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

Options

A. `2muF`
B. `15.5 muF`
C. `26.6muF`
D. `32muF`

Answer

C

Explanation

`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`