JEE Physics Electrostatic Potential and Capacitance MCQs Set 18

Answer: (a) 2 and 3 only

Explanation:
1. Between the plates the field is uniform, \( E=\frac{\sigma}{\epsilon_0} \), so the force is the same at B and C (statement 3) and not zero at C.
2. Outside the plates the fields of the two equal and opposite plates cancel, so the force at A is zero (statement 2).
3. Statements 1 and 4 are false.

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) 2 and 3 only

Explanation:
1. Only the facing (inner) charges set up the field between the plates, and for an isolated pair \( Q_3=-Q_2 \).
2. So \( V=\left|\frac{Q_2}{C}\right|=\left|\frac{Q_3}{C}\right| \).
3. The total charge of one plate divided by C, \( \frac{Q_1+Q_2}{C} \), is not in general the potential difference (it includes the outer-face charge).

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) 1, 2 and 3

Explanation:
1. Charges on the facing surfaces are \( \pm\frac{2Q-(-Q)}{2}=\pm\frac{3Q}{2} \) and each outer surface carries \( \frac{2Q+(-Q)}{2}=\frac Q2 \), so there is a field outside (statement 1 true).
2. Potential difference \( V=\frac{3Q/2}{C}=\frac{3Q}{2C} \) and energy \( \frac12CV^2=\frac{9Q^2}{8C} \) (statements 2 and 3 true).
3. The force between plates depends on the plate area, \( \frac{(3Q/2)^2}{2\epsilon_0A} \); the given expression in d is not valid, so statement 4 is false.

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) 1, 2 and 3

Explanation:
1. Initial charges: 360 \( \mu C \) on the 3 \( \mu F \) and 300 \( \mu C \) on the 2 \( \mu F \) capacitor. Let \( \Delta \) be the charge moved through the 1.5 \( \mu F \) capacitor loop.
2. Loop condition: \( \frac{300-\Delta}{2}-\frac{-360+\Delta}{3}=\frac{\Delta}{1.5} \) gives \( 270-\frac{5\Delta}{6}=\frac{2\Delta}{3} \), so \( \Delta=180\,\mu C \).
3. So the 1.5 \( \mu F \) capacitor holds 180 \( \mu C \) and the 2 \( \mu F \) capacitor holds \( 300-180=120\,\mu C \). The bottom plates change from +360 and -300 to +180 and -120, so 180 \( \mu C \) of positive charge goes through A from left to right.

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 and 3 only

Explanation:
1. A row of \( 2^{n-1} \) capacitors in series has capacitance \( \frac{2}{2^{n-1}} \) F: 2, 1, 0.5, ... F, summing to \( 2+1+\frac12+\dots=4 \) F (statement 1 true).
2. Each row has the same voltage 10 V, so the charge is \( C_{row}V \): 20 C, 10 C, 5 C, ... which is largest in the first row (statement 3 true, statement 2 false).
3. Total energy \( \frac12\times4\times10^2=200 \) J, not 50 J.

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) 1, 2, 3 and 4

Explanation:
1. Charge conservation at the isolated nodes B and D: \( 4(V_B-20)+2V_B+2(V_B-V_D)=0 \) and \( 2(V_D-20)+4V_D+2(V_D-V_B)=0 \).
2. Solving: \( V_B=12 \) V, \( V_D=8 \) V.
3. Check: statement 2: \( 4(8)+2(-4)=24=2\times12 \); statement 3: \( 2(12)+2(4)=32=4\times8 \); statement 4: \( 20=12+8 \). All four hold.

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) 1 and 3 only

Explanation:
1. The final charge is \( q_{max}=CE \) for each circuit and the curves reach the same value, so statement 1 is true.
2. This only fixes the product \( C_1E_1=C_2E_2 \), so the emfs may differ (statement 3) and nothing shows which is larger (statements 2 and 4 are not implied).

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 and 3 only

Explanation:
1. Slower decay means a larger time constant: \( R_1C_1>R_2C_2 \) (statement 1). Statement 2 says the opposite, \( R_1C_1<R_2C_2 \).
2. The initial charges are equal, \( C_1E_1=C_2E_2 \); if \( E_1=E_2 \) then \( C_1=C_2 \), so statement 4 is false.
3. With equal C, \( R_1C>R_2C \) gives \( R_1>R_2 \) (statement 3).

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) 2 and 4 only

Explanation:
1. Both capacitors are charged to the battery emf V, so each starts discharging with \( I_0=\frac VR \): equal and non-zero (statement 2).
2. The time to lose half the charge is \( t=RC\ln2 \), which is shorter for the smaller capacitor \( C_1 \) (statement 4).

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) 2, 3 and 4

Explanation:
1. After a long time no current flows in the capacitor branch (statement 4), so the battery current flows through \( R_1,R_3,R_2 \) in series: \( I=\frac{E}{R_1+R_2+R_3} \) (statement 2).
2. The capacitor branch is across \( R_3+R_2 \), so \( V_C=\frac{(R_2+R_3)E}{R_1+R_2+R_3} \), not E (statement 1 false).
3. Energy \( \frac12CV_C^2 \) gives statement 3.

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) 1, 2 and 3

Explanation:
1. The battery holds V fixed, so \( E=\frac Vd \) is unchanged (1).
2. Capacitance becomes \( KC \) (2) and the charge \( Q=KCV \) increases (3).
3. Energy \( \frac12KCV^2 \) increases, so statement 4 is false.

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: (d) 1, 3 and 4

Explanation:
1. The battery is disconnected, so the charge stays \( Q=C_0V=\frac{\epsilon_0AV}{d} \) (statement 1; statement 2 is false).
2. The field becomes \( \frac{E_0}{K}=\frac{V}{Kd} \) (statement 3).
3. Energy falls from \( \frac{Q^2}{2C_0} \) to \( \frac{Q^2}{2KC_0} \), so \( W=\frac{Q^2}{2C_0}\left(1-\frac1K\right)=\frac{\epsilon_0AV^2}{2d}\left(1-\frac1K\right) \) (statement 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.

Answer: (a) 1 and 4 only

Explanation:
1. With V fixed: \( C\to KC \), so \( U=\frac12CV^2\to KU \) and \( Q=CV\to KQ \) (statements 1 and 4).
2. The field \( V/d \) is unchanged (statement 2 false).
3. The force \( F=\frac12V^2\left|\frac{dC}{dx}\right| \) also scales with C, so it becomes K times, not \( K^2 \) (statement 3 false).

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) 1, 2, 3 and 4

Explanation:
1. The capacitance varies as \( C=\frac{\epsilon_0w}{d}\left[K(L-x)+x\right] \), linear in the length x of slab inside, hence linear in t (statement 1).
2. Charge is fixed: \( V=\frac QC \) (statement 4), which rises as C falls but not linearly (statement 2).
3. Energy \( U=\frac12QV \) is proportional to V (statement 3).

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) 1 and 4 only

Explanation:
1. The battery fixes the potential difference, so \( V=V_0 \) and \( E=\frac Vd=E_0 \).
2. The capacitance increases, so \( Q=CV>Q_0 \) and \( U=\frac12CV^2>U_0 \).

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: (d) 2, 3 and 4

Explanation:
1. PSQ: the capacitor charges with the slab, \( Q=KCV \); PQS: charged without the slab, \( Q=CV \) (statement 2 true).
2. SPQ: \( U=\frac12KCV^2 \); PQS: \( U=\frac{(CV)^2}{2KC}=\frac{CV^2}{2K} \) (statement 3 true).
3. PS and SP both end connected to the battery with the slab, \( E=\frac Vd \) (statement 4 true).
4. In PQR the capacitor discharges and recharges in the opposite sense through the circuit, so heat is produced (statement 1 false).

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.