The figure shows a RC circuit with a parallel plate capacitor. Before switching on the circuit, plate A of the capacitor has a charge $- Q_{0}$ while plate B has no net charge. Now, at t=0, the circuit is switched on. How much time (in second) will elapse before the net charge on plate A becomes zero.
$(G i v e n C = 1 μ F, Q_{0} = 1 m C, ε = 1000 V a n d R = \frac{2 \times 10^{6}}{\ln 3} Ω)$

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answer is 2.

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Detailed Solution

Let at any time t charge flown through the plate B to plate A is q and instantaneous current is l.
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from loop theorem $(\frac{2 q - Q_{0}}{2 C}) + I R - ε = 0$
$\Rightarrow R \frac{d q}{d t} = \frac{2 q + 2 ε C + Q_{0}}{2 C} \Rightarrow \frac{d q}{2 ε C + Q_{0} - 2 q} = \frac{d t}{2 R C}$

Now for charge on plate A to be zero $q = Q_{0}$ .
Integrating $\int_{0}^{Q_{0}} \frac{d p}{2 ε C + Q_{0} - 2 q} = \int_{0}^{t} \frac{d t}{2 R C}$
$= t = R C \ln [\frac{2 ε C + Q_{0}}{2 ε C - Q_{0}}]$
Putting the value of $C, Q_{0}, ε a n d R$
We get t=2 seconds.