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 :

Capacitor PYQ — JEE Main 2022 · 26 Jun · Shift 1 · Q50 with Solution

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Correct answer: A

Initially, capacitor $C_1$ is charged by a battery of voltage $V$ . So its initial charge is $Q_i = C_1 V$

Capacitor $C_2$ is initially uncharged, so $Q_{2i}=0$

Now $C_1$ and $C_2$ are connected together. At equilibrium, both capacitors will have the same potential difference, say $V_f$ .

Since the battery is removed, total charge is conserved. Thus, $Q_i = Q_f$ $C_1V = (C_1+C_2)V_f$

So, $V_f = \frac{C_1V}{C_1+C_2}$

The final charge on $C_2$ is $Q_2 = C_2 V_f$ Substituting $V_f$ , $Q_2 = C_2\left(\frac{C_1V}{C_1+C_2}\right)$ $Q_2 = \frac{C_1C_2}{C_1+C_2}V$

Option A is $\frac{C_1C_2}{C_1+C_2}V$ which matches our result.

So the correct option is A.