A parallel-plate air capacitor has a capacity ' C ', distance with a separation ' x ', a dielectric material with a dielectric constant ' k ' between plates, and a potent

Question

A parallel-plate air capacitor has a capacity ' C ', distance with a separation ' x ', a dielectric material with a dielectric constant ' k ' between plates, and a potential difference with a ' V ' volt applied between plates. The force of attraction between the plates is

TestHub · Accepted Answer

The capacitance of the air capacitor is given by $C = \frac{ϵ _{0} A}{x}$ , where $A$ is the area of the plates.

When a dielectric material of dielectric constant $k$ is introduced between the plates, the new capacitance becomes $C^{'} = k C$ .

The charge on the capacitor plates when a potential difference $V$ is applied is $Q = C^{'} V = k C V$ .

The magnitude of the force of attraction between the plates of the capacitor is given by $F = \frac{Q ^{2}}{2 k ϵ _{0} A}$ .

Substituting $Q = k C V$ and $ϵ_{0} A = C x$ , we get:

$F = \frac{( k C V ) ^{2}}{2 k ( C x )}$

$F = \frac{k ^{2} C ^{2} V ^{2}}{2 k C x}$

$F = \frac{k C V ^{2}}{2 x}$

Answer: $\frac{k C V ^{2}}{2 x}$

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