$\Rightarrow$ they are connected in parallel

$\therefore C_{e f f}=C_{p}=C_{1}+C_{2}$

$C_{p}=\frac{A_{\epsilon_{0}}}{d}+\frac{A_{\epsilon_{0}}}{2 d}$

$C_{p}=\frac{3 A \epsilon_{0}}{2 d}$

Let $V$ be the potential difference across the capacitor which is equal to potential of

plate $Q.$

Then $\mathrm{C}=\frac{q}{V}$

$C_{e f f}=\frac{q}{V}$

$\frac{3 A \epsilon_{0}}{2 d}=\frac{8.85 \times 10^{-8}}{V}$

$\frac{3 \times 2 \times 8.84 \times 10^{-12}}{2 \times 2 \times 10^{-3}}=\frac{8.85 \times 10^{-8}}{V}$

$V=\frac{2}{3} \times 10^{-8-3+12}=\frac{20}{3}=6.67$