Electrostatic Potential and Capacitance, on the exam screen.

The four 10 $\mu F$ capacitors are in series, giving $\frac{10}{4} = 2.5\ \mu F$, and this branch is in parallel with $C_5 = 2.5\ \mu F$, so $C_{eq} = 2.5 + 2.5 = 5\ \mu F$. Each branch has 50 V across it, so the series branch carries $q = 2.5 \times 50 = 125\ \mu C$ (the same charge flows through $C_1$ to $C_4$) and $C_5$ also holds $2.5 \times 50 = 125\ \mu C$. Every capacitor therefore stores 125 $\mu C$, option (2).

When a charged capacitor shares its charge with an identical uncharged one, the energy dissipated is $\dfrac{1}{2}\dfrac{C_1 C_2}{C_1 + C_2}V^2$. With $C_1 = C_2 = 200\ \text{pF}$ and $V = 100\ \text{V}$, this equals $\dfrac{1}{2}(100\ \text{pF})(100)^2 = 0.5 \times 10^{-6}\ \text{J}$ (option (1)). The loss is real and independent of the connecting-wire resistance; answers in whole joules mishandle the pico-farad factor.

In electrostatic equilibrium, the electric field inside a conductor is zero, and any excess charge lies on its surface. The field just outside is normal to the surface and depends on the local surface charge density, $E = \dfrac{\sigma}{\varepsilon_0}\hat{n}$. The potential is constant inside but need not be zero. Therefore internal statements I, III and IV hold; II and V do not.

$P$ and $Q$ form a series branch of equivalent capacitance $C/2$, and that branch sits in parallel with $S$ across the battery, so $C_{eq} = \frac{C}{2} + C = \frac{3C}{2}$ and $U_T = \frac{1}{2}\left(\frac{3C}{2}\right)V^2 = \frac{3}{4}CV^2$. Identical capacitors in series split the voltage equally, so $P$ holds $V/2$ and $U_P = \frac{1}{2}C\left(\frac{V}{2}\right)^2 = \frac{CV^2}{8}$. The ratio is $\frac{CV^2/8}{3CV^2/4} = \frac{1}{6}$. NCERT Class 12, Electrostatic Potential and Capacitance.

Why the other options are wrong:

- (1) treats all three capacitors as being in parallel, so each stores an equal third and $P$ plus $Q$ gives 2/3
- (2) assumes the three capacitors share the energy equally, ignoring that $P$ sees only half the battery voltage
- (3) computes the ratio of $U_P$ to the energy of the series branch alone, forgetting to include $S$ in the total

Work done in moving a charge slowly is path-independent and depends only on the endpoint potentials, so the detour through the tube is a red herring. Both $A$ and $B$ lie OUTSIDE the sphere, where $V(r) = \frac{Q}{4\pi\varepsilon_0 r}$ with $Q = \frac{4}{3}\pi R^3\rho$. Then $|W| = |V_B - V_A| = \frac{Q}{4\pi\varepsilon_0 R}\left|\frac{1}{3} - \frac{1}{2}\right| = \frac{\rho R^2}{18\varepsilon_0}$, giving $n = 18$. NCERT Class 12, Electrostatic Potential and Capacitance.

Why the other options are wrong:

- (1) uses only the $\frac{1}{2}$ from position $A$ and drops the $\frac{4}{3}\pi R^3$ volume factor
- (2) computes $\frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ but forgets the $\frac{1}{3}$ carried by the sphere's total charge
- (3) integrates the internal field along the tube as if the endpoints were inside the sphere

Outside the sphere the potential energy of the attracting pair is $U(r) = -\frac{Qq}{4\pi\varepsilon_0 r}$. Energy conservation from $r = 3R$ (at rest) to $r = R$ gives $\frac{1}{2}mv^2 = \frac{Qq}{4\pi\varepsilon_0 R}\left(1 - \frac{1}{3}\right) = \frac{Qq}{6\pi\varepsilon_0 R}$. Solving, $v = \sqrt{\frac{Qq}{3\pi\varepsilon_0 mR}}$. NCERT Class 12, Electrostatic Potential and Capacitance.

Electric field is the negative spatial gradient of potential. A constant 5 V everywhere in the region has zero gradient, so the field is zero. Chapter: Electrostatic Potential and Capacitance.

The electric potential due to a dipole is given by $V = \frac{1}{4\pi\epsilon_0} \frac{p \cos \theta}{r^2}$. Substituting the values, $V = \frac{9 \times 10^9 \times 16 \times 10^{-9} \times \cos 60^\circ}{(0.6)^2} = \frac{9 \times 16 \times 0.5}{0.36} = 200\ V$. Chapter: Electrostatic Potential and Capacitance.

The dielectric constant is 30/6 = 5. Therefore ε = 5ε$_{0}$ = $4.425 \times 10^{-11}$ = $0.4425 \times 10^{-10}$ C$^{2}$ N$^{-1}$ m$^{-2}$, matching option (2). Chapter: Electrostatic Potential and Capacitance.