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 d/2, 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 capictor is

$C_0=\frac{{\epsilon }_{0}A}{d}$

Where, A is the area of parallel plates

Suppose that the capacitor is connected to a battery, an electric field $E_0$ is produced

now if we insert the dielectric slab of thickness t=d/2 the electric field reduce to E

Eis producedow, if we insert the dielectric slab of thickness t=
t=d2the electric field reduces to ENow, the gap between plates is divided in two parts, for distance there is electric field E and for the remaining distance
(d-t) the electric field is $E_0$

0
ε0 If V be the potential difference between the plates of the capacitor, then

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

$V=\frac{d}{2}(\frac{E_0}{K}+E_0)=\frac{d{E}_0}{2K}(K+1)(as\frac{E_0}{E}=K)$

$now{E}_0=\frac{\sigma }{{\epsilon }_0}=\frac{q}{{\epsilon }_{0}A}\Rightarrow V=\frac{d}{2K}\frac{q}{{\epsilon }_{0}A}(K+1)weknowC=\frac{q}{V}=\frac{2K{\epsilon }_{0}A}{(K+1)d}$