Question

The figure shows a RC circuit with a parallel plate capacitor. Before switching on the circuit, plate A of the capacitor was given a charge $-Q_0$ while plate B had no net charge. Now, at t = 0, the circuit is switched on. Time (in seconds) will elapse before the net charge on plate A becomes zero is K. Value of 125K is

Take C = 1 µF, $Q_0$ = 1 mC, $\epsilon$ = 1000 V and R = $\frac{2 \times 10^6}{ln(3)} \Omega$

• A. 100
• B. 150
• C. 200
• D. 250

Answer 1

Answer: (D)

250 seconds

Answer 2

Solution:

Let $q(t)$ be the charge delivered by the battery by time $t$. Initially, plate A has –Q₀ and plate B has 0. As charge $q$ flows, the net charge on plate A becomes (–Q₀ + q). The capacitor voltage is given by

$V_C = \frac{[ (\text{charge on B}) – (\text{charge on A}) ]}{2C} = \frac{[ q – (–Q₀ + q) ]}{2C} = \frac{Q₀}{2C}$.

Applying KVL in the loop (with resistor R and battery ε):

ε – I·R – V_C = 0, and I = dq/dt.

However, note that the capacitor voltage here is evolving as the charge q changes on the plates. A more careful derivation (as done in the similar problem) shows that the correct KVL leads to the differential equation:

ε – R (dq/dt) – [ –Q₀ + 2q ]/(2C) = 0.

Substitute the given values:

C = 1×10⁻⁶ F, Q₀ = 1×10⁻³ C, ε = 1000 V, and R = (2×10⁶)/(ln3) Ω:

1. Write the equation:

   1000 – R (dq/dt) – [–(1×10⁻³) + 2q]⁄(2×10⁻⁶) = 0.

2. Note that (–1×10⁻³)/(2×10⁻⁶) = –500 V and (2q)/(2×10⁻⁶) = (q)/(1×10⁻⁶) = 10⁶ q.

3. So the equation becomes:

   1000 – R (dq/dt) + 500 – 10⁶ q = 0 ⟹ 1500 – R (dq/dt) – 10⁶ q = 0.

4. Rearranging:

   R (dq/dt) = 1500 – 10⁶ q ⟹ dq/dt = (1500 – 10⁶ q)/R.

5. Substitute R = (2×10⁶)/(ln3):

   dq/dt = (1500 – 10⁶ q)·(ln3)/(2×10⁶).

This is a first‐order linear ODE with steady state solution:

qₛₛ = (1500 ln3/(2×10⁶)) / ((ln3)/2) = 1500/10⁶ = 1.5×10⁻³ C.

The general solution is:

q(t) = 1.5×10⁻³ [1 – exp(–(ln3/2)·t)].

Plate A will be neutral when its net charge becomes zero, i.e.,

–Q₀ + q(t) = 0 ⟹ q(t) = Q₀ = 1×10⁻³ C.

Set:

1.5×10⁻³ [1 – exp(–(ln3/2)·t)] = 1×10⁻³

⟹ 1 – exp(–(ln3/2)·t) = (1×10⁻³)/(1.5×10⁻³) = 2/3

⟹ exp(–(ln3/2)·t) = 1/3

Taking natural logarithm:

–(ln3/2)·t = ln(1/3) = –ln3

⟹ t = (2 ln3)/(ln3) = 2 seconds.

Thus, K = 2 seconds. The question asks for 125K, so:

125K = 125 × 2 = 250 seconds.