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A parallel plate capacitor of capacitanc...

A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is connected to another battery and is charged to potential difference 2V. The charging batteries are now disconnected and the capacitor are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

A

zero

B

`3/2CV^(2)`

C

`25/6CV^(2)`

D

`9/2CV^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
The digramatic repensentation of given problem is shown in figure .
The net charge shared between the two capacitors is
`Q '=Q_2-Q_1=4CV-CV=3CV`
The two capacitor will have the same potential say V'
The net capacitance of the parallel combinations of the two capacitors will be
`C' =C_1+C_2=C+ 2C=3C`
The potential difference across the capacitors will be
`V '(Q ')/(C ')=(3 CV)/(3C)=V`
The electrostatic energy of the capacitor will be
`U '=1/2C ' V '=1/2(3C) V^(2)=3/2CV^(2)`
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Knowledge Check

  • Two idential capacitors are joined in parallel, charged to a potential V and then separated and then connected in series i.e. the positive plate of one is connected to negative of the other

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    The charges on the free plated connected together are destoyed.
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    D
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