Question

Two capacitors having capacitors $C_1\text{and}{C}_2$ respectively are connected as shown in figure. Initially, capacitor $C_1$ is charged to a potential difference $\text{V}$ volt by a battery. The battery is then removed and the charged capacitor $C_1$ is now connected to uncharged capacitor $C_2$ by closing the switch $S$. The amount of charge on the capacitor $C_2$, after equilibrium is

• A. $(C_1+C_2)V$
• B. $\frac{C_1{C}_2}{(C_1+C_2)}V$
• C. $(C_1-C_2)V$
• D. $\frac{(C_1+C_2)}{C_1{C}_2}V$

Answer 1

The correct option is B $\frac{C_1{C}_2}{(C_1+C_2)}V$

$V_{common}=\frac{C_{1}V}{C_1+C_2}$

$\Rightarrow$ Charge on capacitor $C_2$

$=C_2{V}_{common}$

$=\frac{C_1{C}_{2}V}{C_1+C_2}$

$\Rightarrow$ Option (a) si correct.