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 capac

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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 4 μ F . then its new capacity is

TestHub · Accepted Answer

The capacitance of a parallel plate air capacitor is given by
$C_{0} = \frac{ε_{0} A}{d}$
Where, $ε_{0}$ =permittivity of the medium,
A = area of plates
and d = distance between the plates
When the distance between plates is reduced and a dielectric slab is introduced, then the capacitance becomes
$C = \frac{K ε_{0} A}{d_{1}}$ … (ii)
Where, K = dielectric constant of medium.
Here, K = 2, $d_{1} = \frac{d}{4}$ and $C_{0} = 4 μ F$ = $4 \times 10^{- 6} F$
From Eq. (ii), we get
$C = 4 (\frac{K ε_{0} A}{d_{1}}) = 4 K (\frac{ε_{0} A}{d})$
$= 4 K C_{0}$ …. (iii) [From Eq. (i)]
Substituting given values in Eq. (iii), we get
$C = 4 \times 2 \times 4 = 32 μ F$

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