Question

The two plates of a capacitor of capacitance C are given charges $Q_1$ and $Q_2$. This capacitor is connected across a resistance $R$ as shown key is closed at $t=0$. Find the charges on the plates after time $t$.

• A. Total charge of the right plate $q_2=\frac{Q_1+Q_2}{2}-\left(\frac{Q_1-Q_2}{2}\right)e^{-\frac{t}{RC}}$
• B. Total charge on the left plate $q=\frac{Q_1+Q_2}{2}+\left(\frac{Q_1-Q_2}{2}\right)e^{-\frac{t}{RC}}$
• C. Initial potential difference across the plates is given by $\frac{Q_1-Q_2}{2C}$
• D. Initial potential difference across the plates is given by $\frac{Q_1+Q_2}{2C}$

Answer 1

Answer: (A), (B), (C)

Initial potential difference across the plates is given by $\frac{Q_1-Q_2}{2C}$

Initially the charges on the plates will be distributed as shown.

Potential difference across the plates is given by $\frac{Q_1-Q_2}{2C}$ and is independent of the charge on the outer surfaces. Thus, when capacitor is discharged, the charges on the outer surface do not change. Charge on the inner surface decreases according to

the equation $q=\left(\frac{Q_1-Q_2}{2}\right)e^{-\frac{t}{RC}}$
Total charge on the left plate is
$=\frac{Q_1+Q_2}{2}+\left(\frac{Q_1-Q_2}{2}\right)e^{-\frac{t}{RC}}$
And total charge of the right plate is $q_2$
$=\frac{Q_1+Q_2}{2}-\left(\frac{Q_1-Q_2}{2}\right)e^{-\frac{t}{RC}}$