JEE Physics Electrostatic Potential and Capacitance MCQs Set 19

Answer: (b) 8 \( \mu F \)

Explanation:
1. In series the voltage divides inversely: \( V_1=V\frac{C_2+C_3}{C_1+C_2+C_3} \) with \( V=10 \) V.
2. At \( C_3=0 \): \( \frac{C_2}{C_1+C_2}=0.2 \) so \( C_1=4C_2 \). At \( C_3=10 \): \( \frac{C_2+10}{C_1+C_2+10}=0.6 \) gives \( 0.4C_2+4=0.6C_1=2.4C_2 \), so \( C_2=2 \) and \( C_1=8\,\mu F \).
3. Check at \( C_3=6 \): \( \frac{8}{16}\times10=5 \) V.

Teacher's Note:
1. Reduce the network step by step, or write charge conservation at isolated nodes.
2. Check the answer by testing a simple limiting case.

Answer: (c) 2 \( \mu F \)

Explanation:
1. From the first reading, \( \frac{C_2}{C_1+C_2}=\frac{2}{10} \) gives \( C_1=4C_2 \).
2. The reading at \( C_3=10\,\mu F \) then gives \( C_2=2\,\mu F \) and \( C_1=8\,\mu F \) (check: at \( C_3=6 \), \( V_1=\frac{2+6}{8+2+6}\times10=5 \) V).

Teacher's Note:
1. Reduce the network step by step, or write charge conservation at isolated nodes.
2. Check the answer by testing a simple limiting case.

Answer: (a) \( \frac{E}{\sqrt e\,R}\left(1-\frac1e\right) \)

Explanation:
1. At t = RC the charge is \( CE(1-e^{-1}) \) and the discharge begins; at t = 1.5RC, half a time constant later.
2. The charge is \( CE(1-e^{-1})e^{-1/2} \) and the discharge current is \( \frac{q}{RC}=\frac{E}{R}\left(1-\frac1e\right)e^{-1/2}=\frac{E}{\sqrt e\,R}\left(1-\frac1e\right) \).

Teacher's Note:
1. The time constant of an RC circuit is the product of the effective R and C.
2. Charging and discharging both follow exponentials in t over RC.

Answer: (c) 80%

Explanation:
1. Initial energy \( \frac12C_1V^2 \) with charge \( Q=C_1V \).
2. After sharing: \( U_f=\frac{Q^2}{2(C_1+C_2)} \), so \( \frac{U_f}{U_i}=\frac{C_1}{C_1+C_2}=\frac{2}{10} \).
3. So 80% is dissipated.

Teacher's Note:
1. Reduce the network step by step, or write charge conservation at isolated nodes.
2. Check the answer by testing a simple limiting case.

Answer: (d) +48 \( \mu C \)

Explanation:
1. The node joining the lower plate of the 4 \( \mu F \) and the upper plates of the 2 and 3 \( \mu F \) capacitors is isolated and neutral: \( -80+q_2+q_3=0 \).
2. The 2 and 3 \( \mu F \) capacitors are in parallel at the same voltage, so \( q_2:q_3=2:3 \).
3. Hence \( q_2+q_3=80 \) gives \( q_3=48\,\mu C \).

Teacher's Note:
1. Reduce the network step by step, or write charge conservation at isolated nodes.
2. Check the answer by testing a simple limiting case.

Answer: (a) 2 and 4 only

Explanation:
1. After S1: C1 holds \( 2CV_0 \) on its upper plate. Pressing S2 shares it equally with C2: each upper plate carries \( CV_0 \) (statement 2 true).
2. S2 is released, so C1 is isolated at \( CV_0 \).
3. S3 connects C2 across a cell of emf \( V_0 \) with reversed polarity, so the upper plate of C2 ends at \( -CV_0 \) (statement 4 true).

Teacher's Note:
1. Reduce the network step by step, or write charge conservation at isolated nodes.
2. Check the answer by testing a simple limiting case.

Answer: (b) 1 and 4 only

Explanation:
1. The two portions are in parallel, so they have the same potential difference and the same gap: \( E_1=E_2=\frac Vd \) (statement 1 true, 2 false).
2. With area A: \( C_1=\frac{K\epsilon_0A/3}{d} \) and the rest \( \frac{2\epsilon_0A/3}{d} \), so \( \frac{Q_1}{Q_2}=\frac{K}{2} \) (statement 3 false).
3. Total \( C=\frac{\epsilon_0A(K+2)}{3d} \) so \( \frac{C}{C_1}=\frac{K+2}{K} \) (statement 4 true).

Teacher's Note:
1. Decide first whether the battery stays connected (V fixed) or is removed (Q fixed).
2. Then write the energy as 1/2 CV squared or Q squared over 2C, whichever variable is held.

Answer: (c) \( \frac73 \)

Explanation:
1. The two halves of the area are in parallel. Left half (area S/2): full thickness d of \( \epsilon_1=2 \): \( C_a=\frac{2\epsilon_0(S/2)}{d}=\frac{\epsilon_0S}{d} \).
2. Right half: two layers of thickness d/2 in series, \( \frac{2\epsilon_0S}{d} \) and \( \frac{4\epsilon_0S}{d} \): \( C_b=\frac{2\times4}{2+4}\frac{\epsilon_0S}{d}=\frac{4\epsilon_0S}{3d} \).
3. Total \( C_2=\frac{\epsilon_0S}{d}\left(1+\frac43\right)=\frac73\frac{\epsilon_0S}{d} \) and \( C_1=\frac{\epsilon_0S}{d} \), so \( \frac{C_2}{C_1}=\frac73 \).

Teacher's Note:
1. Reduce the network step by step, or write charge conservation at isolated nodes.
2. Check the answer by testing a simple limiting case.

Answer: (d) 1, 2, 3 and 4

Explanation:
1. At t = 0 the uncharged capacitors act as short circuits: the top node is at 5 V and the bottom node at 0 V, so the reading (bottom minus top) is -5 V. After a long time no current flows, the top node returns to the potential of the right corner (0) and the bottom node to that of the left corner (5 V): +5 V.
2. Both branches have \( RC=25\text{ k}\Omega\times40\,\mu F=50\text{ k}\Omega\times20\,\mu F=1 \) s, so \( V_{top}=5e^{-t} \), \( V_{bottom}=5(1-e^{-t}) \) and the reading is \( 5(1-2e^{-t}) \), zero at \( t=\ln2 \) s.
3. Each branch current is proportional to \( e^{-t} \), so the ammeter current is 1/e of its initial value at t = 1 s and tends to zero.

Teacher's Note:
1. The time constant of an RC circuit is the product of the effective R and C.
2. Charging and discharging both follow exponentials in t over RC.

Answer: (b) \( E_C=E_D \)

Explanation:
1. The source supplies \( Q V_0=CV_0^2 \).
2. The capacitor stores \( E_C=\frac12CV_0^2 \), so the heat is \( E_D=CV_0^2-\frac12CV_0^2=\frac12CV_0^2 \).
3. Hence \( E_C=E_D \).

Teacher's Note:
1. The time constant of an RC circuit is the product of the effective R and C.
2. Charging and discharging both follow exponentials in t over RC.

Answer: (d) 1.5

Explanation:
1. After S1: C3 holds \( 1.0\times8=8\,\mu C \).
2. Then C3 is connected in parallel with the series branch of C1 (\( \epsilon_r\ \mu F \)) and C2 (1 \( \mu F \)). C3 has 5 \( \mu C \) so the common voltage is 5 V, and by conservation the series branch holds \( 8-5=3\,\mu C \).
3. Series capacitance \( \frac{3}{5}=0.6\,\mu F=\frac{\epsilon_r}{1+\epsilon_r} \) gives \( \epsilon_r=1.5 \).

Teacher's Note:
1. Reduce the network step by step, or write charge conservation at isolated nodes.
2. Check the answer by testing a simple limiting case.

Answer: (a) 4

Explanation:
1. Let the equivalent capacitance be \( C \). Adding one more section does not change it: \( C=\frac{6(8+C)}{6+8+C} \).
2. So \( C^2+8C-48=0 \), giving \( C=\frac{-8+16}{2}=4\,\mu F \).

Teacher's Note:
1. Reduce the network step by step, or write charge conservation at isolated nodes.
2. Check the answer by testing a simple limiting case.

Answer: (b) 20

Explanation:
1. Take \( V_A=1 \), \( V_B=0 \) and use charge conservation at X and Y (in \( \mu C \)): \( 2(V_X-1)+4V_X+4(V_X-V_Y)=0 \) and \( 4(V_Y-1)+2V_Y+4(V_Y-V_X)=0 \).
2. These give \( V_X=\frac37 \) and \( V_Y=\frac47 \).
3. Charge from A: \( 2\left(1-\frac37\right)+4\left(1-\frac47\right)=\frac87+\frac{12}7=\frac{20}{7} \), so \( C_{eq}=\frac{20}{7}\,\mu F \) and \( n=20 \).

Teacher's Note:
1. Reduce the network step by step, or write charge conservation at isolated nodes.
2. Check the answer by testing a simple limiting case.

Answer: (d) 1

Explanation:
1. The two capacitors are in series in the loop: \( C_{eff}=\frac{C\cdot2C}{3C}=\frac{2C}{3} \), so the time constant is \( \frac{2RC}{3} \) and the exponent is \( \frac{3t}{2RC} \), so \( \beta=2 \).
2. Final charge on 2C: total charge CV shares in the ratio of capacitances, so \( \frac23CV=\frac{\alpha CV}{3} \) and \( \alpha=2 \).
3. Hence \( \alpha/\beta=1 \).

Teacher's Note:
1. The time constant of an RC circuit is the product of the effective R and C.
2. Charging and discharging both follow exponentials in t over RC.

Answer: (a) 5

Explanation:
1. Minimum spacing \( d=\frac{V}{E_{max}}=\frac{4000}{18\times10^6}=2.22\times10^{-4} \) m.
2. Area \( A=\frac{Cd}{K\epsilon_0}=\frac{7\times10^{-8}\times2.22\times10^{-4}\times36\pi}{2.8\times10^{-9}}=0.2\pi \) m\( ^2 \).
3. So \( \frac\pi n=\frac\pi5 \) and \( n=5 \).

Teacher's Note:
1. Decide first whether the battery stays connected (V fixed) or is removed (Q fixed).
2. Then write the energy as 1/2 CV squared or Q squared over 2C, whichever variable is held.

Answer: (b) 4

Explanation:
1. When the plates are lowered a distance x into the oil, \( C=\frac{\epsilon_0w}{d}[(L-x)+Kx] \), so \( \frac{dC}{dt}=\frac{\epsilon_0w(K-1)}{d}v \).
2. Current \( I=V\frac{dC}{dt}=500\times\frac{8\times10^{-12}\times1\times10}{0.01}\times0.001=4\times10^{-9} \) A.
3. So \( n=4 \).

Teacher's Note:
1. Decide first whether the battery stays connected (V fixed) or is removed (Q fixed).
2. Then write the energy as 1/2 CV squared or Q squared over 2C, whichever variable is held.