In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C . The…

In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch S1 is pressed first to fully charge the capacitor C1 and the released. The switch S2 is then pressed to charge the capacitor C2. After some time, S2 is released and then is S3 pressed. After some time,

  1. the charge on the upper plate of C1 is 2CV0 . 
  2. the charge on the upper plate of C1 is CV0 . 
  3. the charge on the upper plate of C2 is 0. 
  4. the charge on the upper plate of C2 is - CV0 . 

Solution

When S1 is closed,
charge on  C 1 = 2 V 0 C
After S2 is closed, half of the change will move to C2
∴   charge on C 1 = 1 2 2 V 0 C
when S2 is closed,
charge on C 2 = V 0 C but with reversed polarity,
∴   Charge on upper plate = - V 0 C .

Asked in: JEE Advanced 2013 (Paper 1)

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