Ncert Exercises & Solutions

A parallel plate capacitor with air between the plates has a capacitance of $8~\text{pF}~(1~\text{pF}=10^{-12}~\text F).$ What will be the capacitance if the distance between the plates is reduced by half, and the space between¹ them is filled with a substance of dielectric constant $6?$
1. $24~\text{pF}$
2. $48~\text{pF}$
3. $96~\text{pF}$
4. $192~\text{pF}$

Given,

Capacitance, C = 8pF.

In first case the parallel plates are at a distance ‘d’ and is filled with air.

Air has dielectric constant, k = 1

Capacitance, $C = \frac{k \times ϵ_{o} \times A}{d} = \frac{ϵ_{o} \times A}{d}$

Here,

A = area of each plate

$ϵ_{o}$ = permittivity of free space.

Now, if the distance between the parallel plates is reduced to half, then d₁ = d/2

Given, dielectric constant of the substance, k₁ = 6

Hence, the capacitance of the capacitor,

$C_{1} = \frac{k_{1} \times ϵ_{o} \times A}{d_{1}} = \frac{6 ϵ_{o} \times A}{d / 2} = \frac{12 ϵ_{o} A}{d} \dots$

Taking ratios of eqns. (1) and (2), we get,

C₁ = 2 x 6 C = 12 C = 12 x 8 pF = 96pF.

Hence, capacitance between the plates is 96pF.

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