JEE Physics Electrostatic Potential and Capacitance MCQs Set 35

Answer: (c) \( 2Q/V \)

Explanation:
1. Identical capacitors in series share the voltage equally: each has \( V/2 \).
2. Capacitance \( C=Q/(V/2)=2Q/V \).

Teacher's Note:
1. Equal capacitors in series divide the voltage equally.
2. Charge on each series capacitor is the same.

Answer: (d) \( 8C \)

Explanation:
1. Let each capacitor be \( C_0 \). In series: \( C_0/2=2C \), so \( C_0=4C \).
2. In parallel: \( 2C_0=8C \).

Teacher's Note:
1. Series of two identical: \( C0/2 \). Parallel: \( 2 C0 \).
2. Parallel = \( 4 x \) series for identical capacitors.

Answer: (c) The electric field between the plates remains the same

Explanation:
1. With the battery disconnected, charge \( Q \) is constant.
2. Field between the plates \( E=\sigma/\epsilon_0=Q/(A\epsilon_0) \) does not depend on separation, so it stays the same.
3. Potential difference \( V=Ed \) increases.

Teacher's Note:
1. Isolated capacitor: \( Q \) constant.
2. Energy increases because work is done in pulling the plates apart.

Answer: (c) 8

Explanation:
1. Charge is constant, so \( V=Q/C \).
2. Potential becomes 1/8, so the capacitance becomes 8 times: \( C'=KC \).
3. Hence \( K=8 \).

Teacher's Note:
1. With dielectric: \( C \) multiplied by \( K \).
2. Constant \( Q \): \( V \) is inversely proportional to \( C \).

Answer: (c) in the field between the plates

Explanation:
1. The energy of a capacitor is stored in the electric field between the plates.
2. Energy density is \( \tfrac12\epsilon_0E^2 \) in that region.

Teacher's Note:
1. Energy is stored in the field, not on the charges.
2. Energy density \( u = (1/2) \) epsilon-0 \( E^2 \).

Answer: (c) 2 times

Explanation:
1. Volume is conserved: \( \frac43\pi R^3=8\cdot\frac43\pi r^3 \), so \( R=2r \).
2. Capacitance of a sphere \( C=4\pi\epsilon_0r \) is proportional to radius, so it becomes 2 times.

Teacher's Note:
1. Conserve volume to get the new radius.
2. Capacitance of an isolated sphere is proportional to radius.

Answer: (a) \( \frac{Q}{A\epsilon_0}\left(\frac{t_1}{k_1}+\frac{t_2}{k_2}\right) \)

Explanation:
1. Field in a dielectric of constant \( k \): \( E=\frac{Q}{Ak\epsilon_0} \).
2. Potential difference \( V=E_1t_1+E_2t_2 \).
3. So \( V=\frac{Q}{A\epsilon_0}\left(\frac{t_1}{k_1}+\frac{t_2}{k_2}\right) \).

Teacher's Note:
1. Dielectric slabs in series add their \( t/k \) contributions.
2. \( V \) = sum of \( E x \) thickness.

Answer: (b) 4

Explanation:
1. Initially \( C_0=\epsilon_0A/d=1\ \text{pF} \).
2. After: \( C=\frac{K\epsilon_0A}{2d}=\frac K2C_0=2 \) pF.
3. So \( K=4 \).

Teacher's Note:
1. Doubling the separation halves \( C \); the dielectric multiplies by \( K \).
2. \( C = K \) epsilon-0 \( A / d \).

Answer: (b) \( 9\times10^{-4}\ \text{J} \)

Explanation:
1. Energy \( U=\tfrac12CV^2 \).
2. Increase \( \Delta U=\tfrac12(6\times10^{-6})(20^2-10^2)=3\times10^{-6}\times300=9\times10^{-4}\ \text{J} \).

Teacher's Note:
1. Use the difference of energies, not \( (1/2) C (\delta V)^2 \).
2. \( V^2 \): \( 400 - 100 = 300 \).

Answer: (d) the electrostatic energy stored in the capacitor increases

Explanation:
1. Charge is constant (isolated).
2. Capacitance decreases as \( C\propto1/d \).
3. Voltage \( V=Q/C \) increases and energy \( U=Q^2/2C \) increases (the work done by the external agent).

Teacher's Note:
1. \( Q \) constant, \( C \) falls, \( V \) and \( U \) rise.
2. Energy is supplied by whoever pulls the plates apart.

Answer: (b) \( \frac12nCV^2 \)

Explanation:
1. Parallel capacitors add: \( C_{eq}=nC \).
2. Energy \( U=\tfrac12C_{eq}V^2=\tfrac12nCV^2 \).

Teacher's Note:
1. Parallel: \( C \) adds, \( V \) same.
2. Series: \( 1/C \) adds, \( Q \) same.

Answer: (a) \( 32\times10^{-32}\ \text{J} \)

Explanation:
1. Work \( W=\frac{Q^2}{2C} \).
2. Substituting: \( W=\frac{(8\times10^{-18})^2}{2\times10^{-4}}=\frac{64\times10^{-36}}{2\times10^{-4}}=32\times10^{-32}\ \text{J} \).

Teacher's Note:
1. Energy stored \( U = Q^2/2C \).
2. 100 micro-farad = \( 10^{-4} F \).

Answer: (d) The potential difference between the free plates becomes 2V

Explanation:
1. Each capacitor carries charge \( q=CV \) with potential difference \( V \).
2. On connecting the positive plate of one to the negative plate of the other, the charges on the connected plates neutralise (the charge on the junction is conserved and redistributes, no change in the free-plate charges).
3. The potential difference across the free plates becomes \( V+V=2V \).

Teacher's Note:
1. Series connection with the same orientation adds the \( p.d.s \).
2. Free plates keep their charges.

Answer: (b) \( Q_1=Q_2+Q_3 \) and \( V=V_1+V_2 \)

Explanation:
1. The charge on \( C_1 \) flows to the parallel pair and divides: \( Q_1=Q_2+Q_3 \).
2. The \( p.d \). across the parallel pair is common: \( V_2=V_3 \).
3. The battery voltage is divided between \( C_1 \) and the pair: \( V=V_1+V_2 \).

Teacher's Note:
1. Series: same charge, voltages add. Parallel: same voltage, charges add.
2. Identify the series and parallel parts first.

Answer: (c) \( \frac14C(V_1-V_2)^2 \)

Explanation:
1. Initial energy \( U_i=\tfrac12C(V_1^2+V_2^2) \).
2. Common potential after connection \( V=\frac{V_1+V_2}2 \); final energy \( U_f=\tfrac12(2C)V^2=\tfrac14C(V_1+V_2)^2 \).
3. Decrease \( =\tfrac14C(2V_1^2+2V_2^2-V_1^2-2V_1V_2-V_2^2)=\tfrac14C(V_1-V_2)^2 \).

Teacher's Note:
1. Charge is conserved; the common potential is the average for equal capacitances.
2. Energy lost as heat in connecting wires.

Answer: (c) \( 2\pi\epsilon_0R \)

Explanation:
1. Charges \( +Q \) and \( -Q \): potentials \( +\frac{Q}{4\pi\epsilon_0R} \) and \( -\frac{Q}{4\pi\epsilon_0R} \) (far apart).
2. Potential difference \( \frac{2Q}{4\pi\epsilon_0R} \).
3. Capacitance \( C=\frac{Q}{\Delta V}=2\pi\epsilon_0R \).

Teacher's Note:
1. Two isolated spheres of opposite charge act as capacitors in series.
2. Each sphere alone has capacitance \( 4 \pi \) epsilon-0 \( R \).

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

Explanation:
1. Initial energy \( 2\times\tfrac12CV^2=CV^2 \).
2. After connection the total charge is zero, so the common potential is zero and final energy is zero.
3. Loss \( CV^2 \).

Teacher's Note:
1. Charge conservation gives common potential \( (Q1+Q2)/(C1+C2) = 0 \).
2. The lost energy appears as heat and radiation.

Answer: (c) 0.01 J

Explanation:
1. Stored energy \( U=\tfrac12CV^2=\tfrac12(2\times10^{-6})(100)^2=0.01\ \text{J} \).
2. All of it is dissipated as heat when the plates are shorted.

Teacher's Note:
1. Heat = energy stored.
2. 2 micro-farad = \( 2 x 10^{-6} F \).

Answer: (b) 90 kW

Explanation:
1. Energy \( U=\tfrac12CV^2=\tfrac12(40\times10^{-6})(3000)^2=180\ \text{J} \).
2. Power \( P=U/t=180/(2\times10^{-3})=9\times10^4\ \text{W}=90\ \text{kW} \).

Teacher's Note:
1. Power = energy / time.
2. Defibrillator capacitors are of the order of tens of micro-farads.

Answer: (c) \( 3V/(k+2) \)

Explanation:
1. Total charge (conserved) \( Q=3CV \).
2. New capacitances: \( kC \) and \( 2C \) in parallel: \( (k+2)C \).
3. New potential difference \( V'=\frac{3CV}{(k+2)C}=\frac{3V}{k+2} \).

Teacher's Note:
1. Isolated parallel combination: total charge is constant.
2. Parallel capacitors share a common potential.

Answer: (a) 2 V

Explanation:
1. Equivalent capacitance: \( \frac1{C_{eq}}=\frac1C+\frac1C+\frac1{2C}=\frac{5}{2C} \), so \( C_{eq}=0.4C \).
2. Charge on each \( Q=0.4C\times10=4C \) (units V-times \( C \)).
3. Potential of the junction = voltage across the \( 2C \) capacitor (other end earthed) \( =\frac{4C}{2C}=2 V \).

Teacher's Note:
1. Series capacitors carry the same charge; voltage divides inversely with capacitance.
2. The earthed end is at \( 0 V \).