Question

The figure shows an $\text{RC}$ circuit with a parallel plate capacitor. Before switching on circuit, plate $\text{A}$ of capacitor has a charge $-Q_0$ while plate $\text{B}$ has no net charge. Now at $t=0$, the circuit is switched on. Find How much time will elapse before the net charge on plate $A$ (in $\text{sec}$) becomes zero.

$[$ Given: $C=1\mu \text{F},Q_0=1\text{mC},E=1000\text{V}$ and $R=\frac{2\times {10}^6}{\ln 3}\Omega ]$

• A. $5\text{sec}$
• B. $4\text{sec}$
• C. $2\text{sec}$
• D. $7\text{sec}$

Answer 1

The correct option is C $2\text{sec}$

Let at any time $t$ , charge flown through the plate $\text{B}$ to plate $\text{A}$ be $q$ and the instantaneous current be $I$.


Applying charge distribution, the charge on outer surfaces of capacitor will be,

$q_{outer}=\frac{(-Q_0+q)-q}{2}=\frac{-Q_0}{2}$

Charge on inner plates:

$q_{inner}=-Q_0+q-\left(\frac{-Q_0}{2}\right)=\frac{-Q_0}{2}+q$

$\Rightarrow q_{inner}=\frac{2q-Q_0}{2}$

Thus charge on capacitor is,

$Q=\frac{2q-Q_0}{2}$

Applying the $KVL$ at any time $t$ gives,

$-\left(\frac{Q}{C}\right)+E-IR=0$

$\Rightarrow \frac{Q}{C}+IR-E=0$

Substituting, $I=\frac{dq}{dt}$ and value of $Q$,

$\Rightarrow \frac{2q-Q_0}{2C}+R\left(\frac{dq}{dt}\right)=E$

$\Rightarrow R\left(\frac{dq}{dt}\right)=E-\frac{(2q-Q_0)}{2C}$

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

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

For the charge on plate $A$ to be zero.

$-Q_0+q=0$

$\therefore q=Q_0$

Thus, integrating by putting limits $q=0$ to $q=Q_0$

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

$\Rightarrow {\left[\frac{\ln [2CE-2q+Q_0]}{-2}\right]}_0^{Q_0}={\left[\frac{t}{2RC}\right]}_0^t$

$\Rightarrow \ln (2CE-Q_0)-\ln (2CE+Q_0)=\frac{-t}{RC}$

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

$\Rightarrow \ln \left[\frac{2CE+Q_0}{2CE-Q_0}\right]=\frac{t}{RC}$

$\Rightarrow t=RC\ln \left[\frac{2CE+Q_0}{2CE-Q_0}\right]$

Now, substituting the values,

$\Rightarrow t={10}^{-6}\left(\frac{2\times {10}^6}{\ln 3}\right)\ln \left[\frac{2\times {10}^{-6}\times 1000+{10}^{-3}}{2\times {10}^{-6}\times 1000-{10}^{-3}}\right]$

$\Rightarrow t={10}^{-6}\left(\frac{2\times {10}^6}{\ln 3}\right)\ln \left[\frac{3\times {10}^{-3}}{1\times {10}^{-3}}\right]$

$\therefore t=\frac{2}{\ln 3}\times \ln 3=2\text{sec}$

Hence, option (c) is the correct answer.