Physics - Electrostatics Question with Solution | TestHub

Three forces $F_{41},F_{42}$ and $f_{43}$ acting on $Q$ are shown Resultant of $F_{41}+F_{43}$ $$\begin{array}{rl}& =\sqrt{2}{F}_{\text{each }}\\ & =\sqrt{2}\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{d^2}\end{array}$$ Resultant on $Q$ becomes zero only when ' $q$ ' charges are of negative nature. $$\begin{array}{rl}& F_{4,2}=\frac{1}{4\pi {\epsilon }_0}\frac{Q\times Q}{(\sqrt{2}d{)}^2}\\ & \Rightarrow \sqrt{2}\frac{dQ}{d^2}=\frac{Q\times Q}{2d^2}\\ & \Rightarrow \sqrt{2}\times q=\frac{Q\times Q}{2}\\ & \therefore q=-\frac{Q}{2\sqrt{2}}\text{ or }\frac{Q}{q}=-2\sqrt{2}\end{array}$$