Two capacitors having capacitance C 1 and C 2 respectively are connected as shown in figure. Initially,…

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

  1. C1C2C1+C2V
  2. C1+C2C1C2V
  3. C1+C2V
  4. C1-C2V

Solution

The initial charge on the capacitor C1 will be Q=C1V. If after closing the  switch the charge flown through the circuit is q, the charge on C2 will be q and C1 will be Q-q. The potential across the capacitors will be same. Hence,

Q-qC1=qC2⇒q=C2QC1+C2⇒q=C1C2C1+C2V

Asked in: JEE Main 2022 (26 Jun Shift 1)

Practice more Electrostatics questions on Aicharya