Question

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 4μF then its new capacity is ______.

Options

• 32μF

• 18μF

• 8μF

• 44μF

Answer 1

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 4μF then its new capacity is 32μF.

Explanation:

The capacitance of a parallel plate air capacitor is given by

`"C"_0 = (epsilon_0"A")/"d"`    ....(i)

where, ε₀ = 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 = `("K" epsilon_0"A")/"d"_1`    ....(ii)

where, K = dielectric constant of medium.

Here, K = 2, `"d"_1 = "d"/4` and C₀ = 4μ F = 4 × 10^-6 F

From Eq. (ii), we get

C = `((4("K"epsilon_0"A"))/"d"_1) = "4K"((epsilon_0"A")/"d")`

= 4KC₀        ....(iii)[From Eq. (i)]

Substituting given values in Eq. (iii), we get

C = 4 × 2 × 4 = 32 μF