JEE Physics Electrostatic Potential And Capacitance MCQs Set 08

Answer: (b) Zero

Explanation:
1. Electric field \( E=-\frac{dV}{dx} \).
2. Between x = 1 m and 3 m the potential is constant, so \( \frac{dV}{dx}=0 \) and the field is zero.

Teacher's Note:
1. Field is the negative slope of the V-x graph.
2. A flat portion of the graph means zero field.

Answer: (a) \( \frac{qQ}{4\pi\varepsilon_0}\left(\frac1a-\frac1b\right) \)

Explanation:
1. Work done by the agent = change in potential energy: \( W=q(V_A-V_B) \).
2. With \( V=\frac{Q}{4\pi\varepsilon_0r} \): \( W=\frac{qQ}{4\pi\varepsilon_0}\left(\frac1a-\frac1b\right) \).

Teacher's Note:
1. Work done against the field equals the increase in potential energy.
2. Potential of a point charge goes as 1/r (not 1/r²).

Answer: (b) \( rE \)

Explanation:
1. Field due to one sphere at distance r/2: \( E=\frac{kq}{(r/2)^2} \).
2. Potential due to one sphere \( =\frac{kq}{r/2}=E\frac r2 \).
3. Both spheres contribute equally: \( V=2\times\frac{rE}{2}=rE \).

Teacher's Note:
1. Potential is a scalar, so the two contributions add directly.
2. V = E × distance for a point charge at distance d only when you use E = kq/d², so V = Ed.

Answer: (d) increases

Explanation:
1. \( U=qV \) with q positive, so the potential energy increases when V increases.

Teacher's Note:
1. Potential energy = charge × potential.
2. For a negative charge the result would be reversed.

Answer: (c) 90 m/s

Explanation:
1. Loss of potential energy \( =kq_1q_2\left(\frac11-\frac1{10}\right)=9\times10^9\times10^{-6}\times10^{-3}\times0.9=8.1\ \text{J} \).
2. \( \tfrac12(0.002)v^2=8.1\Rightarrow v=\sqrt{8100}=90\ \text{m/s} \).

Teacher's Note:
1. Convert μC and mC to coulombs.
2. Electrostatic potential energy becomes kinetic energy.

Answer: (b) decreases then increases

Explanation:
1. Near each positive charge the potential is high.
2. At the midpoint the potential is a minimum (the sum is \( \frac{2kq}{d/2} \) there, lower than close to either charge).
3. So the potential decreases and then increases.

Teacher's Note:
1. Potential near a positive charge is large and positive.
2. The midpoint is the farthest from both charges.

Answer: (a) \( n^{-2/3}V \)

Explanation:
1. Volume is conserved: \( nr^3=R^3\Rightarrow r=Rn^{-1/3} \). Each small drop carries charge \( q'=\frac qn \).
2. \( V'=\frac{kq'}{r}=\frac{kq/n}{Rn^{-1/3}}=n^{-2/3}V \).

Teacher's Note:
1. Volume and charge are both conserved.
2. The radius falls as n^(−1/3).

Answer: (d) \( \frac{CV}{2}\left(\frac{K-1}{K+1}\right) \)

Explanation:
1. Initially the series capacitance is \( \frac C2 \), charge \( \frac{CV}2 \).
2. After: \( C_{eq}=\frac{C\cdot KC}{C+KC}=\frac{KC}{K+1} \), charge \( \frac{KCV}{K+1} \).
3. Charge through the battery \( =\frac{KCV}{K+1}-\frac{CV}2=\frac{CV}{2}\cdot\frac{K-1}{K+1} \).

Teacher's Note:
1. The battery supplies the change in charge.
2. Capacitance rises by a factor K when the dielectric fills the capacitor.

Answer: (c) \( Q_0,\ \frac{V_0}3,\ \frac{E_0}3 \)

Explanation:
1. The capacitor is isolated, so the charge stays \( Q_0 \).
2. Capacitance becomes \( \frac93=3 \) times larger, so the voltage \( V=\frac QC \) falls to \( \frac{V_0}{3} \).
3. The plate separation is unchanged, so the field \( E=\frac Vd \) becomes \( \frac{E_0}3 \).

Teacher's Note:
1. With the battery removed the charge cannot change.
2. E = V/d with d fixed.

Answer: (b) 3.5 mm

Explanation:
1. Original capacity corresponds to an effective separation of 3 mm.
2. With the layer, the effective separation is \( (d'-1)+\frac12=d'-0.5 \) mm.
3. Set \( d'-0.5=3\Rightarrow d'=3.5\ \text{mm} \).

Teacher's Note:
1. A dielectric slab of thickness t and constant K reduces the effective gap by t(1 − 1/K).
2. Match the effective gap to the original.

Answer: (a) \( \frac{K}{K+1}V_0 \)

Explanation:
1. Capacitances in series: \( C \) and \( KC \), so \( C_{eq}=\frac{KC}{K+1} \).
2. Charge \( Q=\frac{KCV_0}{K+1} \).
3. Voltage across the other capacitor \( =\frac QC=\frac{K}{K+1}V_0 \).

Teacher's Note:
1. In series, the same charge flows on both capacitors.
2. The unchanged capacitor's voltage is Q/C.

Answer: (d) 72 component capacitors

Explanation:
1. To withstand 3000 V with 500 V per capacitor, put \( \frac{3000}{500}=6 \) capacitors in series: each row has \( \frac{1}{6}\ \mu\text{F} \).
2. To get 2 μF we need \( \frac{2}{1/6}=12 \) such rows in parallel.
3. Total \( 6\times12=72 \) capacitors.

Teacher's Note:
1. Series reduces the voltage across each capacitor; parallel increases capacitance.
2. First satisfy the voltage limit, then the capacitance.

Answer: (c) \( \frac{9Q}{2} \)

Explanation:
1. Series capacitors carry the same charge, so C1 also has charge Q.
2. Voltage across the series pair: \( V=\frac QC+\frac Q{2C}=\frac{3Q}{2C} \).
3. C3 is across the same voltage: \( Q_3=3C\times\frac{3Q}{2C}=\frac{9Q}{2} \).

Teacher's Note:
1. The charge on capacitors in series is equal.
2. Parallel branches have the same voltage.

Answer: (b) \( 7.4\times10^{-5}\ \text{A} \)

Explanation:
1. Time constant \( RC=10^4\times2\times10^{-6}=0.02\ \text{s} \).
2. Initial current \( I_0=\frac VR=2\times10^{-4}\ \text{A} \).
3. \( I=I_0e^{-t/RC}=2\times10^{-4}e^{-1}=7.4\times10^{-5}\ \text{A} \).

Teacher's Note:
1. Discharge current decays exponentially.
2. Here t equals one time constant.

Answer: (a) \( KU_0 \)

Explanation:
1. With the battery connected, V is constant.
2. Capacitance becomes KC, so \( U=\tfrac12(KC)V^2=KU_0 \).

Teacher's Note:
1. Constant voltage: energy is proportional to C.
2. With the battery disconnected, the result would be U0/K.

Answer: (b) \( \frac V{2k}(R_1+R_2) \)

Explanation:
1. After joining, both are at potential V: \( q_1'=\frac{VR_1}{k},\ q_2'=\frac{VR_2}{k} \).
2. Total charge is conserved: \( 2q=\frac Vk(R_1+R_2) \).
3. \( q=\frac{V}{2k}(R_1+R_2) \).

Teacher's Note:
1. Charge flows until the potentials are equal.
2. The total charge is unchanged.

Answer: (a) C

Explanation:
1. The foil forms two capacitors in series, each with separation \( \frac d2 \) and capacitance \( 2C \).
2. Series: \( \frac{2C\times2C}{2C+2C}=C \).

Teacher's Note:
1. Foil of negligible thickness removes no gap in total.
2. Each half has twice the capacitance, and two halves in series return the original.

Answer: (a) 2C

Explanation:
1. Joining the foil to a plate makes that plate effectively extend to the foil.
2. The gap becomes \( \frac d2 \), so \( C'=\frac{\varepsilon_0A}{d/2}=2C \).

Teacher's Note:
1. A conductor connected to a plate has the same potential as that plate.
2. The effective separation halves.

Answer: (c) 9.0 μC

Explanation:
1. Parallel pair: \( 2+5=7\ \mu\text{F} \). In series with 3 μF: \( \frac{3\times7}{3+7}=2.1\ \mu\text{F} \).
2. Charge on this branch \( Q=2.1\times6=12.6\ \mu\text{C} \), and the voltage across the parallel pair is \( \frac{12.6}{7}=1.8\ \text{V} \).
3. \( Q_5=5\times1.8=9.0\ \mu\text{C} \).

Teacher's Note:
1. The 4 μF branch is independent because it is directly across the battery.
2. Parallel capacitors share the same voltage.

Answer: (d) 5 : 2 : 3

Explanation:
1. B and C in parallel: \( 20\ \mu\text{F} \). In series with A: \( \frac{5\times20}{25}=4\ \mu\text{F} \).
2. Charge on A \( =4\times25=100\ \mu\text{C} \); voltage across B and C \( \frac{100}{20}=5\ \text{V} \).
3. Charges: B \( 8\times5=40 \), C \( 12\times5=60 \). Ratio \( 100:40:60=5:2:3 \).

Teacher's Note:
1. Series branches have equal charge; parallel branches equal voltage.
2. Charge on the parallel pair equals the charge on A.

Answer: (c) \( \frac{C_1+C_2}{C_1} \)

Explanation:
1. Initial energy \( U_i=\tfrac12C_1V_0^2 \).
2. Common potential \( V=\frac{C_1V_0}{C_1+C_2} \), final energy \( U_f=\tfrac12(C_1+C_2)V^2=\frac{C_1^2V_0^2}{2(C_1+C_2)} \).
3. \( \frac{U_i}{U_f}=\frac{C_1+C_2}{C_1} \).

Teacher's Note:
1. Charge is conserved on connection, and some energy is lost as heat.
2. Compare the energies before and after.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A negative charge has lower potential energy at higher potential, \( U=qV \), so it tends to move towards higher potential.
2. So both statements are true and II explains I.

Teacher's Note:
1. Negative charges move opposite to the electric field.
2. Positive charges move towards lower potential.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The charge induces an opposite charge on the grounded plane, which attracts it (image charge effect).
2. The force is due to this induced charge, so II explains I.

Teacher's Note:
1. The force equals that from an image charge of opposite sign at the mirror position.
2. The plane is held at zero potential by the earth connection.

Answer: (c) A is true, but R is false.

Explanation:
1. Electrostatic forces are conservative, so the work done depends only on the end points: A is true.
2. R is false.

Teacher's Note:
1. Conservative forces have path-independent work.
2. Electrostatic force is conservative.

Answer: (c) A is true, but R is false.

Explanation:
1. The force in a medium is \( F=\frac{1}{4\pi\varepsilon_0\varepsilon_r}\frac{q_1q_2}{r^2} \), reduced by the dielectric constant (about 80 for water): A is true.
2. R is false.

Teacher's Note:
1. Dielectric media reduce the electrostatic force.
2. Water has a high dielectric constant.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Introducing a slab (conducting or insulating) reduces the field in the gap, which reduces the potential difference for the same charge.
2. Since \( C=\frac QV \), a smaller V means a larger C, so II explains I.

Teacher's Note:
1. Capacitance is charge per unit potential difference.
2. A conducting slab reduces the effective gap; an insulating slab reduces the field by K.