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 $d$, 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 plate 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 $t=\frac{d}{2}$, 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 $E_0$.
If $V$ is the potential difference between the plates of the capacitor, then $V=Et+E_0(d-t)$
$V=\frac{Ed}{2}+\frac{E_{0}d}{2}=\frac{d}{2}(E+E_0)$ $[\because t=\frac{d}{2}]$
$V=\frac{d}{2}(\frac{E_0}{K}+E_0)$ $[\because \frac{E_0}{E}=K]$
$=\frac{d{E}_0}{2K}(K+1)$
Now, $E_0=\frac{\sigma }{{\epsilon }_0}=\frac{q}{{\epsilon }_{0}A}$

$\Rightarrow V=\frac{d}{2K}\frac{q}{{\epsilon }_{0}A}$$(K+1)$
We know,
$C$ = $\frac{q}{V}=\frac{2K{\epsilon }_{0}A}{d(K+1)}$