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 seconds) will elapse before the net charge on plate A becomes zero? $(GivenC=1\mu F,Q_0=1mc,\epsilon =1000V$ and $R=\frac{2\times {10}^6}{ln3}\Omega )$

• A. 2
• B. 3
• C. 4
• D. 9

Answer 1

The correct option is A 2

At any time $t_0$ after $t=0$
Let $q=Q_0$ at $t=t_0$
$\therefore Q_{+}=0$
$V_{capacitor}=\frac{Q_{+}-Q_{-}}{2C}$ (Capacitance)
$\therefore$ By KVL
$E-\frac{dq}{dt}R-\frac{((-Q_0+q)-(-q))}{2C_{AB}}=0$
$E-\frac{2\times {10}^6}{\ln 3}\frac{dq}{dt}-\left[\frac{-Q_0+2q}{2\times 1\times {10}^{-6}}\right]=0$
By substituting given values
$\therefore [1000+\frac{{10}^{-3}}{2\times {10}^{-6}}]-\frac{2\times {10}^6}{\ln 3}\frac{dq}{dt}-\frac{q}{{10}^{-6}}=0$
$\therefore 1500-\frac{2\times {10}^6}{\ln 3}\frac{dq}{dt}-\frac{q}{{10}^{-6}}=0$
$\therefore \frac{dq}{dt}=\left[\frac{1500\times {10}^{-6}-q}{2}\right]\ln 3$
$\therefore {\int }_0^{Q_0}\frac{dq(1\times 2)}{(1500\times {10}^{-6}-9)\ln 3}={\int }_0^{t_0}dt$
${\int }_0^{{10}^{-3}}\frac{dq\left(2\right)}{(1500\times {10}^{-6}-q)\ln 3}=t_0=2.19sec\simeq 2$

Thus option A