Question

Two capacitors having capacitance C1 and C2 respectively are connected as shown in figure. Initially, capacitor C1 is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C1 is now connected to uncharged capacitor C2 by closing the switch S. The amount of charge on the capacitor C2, after equilibrium, is

• A. $\frac{C_1 C_2}{C_1 + C_2}V$
• B. $\frac{C_1+C_2}{C_1C_2}V$
• C. $(C_1+C_2)V$
• D. $(C_1-C_2)V$

Answer 1

Answer: (A)

$\frac{C_1 C_2}{C_1 + C_2}V$

Answer 2

$V_{\text{common}} = \frac{C_1V}{C_1 + C_2}$
$⇒\text{Charge on Capacitor} C_2$
$=$ $C_2V_{\text{common}}$

$=$ $\frac{C_1C_2V}{C_1+C_2}$
So, the correct option is (A)