Electrostatic Potential and Capacitance: NEET and JEE questions with solutions

Answer: (D) $1/6$

$P$ and $Q$ are in series (equivalent $C/2$), and this combination is in parallel with $S$ across the battery.

Charge on $P$ (and $Q$): $q = \dfrac{C}{2}V$, so
$$U_P = \frac{q^2}{2C} = \frac{C^2V^2/4}{2C} = \frac{CV^2}{8}$$

Equivalent capacitance: $C_{eq} = \dfrac{C}{2} + C = \dfrac{3C}{2}$, so
$$U_T = \frac{1}{2}\cdot\frac{3C}{2}V^2 = \frac{3CV^2}{4}$$

$$\frac{U_P}{U_T} = \frac{1/8}{3/4} = \frac{1}{6}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $18$

The electric field is conservative, so the work done in moving the charge slowly depends only on the end points A and B, not on the path through the sphere.

Total charge: $Q = \dfrac{4}{3}\pi R^3 \rho$. Outside the sphere,
$$V(r) = \frac{Q}{4\pi\epsilon_0 r} = \frac{\rho R^3}{3\epsilon_0 r}$$

$$V_A = V(2R) = \frac{\rho R^2}{6\epsilon_0}, \qquad V_B = V(3R) = \frac{\rho R^2}{9\epsilon_0}$$

Work done by the external agent on the unit charge:
$$|W| = |V_B - V_A| = \frac{\rho R^2}{\epsilon_0}\left(\frac{1}{6} - \frac{1}{9}\right) = \frac{\rho R^2}{18\epsilon_0}$$

So $n = 18$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\sqrt{\dfrac{Qq}{3\pi\epsilon_0 mR}}$

Loss in potential energy = gain in kinetic energy:
$$\frac{1}{2}mv^2 = \frac{Qq}{4\pi\epsilon_0}\left(\frac{1}{R} - \frac{1}{3R}\right) = \frac{Qq}{4\pi\epsilon_0}\cdot\frac{2}{3R}$$

$$v^2 = \frac{4Qq}{3 \cdot 4\pi\epsilon_0 mR} = \frac{Qq}{3\pi\epsilon_0 mR}$$

$$v = \sqrt{\frac{Qq}{3\pi\epsilon_0 mR}}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $2.66$

The slabs fill $\dfrac{3d}{8} + \dfrac{d}{2} = \dfrac{7d}{8}$, leaving an air gap of $\dfrac{d}{8}$. For slabs stacked between the plates:
$$C = \frac{\epsilon_0 A}{\dfrac{d}{8} + \dfrac{3d}{8K_1} + \dfrac{d}{2K_2}}$$

$C = 2C_0 = \dfrac{2\epsilon_0 A}{d}$, so the denominator equals $\dfrac{d}{2}$:
$$\frac{1}{8} + \frac{3}{8K_1} + \frac{1}{2K_2} = \frac{1}{2} \;\Rightarrow\; \frac{3}{8K_1} + \frac{1}{2K_2} = \frac{3}{8}$$

With $K_2 = \dfrac{K_1}{1.25}$: $\dfrac{1}{2K_2} = \dfrac{1.25}{2K_1} = \dfrac{5}{8K_1}$.
$$\frac{3}{8K_1} + \frac{5}{8K_1} = \frac{1}{K_1} = \frac{3}{8} \;\Rightarrow\; K_1 = \frac{8}{3} \approx 2.66$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1.0$ J

$U = -pE\cos\theta$, so
$$\Delta U = pE(\cos 0° - \cos 60°) = 5 \times 10^{-6} \times 4 \times 10^5 \times \left(1 - \frac{1}{2}\right) = 2 \times 0.5 = 1.0\ \text{J}$$

Solution by Sreeraj P, M.Sc Physics