JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Capacitance

Two capacitors having capacitance

C_{1}

and

C_{2}

respectively are connected as shown in figure. Initially, capacitor

C_{1}

is charged to a potential difference 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 :

Answer: (A)

Solution

Initial charge on

C_{1}

Q = C_{1} V

. When connected to uncharged

C_{2}

, charge redistributes. By charge conservation, total charge remains

C_{1} V

. The capacitors are in parallel, so they reach the same potential

V^{'}

. Total capacitance is

C_{1} + C_{2}

, so

V^{'} = \frac{C _{1} V}{C _{1} + C _{2}}

. Charge on

C_{2}

Q_{2} = C_{2} V^{'} = \frac{C _{1} C _{2} V}{C _{1} + C _{2}}

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Single correct option