Question

A capacitor of capacity $C_1$, is charged by connecting it across a battery of e.m.f. $V_0$. The battery is then removed and the capacitor is connected in parallel with an uncharged capacitor of capacity $C_2$. The potential difference across this combination is :

• A. $\frac{C_2}{C_1+C_2}.V_0$
• B. $\frac{C_1}{C_1+C_2}.V_0$
• C. $\frac{C_1+C_2}{C_2}.V_0$
• D. $\frac{C_1+C_2}{C_1}.V_0$

Answer 1

The correct option is B $\frac{C_1}{C_1+C_2}.V_0$

The charge on the first capacitor will be $Q_1=C_1{V}_0$

When battery is disconnected, the charge on the first capacitor remains constant.

Let $V_c$ be the common potential for connecting charged capacitor $C_1$ with an uncharged capacitor $C_2$.

Net capacitance in parallel connection $C_{net}=C_1+C_2$

So, $V_c=\frac{Q_{net}}{C_{net}}=\frac{Q_1+Q_2}{C_1+C_2}=\frac{C_1{V}_0+0}{C_1+C_2}=\frac{C_1}{C_1+C_2}{V}_0$