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The plates of a parallel plate capacitor...

The plates of a parallel plate capacitor of capacity `C_(1)` are moved closer together until they are half their original separation. What is the new capacitance?

A

`C_(2)=C_(1)`

B

`C_(2)=2C_(1)`

C

`C_(2)=(C_(1))/(2)`

D

`C_(2)=3C`

Text Solution

Verified by Experts

The correct Answer is:
B
`C=(KA epsi_(0))/(d)`, C is doubled if d is made halff.
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