CBSE Class 12 · 2018 · ElectrostaticsShort Answer

(a) A charge $Q$ (say at a corner) is acted on by the two adjacent $q$ charges (at distance $a$) and the opposite $Q$ (at distance $a\sqrt{2}$). Each adjacent force is $F_1=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{a^2}$; their resultant along the diagonal is $\sqrt{2}{F}_1$. The diagonal force from the opposite $Q$ is $F_2=\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{2a^2}$. Adding along the diagonal, $F=\frac{Q}{4\pi {\epsilon }_0{a}^2}\left(\sqrt{2}q+\frac{Q}{2}\right)$. (b) Summing over the $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ pairs: four edge pairs ($Qq$ at $a$) and two diagonal pairs ($QQ$ and $qq$ at $a\sqrt{2}$): $U=\frac{1}{4\pi {\epsilon }_0}\left[\frac{4Qq}{a}+\frac{Q^2}{a\sqrt{2}}+\frac{q^2}{a\sqrt{2}}\right]=\frac{1}{4\pi {\epsilon }_{0}a}\left(4Qq+\frac{Q^2+q^2}{\sqrt{2}}\right)$.