JEE Physics Electrostatic Potential and Capacitance MCQs Set 15

Answer: (c) \( C_0\left[\left(\frac{V_0}{V}\right)^{1/n}-1\right] \)

Explanation:
1. Each connection multiplies the voltage by \( \frac{C_0}{C_0+C} \).
2. After n times: \( V=V_0\left(\frac{C_0}{C_0+C}\right)^n \).
3. Solve: \( C=C_0\left[\left(\frac{V_0}{V}\right)^{1/n}-1\right] \).

Teacher's Note:
1. Charge sharing multiplies the voltage by a fixed ratio each time.
2. Take the n-th root to isolate C.

Answer: (a) \( 4\pi\epsilon_0r_2 \)

Explanation:
1. The wire makes both one conductor; all the charge sits on the outer surface of the shell.
2. So the system behaves like an isolated sphere of radius \( r_2 \): \( C=4\pi\epsilon_0r_2 \).

Teacher's Note:
1. The spherical-capacitor formula needs different potentials on the two spheres.
2. Joined by a wire, they form a single conductor.

Answer: (c) \( \frac{C(V_0-V)}{V} \)

Explanation:
1. Charge is conserved: \( CV_0=(C+C_x)V \).
2. So \( C_x=\frac{CV_0}{V}-C=\frac{C(V_0-V)}{V} \).

Teacher's Note:
1. Conservation of charge fixes the final voltage when capacitors are joined.
2. V is smaller than V0, so Cx is positive.

Answer: (a) 1/2 V

Explanation:
1. Charge is fixed and capacitance becomes 8 times: \( V'=\frac{V}{K}=\frac48=0.5\ \text{V} \).

Teacher's Note:
1. Disconnected means Q is constant.
2. V = Q/C falls by the dielectric constant.

Answer: (c) \( 12\times10^3 \) V/m

Explanation:
1. The battery holds the voltage at 300 V, so \( E=\frac Vd=\frac{300}{2.5\times10^{-2}}=12\times10^3\ \text{V/m} \).

Teacher's Note:
1. With the battery connected, V stays fixed and the field changes with d.
2. Use d in metres.

Answer: (d) \( \frac{4\epsilon_r}{3\epsilon_r+1} \)

Explanation:
1. The gap has thickness \( \frac d4 \) with the dielectric and \( \frac{3d}{4} \) of air, effectively in series: \( C=\frac{\epsilon_0A}{\frac{3d}{4}+\frac{d}{4\epsilon_r}} \).
2. \( \frac C{C_0}=\frac{1}{\frac34+\frac1{4\epsilon_r}}=\frac{4\epsilon_r}{3\epsilon_r+1} \).

Teacher's Note:
1. A partly filled gap behaves like series layers: thickness divided by K.
2. Check: for K = 1 the ratio is 1.

Answer: (b) \( \frac14 \)

Explanation:
1. Charge: \( q=q_0e^{-t/\tau} \). Energy \( \propto q^2=e^{-2t/\tau} \).
2. Energy halves: \( e^{-2t_1/\tau}=\frac12 \), \( t_1=\frac{\tau\ln2}{2} \).
3. Charge becomes 1/4: \( t_2=\tau\ln4=2\tau\ln2 \).
4. \( \frac{t_1}{t_2}=\frac{1/2}{2}=\frac14 \).

Teacher's Note:
1. Energy depends on the square of charge, so it decays twice as fast.
2. Take logarithms on both sides.

Answer: (b) \( 5.06\times10^{-8} \) J (increase)

Explanation:
1. \( Q=C_1V=4.5\ \text{pF}\times300=1.35\ \text{nC} \).
2. \( \Delta U=\frac{Q^2}{2}\left(\frac1{C_2}-\frac1{C_1}\right)=\frac{(1.35\times10^{-9})^2}{2}(2.78\times10^{11}-2.22\times10^{11})\approx5.06\times10^{-8}\ \text{J} \).

Teacher's Note:
1. At constant charge the energy is inversely proportional to C, so it increases when C falls.
2. The extra energy equals the work done by the person pulling the plates apart.

Answer: (c) 96 µJ

Explanation:
1. Initial energy \( \frac12\times2\times144=144\ \mu\text{J} \); final energy 48 µJ.
2. Heat = 144 − 48 = 96 µJ. Check: \( \frac12\frac{C_1C_2}{C_1+C_2}V^2=\frac12\times\frac{8}{6}\times144=96 \).

Teacher's Note:
1. The lost energy appears as heat in the wire and as radiation.
2. The formula with the product over the sum gives the same result.

Answer: (b) 2 nF and 12.1 µJ

Explanation:
1. Without the slab \( C_0=\frac{\epsilon_0A}{d}=\frac{8.85\times10^{-12}\times0.04}{8.85\times10^{-4}}=0.4\ \text{nF} \).
2. The two halves are in parallel: \( C=\frac{C_0}{2}(9+1)=2\ \text{nF} \).
3. \( U=\frac12(2\times10^{-9})(110)^2=12.1\ \mu\text{J} \).

Teacher's Note:
1. A slab covering part of the area gives capacitors in parallel.
2. Each half has half the area.

Answer: (a) \( 6\times10^{-4} \) C

Explanation:
1. Net charge \( =900-400=500\ \mu\text{C} \), so \( V=\frac{500}{5}=100\ \text{V} \).
2. Final charge on the 3 µF capacitor is 300 µC, so the charge that flowed is \( 900-300=600\ \mu\text{C}=6\times10^{-4}\ \text{C} \).

Teacher's Note:
1. Opposite polarity: subtract the charges.
2. Flow equals the change of charge on either capacitor (here 900 to 300 microcoulomb).

Answer: (b) double

Explanation:
1. A conductor slab removes its own thickness from the effective gap: \( d_{eff}=d-\frac d2=\frac d2 \).
2. \( C'=\frac{\epsilon_0A}{d/2}=2C \).

Teacher's Note:
1. The field is zero inside a conductor, so that thickness contributes no potential drop.
2. The result does not depend on where the slab sits.

Answer: (a) 25 µC from the smaller to the bigger sphere

Explanation:
1. Joined spheres reach the same potential, so charge divides in proportion to radius: \( Q_1:Q_2=5:10=1:2 \).
2. Total charge is 150 µC, so the final charges are 50 µC and 100 µC.
3. The smaller sphere goes from 75 µC to 50 µC, so 25 µC moves from the smaller to the bigger sphere.

Teacher's Note:
1. Capacitance of a sphere is proportional to its radius, so the bigger sphere holds more charge at a common potential.
2. Always compare final charges with initial charges to get the direction.

Answer: (d) 2.5 s

Explanation:
1. Half-time \( t=RC_{eq}\ln2 \).
2. Parallel: \( C_{eq}=2C \); series: \( C_{eq}=\frac C2 \), which is one fourth.
3. So the series time is \( \frac{10}{4}=2.5\ \text{s} \).

Teacher's Note:
1. The half-time is proportional to the equivalent capacitance.
2. Parallel to series changes C_eq by a factor of 4.

Answer: (d) 4

Explanation:
1. Two capacitors in parallel give 4 µF and two in series give 1 µF; putting these in parallel gives 5 µF with 4 capacitors.
2. Three capacitors can only give 2/3, 1, 4/3, 2, 3, 4 or 6 µF, never 5 µF.

Teacher's Note:
1. List what small groups can produce before choosing.
2. Series pieces give fractions; parallel pieces add.

Answer: (a) \( \frac{3W}{4} \)

Explanation:
1. \( W=\frac12C(40^2-20^2)=600C \).
2. Second work \( =\frac12C(50^2-40^2)=450C \).
3. \( \frac{450C}{600C}=\frac34 \), so the second work is 3W/4.

Teacher's Note:
1. Work equals the change in stored energy, one half C V squared.
2. Use differences of squares, not differences of voltages.

Answer: (b) 10 J/s

Explanation:
1. The rate of energy storage is \( P=vi=E(1-e^{-t/\tau})\frac ERe^{-t/\tau} \).
2. With \( x=e^{-t/\tau} \), \( P=\frac{E^2}{R}x(1-x) \), maximum at \( x=\frac12 \).
3. \( P_{max}=\frac{E^2}{4R}=\frac{400}{40}=10\ \text{J/s} \).

Teacher's Note:
1. The capacitor absorbs maximum power when its voltage is half of E.
2. This is the maximum power transfer result for a resistor-capacitor charging circuit.

Answer: (d) 4 V

Explanation:
1. The battery keeps the voltage fixed at 4 V; only the charge changes (it becomes 8 times larger).

Teacher's Note:
1. With the battery connected the voltage is set by the battery.
2. Charge on the plates changes instead.

Answer: (d) 0.5 nC and 10 V

Explanation:
1. The charge on the middle plate spreads equally on its two faces, 0.5 nC each, inducing −0.5 nC on the facing inner surfaces of the neutral outer plates; each outer surface then carries +0.5 nC.
2. Field between plates \( E=\frac{0.5\times10^{-9}}{\epsilon_0A}=\frac{0.5\times10^{-9}}{8.85\times10^{-12}\times0.05}=1.13\times10^3\ \text{V/m} \).
3. \( V=Ed=1130\times8.85\times10^{-3}=10\ \text{V} \).

Teacher's Note:
1. For any set of parallel plates, the outer surfaces share the total charge equally.
2. The inner surfaces are fixed by the induced charges.

Answer: (b) \( \frac{4\pi\epsilon_0}{\frac1a+\frac1b-\frac2d} \)

Explanation:
1. For charges +q on a and −q on b: \( V_a=\frac{kq}{a}-\frac{kq}{d} \) and \( V_b=-\frac{kq}{b}+\frac{kq}{d} \).
2. \( V_a-V_b=kq\left(\frac1a+\frac1b-\frac2d\right) \), and \( C=\frac{q}{V_a-V_b} \).

Teacher's Note:
1. Add the potential due to the other sphere's charge, using d as its distance.
2. Distant spheres: treat each charge as a point at the centre.