Physics - Capacitor Question with Solution | TestHub

If $d$ is the initial distance between the plates then capacitance is given by $\mathrm{C}=\frac{\mathrm{kA}{\in }_0}{\mathrm{d}}$ When a plate of thickness $\mathrm{t}=2\mathrm{mm}$ is inserted between the plates the capacitance, $$C_1=\frac{kA{ϵ}_0}{\left(d-t+\frac{t}{k}\right)}$$ If distance between the plates is increased by $\mathrm{x}=1.6\mathrm{mm}$ Capacitance become, ${\mathrm{C}}_2=\frac{\mathrm{kA}{\in }_0}{\left(\mathrm{d}+\mathrm{x}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{k}}\right)}$ But ${\mathrm{C}}_2=\mathrm{C}$ $\begin{array}{rl}& \therefore \mathrm{d}+\mathrm{x}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{k}}=\mathrm{d}\\ & \therefore \mathrm{t}-\mathrm{x}=\frac{\mathrm{t}}{\mathrm{k}}\\ & \therefore 2-1.6=\frac{2}{\mathrm{k}}\\ & \therefore 0.4=\frac{2}{\mathrm{k}}\therefore \mathrm{k}=5\end{array}$