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. 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 $C=1\mu F,Q_0=1mC,E=1000V$ and $R=\frac{2\times {10}^6}{ln3}\Omega$ )

• A. $2$
• B. $3.33$
• C. $1.33$
• D. $1.67$

Answer 1

The correct option is A $2$

Let at any time t charge flown through the plate B to plate A is q and instantaneous current is I.From loop theorem $\left(\frac{2q-Q_0}{2C}\right)+IR-E=0$

$\Rightarrow R\frac{dq}{dt}=\frac{-2q+2EC+Q_0}{2C}$

$\Rightarrow \frac{dq}{2EC+Q_0-2q}=\frac{dt}{2RC}$

Now for change on plate A to be zero $q=Q_0$ .Integrating

${\int }_0^{Q_0}\frac{dq}{2EC+Q_0-2q}={\int }_0^t\frac{dt}{2RC}$

$=t=RCln\left[\frac{2EC+Q_0}{2EC-Q_0}\right]$

Putting the value of $C,Q_0,EandR$ We get $t=2$ seconds.