Correct Option: (b) $\frac{b}{a}$

When connected by a wire, both spheres reach the same electric potential ($V_a = V_b = V$).

Since potential is $V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$ and surface electric field is $E = \frac{V}{r}$:

$$E_a = \frac{V}{a}, \quad E_b = \frac{V}{b}$$ $$\frac{E_a}{E_b} = \frac{V/a}{V/b} = \frac{b}{a}$$

Correct Option: (c) Both a net force and a torque

In a non-uniform field, the electric field strength varies across space ($\vec{E}_+ \neq \vec{E}_-$), so the forces on $+q$ and $-q$ do not cancel ($\vec{F}_{\text{net}} \neq 0$). Simultaneously, non-collinear forces create a net torque ($\vec{\tau} \neq 0$).

Correct Option: (a) $\frac{2q}{4\pi\varepsilon_0 L}\left(1 - \frac{1}{\sqrt{5}}\right)$

Distance from $A$ to the two adjacent charges $+q$ is $L$. Distance to the opposite charges $-q$ is $\sqrt{(2L)^2 + L^2} = \sqrt{5}L$.

$$V_A = 2 \times \frac{kq}{L} + 2 \times \frac{k(-q)}{\sqrt{5}L} = \frac{2kq}{L}\left(1 - \frac{1}{\sqrt{5}}\right) = \frac{2q}{4\pi\varepsilon_0 L}\left(1 - \frac{1}{\sqrt{5}}\right)$$

The sphere centered at origin has radius $r = 3a$.

• Charge $-2Q$ at $(a, 0)$ lies inside the sphere since $a < 3a$.
• Charge $+Q$ at $(4a, 0)$ lies outside the sphere since $4a > 3a$.

Applying Gauss's Law:

$$\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{-2Q}{\varepsilon_0}$$

Initial attractive force magnitude:

$$F = \frac{k |q_1 q_2|}{r^2} = \frac{k(10 \times 20)}{r^2} = \frac{200k}{r^2}$$

When touched, the charges redistribute equally across both identical spheres:

$$q' = \frac{q_1 + q_2}{2} = \frac{10 - 20}{2} = -5\,\mu\text{C}$$

The new distance is $r' = \frac{r}{2}$. The new force is:

$$F' = \frac{k (q')^2}{(r/2)^2} = \frac{k(-5)^2}{\frac{r^2}{4}} = \frac{4 \times 25k}{r^2} = \frac{100k}{r^2}$$

Comparing $F'$ to $F$:

$$\frac{F'}{F} = \frac{100k/r^2}{200k/r^2} = \frac{1}{2} \implies F' = \frac{F}{2}$$

Since both spheres now carry negative charges ($-5\,\mu\text{C}$), the force is repulsive.

The total electrostatic potential energy of a 3-charge system is:

$$U = \frac{1}{4\pi\varepsilon_0}\left[\frac{q_1 q_2}{r} + \frac{q_2 q_3}{r} + \frac{q_1 q_3}{r}\right]$$

Substitute $r = 2.0\text{ m}$ and the charge values:

$$U = \frac{9 \times 10^9}{2.0} \times 10^{-12} \left[(4)(-4) + (-4)(2) + (4)(2)\right]$$ $$U = 4.5 \times 10^{-3} \times [-16 - 8 + 8] = 4.5 \times 10^{-3} \times (-16) = -7.2 \times 10^{-2}\text{ J}$$

Area of circular faces: $A = 25\text{ cm}^2 = 25 \times 10^{-4}\text{ m}^2$.

(i) Net Flux:

• Left face ($x = 1\text{ m}$): $\vec{E}_L = 50(1)\hat{\imath} = 50\hat{\imath}\text{ N/C}$, normal points along $-\hat{\imath}$.
  $$\Phi_L = \vec{E}_L \cdot \Delta\vec{A}_L = (50\hat{\imath}) \cdot (-25 \times 10^{-4}\hat{\imath}) = -0.125\text{ N}\cdot\text{m}^2/\text{C}$$
• Right face ($x = 2\text{ m}$): $\vec{E}_R = 50(2)\hat{\imath} = 100\hat{\imath}\text{ N/C}$, normal points along $+\hat{\imath}$.
  $$\Phi_R = \vec{E}_R \cdot \Delta\vec{A}_R = (100\hat{\imath}) \cdot (+25 \times 10^{-4}\hat{\imath}) = +0.250\text{ N}\cdot\text{m}^2/\text{C}$$
• Curved surface: Electric field is parallel to the surface ($\vec{E} \perp \hat{n}$), so $\Phi_{\text{curved}} = 0$.

$$\Phi_{\text{net}} = \Phi_L + \Phi_R = -0.125 + 0.250 = +0.125\text{ N}\cdot\text{m}^2/\text{C}$$

(ii) Enclosed Charge:

$$q = \varepsilon_0 \Phi_{\text{net}} = (8.854 \times 10^{-12}) \times 0.125 \approx 1.11 \times 10^{-12}\text{ C}$$

Let point $P$ be at perpendicular distance $r$ on the equatorial line from the center $O$. The distance from each charge to $P$ is $\sqrt{r^2 + a^2}$.

Magnitudes of individual fields:

$$E_{+q} = E_{-q} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2 + a^2}$$

Components perpendicular to the dipole axis ($E\sin\theta$) are equal and opposite, cancelling out. Components parallel to the dipole axis ($E\cos\theta$) add up in the direction opposite to the dipole moment $\vec{p}$:

$$E_{\text{net}} = 2 E_{+q} \cos\theta$$

From the geometry of the triangle, $\cos\theta = \frac{a}{\sqrt{r^2 + a^2}}$:

$$E_{\text{net}} = 2 \left(\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2 + a^2}\right)\left(\frac{a}{\sqrt{r^2 + a^2}}\right) = \frac{1}{4\pi\varepsilon_0}\frac{q(2a)}{(r^2 + a^2)^{3/2}}$$

Using dipole moment $p = q(2a)$:

$$\vec{E}_{\text{eq}} = -\frac{1}{4\pi\varepsilon_0}\frac{\vec{p}}{(r^2 + a^2)^{3/2}}$$

For a short dipole ($r \gg a$):

$$\vec{E}_{\text{eq}} \approx -\frac{1}{4\pi\varepsilon_0}\frac{\vec{p}}{r^3}$$

(i) Electric field expression:

Construct a coaxial cylindrical Gaussian surface of radius $r$ and length $l$.

• Flux through circular flat caps: $\vec{E} \perp \hat{n} \implies \Phi_{\text{caps}} = 0$.
• Flux through curved surface: $\vec{E} \parallel \hat{n} \implies \Phi_{\text{curved}} = E(2\pi rl)$.

Charge enclosed: $q_{\text{enc}} = \lambda l$. By Gauss's Law:

$$\oint \vec{E}\cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0} \implies E(2\pi rl) = \frac{\lambda l}{\varepsilon_0}$$ $$E = \frac{\lambda}{2\pi\varepsilon_0 r}$$

(ii) Kinetic Energy of revolving electron:

The electrostatic attraction provides the required centripetal force for circular motion:

$$\frac{m v^2}{r} = eE = e\left(\frac{\lambda}{2\pi\varepsilon_0 r}\right)$$ $$m v^2 = \frac{e\lambda}{2\pi\varepsilon_0}$$

Therefore, the kinetic energy is:

$$K = \frac{1}{2}m v^2 = \frac{e\lambda}{4\pi\varepsilon_0}$$

(Notice that the kinetic energy is independent of the orbital radius $r$).

The restoring torque acting on the dipole at angle $\theta$ is $\tau = pE\sin\theta$.

Work done by an external torque in rotating the dipole from $\theta_0$ to $\theta_1$ without acceleration:

$$W = \int_{\theta_0}^{\theta_1} \tau_{\text{ext}} d\theta = \int_{\theta_0}^{\theta_1} pE\sin\theta d\theta = pE [-\cos\theta]_{\theta_0}^{\theta_1} = pE(\cos\theta_0 - \cos\theta_1)$$

Taking the reference zero potential energy at $\theta_0 = 90^\circ$ ($\frac{\pi}{2}$):

$$U(\theta) = -pE\cos\theta = -\vec{p}\cdot\vec{E}$$

Special Cases:

• Minimum (Stable): At $\theta = 0^\circ$, $U_{\text{min}} = -pE$ (dipole aligns with field).
• Maximum (Unstable): At $\theta = 180^\circ$, $U_{\text{max}} = +pE$ (dipole opposite to field).

Correct Option: (b) $\sqrt{\frac{m_P}{m_Q}}$

The electrostatic acceleration of a charged particle in a uniform field is $a = \frac{qE}{m}$.

Starting from rest ($u = 0$), distance travelled is $s = \frac{1}{2}at^2 = \frac{1}{2}\left(\frac{qE}{m}\right)t^2$.

$$t = \sqrt{\frac{2ms}{qE}} \implies t \propto \sqrt{m}$$ $$\frac{t_P}{t_Q} = \sqrt{\frac{m_P}{m_Q}}$$

Correct Option: (a) Spherical surfaces

At very large distances from a localized group of charges, the spatial separation between the individual charges becomes negligible, and the system behaves as a single point charge having total charge $Q = \sum q_i$. The equipotential surfaces of an isolated point charge are concentric spheres.

By convention, incoming electric flux is negative ($-\phi_1$) and outgoing electric flux is positive ($+\phi_2$):

$$\Phi_{\text{net}} = \phi_2 - \phi_1$$

Applying Gauss's Law, $\Phi_{\text{net}} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$:

$$q_{\text{enclosed}} = \varepsilon_0(\phi_2 - \phi_1)$$

Correct Pair: Pair IV

Electric field lines always point in the direction of steepest decreasing potential ($E_x = -\frac{dV}{dx}$):

• For Pair IV: Left plate is at $-100\text{ V}$, right plate is at $-400\text{ V}$. Potential decreases from left to right, meaning the electric field points along $+\hat{\imath}$.
• For Pairs I, II, and III: Potential increases towards the right, so the electric field points along $-\hat{\imath}$.

Correct Option: (d) Zero

Electric flux is $\Phi = \vec{E} \cdot \vec{A} = EA\cos\theta$. The normal vector $\hat{n}$ to the plane of the paper is perpendicular to the paper ($\theta = 90^\circ$). Because the electric field lies in the plane of the paper, no lines cross through the surface ($\cos 90^\circ = 0$).

Area vector of the sheet parallel to $y-z$ plane: $\vec{A} = a^2 \hat{\imath}$.

Electric field at position $x = c$: $\vec{E} = C c \hat{\imath}$.

$$\Phi = \vec{E} \cdot \vec{A} = (C c \hat{\imath}) \cdot (a^2 \hat{\imath}) = C c a^2$$

Let point $P$ be at distance $r$ from $+1\,\mu\text{C}$. For net zero field ($E_1 = E_2$):

$$\frac{k q_1}{r^2} = \frac{k q_2}{(30 - r)^2} \implies \frac{1}{r^2} = \frac{4}{(30 - r)^2}$$

Taking square root on both sides:

$$\frac{1}{r} = \frac{2}{30 - r} \implies 30 - r = 2r \implies 3r = 30 \implies r = 10\text{ cm}$$

The negative point charge $-Q$ induces positive charges on the near face of the uncharged conducting plate and negative charges on the far face. Field lines emerge normally from the induced positive charges on the plate and terminate radially inward on $-Q$.

Area of square: $A = (0.1\text{ m})^2 = 0.01\text{ m}^2 = 10^{-2}\text{ m}^2$.

Case (i): Plane parallel to $y-z$ plane:
The normal to the plane is along the $x$-axis ($\hat{n} = \hat{\imath}$, so $\theta = 0^\circ$).

$$\Phi = EA\cos 0^\circ = (5 \times 10^3) \times 0.01 \times 1 = 50\text{ N}\cdot\text{m}^2/\text{C}$$

Case (ii): Plane makes an angle of $30^\circ$ with $x$-axis:
The angle $\theta$ between the normal $\hat{n}$ and the electric field $\vec{E}$ is $\theta = 90^\circ - 30^\circ = 60^\circ$.

$$\Phi = EA\cos 60^\circ = (5 \times 10^3) \times 0.01 \times \frac{1}{2} = 25\text{ N}\cdot\text{m}^2/\text{C}$$

Face area: $A = a^2 = (0.1)^2 = 0.01\text{ m}^2$. Field only has an $x$-component.

• Left face ($x = 0.1\text{ m}$): $E_L = 500(0.1) = 50\text{ N/C}$, normal $\hat{n} = -\hat{\imath}$.
  $$\Phi_L = -(50) \times (0.01) = -0.5\text{ N}\cdot\text{m}^2/\text{C}$$
• Right face ($x = 0.1 + 0.1 = 0.2\text{ m}$): $E_R = 500(0.2) = 100\text{ N/C}$, normal $\hat{n} = +\hat{\imath}$.
  $$\Phi_R = +(100) \times (0.01) = +1.0\text{ N}\cdot\text{m}^2/\text{C}$$
• Other four faces: Field is parallel to faces ($\Phi = 0$).

(i) Net outward flux:

$$\Phi_{\text{net}} = \Phi_L + \Phi_R = -0.5 + 1.0 = +0.5\text{ N}\cdot\text{m}^2/\text{C}$$

(ii) Net enclosed charge:

$$q = \varepsilon_0 \Phi_{\text{net}} = (8.854 \times 10^{-12}) \times 0.5 \approx 4.4 \times 10^{-12}\text{ C}$$

(i) At an external point ($r > R$):

Consider a concentric spherical Gaussian surface of radius $r > R$ passing through point $P$.

By symmetry, electric field $\vec{E}$ is directed radially outward and has identical magnitude over the entire Gaussian surface.

$$\oint \vec{E} \cdot d\vec{A} = E \oint dA = E(4\pi r^2)$$

The total charge enclosed by the Gaussian surface is $q_{\text{enc}} = q = \sigma(4\pi R^2)$. Applying Gauss's Law:

$$E(4\pi r^2) = \frac{q}{\varepsilon_0} \implies E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$$

In terms of surface charge density $\sigma$:

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

(ii) At an internal point ($r < R$):

Construct a concentric spherical Gaussian surface of radius $r < R$.

Since all the charge resides on the outer surface of the shell, the charge enclosed inside this Gaussian sphere is zero ($q_{\text{enc}} = 0$).

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

Correct Option: (c) $2E$

Electric field due to a point charge $2q$ at distance $r$ is:

$$E = \frac{1}{4\pi\varepsilon_0}\frac{2q}{r^2}$$

For an external point ($\frac{r}{2} \gg R$), a spherical shell of charge $q$ behaves as a point charge situated at its center:

$$E' = \frac{1}{4\pi\varepsilon_0}\frac{q}{(r/2)^2} = \frac{1}{4\pi\varepsilon_0}\frac{4q}{r^2} = 2 \left(\frac{1}{4\pi\varepsilon_0}\frac{2q}{r^2}\right) = 2E$$

Correct Option: (c) Neutral or positively charged

Since $X$ is negatively charged and repels $Y$, $Y$ must also be negatively charged. A negatively charged body attracts both oppositely charged bodies (positive) and uncharged neutral bodies (via electrostatic induction).

Correct Option: (d) Both (A) and (R) are false

Electrostatic force on a charge $q$ is $\vec{F} = q\vec{E}$. For a negative charge ($q = -e$), the force is directed opposite to the electric field ($\vec{F} = -e\vec{E}$). Consequently, it tends to accelerate in the direction opposite to $\vec{E}$.

Correct Option: (d)

According to Gauss's Law, only charges enclosed within the surface determine the net flux ($\Phi = \frac{q_{\text{enc}}}{\varepsilon_0}$), which is independent of the size or shape of the surface. For practical evaluation using standard symmetry, the electric field over the Gaussian surface is continuous and uniform.

The electrostatic Coulomb force in air is:

$$F_0 = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{r^2}$$

When a dielectric plastic sheet of constant $K$ is introduced, the force decreases to:

$$F = \frac{F_0}{K}$$

Because $K > 1$, the repulsive force between the spheres decreases. As a result, the restoring tension brings the balls closer to each other, decreasing their separation distance.

An electric dipole consists of two equal and opposite charges, $+q$ and $-q$.

The total net charge enclosed within the cube is:

$$q_{\text{enc}} = (+q) + (-q) = 0$$

By Gauss's Law:

$$\Phi = \frac{q_{\text{enc}}}{\varepsilon_0} = \frac{0}{\varepsilon_0} = 0$$

Gauss's law states that the net outward flux depends only on the charge enclosed inside the surface:

$$q_{\text{enc}} = q_2 - q_1$$ $$\Phi = \frac{q_{\text{enc}}}{\varepsilon_0} = \frac{q_2 - q_1}{\varepsilon_0}$$

The external charge $+q_3$ contributes zero net flux through the closed surface.

The area vector is $\vec{A} = A\hat{n} = 10^{-4}(0.8\hat{\imath} + 0.6\hat{k})\text{ m}^2$.

$$\Phi_E = \vec{E} \cdot \vec{A} = (100\hat{\imath}) \cdot \left[10^{-4}(0.8\hat{\imath} + 0.6\hat{k})\right]$$ $$\Phi_E = 100 \times 0.8 \times 10^{-4} = 8.0 \times 10^{-3}\text{ N}\cdot\text{m}^2/\text{C}$$

Dimensional Formula of Electric Flux:

$$[\Phi_E] = [E][A] = \left[\text{M L T}^{-3}\text{ A}^{-1}\right]\left[\text{L}^2\right] = \left[\text{M L}^3\text{ T}^{-3}\text{ A}^{-1}\right]$$

(i) Electric field magnitude:

$$E = \frac{\sigma}{2\varepsilon_0} = \frac{1.0 \times 10^{-18}\text{ C/m}^2}{2 \times 8.854 \times 10^{-12}\text{ C}^2/(\text{N}\cdot\text{m}^2)} \approx 5.65 \times 10^{-8}\text{ N/C}$$

The field is directed normally outwards from the positive sheet.

(ii) Comparison among points $P, Q,$ and $R$:
Point $Q$ lies directly along the central normal axis opposite to point $A$. For a plane sheet of finite dimensions ($10\text{ cm} \times 10\text{ cm}$ at $50\text{ cm}$), edge-fringing affects the peripheral points $P$ and $R$. The central symmetry around point $Q$ matches that of $A$, keeping the electric field magnitude equal.

As derived earlier, the electrostatic force provides the centripetal force:

$$\frac{m v^2}{r} = eE = e\left(\frac{\lambda}{2\pi\varepsilon_0 r}\right) \implies m v^2 = \frac{e\lambda}{2\pi\varepsilon_0}$$

Kinetic energy:

$$K = \frac{1}{2}m v^2 = \frac{e\lambda}{4\pi\varepsilon_0}$$

Because $K \propto \lambda$, the graph is a straight line passing through the origin with positive slope $\frac{e}{4\pi\varepsilon_0}$.

(i) Electric field:
Yes. At the midpoint $x = \frac{a}{2}$, the electric field vectors produced by both charges have equal magnitude $\frac{kq}{(a/2)^2}$ but point in opposite directions:

$$\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2 = \frac{kq}{(a/2)^2}\hat{\imath} - \frac{kq}{(a/2)^2}\hat{\imath} = 0$$

(ii) Electric potential:
No. Electric potential is a scalar quantity. Since both charges are positive, the total potential at any point between them is the sum of two positive numbers:

$$V = \frac{kq}{x} + \frac{kq}{a - x} > 0 \quad (\text{for } 0 < x < a)$$

Because the sum of two positive terms cannot be zero, the electric potential is non-zero everywhere along the finite line segment joining the charges.

Correct Option: (a) No net charge is enclosed by the surface

According to Gauss's Law, $\Phi = \frac{q_{\text{net, enclosed}}}{\varepsilon_0}$. When the net outward flux is zero ($\Phi = 0$), the algebraic sum of all enclosed charges must be zero ($\sum q_{\text{enc}} = 0$). This allows for equal amounts of positive and negative charges to be present inside (such as an electric dipole), so charge is not necessarily absent.

The electric flux remains unchanged.

By Gauss's Law, the total electric flux depends solely on the magnitude of the enclosed charge ($q_{\text{enc}}$) and the medium's permittivity:

$$\Phi = \frac{q_{\text{enc}}}{\varepsilon_0}$$

The total flux is completely independent of the shape, size, or radius of the Gaussian surface.

Definition: The electric dipole moment of an electric dipole is defined as the product of the magnitude of either charge ($q$) and the separation distance ($2a$) between them:

$$\vec{p} = q(2\vec{a})$$

• Nature: It is a vector quantity.
• Direction: Directed along the dipole axis from the negative charge ($-q$) to the positive charge ($+q$).
• SI Unit: Coulomb-meter ($\text{C}\cdot\text{m}$).

(i) Stable Equilibrium ($\theta = 0^\circ$):
The dipole moment vector $\vec{p}$ points in the same direction as the external electric field $\vec{E}$.

$$\tau = pE\sin(0^\circ) = 0$$ $$U_{\text{min}} = -pE\cos(0^\circ) = -pE$$

(ii) Unstable Equilibrium ($\theta = 180^\circ$):
The dipole moment vector $\vec{p}$ is anti-parallel to the electric field $\vec{E}$.

$$\tau = pE\sin(180^\circ) = 0$$ $$U_{\text{max}} = -pE\cos(180^\circ) = +pE$$

A negative electrostatic potential energy ($U < 0$) indicates that the system is bound and stable under attractive forces. It signifies that work was done by the attractive electrostatic field when assembling the charges from infinity, and external positive work must be supplied to overcome the binding forces and separate the charges back to infinite distance.

The relation between electric field intensity and electric potential gradient is:

$$\vec{E} = -\frac{dV}{dx}\hat{\imath}$$

Given $V = 6x$:

$$\vec{E} = -\frac{d}{dx}(6x)\hat{\imath} = -6\hat{\imath}\text{ V/m}$$

Therefore, the electric field has a constant magnitude of $6\text{ V/m}$ and is directed along the negative $x$-axis ($-\hat{\imath}$).