JEE Physics Electrostatic Potential and Capacitance MCQs Set 26

Answer: (a) of the same magnitude but opposite in nature

Explanation:
1. The two plates form a pair: the field of one induces an equal and opposite charge on the facing surface of the other.
2. So the magnitudes are equal while the signs are opposite; only the charge density differs.

Teacher's Note:
1. Capacitor plates always carry equal and opposite charges (isolated pair).
2. Different areas change the surface charge density, not the charge.

Answer: (d) 40 μC

Explanation:
1. \( C_{eq} = \frac{6\times12}{18} = 4\ \mu\text{F} \).
2. Charge on each in series \( = C_{eq}V = 4\times10 = 40\ \mu\text{C} \).

Teacher's Note:
1. Capacitors in series carry the same charge.
2. Total voltage across the series pair is \( 10 V \).

Answer: (b) \( \frac{\epsilon_0A_2}{d} \)

Explanation:
1. Capacitance depends on the overlapping area of the facing plates.
2. Overlap is the smaller area, so \( C = \frac{\epsilon_0A_2}{d} \).

Teacher's Note:
1. Only facing areas contribute.
2. Edge effects are neglected.

Answer: (b) 80 μC

Explanation:
1. Series of \( 2 \muF \) and \( 1 \muF \): \( \frac{2\times1}{3} = \frac23\ \mu\text{F} \) across \( 120 V \).
2. Charge \( = \frac23\times120 = 80\ \mu\text{C} \), the same on each in series, including the \( 2 \muF \).

Teacher's Note:
1. Parallel branches each have the full source voltage.
2. Series capacitors share the same charge.

Answer: (c) 24 μC

Explanation:
1. Parallel pair \( = 1+5 = 6\ \mu\text{F} \). Series with \( 4 \muF \): \( \frac{4\times6}{10} = 2.4\ \mu\text{F} \).
2. Charge on the branch \( = 2.4\times10 = 24\ \mu\text{C} \), which is the charge on the \( 4 \muF \) capacitor.

Teacher's Note:
1. Reduce the parallel part first, then the series chain.
2. The \( 3 \muF \) capacitor is across the battery directly and does not affect this branch.

Answer: (a) E/4

Explanation:
1. Charge \( Q \) divides equally, each capacitor having \( Q/2 \).
2. Energy in \( C_{2} = \frac{(Q/2)^2}{2C} = \frac14\cdot\frac{Q^2}{2C} = \frac E4 \).

Teacher's Note:
1. Half the charge gives a quarter of the energy because \( U ∝ Q^{2} \).
2. Half of the original energy is lost as heat.

Answer: (d) increase

Explanation:
1. \( C_B \) increases, so the series capacitance increases and, with \( V \) fixed, the charge \( Q = C_{eq} V \) increases.
2. For \( A \): \( U_A = \frac{Q^2}{2C_A} \) increases because \( Q \) increases and \( C_A \) is unchanged.

Teacher's Note:
1. Battery stays connected, so voltage is fixed, not charge.
2. Both capacitors in series carry the new, larger charge.

Answer: (b) 30 μF

Explanation:
1. The two parts are in parallel. Each part has air capacitance \( 5 \muF \).
2. \( C = 5\times2 + 5\times4 = 30\ \mu\text{F} \).

Teacher's Note:
1. Side-by-side regions across the same plates are in parallel.
2. Dielectric multiplies the capacitance by \( K \).

Answer: (d) the same

Explanation:
1. The battery maintains the potential difference \( V \), so \( E = V/d \) remains the same when the slab fills the gap.
2. The charge on the plates increases instead.

Teacher's Note:
1. With the battery connected, \( V \) is fixed.
2. With the battery disconnected, \( E \) would reduce by \( K \).

Answer: (a) \( \tau\ln\frac{10}{9} \)

Explanation:
1. \( q = q_0(1-e^{-t/\tau}) = 0.1q_0 \Rightarrow e^{-t/\tau} = 0.9 \).
2. \( t = -\tau\ln0.9 = \tau\ln\frac{10}{9} \).

Teacher's Note:
1. Charging follows \( q = q_{0}(1 - e^{-t/\tau }) \).
2. \( -\ln 0.9 \) equals \( \ln(10/9) \).

Answer: (c) 0.45 m²

Explanation:
1. Minimum gap: \( d = \frac{V}{E_{max}} = \frac{1200}{3\times10^6} = 4\times10^{-4} m \). \( C = \frac QV = 10^{-8} F \).
2. \( A = \frac{Cd}{\epsilon_0} = \frac{10^{-8}\times4\times10^{-4}}{8.85\times10^{-12}} \approx 0.45\ \text{m}^2 \).

Teacher's Note:
1. Dielectric strength sets the smallest safe plate separation.
2. Then area follows from \( C \) = ε₀A/d.

Answer: (c) 1 V

Explanation:
1. On an isolated pair of plates the total charge \( 20 \muC \) divides so that the facing surfaces carry \( +10 \muC \) and \( -10 \muC \).
2. \( V = \frac{10\ \mu\text{C}}{10\ \mu\text{F}} = 1 V \).

Teacher's Note:
1. The charge that matters is the charge on the inner (facing) surfaces.
2. Half of the given charge ends on the outer surface.

Answer: (a) E is uniform in each layer, with a larger value in the K = 2 layer than in the K = 4 layer

Explanation:
1. Displacement \( D \) is continuous across the layers, so \( E = \frac{D}{K\epsilon_0} \propto \frac1K \).
2. Hence \( E \) is constant within each layer and the \( K = 2 \) layer has twice the field of the \( K = 4 \) layer.

Teacher's Note:
1. Normal component of \( D \) is continuous at a dielectric boundary.
2. No free charge between the plates, so \( E \) does not vary within a uniform layer.

Answer: (d) 10 kW

Explanation:
1. \( U = \tfrac12CV^2 = \tfrac12\times20\times10^{-6}\times(2000)^2 = 40 J \).
2. \( P = \frac{40}{4\times10^{-3}} = 10^4 W = 10 \) kW.

Teacher's Note:
1. Energy stored is ½CV².
2. Power is energy per unit time.