In a parallel plate air capacitor the distance between plates is reduced to one fourth and the space between them is filled with a dielectric medium of constant 2. If the initial capacity of the capacitor is `4muF`. then its new capacity is .

Options

A. `32 muF`
B. `18 muF`
C. `8 muF`
D. `44 muF`

Answer

A

Explanation

The capacitance of parallel plate air capacitor is given by
`C_(0) = (epsi_(0)A)/(d)" "……..(i)`
Where `,epsi_(0)` = permittivity of the medium.
A = area of plates
and d= distance between the plates
When the distance between plates in reduced and a dielectric slab is introduced , then capacitance becomes .
`C= (k epsi_(0)A)/(d_(1)) " "........(ii)`
where , k = dielectric constant of medium .
Here`k= 2, d_(1) = (d)/(4) and c_(0) = 4 mu F = 4 xx 10^(-6)F`
From Eq.(ii) we get
`C = 4 ((k epsi_(0) A)/(d_(1))) = 4k((epsi_(0)A)/(d))`
` = 4kC_(0)" "....(iii)` [from Eq.(i)]
Substituting given values in Eq. (iii) we , get
`C = 4xx 2 xx 4 = 32 muF`