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In a capacitor of capacitance 20 muF,the...

In a capacitor of capacitance `20 muF`,the distance between the plates is `2 mm`. If a dielectric slab of width `1 mm` and dielectric constant `12` is inserted between the plates, then the new capacitance is

A

`2muF`

B

`15.5 muF`

C

`26.6muF`

D

`32muF`

Text Solution

Verified by Experts

The correct Answer is:
C
`C=20 muF`
`C.=(Aepsilon_0)/([d-t+t/k])=(epsilon_0A)/([1-t/d+t/(dk)])=(20 xx10^(-6))/([1-1/2+1/(2k)])`
`C.=(4xx20 xx10^(-6))/3=26.67muF`
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