A parallel plate capacitor of capacitance C 0 has plates of area ( A ) with a se

Question

A parallel plate capacitor of capacitance C0 has plates of area ( A ) with a separation ( d ) between them. When it is connected to a battery of voltage ( V0 ) it has charge of magnitude ( Q0 ) on its plates. it is then disconnected from the battery and the plates are pulled apart to a separation of ( 2d ) without discharging them. after the plates are ( 2d ) apart, the new capacitance and the potential difference between the plates are?

A) ( 1/2)C0, (1/2)V0

B) ( 1/2)C0, V0

C) ( 1/2)C0, (2)V0

D) C0, (2)V0

E) ( 2)C0, (2)V0

Please help,

Thank you!

Explanation / Answer

c option is correct and ( 1/2)C0, (2)V0

As C=Ar0/d

For vacuum r=1;

so C=C0=A0/d,

When the plates are 2d apart,the new capacitance will be

C=A0/(2d)=C0/2

Also,qo=CoVo

and V=q/C, now if the capacitance becomes half, the voltage or potential difference will be doubled.

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