Correct Option: (c) For a uniform electric field, they are concentric spheres.

In a uniform electric field, equipotential surfaces are parallel equidistant planes oriented perpendicular to the electric field lines, not concentric spheres.

Correct Option: (d) both (a) and (b) are correct.

From the relation $E = -\frac{dV}{dr}$, the spacing between equipotential surfaces is inversely proportional to the electric field intensity ($dr \propto \frac{1}{E}$). Since surface charge density $\sigma$ is highest at sharp corners, the electric field is stronger there, making equipotential surfaces more crowded.

Correct Option: (d) Zero

The potential of an electric dipole at any point $(r, \theta)$ is given by:

$$V = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2}$$

For any point on the equatorial line, $\theta = 90^\circ$. Since $\cos 90^\circ = 0$, $V = 0$.

Correct Option: (d) no work is done

Work done in moving a test charge $q_0$ between two points on an equipotential surface is $W = q_0(V_B - V_A)$. Since potential is identical everywhere on an equipotential surface ($V_A = V_B$), $W = 0$.

Correct Option: (a)

Electric field is related to potential gradient by $E = -\frac{dV}{dX}$:

• For $0 \le X \le 2$: Slope $\frac{dV}{dX} > 0$ and constant $\implies E = -E_0$ (constant negative).
• For $2 \le X \le 4$: Slope $\frac{dV}{dX} = 0 \implies E = 0$.
• For $4 \le X \le 6$: Slope $\frac{dV}{dX} < 0$ and constant $\implies E = +E_0$ (constant positive).

Correct Option: (a) $6\text{ V/m}$ along $(-x)$-axis

The components of the electric field are:

$$E_x = -\frac{\partial V}{\partial x} = -\frac{d}{dx}(3x^2) = -6x$$ $$E_y = -\frac{\partial V}{\partial y} = 0, \quad E_z = -\frac{\partial V}{\partial z} = 0$$

At $x = 1\text{ m}$:

$$\vec{E} = -6(1)\hat{\imath} = -6\hat{\imath}\text{ V/m}$$

Thus, the field is $6\text{ V/m}$ directed along the negative $x$-axis.

Correct Option: (b) $pE$

Work done in rotating a dipole from angle $\theta_1$ to $\theta_2$ in a uniform electric field:

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

Given $\theta_1 = 0^\circ$ and $\theta_2 = 90^\circ$:

$$W = pE(\cos 0^\circ - \cos 90^\circ) = pE(1 - 0) = pE$$

Correct Option: (c) $E \neq 0$ and $V = 0$

Each vertex is at the same distance $r$ from the centroid $G$. The electric potential is scalar:

$$V_G = \frac{k(2q)}{r} + \frac{k(-q)}{r} + \frac{k(-q)}{r} = \frac{k}{r}(2q - q - q) = 0$$

For the electric field: The field due to $+2q$ points downward away from the top vertex, and the vector sum of fields from both $-q$ charges also points downward. Hence, the net field vectors reinforce each other and $E_{\text{net}} \neq 0$.

Correct Option: (d) Both the electric potential and electric field achieve a maximum magnitude at $B$.

Inside the conductor ($r < R$), the electric field is zero. At the surface $r = R$ (point $B$), the electric field jumps to its peak magnitude $E_{\text{max}} = \frac{kQ}{R^2}$. The electric potential throughout the solid conductor is uniform and equal to its maximum surface value at $r = R$.

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

Distance from $A$ to each adjacent charge $+q$ is $L$. Distance from $A$ to each opposite charge $-q$ is $\sqrt{(2L)^2 + L^2} = \sqrt{5}L$:

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

Correct Option: (b) $102.1\text{ J}$

The potential associated with the radial electric field $E = \frac{B}{r^2}$ is $V(r) = \frac{B}{r}$:

$$V_1 = V_2 = \frac{1.2 \times 10^6}{0.12\text{ m}} = 10^7\text{ V}$$

The total electrostatic potential energy is the sum of interaction energy and external potential energies:

$$U = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}} + q_1 V_1 + q_2 V_2$$

Separation between charges $r_{12} = 12 - (-12) = 24\text{ cm} = 0.24\text{ m}$:

$$U_{12} = \frac{(9 \times 10^9) \times (14 \times 10^{-6}) \times (-4 \times 10^{-6})}{0.24} = -2.1\text{ J}$$ $$q_1 V_1 + q_2 V_2 = (14 \times 10^{-6} - 4 \times 10^{-6}) \times 10^7 = 10 \times 10^{-6} \times 10^7 = 100\text{ J}$$ $$U_{\text{total}} = 100 - 2.1 = 102.1\text{ J}$$

Correct Option: (a) spheres

At distances much larger than the geometric dimensions of the charge distribution, the detailed arrangement becomes negligible and the entire system acts as an isolated point charge ($Q_{\text{net}} = \sum q_i \neq 0$). The equipotentials of a point charge are concentric spheres.

Correct Option: (a)

Because the electrostatic force is conservative, the line integral of the electric field around any closed loop vanishes ($\oint \vec{E}\cdot d\vec{l} = 0$). Hence, work done over any closed path is zero.

From the geometry:

• $AC = L$ and $BC = L$.
• $BD = BC = L$ (radius of semicircle). Distance $AD = AB + BD = 2L + L = 3L$.

Potential at initial point $C$:

$$V_C = \frac{kQ}{AC} + \frac{k(-Q)}{BC} = \frac{kQ}{L} - \frac{kQ}{L} = 0$$

Potential at final point $D$:

$$V_D = \frac{kQ}{AD} + \frac{k(-Q)}{BD} = \frac{kQ}{3L} - \frac{kQ}{L} = -\frac{2kQ}{3L}$$

Work done in moving charge $+q$ from $C$ to $D$:

$$W = q(V_D - V_C) = q\left(-\frac{2kQ}{3L} - 0\right) = -\frac{2kQq}{3L} = -\frac{Qq}{6\pi\varepsilon_0 L}$$

Let point $P$ lie at distance $x$ (in metres) from the positive charge $q_1 = 5 \times 10^{-8}\text{ C}$. Distance from $q_2$ is $(0.16 - x)\text{ m}$.

$$V = \frac{1}{4\pi\varepsilon_0}\left[\frac{q_1}{x} + \frac{q_2}{0.16 - x}\right] = 0$$ $$\frac{5 \times 10^{-8}}{x} = \frac{3 \times 10^{-8}}{0.16 - x}$$ $$5(0.16 - x) = 3x \implies 0.80 - 5x = 3x \implies 8x = 0.80 \implies x = 0.10\text{ m} = 10\text{ cm}$$

The electric potential is zero at a point $10\text{ cm}$ from the positive charge between the charges.

(i) Electric field:

$$\vec{E} = -\frac{dV}{dx}\hat{\imath} = -\frac{d}{dx}(10x + 5)\hat{\imath} = -10\hat{\imath}\text{ N/C}$$

(ii) Net Electric Flux:
Since $\vec{E} = -10\hat{\imath}\text{ N/C}$ is uniform, constant in magnitude, and directed along $-\hat{\imath}$, the electric flux entering the right face is equal to the electric flux leaving the left face ($\Phi_{\text{in}} = \Phi_{\text{out}}$):

$$\Phi_{\text{net}} = 0$$

Let point $P$ lie along the dipole axis at distance $r$ from center $O$.

• Distance from $+q$ to $P$ is $(r - a)$.
• Distance from $-q$ to $P$ is $(r + a)$.

$$V_P = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{r - a} + \frac{-q}{r + a}\right] = \frac{q}{4\pi\varepsilon_0}\left[\frac{(r + a) - (r - a)}{r^2 - a^2}\right]$$ $$V_P = \frac{q(2a)}{4\pi\varepsilon_0(r^2 - a^2)}$$

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

$$V_P = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^2 - a^2}$$

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

$$V_P \approx \frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}$$

(i) Net force on charge $q$ at $A$:

• Attractive force due to $-4q$ along $\vec{AB}$: $F_B = \frac{k(q)(4q)}{l^2} = \frac{4kq^2}{l^2}$.
• Repulsive force due to $2q$ along $\vec{CA}$ produced: $F_C = \frac{k(q)(2q)}{l^2} = \frac{2kq^2}{l^2}$.
• Angle between $\vec{F}_B$ and $\vec{F}_C$ is $\theta = 180^\circ - 60^\circ = 120^\circ$.

$$F_{\text{net}} = \sqrt{F_B^2 + F_C^2 + 2F_B F_C \cos 120^\circ}$$ $$F_{\text{net}} = \frac{kq^2}{l^2}\sqrt{4^2 + 2^2 + 2(4)(2)(-1/2)} = \frac{kq^2}{l^2}\sqrt{16 + 4 - 8} = \frac{\sqrt{12}kq^2}{l^2} = \frac{2\sqrt{3}kq^2}{l^2}$$

(ii) Work done to separate charges to infinity:

$$W = U_f - U_i = 0 - U_i = -U_i$$ $$U_i = \frac{k}{l}\left[q(-4q) + (-4q)(2q) + (q)(2q)\right] = \frac{k}{l}[-4q^2 - 8q^2 + 2q^2] = -\frac{10kq^2}{l}$$ $$W = -\left(-\frac{10kq^2}{l}\right) = +\frac{10kq^2}{l} = \frac{10q^2}{4\pi\varepsilon_0 l}$$

Correct Option: (a) decreases $K$ times [1]

The electrostatic force in vacuum/air is $F_0 = \frac{1}{4\pi\varepsilon_0}\frac{Q_1 Q_2}{d^2}$ [1]. In a dielectric medium of constant $K$, the force becomes:

$$F = \frac{1}{4\pi\varepsilon_0 K}\frac{Q_1 Q_2}{d^2} = \frac{F_0}{K}$$

Because $K > 1$, the electrostatic force decreases $K$ times [1, 2].

Correct Answer: $4\,\mu\text{F}$ [1, 2]

The circuit forms a balanced Wheatstone bridge since $\frac{C_1}{C_2} = \frac{C_3}{C_4} = \frac{4\,\mu\text{F}}{4\,\mu\text{F}} = 1$ [1]. Hence, no charge flows through the central $8\,\mu\text{F}$ capacitor, and it can be removed from the calculation [1]:

• Upper branch: two $4\,\mu\text{F}$ capacitors in series $\implies C_{\text{top}} = \frac{4 \times 4}{4 + 4} = 2\,\mu\text{F}$ [1].
• Lower branch: two $4\,\mu\text{F}$ capacitors in series $\implies C_{\text{bottom}} = \frac{4 \times 4}{4 + 4} = 2\,\mu\text{F}$ [1].

These two branches are in parallel across $AB$ [1]:

$$C_{AB} = C_{\text{top}} + C_{\text{bottom}} = 2\,\mu\text{F} + 2\,\mu\text{F} = 4\,\mu\text{F}$$

Correct Option: (c) $2 : 1$ [1]

Let $C_A = C_0$. For capacitor $B$, $C_B = K C_0 = 2C_0$ [1].

In a series circuit, both capacitors store the same charge ($Q_A = Q_B = Q$) [1]. Stored energy in terms of charge is $U = \frac{Q^2}{2C}$ [1]:

$$\frac{U_A}{U_B} = \frac{Q^2 / (2C_A)}{Q^2 / (2C_B)} = \frac{C_B}{C_A} = \frac{2C_0}{C_0} = \frac{2}{1}$$

Correct Option: (c) $\frac{V}{K}$ [1, 3]

When the battery is disconnected, the charge $Q$ on the plates remains constant ($Q = CV$) [3]. Inserting a dielectric slab increases the capacitance to $C' = KC$ [3]. The new potential difference is [3]:

$$V' = \frac{Q}{C'} = \frac{CV}{KC} = \frac{V}{K}$$

Correct Option: (a) $1\,\mu\text{F}$ [3, 4]

The voltage remains fixed at $V = 200\text{ V}$. The change in stored energy is [3]:

$$\Delta U = U_i - U_f = \frac{1}{2} C_i V^2 - \frac{1}{2} C V^2 = \frac{1}{2} (C_i - C) V^2$$ $$2 \times 10^{-2}\text{ J} = \frac{1}{2} (2 \times 10^{-6} - C) (200)^2$$ $$2 \times 10^{-2} = \frac{1}{2} (2 \times 10^{-6} - C) \times 40000 = 20000 (2 \times 10^{-6} - C)$$ $$2 \times 10^{-6} - C = \frac{2 \times 10^{-2}}{2 \times 10^4} = 10^{-6}\text{ F} = 1\,\mu\text{F}$$ $$C = 2\,\mu\text{F} - 1\,\mu\text{F} = 1\,\mu\text{F}$$

Initial charge on capacitor $A$: $Q_0 = CV$ [5].

Common potential $V'$ when connected in parallel [5]:

$$V' = \frac{Q_{\text{total}}}{C_A + C_B} = \frac{CV}{C + 2C} = \frac{V}{3}$$

(i) Ratio of final charges: [5, 6]

$$q_A = C V' = C\left(\frac{V}{3}\right) = \frac{CV}{3}$$ $$q_B = (2C) V' = 2C\left(\frac{V}{3}\right) = \frac{2CV}{3}$$ $$\frac{q_A}{q_B} = \frac{CV/3}{2CV/3} = \frac{1}{2}$$

(ii) Ratio of stored electrostatic energies: [5, 6]

Initial energy stored in $A$: $U_i = \frac{1}{2}CV^2$ [5].

Final total energy stored in combination: [5]

$$U_f = \frac{1}{2}(C_A + C_B)(V')^2 = \frac{1}{2}(3C)\left(\frac{V}{3}\right)^2 = \frac{1}{2}(3C)\left(\frac{V^2}{9}\right) = \frac{1}{6}CV^2$$ $$\frac{U_f}{U_i} = \frac{\frac{1}{6}CV^2}{\frac{1}{2}CV^2} = \frac{1}{3}$$

Case 1: Capacitance with a single slab of thickness $d$ [7, 8]:

$$C = \frac{K\varepsilon_0 A}{d}$$

Case 2: Two dielectric slabs in series, each of thickness $\frac{d}{2}$ [8]:

$$C_1 = \frac{K_1 \varepsilon_0 A}{d/2} = \frac{2K_1 \varepsilon_0 A}{d}, \quad C_2 = \frac{K_2 \varepsilon_0 A}{d/2} = \frac{2K_2 \varepsilon_0 A}{d}$$

For series combination [8]:

$$\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{2\varepsilon_0 A}\left(\frac{1}{K_1} + \frac{1}{K_2}\right) = \frac{d}{2\varepsilon_0 A}\left(\frac{K_1 + K_2}{K_1 K_2}\right)$$ $$C' = \frac{2\varepsilon_0 A}{d}\left(\frac{K_1 K_2}{K_1 + K_2}\right)$$

Equating $C = C'$ [7, 8]:

$$\frac{K\varepsilon_0 A}{d} = \frac{2\varepsilon_0 A}{d}\left(\frac{K_1 K_2}{K_1 + K_2}\right) \implies K = \frac{2K_1 K_2}{K_1 + K_2}$$ $$\text{or equivalently: } \frac{1}{K} = \frac{1}{2}\left(\frac{1}{K_1} + \frac{1}{K_2}\right)$$

Let two parallel conducting plates, each of cross-sectional area $A$, carry charges $+Q$ and $-Q$ so that surface charge density is $\sigma = \frac{Q}{A}$ [9].

The electric field between the plates by superposition is [9]:

$$E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$$

The potential difference between the plates separated by distance $d$ is [9]:

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

Therefore, the capacitance is [9]:

$$C = \frac{Q}{V} = \frac{Q}{\frac{Q d}{\varepsilon_0 A}} = \frac{\varepsilon_0 A}{d}$$

1. $C_2$ and $C_3$ are in series [9]:

$$C_{23} = \frac{200 \times 200}{200 + 200} = 100\text{ pF}$$

2. $C_1$ is in parallel with $C_{23}$ [9]:

$$C_{123} = C_1 + C_{23} = 100\text{ pF} + 100\text{ pF} = 200\text{ pF}$$

3. $C_{123}$ is in series with $C_4$ ($100\text{ pF}$) across the $300\text{ V}$ supply [9]:

$$C_{\text{eq}} = \frac{C_{123} \times C_4}{C_{123} + C_4} = \frac{200 \times 100}{200 + 100} = \frac{200}{3}\text{ pF} \approx 66.7\text{ pF}$$

Charges: [9, 10]

• Total charge through $C_4$: $Q_4 = C_{\text{eq}} V = \left(\frac{200}{3} \times 10^{-12}\right) \times 300 = 2 \times 10^{-8}\text{ C}$ [9].
• Potential across $C_{123}$: $V_{123} = 300\text{ V} - \frac{Q_4}{C_4} = 300 - \frac{2 \times 10^{-8}}{100 \times 10^{-12}} = 100\text{ V}$ [10].
• Charge on $C_1$: $Q_1 = C_1 V_{123} = 100\text{ pF} \times 100\text{ V} = 1 \times 10^{-8}\text{ C}$ [10].
• Charge on $C_2$ and $C_3$: $Q_2 = Q_3 = C_{23} V_{123} = 100\text{ pF} \times 100\text{ V} = 1 \times 10^{-8}\text{ C}$ [10].

Given: $A = 6 \times 10^{-3}\text{ m}^2$, $d = 3\text{ mm} = 3 \times 10^{-3}\text{ m}$, $V = 100\text{ V}$ [9].

(i) Capacitance in air:

$$C_0 = \frac{\varepsilon_0 A}{d} = \frac{(8.854 \times 10^{-12}) \times (6 \times 10^{-3})}{3 \times 10^{-3}} = 17.7 \times 10^{-12}\text{ F} = 17.7\text{ pF}$$

(ii) Charge on plates:

$$Q_0 = C_0 V = (17.7 \times 10^{-12}\text{ F}) \times 100\text{ V} = 1.77 \times 10^{-9}\text{ C} = 1.77\text{ nC}$$

(iii) Mica sheet inserted with battery connected ($V = \text{constant}$):

Since the voltage supply remains connected, the capacitance increases to $C' = K C_0$ [9]:

$$Q' = C' V = K (C_0 V) = 6 \times (1.77 \times 10^{-9}\text{ C}) = 1.062 \times 10^{-8}\text{ C}$$

The charge on the plates increases by a factor of 6 [9].

(i) New Capacitance:

Initial capacitance $C_0 = \frac{\varepsilon_0 A}{d}$ [10]. When distance is halved and dielectric $K = 10$ is placed [10]:

$$C' = \frac{K \varepsilon_0 A}{d'} = \frac{10 \varepsilon_0 A}{d / 2} = 20 \left(\frac{\varepsilon_0 A}{d}\right) = 20 C_0$$

The capacitance increases 20 times.

(ii) Electric Field:

Because the battery remains connected, the potential difference $V$ across the plates remains constant:

$$E' = \frac{V}{d'} = \frac{V}{d/2} = 2\left(\frac{V}{d}\right) = 2E_0$$

The electric field between the plates doubles.

(i) Force on the charge at centre $C$:
Due to electrostatic shielding by the conducting metallic shell, the external charge $+2Q$ induces charges on the outer surface of the shell which cancel out inside [11]. The electric field at the centre due to external charges is zero, so:

$$F_C = 0$$

(ii) Force on the charge at $A$ ($+2Q$):
For any point outside the shell ($x > R$), the shell and the internal charge behave as a single point charge located at the centre $C$ [11]:

$$q_{\text{eff}} = Q + \frac{Q}{2} = \frac{3Q}{2}$$ $$F_A = \frac{1}{4\pi\varepsilon_0}\frac{q_{\text{eff}}(2Q)}{x^2} = \frac{1}{4\pi\varepsilon_0}\frac{(3Q/2)(2Q)}{x^2} = \frac{1}{4\pi\varepsilon_0}\frac{3Q^2}{x^2}$$

(iii) Electric flux through the shell:
By Gauss's Law, the flux through a Gaussian surface just inside the outer boundary depends on the enclosed charge $+Q/2$ [11]:

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

Because of the connecting wires in parallel across the pairs $(C_1, C_2)$ and $(C_4, C_5)$, both pairs are short-circuited [12]. The only operational capacitor connected across terminals $A$ and $B$ is $C_3$ [12]:

$$C_{\text{eq}} = C_3 = 2\,\mu\text{F}$$

(ii) Total charge drawn from battery:

$$Q = C_{\text{eq}} V = (2 \times 10^{-6}\text{ F}) \times 5\text{ V} = 10 \times 10^{-6}\text{ C} = 10\,\mu\text{C}$$

(iii) Total energy stored in network:

$$U = \frac{1}{2} C_{\text{eq}} V^2 = \frac{1}{2} \times (2 \times 10^{-6}\text{ F}) \times (5\text{ V})^2 = 25 \times 10^{-6}\text{ J} = 25\,\mu\text{J}$$

Case I: In Series ($C_1 = C_2 = 12\text{ pF}$, $V = 50\text{ V}$): [12]

$$C_s = \frac{12 \times 12}{12 + 12} = 6\text{ pF}$$ $$U_s = \frac{1}{2} C_s V^2 = \frac{1}{2} \times (6 \times 10^{-12}) \times (50)^2 = 7.5 \times 10^{-9}\text{ J}$$ $$Q_s = C_s V = (6 \times 10^{-12}\text{ F}) \times 50\text{ V} = 300 \times 10^{-12}\text{ C} = 300\text{ pC}$$

Case II: In Parallel: [12]

$$C_p = 12\text{ pF} + 12\text{ pF} = 24\text{ pF}$$ $$U_p = \frac{1}{2} C_p V^2 = \frac{1}{2} \times (24 \times 10^{-12}) \times (50)^2 = 3.0 \times 10^{-8}\text{ J}$$ $$Q_p = C_p V = (24 \times 10^{-12}\text{ F}) \times 50\text{ V} = 1200\text{ pC} = 1.2\text{ nC}$$

Correct Option: (d) $5.0 \times 10^{-2}\text{ J}$ [13, 14]

For 10 identical capacitors in parallel:

$$C_p = 10 \times (1\,\mu\text{F}) = 10\,\mu\text{F} = 10 \times 10^{-6}\text{ F}$$

Total electrostatic energy stored:

$$U = \frac{1}{2} C_p V^2 = \frac{1}{2} \times (10 \times 10^{-6}\text{ F}) \times (100\text{ V})^2 = \frac{1}{2} \times 10^{-5} \times 10^4 = 0.05\text{ J} = 5.0 \times 10^{-2}\text{ J}$$