Question

In a parallel plate capacitor set up, the plate area of capacitor is $2{\text{m}}^2$ and the plates are separated by $1\text{m}$. If the space between the plates are filled with a dielectric material of thickness $0.5\text{m}$ and area $2{\text{m}}^2$ (see figue) the capacitance of the set-up will be $n{ϵ}_0$. The value of $n$ is
(Dielectric constant of the material $=3.2$) (Round off to the Nearest Integer)

Answer 1

This capacitor can be thought as two capacitor connected in parallel. one with dilectric and another without dielectric.

$C_1=\frac{K{\epsilon }_{0}A}{d/2},C_2=\frac{{\epsilon }_{0}A}{d/2}$
Equivalent capacitance can be calculated as:
$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{d}{2K{\epsilon }_{0}A}+\frac{d}{2{\epsilon }_{0}A}$

$\frac{1}{C_{eq}}=\frac{d}{2{\epsilon }_{0}A}\left(\frac{K+1}{K}\right)$

$C_{eq}=\frac{2{\epsilon }_{0}AK}{d(K+1)}=\frac{2\times 2\times 3.2}{1\times 4.2}{\epsilon }_0=3.04{\epsilon }_0$