
$\begin{aligned}
& \mathrm{C}_{\mathrm{eq}}=\mathrm{C}_1+\mathrm{C}_2 \
& \mathrm{C}_1=\frac{2 \epsilon_0 \mathrm{A}}{2 \times \mathrm{d}}=10 \mu \mathrm{F} \
& \mathrm{C}_2=\frac{3 \epsilon_0 \mathrm{A}}{2 \mathrm{~d}}=15 \mu \mathrm{F} \
& \mathrm{C}_{\mathrm{eq}}=25 \mu \mathrm{F}
\end{aligned}$
Now the charge on $\begin{aligned}
& \mathrm{C}_1=10 \mathrm{V} \mu \mathrm{c} \
& \mathrm{C}_2=1.5 \mathrm{V} \mu \mathrm{C} .
\end{aligned}$
Now force between the plates [F=2A∈0Q2] $\begin{aligned}
& \frac{100 \mathrm{V}^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \epsilon_0}+\frac{225 \mathrm{V}^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \times \epsilon_0}=8 \
& 325 \mathrm{V}^2=8 \times 4 \times 10^{-4} \times 8.85 \
& \mathrm{V}^2=\frac{32 \times 8.85 \times 10^{-4}}{325} \
& \therefore \mathrm{V}=\sqrt{\frac{283.2 \times 10^{-4}}{325}} \
& \mathrm{~V}=0.93 \times 10^{-2}
\end{aligned}$
