JEE Physics Electrostatic Potential and Capacitance MCQs Set 14

Answer: (b) \( \frac{6C}{5} \)

Explanation:
1. Use symmetry: the three edges from the first corner are at one potential level, the six face-diagonal-linked edges form a middle group, and the last three edges meet the far corner.
2. Parallel groups: 3C, then 6C, then 3C, joined in series: \( \frac1{C_{eq}}=\frac1{3C}+\frac1{6C}+\frac1{3C}=\frac{5}{6C} \).
3. So \( C_{eq}=\frac{6C}{5} \) (capacitors combine like conductances, so the cube gives 6C/5 where resistors R would give 5R/6).

Teacher's Note:
1. Capacitors in parallel add, so this is the reverse of the resistance result.
2. Do not copy the 5R/6 value without inverting.

Answer: (d) \( \frac{16}{3}e^{-1/3} \) µJ/s

Explanation:
1. Source power \( =Ei=4\times\frac43e^{-1/3}=\frac{16}{3}e^{-1/3} \) µW.

Teacher's Note:
1. Power delivered by a source is its emf times the current.
2. The current at time t is (E/R) e^(-t/tau).

Answer: (a) \( 3C_1=5C_2 \)

Explanation:
1. Charges must cancel: \( 120C_1=200C_2 \).
2. Divide by 40: \( 3C_1=5C_2 \).

Teacher's Note:
1. Zero final potential means the total charge is zero, so the connection is with opposite polarities.
2. The charges are C V in each.

Answer: (c) 2000 µC

Explanation:
1. Initial charge \( 200\times20=4000\ \mu\text{C} \).
2. In steady state the capacitor voltage equals the battery emf: \( q=200\times10=2000\ \mu\text{C} \).

Teacher's Note:
1. The new battery fixes the final voltage.
2. The extra charge flows back through that battery.

Answer: (a) 10 V

Explanation:
1. Total charge 1.0 nC divides equally between the two outer surfaces, 0.5 nC each, so the facing surface of the upper plate has 0.5 nC.
2. Field between upper and middle plate \( E=\frac{0.5\times10^{-9}}{\epsilon_0A} \), so \( V=Ed=10\ \text{V} \).

Teacher's Note:
1. The two outer surfaces of a plate stack always hold equal charge.
2. Both gaps then have the same field and same voltage.

Answer: (c) the charge stays constant, the potential difference increases and the capacitance decreases

Explanation:
1. Isolated means the charge cannot change.
2. \( C=\frac{\epsilon_0A}{d} \) falls as d rises.
3. \( V=Q/C \) therefore rises.

Teacher's Note:
1. Decide first whether Q (isolated) or V (battery connected) is held fixed.
2. Then use C proportional to 1/d.

Answer: (b) \( \frac{\epsilon_0AV^2}{2d} \)

Explanation:
1. Charge is fixed, so energy is \( \frac{Q^2}{2C} \) and doubling d doubles the energy.
2. Initial energy \( U_1=\frac12\frac{\epsilon_0A}{d}V^2 \).
3. Work done \( =U_2-U_1=U_1=\frac{\epsilon_0AV^2}{2d} \).

Teacher's Note:
1. At constant charge, energy is proportional to d.
2. The work done equals the increase in energy.

Answer: (a) 2.5 J and 1/2

Explanation:
1. Initial energy \( \frac12(0.1)(10)^2=5\ \text{J} \).
2. The charge divides equally, voltage becomes 5 V across a total of 0.2 F: \( U=\frac12(0.2)(5)^2=2.5\ \text{J} \).
3. Ratio \( \frac{2.5}{5}=\frac12 \).

Teacher's Note:
1. Sharing equally between equal capacitors always halves the energy.
2. The other half is lost as heat.

Answer: (a) 24 V

Explanation:
1. Total charge \( =20\times30+30\times20=1200\ \mu\text{C} \).
2. Total capacitance 50 µF, so \( V=\frac{1200}{50}=24\ \text{V} \).

Teacher's Note:
1. Like plates joined means the charges add.
2. Common voltage is total charge divided by total capacitance.

Answer: (c) 150 µA, 1.1 s

Explanation:
1. With the meter across one resistor, the other resistor has 1.0 V and the parallel pair 0.5 V, so the pair has half of the other resistance: \( R_{pair}=5\ \text{k}\Omega \), hence the meter resistance is \( R_V=10\ \text{k}\Omega \).
2. Capacitor charges to 1.5 V; initial meter current \( \frac{1.5}{10\ \text{k}\Omega}=150\ \mu\text{A} \).
3. Discharge time constant \( RC=10^4\times10^{-4}=1\ \text{s} \); reading 0.5 V when \( e^{-t}=\frac13 \), \( t=\ln3=1.1\ \text{s} \).

Teacher's Note:
1. A voltmeter has finite resistance, which loads the circuit.
2. Use the voltage-divider ratio to find that resistance first.

Answer: (c) \( \ln 2 \) ms

Explanation:
1. Maximum occurs when \( e^{-t/\tau}=\frac12 \), i.e. \( t=\tau\ln2 \).
2. \( \tau=RC=10\times100\times10^{-6}=1\ \text{ms} \), so \( t=\ln2\ \text{ms} \).

Teacher's Note:
1. The capacitor voltage is E/2 at this moment.
2. Tau is RC = 1 ms here.

Answer: (a) 0.09 A

Explanation:
1. Initial current \( I_0=\frac{6}{24}=0.25\ \text{A} \).
2. \( I(\tau)=I_0e^{-1}=0.25/2.718\approx0.09\ \text{A} \).

Teacher's Note:
1. After one time constant the current falls to 1/e (about 37%) of its first value.
2. The first value is E/R because the empty capacitor acts like a wire.

Answer: (a) \( 15\times10^3 \) V/m

Explanation:
1. Charge stays constant, and the field between the plates is \( E=\frac{\sigma}{\epsilon_0}=\frac{Q}{\epsilon_0A} \), independent of d.
2. Initial field \( =\frac{300}{2\times10^{-2}}=15\times10^3\ \text{V/m} \), and it stays the same.

Teacher's Note:
1. For an isolated capacitor E depends only on charge density, so separating the plates does not change it.
2. Voltage rises instead.

Answer: (d) −405 × 10⁻¹⁰ J

Explanation:
1. \( C_1=\frac{\epsilon_0A}{d_1}=\frac{9\times10^{-12}\times10^{-2}}{2\times10^{-2}}=4.5\ \text{pF} \) and \( C_2=3.6\ \text{pF} \).
2. \( \Delta U=\frac12(C_2-C_1)V^2=\frac12(-0.9\times10^{-12})(300)^2=-4.05\times10^{-8}\ \text{J} \).

Teacher's Note:
1. At constant voltage the stored energy is proportional to the capacitance.
2. Moving the plates apart lowers C, so the energy decreases.

Answer: (b) decreases if \( Q_1R_2\neq Q_2R_1 \)

Explanation:
1. Charge flows only if the potentials differ: \( \frac{Q_1}{R_1}\ne\frac{Q_2}{R_2} \), that is \( Q_1R_2\ne Q_2R_1 \).
2. Energy lost as heat is \( \frac12\frac{C_1C_2}{C_1+C_2}(V_1-V_2)^2 \), which is positive when the potentials differ and zero otherwise.

Teacher's Note:
1. Sharing of charge between conductors always wastes energy as heat unless the potentials are already equal.
2. The words 'in every case' fail because equal potentials give no change.

Answer: (d) The energy stored in the capacitor increases.

Explanation:
1. Charge is fixed because the battery is disconnected.
2. C falls, so \( U=\frac{Q^2}{2C} \) rises and \( V=\frac QC \) rises.

Teacher's Note:
1. Extra energy comes from the work done in pulling the plates apart against their attraction.
2. Capacitance depends only on geometry, which reduced it.

Answer: (d) \( \frac1{d^2} \)

Explanation:
1. \( C=\frac{\epsilon_0A}{d} \), so \( \frac{dC}{dt}=-\frac{\epsilon_0A}{d^2}\frac{dd}{dt}=-\frac{\epsilon_0Av}{d^2} \).

Teacher's Note:
1. Differentiate C = e0 A / d using the chain rule.
2. The magnitude falls as the plates move apart.

Answer: (b) 48 µJ

Explanation:
1. \( U=\frac12(6\ \mu\text{F})(4\ \text{V})^2=48\ \mu\text{J} \).

Teacher's Note:
1. Use the common voltage and the total capacitance.
2. Initial energy was 144 µJ, so two thirds is lost.

Answer: (b) \( \frac43e^{-1/3} \) µC/s

Explanation:
1. Time constant \( \tau=RC=3\times10^6\times10^{-6}=3\ \text{s} \).
2. Current \( i=\frac{E}{R}e^{-t/\tau}=\frac43e^{-1/3}\ \mu\text{A} \), which is the rate of increase of charge in µC/s.

Teacher's Note:
1. The charging current decays exponentially from E/R.
2. The rate of change of charge is the current.

Answer: (d) 10 mJ

Explanation:
1. Charge flowing back: 2000 µC into the 10 V battery, so work done on the battery is \( 2000\times10^{-6}\times10=20\ \text{mJ} \).
2. Decrease in capacitor energy \( \frac12(200\ \mu\text{F})(20^2-10^2)=30\ \text{mJ} \).
3. Heat = 30 − 20 = 10 mJ.

Teacher's Note:
1. Energy lost by the capacitor goes partly to the battery (it is being charged) and partly to heat.
2. Check each step separately.