hence:

$$ \begin{aligned} \text{Force by right charge}&:& \vec F_R = -\,k\frac{q^{2}}{(1-x)^{2}}\hat i,\\[6pt] \text{Force by left charge}&:& \vec F_L = +\,k\frac{q^{2}}{(1+x)^{2}}\hat i. \end{aligned} $$ (The negative sign on $$\vec F_R$$ shows that it acts towards the left, while the positive sign on $$\vec F_L$$ shows that it acts towards the right.)

The net force on the particle is therefore

$$ \vec F = \vec F_L + \vec F_R = kq^{2}\left[\frac1{(1+x)^{2}} - \frac1{(1-x)^{2}}\right]\hat i. $$

For very small $$x$$ we may expand each reciprocal square by the binomial approximation $$\displaystyle(1\pm x)^{-2}\approx 1\mp2x+3x^{2}\ldots$$ Retaining only the first-order term in $$x$$ gives

$$ \begin{aligned} \frac1{(1+x)^{2}} &\approx 1 - 2x,\\ \frac1{(1-x)^{2}} &\approx 1 + 2x. \end{aligned} $$

Substituting these into the expression for $$\vec F$$ we obtain

$$ \vec F \approx kq^{2}\big[(1-2x) - (1+2x)\big]\hat i = kq^{2}(-4x)\hat i = -4kq^{2}x\,\hat i. $$

This force is proportional to $$-x$$, i.e. directed opposite to the displacement, which is the hallmark of simple harmonic motion (SHM). Comparing with the standard SHM equation

$$ m\frac{d^{2}x}{dt^{2}} = -\omega^{2}x, $$

we identify

$$ \omega^{2} = \frac{4kq^{2}}{m},\qquad\text{so}\qquad \omega = 2\sqrt{\frac{kq^{2}}{m}}. $$

Now we substitute the numerical values. We are given $$q^{2}=10\,\text{C}^{2}$$, hence we need only $$q^{2}$$, not $$q$$ itself.

$$ \begin{aligned} \omega &= 2\sqrt{\frac{(9\times10^{9})\,(10)}{1\times10^{-6}}}\\[6pt] &= 2\sqrt{9\times10^{10+6}}\\[6pt] &= 2\sqrt{9\times10^{16}}\\[6pt] &= 2\,(3\times10^{8})\\[6pt] &= 6\times10^{8}\ \text{rad s}^{-1}. \end{aligned} $$

The problem asks for the coefficient in

$$ \omega = (\,\text{blank}\,)\times10^{5}\ \text{rad s}^{-1}. $$

Dividing our result by $$10^{5}$$ gives

$$ \frac{6\times10^{8}}{10^{5}} = 6\times10^{3} = 6000. $$

So, the answer is $$6000$$.

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$$^{-1}$$.

The total charge enclosed in a volume is related to the electric flux density $$\vec{D}$$ through Gauss's law in differential form: $$\rho_v = \nabla \cdot \vec{D}$$.

Given $$\vec{D} = e^{-x}\sin y\,\hat{i} - e^{-x}\cos y\,\hat{j} + 2z\,\hat{k}$$, the divergence is:

$$\nabla \cdot \vec{D} = \frac{\partial}{\partial x}(e^{-x}\sin y) + \frac{\partial}{\partial y}(-e^{-x}\cos y) + \frac{\partial}{\partial z}(2z)$$

$$= -e^{-x}\sin y + e^{-x}\sin y + 2 = 2\,\text{C\,m}^{-3}$$

The divergence is constant (equals 2 everywhere, independent of position), so the volume charge density at the origin is $$\rho_v = 2\,\text{C\,m}^{-3}$$.

The total charge enclosed in the incremental volume $$\Delta v = 2 \times 10^{-9}\,\text{m}^3$$ is:

$$Q = \rho_v \cdot \Delta v = 2 \times 2 \times 10^{-9} = 4 \times 10^{-9}\,\text{C} = 4\,\text{nC}$$

Each sphere has mass $$m = 10$$ mg $$= 10 \times 10^{-6}$$ kg $$= 10^{-5}$$ kg, and they are suspended from the same point by threads of length $$L = 0.5$$ m. They repel each other to a separation of $$d = 0.20$$ m.

For equilibrium of each sphere, the forces acting are: tension $$T$$ along the thread, weight $$mg$$ downward, and electrostatic repulsion $$F$$ horizontally. The half-separation is $$\frac{d}{2} = 0.10$$ m, so $$\sin\theta = \frac{0.10}{0.5} = 0.2$$.

From the equilibrium conditions: $$\tan\theta = \frac{F}{mg}$$. Since $$\sin\theta = 0.2$$, we get $$\cos\theta = \sqrt{1 - 0.04} = \sqrt{0.96}$$ and $$\tan\theta = \frac{0.2}{\sqrt{0.96}}$$.

The electrostatic force is $$F = \frac{kq^2}{d^2} = \frac{9 \times 10^9 \cdot q^2}{(0.20)^2} = \frac{9 \times 10^9 \cdot q^2}{0.04}$$.

From $$\tan\theta = \frac{F}{mg}$$: $$\frac{0.2}{\sqrt{0.96}} = \frac{9 \times 10^9 \cdot q^2}{0.04 \times 10^{-5} \times 10}$$.

The denominator is $$0.04 \times 10^{-4} = 4 \times 10^{-6}$$. So $$\frac{0.2}{\sqrt{0.96}} = \frac{9 \times 10^9 \cdot q^2}{4 \times 10^{-6}}$$.

Solving for $$q^2$$: $$q^2 = \frac{0.2 \times 4 \times 10^{-6}}{9 \times 10^9 \times \sqrt{0.96}} = \frac{0.8 \times 10^{-6}}{9 \times 10^9 \times \sqrt{0.96}}$$.

Now $$\sqrt{0.96} \approx 0.98$$, so $$q^2 \approx \frac{8 \times 10^{-7}}{8.82 \times 10^9} = \frac{8}{8.82} \times 10^{-16} \approx 0.907 \times 10^{-16}$$.

Therefore $$q \approx \sqrt{0.907 \times 10^{-16}} \approx 0.953 \times 10^{-8}$$ C.

The answer is given as $$q = \frac{a}{21} \times 10^{-8}$$ C. So $$\frac{a}{21} = 0.953$$, giving $$a = 0.953 \times 21 \approx 20$$.

To verify exactly: with the standard approximation $$\tan\theta \approx \sin\theta$$ (valid for small angles), $$\tan\theta \approx 0.2$$, giving $$q^2 = \frac{0.2 \times 0.04 \times 10^{-5} \times 10}{9 \times 10^9} = \frac{0.2 \times 4 \times 10^{-6}}{9 \times 10^9} = \frac{8 \times 10^{-7}}{9 \times 10^9} = \frac{8}{9} \times 10^{-16}$$.

So $$q = \sqrt{\frac{8}{9}} \times 10^{-8} = \frac{2\sqrt{2}}{3} \times 10^{-8}$$. Now $$\frac{2\sqrt{2}}{3} \approx \frac{2 \times 1.414}{3} = \frac{2.828}{3} = 0.9428$$. Then $$a = 0.9428 \times 21 \approx 19.8 \approx 20$$.

Therefore, the value of $$a$$ is $$20$$.

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 line charge lies along the $$z$$-axis, so the electric field at any point in the $$x$$-$$y$$ plane is radial and has magnitude

$$E(r)=\frac{\lambda}{2\pi\varepsilon_0\,r}\qquad -(1)$$
where $$\lambda=3.0\times10^{-6}\,{\rm C\,m^{-1}}$$ and $$r$$ is the perpendicular distance from the line.

The dipole is on the $$x$$-axis.
Positive charge $$+q$$ is at $$x_1=10\;{\rm mm}=0.01\;{\rm m}$$.
Negative charge $$-q$$ is at $$x_2=12\;{\rm mm}=0.012\;{\rm m}$$.

Using $$-(1)$$, the fields at the two charges are

$$E_1=\frac{\lambda}{2\pi\varepsilon_0\,x_1},\qquad E_2=\frac{\lambda}{2\pi\varepsilon_0\,x_2}\qquad -(2)$$

Direction of the field is $$+x$$ (away from the line) at both points.
Hence the individual forces are

$$F_+=+qE_1\quad\text{(on }+q\text{, toward }+x),$$ $$F_-=-qE_2\quad\text{(on }-q\text{, toward }-x)$$

The net force on the dipole is therefore

$$F=F_+ + F_- = q\bigl(E_1-E_2\bigr)\qquad -(3)$$

Substituting $$-(2)$$ into $$-(3)$$ gives

$$F = q\,\frac{\lambda}{2\pi\varepsilon_0}\left(\frac1{x_1}-\frac1{x_2}\right)\qquad -(4)$$

Given $$F=4\;{\rm N}$$, solve $$-(4)$$ for $$q$$:

First evaluate the bracket:
$$\frac1{x_1}-\frac1{x_2}= \frac1{0.01}-\frac1{0.012} =100-83.333=16.6667\;{\rm m^{-1}}\qquad -(5)$$

Next compute the constant factor:
$$\frac{\lambda}{2\pi\varepsilon_0}= \frac{3.0\times10^{-6}}{2\pi(8.854\times10^{-12})} =5.39\times10^{4}\;{\rm N\,C^{-1}}\qquad -(6)$$

Hence the field difference is
$$E_1-E_2 = (5.39\times10^{4})(16.6667) =8.98\times10^{5}\;{\rm N\,C^{-1}}\qquad -(7)$$

Finally, from $$-(4)$$,
$$q=\frac{F}{E_1-E_2}= \frac{4}{8.98\times10^{5}} =4.46\times10^{-6}\;{\rm C}\qquad -(8)$$

$$4.46\times10^{-6}\;{\rm C}=4.46\;\mu{\rm C}$$

The magnitude of each charge in the dipole is therefore approximately $$4.44\;\mu{\rm C}$$.

Answer: Option D - $$4.44\;\mu{\rm C}$$

A uniform line charge of linear charge density $$\lambda = 8$$ nC/m lies along the $$z$$-axis. The electric field at a perpendicular distance $$r$$ from an infinite line charge is $$E = \frac{\lambda}{2\pi\varepsilon_0 r}$$.

At the point where the $$x = 3$$ m plane intersects the $$x$$-axis, the perpendicular distance from the $$z$$-axis is $$r = 3$$ m.

The surface charge density at this point is defined as $$\sigma = \varepsilon_0 E$$. Substituting the expression for $$E$$: $$\sigma = \varepsilon_0 \times \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{\lambda}{2\pi r}$$.

Computing: $$\sigma = \frac{8 \times 10^{-9}}{2\pi \times 3} = \frac{8 \times 10^{-9}}{6\pi} = \frac{8 \times 10^{-9}}{18.85} = 0.4244 \times 10^{-9}$$ C/m$$^2$$ $$= 0.424$$ nC/m$$^2$$.

We have ten point charges fixed on the circumference of a circle of radius $$R$$. They are equally spaced, so the angular separation between successive charges is

$$\Delta\theta=\frac{2\pi}{10}=36^{\circ}.$$

Charges numbered 1, 3, 5, 7, 9 each carry $$+q$$, while 2, 4, 6, 8, 10 each carry $$-q$$. Thus there are exactly five positive charges and five negative charges.

First we examine the electric potential at the centre. The potential produced at the centre by a single point charge is obtained from the formula

$$V=\frac{1}{4\pi\varepsilon_{0}}\frac{q}{r},$$

where $$r$$ is the distance from the charge to the point of observation. Here that distance is the radius $$R$$ for every charge. Hence the contribution of one $$+q$$ charge is

$$V_{(+)}=\frac{1}{4\pi\varepsilon_{0}}\frac{(+q)}{R},$$

while that of one $$-q$$ charge is

$$V_{(-)}=\frac{1}{4\pi\varepsilon_{0}}\frac{(-q)}{R}.$$

Because there are five charges of each sign, the total potential is

$$V_{\text{total}} =5V_{(+)}+5V_{(-)} =5\left(\frac{1}{4\pi\varepsilon_{0}}\frac{q}{R}\right) +5\left(\frac{1}{4\pi\varepsilon_{0}}\frac{-q}{R}\right) =\frac{5q}{4\pi\varepsilon_{0}R} -\frac{5q}{4\pi\varepsilon_{0}R} =0.$$

So the potential at the centre is zero.

Now we turn to the electric field. The magnitude of the field at a distance $$r$$ from a point charge is given by

$$E=\frac{1}{4\pi\varepsilon_{0}}\frac{|q|}{r^{2}},$$

and its direction is along the line joining the charge and the point: radially inward for a positive charge (because the field lines point away from the charge, towards the centre), and radially outward for a negative charge (field lines point toward the charge).

For our ring, every charge lies the same distance $$R$$ from the centre, so each field contribution has magnitude

$$E_{0}=\frac{1}{4\pi\varepsilon_{0}}\frac{q}{R^{2}}.$$

Let $$\hat{r}_{k}$$ be the unit vector pointing from the centre toward the position of the $$k^{\text{th}}$$ charge. Then the field at the centre produced by that charge is

$$\vec{E}_{k}= \begin{cases} -E_{0}\hat{r}_{k}, & \text{for a } +q \text{ charge (inward)}\\[4pt] +E_{0}\hat{r}_{k}, & \text{for a } -q \text{ charge (outward)} \end{cases}.$$

Let us collect the five positive charges first. They occupy the angles

$$0^{\circ},\;72^{\circ},\;144^{\circ},\;216^{\circ},\;288^{\circ},$$

i.e. every $$72^{\circ}$$, forming a perfect regular pentagon. Their field vectors are the five inward radial vectors directed to the centre from those vertices. The vector sum of the five unit radial vectors of a regular pentagon is well known to be zero because they are symmetrically spaced; therefore

$$\sum_{\text{five }+q}\vec{E}_{k}=-E_{0}\sum_{m=0}^{4}\hat{r}_{m}=0.$$

Exactly the same argument applies to the five negative charges. They lie at the intermediate angles

$$36^{\circ},\;108^{\circ},\;180^{\circ},\;252^{\circ},\;324^{\circ},$$

another regular pentagon, merely rotated by $$36^{\circ}$$. The outward radial vectors at these angles are again symmetrically placed, so

$$\sum_{\text{five }-q}\vec{E}_{k}=+E_{0}\sum_{m=0}^{4}\hat{r}_{m}^{\,'}=0.$$

Adding the two groups together, the total electric field at the centre is

$$\vec{E}_{\text{total}}=0+0=0.$$

We therefore conclude

$$V=0,\qquad \vec{E}=0.$$

Hence, the correct answer is Option C.

First we recall the basic formula for the electric field produced by a point charge. For a charge $$Q$$ situated at a distance $$r$$ from the observation point, the magnitude of the field is

$$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{|Q|}{r^{2}}$$

The direction of the field is • from the charge towards the observation point if the charge is positive, and • from the observation point towards the charge if the charge is negative.

In the present problem every charge lies on the circumference of the circle, so the distance of each charge from the centre $$O$$ is the radius $$d$$. Hence for every charge the magnitude of the field at $$O$$ will be

$$\dfrac{1}{4\pi\varepsilon_0}\dfrac{|q_i|}{d^{2}}$$

where $$q_i$$ is the magnitude of the respective charge.

Let us adopt a convenient coordinate system. We place the centre $$O$$ at the origin and take the line $$OA$$ as the positive $$x$$-axis. Because the triangle $$AOC$$ is equilateral, the angle $$\angle AOC$$ equals $$60^{\circ}$$, so the radius $$OC$$ makes an angle $$60^{\circ}$$ with the positive $$x$$-axis.

The figure supplied in the question shows the charge $$B$$ diametrically opposite to the charge $$C$$. Therefore the radius $$OB$$ forms an angle of $$60^{\circ}+180^{\circ}=240^{\circ}$$ with the positive $$x$$-axis.

Now we examine each charge one by one.

Charge at A: Value $$q_A=-4q$$ (negative). Magnitude of its field at $$O$$:

$$E_A=\dfrac{1}{4\pi\varepsilon_0}\dfrac{4q}{d^{2}}$$

Because the charge is negative, the field points from $$O$$ towards $$A$$, i.e. outward along the radius $$OA$$ or the $$+x$$ direction. Hence in vector form

$$\vec E_A=\dfrac{1}{4\pi\varepsilon_0}\dfrac{4q}{d^{2}}\;\hat i$$

Charge at C: Value $$q_C=-2q$$ (negative). Magnitude:

$$E_C=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2q}{d^{2}}$$

Direction is outward along the radius $$OC$$, that is at $$60^{\circ}$$.

$$\vec E_C=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2q}{d^{2}}\;(\cos60^{\circ}\,\hat i+\sin60^{\circ}\,\hat j)$$

Charge at B: Value $$q_B=+2q$$ (positive). Magnitude:

$$E_B=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2q}{d^{2}}$$

This time the charge is positive, so the field points from the charge towards the centre, i.e. inward along the radius $$BO$$. Since the outward radius $$OB$$ is at $$240^{\circ}$$, the inward direction is opposite to it, namely $$240^{\circ}-180^{\circ}=60^{\circ}$$. Thus $$\vec E_B$$ is also directed at $$60^{\circ}$$.

$$\vec E_B=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2q}{d^{2}}\;(\cos60^{\circ}\,\hat i+\sin60^{\circ}\,\hat j)$$

We now add the three vector contributions.

First collect the $$x$$-components:

$$E_x=\dfrac{1}{4\pi\varepsilon_0\,d^{2}}\Big[4q\;(1)+2q\;(\cos60^{\circ})+2q\;(\cos60^{\circ})\Big]$$

Using $$\cos60^{\circ}= \dfrac12$$, we get

$$E_x=\dfrac{1}{4\pi\varepsilon_0\,d^{2}}\Big[4q+2q\left(\dfrac12\right)+2q\left(\dfrac12\right)\Big] =\dfrac{1}{4\pi\varepsilon_0\,d^{2}}\Big[4q+q+q\Big] =\dfrac{1}{4\pi\varepsilon_0\,d^{2}}\;(6q)$$

So

$$E_x=\dfrac{6q}{4\pi\varepsilon_0 d^{2}}$$

Next the $$y$$-components:

$$E_y=\dfrac{1}{4\pi\varepsilon_0\,d^{2}}\Big[4q\;(0)+2q\;(\sin60^{\circ})+2q\;(\sin60^{\circ})\Big]$$

Because $$\sin60^{\circ}= \dfrac{\sqrt3}{2}$$,

$$E_y=\dfrac{1}{4\pi\varepsilon_0\,d^{2}}\Big[2q\left(\dfrac{\sqrt3}{2}\right)+2q\left(\dfrac{\sqrt3}{2}\right)\Big] =\dfrac{1}{4\pi\varepsilon_0\,d^{2}}\;\big[q\sqrt3+q\sqrt3\big] =\dfrac{1}{4\pi\varepsilon_0\,d^{2}}\;(2\sqrt3\,q)$$

Thus

$$E_y=\dfrac{2\sqrt3\,q}{4\pi\varepsilon_0 d^{2}}$$

The resultant electric field vector is

$$\vec E=\big(E_x\,\hat i+E_y\,\hat j\big)$$

Its magnitude is calculated by Pythagoras:

$$|\vec E|=\sqrt{E_x^{2}+E_y^{2}} =\sqrt{\left(\dfrac{6q}{4\pi\varepsilon_0 d^{2}}\right)^{2}+\left(\dfrac{2\sqrt3\,q}{4\pi\varepsilon_0 d^{2}}\right)^{2}}$$

$$|\vec E|=\dfrac{q}{4\pi\varepsilon_0 d^{2}}\;\sqrt{(6)^{2}+(2\sqrt3)^{2}} =\dfrac{q}{4\pi\varepsilon_0 d^{2}}\;\sqrt{36+12} =\dfrac{q}{4\pi\varepsilon_0 d^{2}}\;\sqrt{48}$$

Since $$\sqrt{48}=4\sqrt3$$, we obtain

$$|\vec E|=\dfrac{q}{4\pi\varepsilon_0 d^{2}}\;(4\sqrt3) =\dfrac{\sqrt3\,q}{\pi\varepsilon_0 d^{2}}$$

Therefore the magnitude of the electric field at the centre is

$$E=\dfrac{\sqrt3\,q}{\pi\varepsilon_0 d^{2}}$$

Hence, the correct answer is Option A.

We are given an electric dipole located at the origin whose dipole moment is

$$\vec p = (-\hat i - 3\hat j + 2\hat k)\; \times 10^{-29}\; \text{C m}.$$

We want the electric field at the point whose position vector is

$$\vec r = +\hat i + 3\hat j + 5\hat k.$$

It is also told that $$\vec p \cdot \vec r = 0,$$ which means the dipole moment is perpendicular to the position vector.

For a point dipole placed at the origin, the electric field at position $$\vec r$$ in free space is given by the standard dipole-field formula

$$\vec E(\vec r)=\frac{1}{4\pi\varepsilon_0}\,\frac{1}{r^{3}}\Bigl[\,3(\vec p\cdot\hat r)\,\hat r-\vec p\Bigr],$$

where $$r=\lVert\vec r\rVert$$ and $$\hat r=\dfrac{\vec r}{r}$$ is the unit vector along $$\vec r.$$

First we note that $$\vec p\cdot\hat r=\dfrac{\vec p\cdot\vec r}{r}=0$$ because $$\vec p\cdot\vec r=0.$$

Hence the first term inside the bracket vanishes, leaving

$$\vec E(\vec r)=\frac{1}{4\pi\varepsilon_0}\,\frac{1}{r^{3}}\,(0-\vec p)= -\;\frac{1}{4\pi\varepsilon_0}\,\frac{\vec p}{r^{3}}.$$

Thus the electric field is directed opposite to $$\vec p,$$ i.e. parallel to $$-\vec p.$$

We now write $$-\vec p$$ explicitly:

$$-\vec p = -\bigl(-\hat i - 3\hat j + 2\hat k\bigr)\times 10^{-29} =(+\hat i + 3\hat j - 2\hat k)\times 10^{-29}.$$

All constant numerical factors (such as $$\tfrac{1}{4\pi\varepsilon_0}$$ and $$\tfrac{1}{r^{3}}$$) merely scale the magnitude and do not affect the direction. Therefore the field is parallel to the vector

$$+\hat i + 3\hat j - 2\hat k.$$

This matches exactly with the vector given in Option C.

Hence, the correct answer is Option C.

We start from the statement of Gauss’s law in integral form

$$\oint_{\text{closed}} \vec{E}\cdot d\vec{A}= \frac{q_{enc}}{\varepsilon_0}.$$

Here $$d\vec{A}$$ is the outward-drawn area element on the chosen Gaussian surface and $$q_{enc}$$ is the total charge enclosed by that surface. The dot product can be written with its magnitude and the angle $$\theta$$ between $$\vec{E}$$ and the outward normal $$\hat{n}$$ to the surface element, giving

$$\oint_{\text{closed}} |\vec{E}|\,\cos\theta \; dA \;=\; \frac{q_{enc}}{\varepsilon_0}.$$

The right side already has no integral, so to convert the left side into the simple product $$|\vec{E}|\,A$$ we must be able to take both $$|\vec{E}|$$ and $$\cos\theta$$ outside the integral sign. This is possible only under the following two simultaneous requirements:

1. $$|\vec{E}|$$ is the same at every point of the Gaussian surface, so that it can be treated as a constant during the integration.

2. The angle $$\theta$$ is the same at every point. The easiest way to secure this is to make $$\theta = 0^{\circ}$$ everywhere, i.e. the electric field is everywhere perpendicular to the surface. When the field is perpendicular everywhere, the scalar potential cannot change along the surface, hence the surface is an equipotential surface.

Under these two conditions we can write

$$\oint_{\text{closed}} |\vec{E}|\,\cos\theta \; dA \;=\; |\vec{E}|\,\cos\theta\, \oint_{\text{closed}} dA \;=\; |\vec{E}|\,(\cos 0^{\circ})\,A \;=\; |\vec{E}|\,A,$$

and Gauss’s law becomes

$$|\vec{E}|\,A = \frac{q_{enc}}{\varepsilon_0}\quad\Longrightarrow\quad |\vec{E}| = \frac{q_{enc}}{\varepsilon_0\,A}.$$

Therefore the compact formula quoted in the question is valid only when (i) the Gaussian surface is an equipotential surface so that $$\vec{E}$$ is normal to it everywhere, and (ii) the magnitude of the field is uniform over the entire surface.

Examining the given options, only Option B states both of these necessary conditions.

Hence, the correct answer is Option B.

First, recall the expression for the electrostatic potential energy of two point dipoles. For dipoles having moments $$\vec p_1$$ and $$\vec p_2$$ separated by the displacement vector $$\vec r$$ (with magnitude $$r$$ and unit vector $$\hat r$$) the formula is

$$U \;=\; \frac1{4\pi\varepsilon_0\,r^{3}}\Bigl[\;\vec p_1\!\cdot\!\vec p_2 \;-\; 3\bigl(\vec p_1\!\cdot\!\hat r\bigr)\bigl(\vec p_2\!\cdot\!\hat r\bigr)\Bigr].$$

We have the two identical dipoles placed on the x-axis a distance $$a$$ apart. Their moments are given as

$$\vec p_1 = +p\hat i, \qquad \vec p_2 = -p\hat i.$$

We choose the displacement vector $$\vec r$$ from dipole 1 to dipole 2 to be along the +x direction, so

$$\vec r = a\hat i, \qquad r = a, \qquad \hat r = \hat i.$$

Now evaluate the scalar products appearing in the energy formula.

1. The direct dot product of the two moments is

$$\vec p_1\!\cdot\!\vec p_2 \;=\; (+p\hat i)\!\cdot\!(-p\hat i) \;=\; -p^{2}.$$

2. The projection of each dipole on the line joining them is

$$\vec p_1\!\cdot\!\hat r \;=\; (+p\hat i)\!\cdot\!\hat i \;=\; +p,$$

$$\vec p_2\!\cdot\!\hat r \;=\; (-p\hat i)\!\cdot\!\hat i \;=\; -p.$$

Multiplying these projections gives

$$\bigl(\vec p_1\!\cdot\!\hat r\bigr)\bigl(\vec p_2\!\cdot\!\hat r\bigr) \;=\; (+p)\times(-p) \;=\; -p^{2}.$$

Substituting all these results into the energy formula:

$$\begin{aligned} U_i &= \frac1{4\pi\varepsilon_0\,a^{3}}\Bigl[\,-p^{2}\;-\;3(-p^{2})\Bigr] \\ &= \frac1{4\pi\varepsilon_0\,a^{3}}\Bigl[-p^{2} + 3p^{2}\Bigr] \\ &= \frac1{4\pi\varepsilon_0\,a^{3}}\,(2p^{2}) \\ &= \frac{2p^{2}}{4\pi\varepsilon_0\,a^{3}} \\ &= \frac{p^{2}}{2\pi\varepsilon_0\,a^{3}}. \end{aligned}$$

This $$U_i$$ is the initial potential energy while both dipoles are at rest, so the initial kinetic energy is zero.

When the dipoles are released they move apart along the x-axis, keeping their orientations fixed. At infinite separation, the dipole-dipole interaction vanishes, so the final potential energy is

$$U_f = 0.$$

By conservation of mechanical energy, the loss in potential energy equals the gain in kinetic energy. Let $$v$$ be the speed acquired by each dipole when they are far apart. Each dipole then has kinetic energy $$\tfrac12 m v^{2}$$, so the total kinetic energy of the pair is

$$K_f = 2\left(\tfrac12 m v^{2}\right) = m v^{2}.$$

Energy conservation gives

$$U_i + 0 = 0 + K_f \quad\Longrightarrow\quad U_i = m v^{2}.$$

Substituting the value of $$U_i$$ calculated earlier, we get

$$\frac{p^{2}}{2\pi\varepsilon_0\,a^{3}} = m v^{2}.$$

Solving for $$v$$:

$$\begin{aligned} v^{2} &= \frac{p^{2}}{2\pi\varepsilon_0\,a^{3} \, m},\\[4pt] v &= \frac{p}{a}\sqrt{\frac{1}{2\pi\varepsilon_0\,m\,a}}. \end{aligned}$$

Thus the speed of each dipole when they are infinitely far apart is

$$v = \frac{p}{a}\sqrt{\frac{1}{2\pi\varepsilon_0\,m\,a}}.$$

Hence, the correct answer is Option B.

We start with Gauss’s law in integral form, stated as

$$\oint_{\text{closed}} \vec E\cdot d\vec A=\frac{q_{\text{enc}}}{\varepsilon_0}.$$

A uniformly charged thin spherical shell of radius $$R$$ carries a total charge $$Q$$ spread evenly over its surface. We place the test charge $$q$$ at a distance $$r$$ from the centre of the shell and look for the electrostatic force $$\vec F$$ acting on it.

Because the shell is spherically symmetric, the electric field at any point depends only on the distance from the centre. We therefore choose a concentric spherical Gaussian surface of radius $$r$$. On this Gaussian surface, the magnitude of the field $$E(r)$$ is the same everywhere and is directed radially. The area element $$d\vec A$$ is also radial, so the dot-product simplifies to $$\vec E\cdot d\vec A = E(r)\,dA$$. The closed-surface integral then becomes

$$\oint_{\text{closed}} \vec E\cdot d\vec A = E(r)\oint_{\text{closed}} dA = E(r)\,(4\pi r^{2}).$$

Now we distinguish the two possible locations of the test charge.

Case 1: $$r < R$$ (inside the shell)

For a point strictly inside the shell, the Gaussian surface of radius $$r$$ encloses no charge, because all the charge $$Q$$ resides on the shell’s surface, which lies outside our Gaussian sphere. Therefore

$$q_{\text{enc}} = 0.$$

Substituting into Gauss’s law gives

$$E(r)\,(4\pi r^{2}) = \frac{0}{\varepsilon_0}=0,$$

so

$$E(r)=0.$$

The force on the charge $$q$$ is obtained from $$\vec F = q\vec E$$. Because the field magnitude is zero, we have

$$F = q\,E(r)=0 \quad\text{for}\; r < R.$$

Case 2: $$r > R$$ (outside the shell)

When the Gaussian surface lies completely outside the shell, it encloses the entire charge $$Q$$. Thus

$$q_{\text{enc}} = Q.$$

Gauss’s law now gives

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

so

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

The corresponding force on the test charge $$q$$ is

$$F = qE(r)=\frac{1}{4\pi\varepsilon_0}\frac{Qq}{r^{2}}\quad\text{for}\; r > R.$$

Summarising the results:

$$F = 0 \quad\text{for}\; r < R,$$

$$F = \frac{1}{4\pi\varepsilon_0}\frac{Qq}{r^{2}} \quad\text{for}\; r > R.$$

We now compare these findings with each option:

Option A asserts a non-zero constant force $$\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qq}{R^{2}}$$ for $$r < R$$, but we have just shown that the force is zero inside, so Option A is false.

Option B claims $$\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qq}{R^{2}} > F > 0$$ for $$r < R$$. Since the true value of $$F$$ inside is exactly zero, the inequality is violated; hence Option B is also false.

Option C states $$F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Qq}{r^{2}}$$ for $$r > R$$, which matches the expression we derived. Therefore, Option C is correct.

Option D maintains the same expression for all values of $$r$$, but we know it fails for $$r < R$$, so Option D is incorrect.

Hence, the correct answer is Option C.

We have an electric field in the $$x$$-direction given to us as

$$E(x)=E_0\,(1-ax^{2}),$$

where $$E_0$$ and $$a$$ are positive constants. A particle of charge $$q$$ and mass $$m$$ is released from rest at the origin $$x=0$$. Because the particle starts from rest, its initial kinetic energy is

$$K_i = 0.$$

While the particle moves, the electric field does work on it. According to the work-energy theorem,

$$\text{Work done by all forces} = K_f - K_i.$$

Here the only force is the electric force $$\mathbf{F}=q\mathbf{E}$$, so the work done by this force as the particle moves from $$x=0$$ to some position $$x$$ is

$$W = \int_{0}^{x} F_x\,dx' = \int_{0}^{x} q\,E(x')\,dx'.$$

We want the kinetic energy at position $$x$$ to again become zero, i.e.

$$K_f = 0.$$

Substituting $$K_i=0$$ and $$K_f=0$$ into the work-energy theorem gives

$$W = 0.$$

So we must find that particular distance $$x$$ (other than $$x=0$$) for which the work done is zero:

$$\int_{0}^{x} q\,E_0\,(1-a{x'}^{2})\,dx' = 0.$$

The charge $$q$$ and the constant $$E_0$$ are common factors and can be taken outside the integral:

$$qE_0 \int_{0}^{x} (1-a{x'}^{2})\,dx' = 0.$$

Because $$qE_0$$ is non-zero, the integral itself must vanish:

$$\int_{0}^{x} (1-a{x'}^{2})\,dx' = 0.$$

Now we evaluate the integral step by step:

$$\int_{0}^{x} 1\,dx' = x,$$

$$\int_{0}^{x} a{x'}^{2}\,dx' = a\int_{0}^{x} {x'}^{2}\,dx' = a\left[\frac{{x'}^{3}}{3}\right]_{0}^{x}=a\left(\frac{x^{3}}{3}-0\right)=\frac{a x^{3}}{3}.$$

Combining these results, we have

$$x-\frac{a x^{3}}{3}=0.$$

Factor out the common term $$x$$:

$$x\left(1-\frac{a x^{2}}{3}\right)=0.$$

This product is zero when either factor is zero. The first possibility is

$$x=0,$$

which simply reproduces the starting point and is therefore not the distance we are looking for. The second possibility is

$$1-\frac{a x^{2}}{3}=0.$$

Solving this equation for $$x^{2}$$ gives

$$\frac{a x^{2}}{3}=1 \quad\Longrightarrow\quad x^{2}=\frac{3}{a}.$$

Taking the positive square root (distance is positive), we obtain

$$x=\sqrt{\frac{3}{a}}.$$

This is the unique position (other than the origin) where the kinetic energy again becomes zero.

Hence, the correct answer is Option C.

Initially sphere $$S_1$$ has radius $$r_1 = \dfrac{2}{3}R$$ and carries charge $$Q_1 = 12\,\mu\text{C}$$, whereas sphere $$S_2$$ has radius $$r_2 = \dfrac{1}{3}R$$ and carries charge $$Q_2 = -3\,\mu\text{C}$$. They are far apart, so each sphere’s potential with respect to infinity is produced only by its own charge, namely $$V = k\dfrac{Q}{r}$$ where $$k = \dfrac{1}{4\pi\varepsilon_0}$$.

When the two spheres are joined by a long, thin conducting wire, charge can flow freely until both spheres reach the same electric potential. Let the final charges be $$q_1$$ on $$S_1$$ and $$q_2$$ on $$S_2$$. Two fundamental conditions now govern the situation:

1. Conservation of charge (no charge is lost to the surroundings):

$$q_1 + q_2 = Q_1 + Q_2 = 12\,\mu\text{C} + (-3\,\mu\text{C}) = 9\,\mu\text{C}.$$

2. Equality of potentials, because the conducting wire forces the potentials to be identical:

$$V_1 = V_2 \quad\Longrightarrow\quad k\dfrac{q_1}{r_1} = k\dfrac{q_2}{r_2}.$$

The constant $$k$$ cancels on both sides, leaving

$$\dfrac{q_1}{r_1} = \dfrac{q_2}{r_2}.$$

Now substitute the given radii:

$$\dfrac{q_1}{\dfrac{2}{3}R} = \dfrac{q_2}{\dfrac{1}{3}R}.$$

To remove the complex denominators multiply numerator and denominator appropriately:

$$\dfrac{3q_1}{2R} = \dfrac{3q_2}{R}.$$

Cancel the common factor $$\dfrac{3}{R}$$ from both sides:

$$\frac{q_1}{2} = q_2.$$

Hence the simple relation between the final charges is

$$q_1 = 2q_2.$$

Insert this relation into the conservation equation $$q_1 + q_2 = 9\,\mu\text{C}$$:

$$2q_2 + q_2 = 9\,\mu\text{C}$$

$$3q_2 = 9\,\mu\text{C}$$

$$q_2 = \dfrac{9\,\mu\text{C}}{3} = 3\,\mu\text{C}.$$