Physics - Capacitor Question with Solution | TestHub

With insertion of dielectric, the system can be considered as combination of two capacitors in series.

Thus, equivalent capacitance of the system is 

$\frac{1}{C_{\mathrm{eq}}}=\frac{1}{C_1}+\frac{1}{C_2}$

$=\frac{1}{\frac{K{\epsilon }_0 A}{t}}+\frac{1}{\frac{{\epsilon }_0 A}{ d-t}}$

$=\frac{t}{K{\epsilon }_0 A}+\frac{d-t}{{\epsilon }_0 A}$

$=\frac{1\times {10}^{-3}}{5{\epsilon }_0\times 40\times {10}^{-4}}+\frac{1\times {10}^{-3}}{{\epsilon }_0\times 40\times {10}^{-4}}$

$\Rightarrow \frac{1}{C_{\mathrm{eq}}}=\frac{1}{20{\epsilon }_0}+\frac{1}{4{\epsilon }_0}$

$\Rightarrow C_{\mathrm{eq}}=\frac{20\times 4{\epsilon }_0}{24}=\frac{10{\epsilon }_0}{3} \mathrm{F}$.