Physics - Capacitor Question with Solution | TestHub

Given:
Plate area $A=40{\text{ cm}}^2=40\times 10^{-4}{\text{ m}}^2$
Plate separation $d=2\text{ mm}=2\times 10^{-3}\text{ m}$
Thickness of dielectric $t=1\text{ mm}=1\times 10^{-3}\text{ m}$
Dielectric constant $K=5$

The capacitance of a parallel plate capacitor partially filled with a dielectric slab is given by:
$C=\frac{{ϵ}_{0}A}{d-t+\frac{t}{K}}$

Substituting the given values:
$C=\frac{{ϵ}_0\times 40\times 10^{-4}}{2\times 10^{-3}-1\times 10^{-3}+\frac{1\times 10^{-3}}{5}}$

$C=\frac{{ϵ}_0\times 40\times 10^{-4}}{10^{-3}\left(1+\frac{1}{5}\right)}$

$C=\frac{{ϵ}_0\times 40\times 10^{-4}}{10^{-3}\times \frac{6}{5}}$

$C=\frac{{ϵ}_0\times 4\times 10^{-3}}{\frac{6}{5}\times 10^{-3}}$

$C=\frac{4{ϵ}_0\times 5}{6}=\frac{20{ϵ}_0}{6}=\frac{10{ϵ}_0}{3}\text{ F}$

Answer: $\frac{10{ϵ}_0}{3}F$