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If q(f) is the free charge on the capaci...

If `q_(f)` is the free charge on the capacitor plates and `q_(b)` is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge `q_(b)` can be expressed as :

A

`q_(b)=q_(f)(1-(1)/(k))`

B

`q_(b)=q_(f)(1+(1)/(k))`

C

`q_(b)=q_(f)(1-(1)/(sqrtk))`

D

`q_(b)=q_(f)(1+(1)/(sqrtk))`

Text Solution

Verified by Experts

The correct Answer is:
A
`sigma_(b)=sigma_(f)(1-(1)/(k))`
`q_(b)=q_(f)(1-(1)/(k))`.
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