A parallel plate air capacitor consists of two metal plates of area A and separation d . A slab of thickness t and dielectric constant 2 k is inserted between the plates with its faces parallel to the

Question

A parallel plate air capacitor consists of two metal plates of area A and separation d . A slab of thickness t and dielectric constant 2 k is inserted between the plates with its faces parallel to the plates. The effective capacitance of the system is ( ϵ 0 ​ = permittivity of free space)

TestHub · Accepted Answer

The system can be considered as two capacitors connected in series.

The first capacitor is formed by the air gap of thickness $d - t$ , and its capacitance is $C_{1} = \frac{ϵ _{0} A}{d - t}$ .

The second capacitor is formed by the dielectric slab of thickness $t$ and dielectric constant $2 k$ , and its capacitance is $C_{2} = \frac{2 k ϵ _{0} A}{t}$ .

The equivalent capacitance $C$ of the series combination is given by:

$\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}}$

Substituting the values of $C_{1}$ and $C_{2}$ :

$\frac{1}{C} = \frac{d - t}{ϵ _{0} A} + \frac{t}{2 k ϵ _{0} A}$

$\frac{1}{C} = \frac{1}{ϵ _{0} A} ((d - t) + \frac{t}{2 k})$

$C = \frac{ϵ _{0} A}{( d - t ) + \frac{t}{2 k}}$

Answer: $\frac{ϵ _{0} A}{( d - t ) + \frac{t}{2 k}}$

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