See the diagram. Area of each plate is \(2.0 \mathrm{~m}^{2}\) and \(d=2 \times 10^{-3} \mathrm{~m}\). A charge of \(8.85 \times 10^{-8} \mathrm{C}\) is given to \(Q\). Then the potential of \(Q\) becomes

In the given arrangement, plate \(Q\) is common for two capacitors which are connected in parallel.
\(\therefore C_{\text {eff }}=C_{P}=C_{1}+C_{2} \)
\( \Rightarrow C_{P}=\frac{\varepsilon_{0} A}{d}+\frac{\varepsilon_{0} A}{2 d}=\frac{2 \varepsilon_{0} A}{2 d}\)
The given arrangement can be shown as
Let \(V\) be the potential difference across the capacitor, which is equal to the potential of the plate \(Q\)

So, \(\quad C=\frac{q}{V}\) i.e., \(C_{\text {eff }}=\frac{q}{V}\)
i.e., \(\quad \frac{3 \varepsilon_{0} A}{2 d}=\frac{8.85 \times 10^{-8}}{V}\)
\(\Rightarrow V =\frac{8.85 \times 10^{-8} \times 2 \times 2 \times 10^{-3}}{3 \times 8.85 \times 10^{-12} \times 2} \)
\( =\frac{2}{3} \times 10=\frac{20}{3}=6.67 \mathrm{~V}\)