JEE Physics Electrostatic Potential and Capacitance MCQs Set 07

Answer: (b) 3 μF

Explanation:
1. The parallel pair gives \( 2+3=5 \) μF.
2. Series: \( \frac1C=\frac1{12}+\frac15+\frac1{20}=\frac{5+12+3}{60}=\frac13 \).
3. \( C=3 \) μF.

Teacher's Note:
1. Parallel capacitances add; series capacitances add reciprocally.
2. The series result is always smaller than the smallest capacitor in the chain.

Answer: (c) 400 V

Explanation:
1. Parallel pair: \( 4+2=6 \) μF in series with 3 μF; both carry the same charge.
2. Voltages divide inversely with capacitance: across the 6 μF pair \( V=1200\times\frac{3}{3+6}=400 \) V.
3. B is earthed, so the potential of P is +400 V.

Teacher's Note:
1. In series the smaller capacitor takes the larger share of the voltage.
2. P sits above earth by the voltage across the pair.

Answer: (d) 2500 × 10⁻⁶ J

Explanation:
1. \( U=\frac12CV^2=\frac12\times2\times10^{-6}\times2500=2.5\times10^{-3} \) J.
2. This equals 2500 × 10⁻⁶ J.

Teacher's Note:
1. Convert μF to F before using the energy formula.
2. Option (b) forgets the factor 10⁻⁶.

Answer: (b) 90 kW

Explanation:
1. \( U=\frac12CV^2=\frac12\times40\times10^{-6}\times(3000)^2=180 \) J.
2. \( P=\frac Ut=\frac{180}{2\times10^{-3}}=9\times10^4 \) W = 90 kW.

Teacher's Note:
1. Power is energy divided by time.
2. Use SI units throughout: μF to F and ms to s.

Answer: (b) 6 V and 4 V

Explanation:
1. \( C_{eq}=\frac{2\times3}{5}=1.2 \) μF, so each carries \( Q=1.2\times10=12 \) μC.
2. \( V_A=\frac{12}{2}=6 \) V and \( V_B=\frac{12}{3}=4 \) V.

Teacher's Note:
1. Equal charge on series capacitors means voltage is inversely proportional to capacitance.
2. The voltages add up to the battery emf: 6 + 4 = 10 V.

Answer: (d) 4

Explanation:
1. \( C=\frac{\varepsilon_0A}{d} \). After the change \( C'=\frac{K\varepsilon_0A}{2d}=\frac K2C \).
2. Given \( C'=2C \), so \( K=4 \).

Teacher's Note:
1. Doubling the gap halves the capacitance; the dielectric multiplies it by K.
2. Combine the two effects into one ratio.

Answer: (c) 30 μF

Explanation:
1. The two halves behave as capacitors in parallel, each with half the area: \( C_1=\frac{K_1C_0}{2}=10 \) μF and \( C_2=\frac{K_2C_0}{2}=20 \) μF.
2. \( C=C_1+C_2=30 \) μF.

Teacher's Note:
1. Media side by side across the plates give parallel capacitors; layers one behind the other give series capacitors.
2. Each half carries half the original area.

Answer: (a) 25 μC from smaller to bigger

Explanation:
1. After joining, the potentials are equal, so the charges share in the ratio of radii (1 : 2). Total charge 150 μC splits into 50 μC and 100 μC.
2. The smaller sphere goes from 75 μC to 50 μC, so 25 μC flows from the smaller to the bigger sphere.

Teacher's Note:
1. Charge flows from higher potential to lower potential. The smaller sphere is at the higher potential initially.
2. Joined spheres share charge in proportion to their capacitance (radius).

Answer: (d) potential or Q/C ratio

Explanation:
1. Charge moves only if there is a potential difference.
2. With equal potential V = Q/C there is no flow.

Teacher's Note:
1. Charge flows to equalise potential, not to equalise charge.
2. This is similar to water flowing until levels are equal.

Answer: (d) metal of the plates

Explanation:
1. \( C=\frac{K\varepsilon_0A}{d} \) depends on area, separation and the medium.
2. It does not depend on which metal the plates are made of.

Teacher's Note:
1. Plate material, charge and applied voltage do not change C.
2. C depends only on geometry and the dielectric.

Answer: (c) 425 V

Explanation:
1. Charges: \( 3\times300=900 \) μC and \( 5\times500=2500 \) μC.
2. \( V=\frac{900+2500}{3+5}=425 \) V.

Teacher's Note:
1. Common potential = total charge / total capacitance.
2. The result lies between the two initial potentials.

Answer: (a) 20 V

Explanation:
1. Total charge on the plates joined at O is zero: \( C_1(V_O-V_A)+C_2(V_O-V_B)+C_3(V_O-V_D)=0 \).
2. \( V_O=\frac{1(10)+2(25)+3(20)}{1+2+3}=\frac{120}{6}=20 \) V.

Teacher's Note:
1. For an initially neutral junction, the sum of charges on the connected plates is zero.
2. This is the capacitor version of the node (Millman) rule.

Answer: (a) 1 : 4

Explanation:
1. Series: equal charge, so \( \frac{V_1}{V_2}=\frac{C_2}{C_1}=\frac14 \).

Teacher's Note:
1. Voltage is inversely proportional to capacitance in series.
2. The actual voltages are 40 V and 160 V.

Answer: (d) 9 kV

Explanation:
1. Maximum charges: \( Q_1=1\times6=6 \) mC and \( Q_2=2\times4=8 \) mC (μF × kV).
2. In series the same charge flows, so the limit is \( Q=6 \) mC.
3. Voltages: \( V_1=6 \) kV and \( V_2=\frac{6}{2}=3 \) kV, total 9 kV.

Teacher's Note:
1. The weaker capacitor, in terms of charge, sets the series limit.
2. Compare charge limits, not voltage limits.

Answer: (b) 31/6 V

Explanation:
1. Charge limits: \( 2\times3=6 \), \( 3\times2=6 \), \( 5\times1=5 \) μC. The series limit is \( Q=5 \) μC.
2. Voltages: \( \frac52+\frac53+\frac55=\frac{15+10+6}{6}=\frac{31}{6} \) V.

Teacher's Note:
1. Find each capacitor's maximum charge first.
2. In series the smallest maximum charge governs.

Answer: (c) becomes four times

Explanation:
1. Battery connected: V is constant. \( E=\frac Vd \) doubles and \( Q=CV \) doubles.
2. Force \( F=\frac{QE}{2} \) becomes \( 2\times2=4 \) times.

Teacher's Note:
1. Force between plates is half of Q times the field due to both plates.
2. With constant Q instead, the force would not change.

Answer: (c) −Qr/R

Explanation:
1. Let the charge on A₁ be q. Its potential is zero: \( \frac{kq}{r}+\frac{kQ}{R}=0 \).
2. \( q=-\frac{Qr}{R} \).

Teacher's Note:
1. The earthed sphere is at zero potential, not zero charge.
2. The potential inside a charged shell equals kQ/R.

Answer: (a) the electric field only

Explanation:
1. Energy is stored in the electric field between the plates, with density \( \frac12\varepsilon_0E^2 \).
2. No current flows in the steady state, so no magnetic energy is stored.

Teacher's Note:
1. Energy density in the field is ½ε₀E².
2. Inductors, not capacitors, store energy in magnetic fields.

Answer: (b) 3W/4

Explanation:
1. \( W=\frac12C(V_2^2-V_1^2) \).
2. First: \( \frac12C(1600-400)=600C \). Second: \( \frac12C(2500-1600)=450C \).
3. Ratio: \( \frac{450}{600}W=\frac{3W}{4} \).

Teacher's Note:
1. Work done equals the change in stored energy.
2. Energy depends on V², not on V.

Answer: (a) \( 7\times10^{-4} \) J

Explanation:
1. Loss \( =\frac12\frac{C_1C_2}{C_1+C_2}V^2=\frac12\times\frac{0.5\times0.2}{0.7}\times10^{-6}\times10^4 \).
2. \( =7.1\times10^{-4} \) J.

Teacher's Note:
1. The formula ½ [C₁C₂/(C₁+C₂)] V² gives the heat lost when a charged capacitor is joined to an uncharged one.
2. Energy is lost as heat and radiation in the connecting wire.

Answer: (c) \( \frac{3CV^2}{2} \)

Explanation:
1. Charges: \( Q_1=CV \) and \( Q_2=2C\times2V=4CV \). Opposite plates joined, so net charge \( =4CV-CV=3CV \).
2. Common potential \( V'=\frac{3CV}{3C}=V \).
3. \( U=\frac12(3C)V^2=\frac{3CV^2}{2} \).

Teacher's Note:
1. When opposite polarities are joined the charges subtract.
2. The final energy uses the total capacitance in parallel.

Answer: (b) 600 μJ

Explanation:
1. Charges: \( 20 \) μC and \( 80 \) μC; opposite plates joined so net charge \( =60 \) μC.
2. Final potential \( V=\frac{60}{6}=10 \) V; \( U_f=\frac12(6)(100)=300 \) μJ.
3. \( U_i=\frac12(2)(100)+\frac12(4)(400)=900 \) μJ. Heat \( =900-300=600 \) μJ.

Teacher's Note:
1. Heat evolved = initial energy − final energy.
2. Check that you subtracted the charges, since opposite polarities are joined.

Answer: (b) double

Explanation:
1. A conducting slab of thickness t effectively reduces the gap: \( C'=\frac{\varepsilon_0A}{d-t}=\frac{\varepsilon_0A}{d/2}=2C \).

Teacher's Note:
1. Inside the conductor the field is zero, so only the remaining air gap counts.
2. The position of the slab inside the gap does not matter.

Answer: (c) 4 V

Explanation:
1. The battery stays connected, so the potential difference stays equal to its emf: 4 V.
2. (Only the charge and stored energy change.)

Teacher's Note:
1. With a battery connected, V is fixed.
2. With the battery disconnected, Q is fixed and V falls by the factor K.