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.

Given $(C = 1 μ F, Q_{0} = 1 m C, ε = 1000 V a n d R = \frac{2 \times 10^{6}}{1 n 3} Ω)$

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Solution

Let at any time t charge flown through the plate B to plate A is q and instantaneous current is I.

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 $\int_{0}^{Q_{0}} \frac{d q}{2 ε C + Q_{0} - 2 q} = \int_{0}^{t} \frac{d t}{2 R C}$

= t = RC 1n $[\frac{2 ε C + Q_{0}}{2 ε C - Q_{0}}]$

Putting the value of C, $Q_{0} ε$ and R

We get t = 2 seconds.

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