Question

The figure shows an $RC$ 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. If at $t=0$ , the switch is closed then after how much time (in seconds) will the net charge on plate A becomes zero?[Given: $C=1\mu F$ , $Q_0=1mC$ , $ϵ=1000V$ and $R=\frac{2\times 10^6}{ln3}\Omega$ ]

Answer 1

Answer: 2

Let at any time $t$ the charge that flows from the plate $B$ to plate $A$ is $q$ and the instantaneous current is $I$ .

$\left(\frac{2q-Q_0}{2C}\right)+IR-ϵ=0$
$\Rightarrow R\frac{dq}{dt}=\frac{-2q+2ϵC+Q_0}{2C}$
$\Rightarrow \frac{dq}{2ϵC+Q_0-2q}=\frac{dt}{2RC}$
Now for the charge on plate $A$ to be zero $q=Q_0$ .
${\int }_0^{Q_0}\frac{dq}{2ϵC+Q_0-2q}={\int }_0^t\frac{dt}{2RC}$
$\Rightarrow t=RCln\left[\frac{2ϵC+Q_0}{2ϵC-Q_0}\right]=2s$