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 the thickness 2d/3, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.

Answer 1

Initially when there is vacuum between the two plates, the capacitance of the two parallel plates is, ,${\displaystyle C_0=\frac{{\epsilon }_{0}A}{d}}$ where, A is the area of parallel plates.

Suppose that the capacitor is connected to a battery, an electric field E₀ is produced.

Now if we insert the dielectric slab of thickness ${\displaystyle t=\frac{2d}{3}}$the electric field reduces to E.

Now the gap between plates is divided in two parts, for distance t there is electric field E and for the remaining distance (d–t) the electric field is E₀.

If V be the potential difference between the plates of the capacitor, then V=Et + E₀(d–t)

${\displaystyle V=\frac{2Ed}{3}+E_0(d-\frac{2d}{3})=\frac{2Ed}{3}+\frac{E_{0}d}{3}=\frac{d}{3}(2E+E_0) (\because t=\frac{d}{2})}$

${\displaystyle \text{or},V=\frac{d}{3}(\frac{2E_0}{K}+E_0)=\frac{d{E}_0}{3K}(K+2) (As,\frac{E_0}{E}=K)}$

${\displaystyle Now,E_0=\frac{\sigma }{{\epsilon }_0}=\frac{q}{{\epsilon }_{0}A} \Rightarrow V=\frac{d}{3K}\frac{q}{{\epsilon }_{0}A}(K+2)}$

${\displaystyle \therefore C=\frac{q}{V} =\frac{3K_{{\epsilon }_{0}A}}{d(K+2)}}$.