Physics - Electrostatics Question with Solution | TestHub

Let $a$ be the side length of the square $ABCD$. $\begin{array}{rl}& AC=BD=\sqrt{a^2+a^2}=a\sqrt{2}\\ & OA=OB=OC=OD=\frac{a\sqrt{2}}{2}=\frac{a}{\sqrt{2}}\end{array}$ Potential is a scalar quantity. Potential at the centre $O$ due to given charge configuration is $\begin{array}{rl}& V=\frac{1}{4\pi {\epsilon }_0}\left[\frac{(-Q)}{\left(\frac{a}{\sqrt{2}}\right)}+\frac{(-q)}{\left(\frac{a}{\sqrt{2}}\right)}+\frac{(2q)}{\left(\frac{a}{\sqrt{2}}\right)}+\frac{2Q}{\left(\frac{a}{\sqrt{2}}\right)}\right]=0\\ & \Rightarrow -Q-q+2q+2Q=0\\ & \text{ or }Q+q=0\text{ or }Q=-q\end{array}$