JEE Physics Electrostatic Potential and Capacitance MCQs Set 33

Answer: (a) \( CE\dfrac{r_2}{r+r_2} \)

Explanation:
1. In steady state no current flows in the capacitor branch, so the current is \( I=\dfrac{E}{r+r_2} \) through \( r_2 \).
2. The capacitor voltage equals the voltage across \( r_2 \): \( V=Ir_2=\dfrac{Er_2}{r+r_2} \), so \( Q=CV=CE\dfrac{r_2}{r+r_2} \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (b) A potential difference appears between the two cylinders when a charge density is given to the inner cylinder

Explanation:
1. Charge on the inner cylinder produces a field in the gap (Gauss's law), hence a potential difference.
2. Charge on the outer cylinder alone gives zero field inside, so no p.d. A line charge on the axis induces opposite charge on the inner cylinder and creates a field in the gap. With equal charge densities on both the field in the gap still exists, so a p.d. appears.

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (c) \( x^2 \)

Explanation:
1. \( U=kqQ\left[\dfrac1{a-x}+\dfrac1{a+x}\right]=\dfrac{2kqQa}{a^2-x^2} \).
2. For small x: \( U\approx\dfrac{2kqQ}{a}\left(1+\dfrac{x^2}{a^2}\right) \), so \( \Delta U\propto x^2 \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (d) \( \dfrac{q(Q_1-Q_2)(\sqrt2-1)}{\sqrt2\,(4\pi\varepsilon_0R)} \)

Explanation:
1. Potential at the centre of ring 1: \( V_1=\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{Q_1}{R}+\dfrac{Q_2}{\sqrt2R}\right) \); at the centre of ring 2: \( V_2=\dfrac1{4\pi\varepsilon_0}\left(\dfrac{Q_2}{R}+\dfrac{Q_1}{\sqrt2R}\right) \).
2. \( W=q(V_2-V_1)\) in magnitude \( =\dfrac{q(Q_1-Q_2)}{4\pi\varepsilon_0R}\left(1-\dfrac1{\sqrt2}\right)=\dfrac{q(Q_1-Q_2)(\sqrt2-1)}{\sqrt2(4\pi\varepsilon_0R)} \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (a) (i) and (iii)

Explanation:
1. The ring attracts P towards O along the axis; P oscillates through O and back, so the motion is periodic for any \( z_0 \) (i).
2. The force on the axis is \( F=-\dfrac{qQz}{4\pi\varepsilon_0(R^2+z^2)^{3/2}} \), linear in z only when \( z\ll R \) (iii). P is attracted back after crossing O, so it does not escape (iv is false).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (b) \( V\left(\dfrac a{3t}\right)^{1/3} \)

Explanation:
1. Charge is conserved: \( Q=4\pi\varepsilon_0aV \). Volume: \( 4\pi a^2t=\dfrac43\pi r^3\Rightarrow r=(3a^2t)^{1/3} \).
2. \( V'=\dfrac{Q}{4\pi\varepsilon_0r}=\dfrac{aV}{(3a^2t)^{1/3}}=V\left(\dfrac a{3t}\right)^{1/3} \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (c) \( H=\dfrac43a \)

Explanation:
1. Energy conservation, with speed zero at both ends: \( mgH=q[V(0)-V(H)]=\dfrac{q\sigma}{2\varepsilon_0}\left(a-\sqrt{H^2+a^2}+H\right) \).
2. Using \( q/m=4\varepsilon_0g/\sigma \): \( H=2\left(a+H-\sqrt{H^2+a^2}\right)\Rightarrow2\sqrt{H^2+a^2}=2a+H \). Squaring: \( 3H^2=4aH \), so \( H=\dfrac{4a}{3} \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (d) \( \dfrac{Q^2R}{8\pi\varepsilon_0r^2} \)

Explanation:
1. At contact the total charge \( Q+q_{n-1} \) divides in proportion to the radii, so \( q_n=\dfrac{R}{R+r}(Q+q_{n-1}) \).
2. In the limit \( q_\infty=\dfrac{R}{R+r}(Q+q_\infty)\Rightarrow q_\infty=\dfrac{QR}{r} \). Energy \( U=\dfrac{q_\infty^2}{8\pi\varepsilon_0R}=\dfrac{Q^2R}{8\pi\varepsilon_0r^2} \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (a) \( \dfrac56\sigma \)

Explanation:
1. Charges: \( Q_1=4\pi R^2\sigma \), \( Q_2=4\pi(2R)^2\sigma=16\pi R^2\sigma \); total \( 20\pi R^2\sigma \).
2. After connection the potentials are equal, so the charge divides in proportion to the radii: bigger sphere gets \( \dfrac23 \) of the total \( =\dfrac{40}{3}\pi R^2\sigma \). New density \( =\dfrac{40\pi R^2\sigma/3}{16\pi R^2}=\dfrac56\sigma \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (b) \( \sqrt{\dfrac{q\lambda}{2\varepsilon_0m}} \)

Explanation:
1. Axial potential: \( V(x)=\dfrac{\lambda R}{2\varepsilon_0\sqrt{R^2+x^2}} \). It is maximum at O: \( V(O)=\dfrac{\lambda}{2\varepsilon_0} \) and \( V(P)=\dfrac{\lambda}{4\varepsilon_0} \).
2. The particle must cross the hill at O: \( \tfrac12mv^2\ge q[V(O)-V(P)]=\dfrac{q\lambda}{4\varepsilon_0} \), so \( v_{min}=\sqrt{\dfrac{q\lambda}{2\varepsilon_0m}} \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (c) 13.86 s

Explanation:
1. \( V_C=3V_R \) and \( V_C+V_R=12 \) give \( V_C=9 \) V. \( \tau=RC=2.5\times10^6\times4\times10^{-6}=10 \) s.
2. \( 9=12(1-e^{-t/\tau})\Rightarrow e^{-t/\tau}=\dfrac14\Rightarrow t=\tau\ln4=10\times1.386=13.86 \) s.

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (d) (ii) and (iv)

Explanation:
1. Both capacitors have the battery voltage V, so the initial currents are \( V/R \): equal and non-zero (ii).
2. Charge decays as \( e^{-t/RC} \); the smaller capacitor has the smaller time constant, so \( C_1 \) loses half its charge sooner (iv).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (a) \( \dfrac{CVR_2}{R_1+R_2} \)

Explanation:
1. In steady state no current flows through C, so the current is \( \dfrac V{R_1+R_2} \).
2. The capacitor voltage equals the drop across \( R_2 \): \( Q_0=C\dfrac{VR_2}{R_1+R_2} \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (b) \( \dfrac{R_1+R_2}{CR_1R_2} \)

Explanation:
1. The capacitor charges through the Thevenin resistance seen by C, which is \( R_1 \) and \( R_2 \) in parallel: \( R_{eq}=\dfrac{R_1R_2}{R_1+R_2} \).
2. \( \alpha=\dfrac1{R_{eq}C}=\dfrac{R_1+R_2}{CR_1R_2} \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (c) \( 0.198\ \mu\text A \)

Explanation:
1. The capacitor leaks like an RC circuit with \( \tau=RC=\dfrac{\varepsilon_0K}{\sigma}=\dfrac{8.85\times10^{-12}\times5}{7.4\times10^{-12}}=5.98 \) s (independent of geometry).
2. \( i=\dfrac{q_0}{\tau}e^{-t/\tau}=\dfrac{8.85}{5.98}e^{-12/5.98}=1.48\times0.134=0.198\ \mu \)A.

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (d) \( E_C=E_D \)

Explanation:
1. Energy stored: \( E_C=\tfrac12CV_0^2 \). The source supplies \( CV_0^2 \) in total.
2. \( E_D=CV_0^2-\tfrac12CV_0^2=\tfrac12CV_0^2 \), so \( E_C=E_D \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (a) \( \dfrac13\left(\dfrac12CV_0^2\right) \)

Explanation:
1. Each step raises the voltage by \( V_0/3 \); the energy dissipated in a step is \( \tfrac12C(\Delta V)^2=\tfrac12C\left(\dfrac{V_0}3\right)^2 \).
2. Three steps: \( E_D=3\times\tfrac12C\dfrac{V_0^2}{9}=\dfrac13\left(\dfrac12CV_0^2\right) \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.