\(V_{A}=\frac{2 k q}{a}-\frac{2 k q}{a \sqrt{5}}\)

\(\Rightarrow V_{A}=\frac{1}{4 \pi E} \frac{2 q}{a}\left(1-\frac{1}{\sqrt{5}}\right)\)

(Here potential due to each \(q=\frac{k q}{a}\) and potential due to each \(-q=\frac{-k q}{a \sqrt{5}}\))

Final potential of the charge

\(V_{B}=0\)

( \(\because\) Point \(B\) is equidistant from all the four charges)

\(\therefore\) Using work energy theorem,

\(\left(W_{A B}\right)_{\text {electric }}=Q\left(V_{A}-V_{B}\right)\)

\(=\frac{2 q Q}{4 \pi E_{0} a}\left[1-\frac{1}{\sqrt{5}}\right]\)

\(=\left(\frac{1}{4 \pi \varepsilon_{0}}\right) \frac{2 Q q}{a}\left[1-\frac{1}{\sqrt{5}}\right]\)