Question

A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has a thickness $\dfrac{d}{2}$, where d is the distance between plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.

Answer 1

We first need to find out the potential difference between the plates after the dielectric slab is introduced between the plates of the capacitor. Then the capacitance is given as the ratio of the charge on the capacitor to the potential difference between plates of the capacitor.
Formula used:
The expression to calculate capacitance are given as

q = CV \\\ $$ The electric field between plates of capacitor is given as $E = \dfrac{\sigma }{{{ \in _0}}} = \dfrac{q}{{{ \in _0}A}}$ **Complete step-by-step solution:** First of all, let us consider the capacitor with vacuum between its plates. In that case, the capacitance can be given as $${C_0} = \dfrac{{{ \in _0}A}}{d}$$ Here ${ \in _0}$ is the permittivity of the vacuum while A is the area of the plates of the capacitor and d is the distance between the plates of the capacitor. Let ${E_0}$ be the electric field between the plates of this capacitor. It can be written as ${E_0} = \dfrac{\sigma }{{{ \in _0}}} = \dfrac{q}{{{ \in _0}A}}$ Here $\sigma $ represents the surface charge density on the plates of the capacitor having total charge q. Now when a dielectric slab of material of dielectric constant K having area A and thickness d/2 is introduced between the plates of this capacitor then the electric field between the plates of the capacitor will reduce. Let the new reduced electric field be E. It is related to the original electric field by the following relation. $\dfrac{{{E_0}}}{E} = K$ As we can see in the diagram that half of the space between the plates of the capacitor is occupied by the dielectric slab while the rest of the half is the vacuum. Now we can write the potential difference between the plates of the capacitor in the following way.  $ V = E \times \dfrac{d}{2} + {E_0}\left( {d - \dfrac{d}{2}} \right) \\\ = \dfrac{{Ed}}{2} + \dfrac{{{E_0}d}}{2} \\\ = \dfrac{{d\left( {E + {E_0}} \right)}}{2} \\\ = \dfrac{d}{2}\left( {\dfrac{{{E_0}}}{K} + {E_0}} \right) \\\ = \dfrac{{d{E_0}}}{{2K}}\left( {1 + K} \right) \\\ = \dfrac{{qd}}{{2{ \in _0}AK}}\left( {1 + K} \right) \\\ $ **Now we can write the capacitance of the capacitor with slab inside it, in the following way. $C = \dfrac{q}{V} = \dfrac{{2K{ \in _0}A}}{{d\left( {K + 1} \right)}}$, This is the required answer for capacitance of the capacitor.** **Note:** It should be noted that the electric field between the plates of the capacitor reduces when a dielectric material is introduced inside it. This happens because the dielectric slab produces an electric field in the direction opposite to that of the original electric field reducing the net electric field between the plates.