Question

Four charges q on a,q on b ,-q on c,q on d are placed at the corner of a square with side length l . Find the force on the charge which is placed at the corner c .

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{force}\mathrm{acting}\mathrm{on}\mathrm{c}\mathrm{due}\mathrm{to}\mathrm{B}\mathrm{and}\mathrm{D}\mathrm{will}\mathrm{be}\mathrm{same} \\ {\mathrm{F}}_{\mathrm{B}}={\mathrm{F}}_{\mathrm{D}}=\frac{9\times {10}^9\times (-{\mathrm{q}}^2)}{{\mathrm{l}}^2}=-\frac{9\times {10}^9\times {\mathrm{q}}^2}{{\mathrm{l}}^2} \\ \mathrm{The}\mathrm{directions}\mathrm{of}\mathrm{these}\mathrm{forces}\mathrm{will}\mathrm{be}\mathrm{along}\mathrm{the}\mathrm{edges},\mathrm{their}\mathrm{sum}\mathrm{will}\mathrm{lie}\mathrm{along}\mathrm{the}\mathrm{diagonal}\mathrm{from}\mathrm{A}\mathrm{to}\mathrm{C}\mathrm{and}\mathrm{have}\mathrm{magnitude}, \\ {\mathrm{F}}_{\mathrm{B}}\cos {45}^0+{\mathrm{F}}_{\mathrm{D}}\cos {45}^0=\frac{2\times [-\frac{9\times {10}^9\times {\mathrm{q}}^2}{{\mathrm{l}}^2}]}{\sqrt{2}}=-\frac{9\sqrt{2}\times {10}^9\times {\mathrm{q}}^2}{{\mathrm{l}}^2} \\ \mathrm{The}\mathrm{remaining}\mathrm{force}{\mathrm{F}}_{\mathrm{A}}\mathrm{is}\mathrm{also}\mathrm{along}\mathrm{this}\mathrm{diagonal}. \\ {\mathrm{F}}_{\mathrm{A}}=\frac{9\times {10}^9\times (-{\mathrm{q}}^2)}{[\mathrm{l}\sqrt{2}{]}^2}=\frac{9\times {10}^9\times (-{\mathrm{q}}^2)}{2{\mathrm{l}}^2} \\ \mathrm{The}\mathrm{resultant}\mathrm{of}\mathrm{all}\mathrm{these}\mathrm{forces}\mathrm{will}\mathrm{point}\mathrm{along}\mathrm{the}\mathrm{diagonal}\mathrm{and}\mathrm{away}\mathrm{from}\mathrm{the}\mathrm{square}. \\ \mathrm{F}=-\frac{9\sqrt{2}\times {10}^9\times {\mathrm{q}}^2}{{\mathrm{l}}^2}-\frac{9\times {10}^9\times {\mathrm{q}}^2}{2{\mathrm{l}}^2} \\ =-\frac{9\times {10}^9\times {\mathrm{q}}^2}{{\mathrm{l}}^2}[\sqrt{2}+\frac{1}{2}] \\ =-\frac{9\times {10}^9\times {\mathrm{q}}^2}{{\mathrm{l}}^2}\left[\frac{2\sqrt{2}+1}{2}\right] \end{gathered}$$