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A capacitor is connected to a battery of...

A capacitor is connected to a battery of voltage V. Now a di-electric slab of di-electric constant k is completely inserted between the plates, then the final charge on the capacitor will be : (If initial charge is `q_(0)`)

A

`(epsi_(0)A)/(d) V`

B

`(kepsi_(0)A)/(d) V`

C

`(epsi_(0)A)/(kd) A`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
B
`q = C_(1)V = ((kepsi_(0)A)/(d))V = Kq_(0)`
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