Explanation:

(a)
\(\mathrm{AC}=\mathrm{BD}=\sqrt{(\sqrt{2})^2+(\sqrt{2})^2}=2 \mathrm{~m}\)
\(\therefore \mathrm{DO}=\mathrm{OB}=\mathrm{AO}=\mathrm{OC}=\frac{2}{2}=\mathrm{Im}\)
\(\therefore\) Potential at the centre \(0, \mathrm{~V}=\mathbf{k} \frac{\mathrm{q}}{\mathrm{r}}\)
\(\begin{aligned}
& V=k\left[\frac{2 \times 10^{-8}}{1}+\frac{-2 \times 10^{-8}}{1}+\frac{-3 \times 10^{-8}}{1}+\frac{6 \times 10^{-8}}{1}\right] \\
& V=k \times 3 \times 10^{-8}=9 \times 10^9 \times 3 \times 10^{-8} \text { volt } \\
& =27 \times 10=270 \mathrm{~V}
\end{aligned}\)