Physics - Capacitor Question with Solution | TestHub

Capacitance of each capacitor 

$C_1=\frac{A3{\epsilon }_0}{{\displaystyle \frac{1}{2}}}=6 A{ϵ}_0$ and $C_2=\frac{A4{ϵ}_0}{1}=4A{\epsilon }_0$

Here, $A$ is area of plate.

Equivalent capacitance is 

$C_{\mathrm{eq}}=\frac{C_1{C}_2}{C_1+C_2}=\frac{6A{\epsilon }_0\times 4A{\epsilon }_0}{6A{\epsilon }_0+4A{\epsilon }_0}=\frac{24}{10}A{ϵ}_0$

Now, net charge stored is $q_{\mathrm{net}}={\mathrm{C}}_{\mathrm{eq}}\left(\mathrm{\Delta V}\right)=\frac{24}{10}\mathrm{A}{\epsilon }_0\left(100\right)=240\mathrm{A}{\epsilon }_0$

Potential drop across $C_2$ is $\Delta V_2=\frac{240A{\epsilon }_0}{4A{\epsilon }_0}=60 \mathrm{V}$

Thus, voltage of the conducting foil is $V_{\mathrm{foil} }=60 \mathrm{V}$