Question

four equal charges q are placed at the four corners A,B,C,D of a square of length a. find the magnitude of the force on another charge q at the midpoint of AB.

Answer 1

Dear student
$$\begin{gathered} ForceonthechargeqduetothechargesatAandBcanceleachother.Forceonthechargeqduetoothertwochargesisshowninfigureis \\ forceonchargeqduetochargeatCoratDis \\ F=\frac{1}{4\pi \in }\frac{q^2}{r^2} \\ verticalcomponentofforceis \\ {F}_v=F\cos \theta \\ andthehorizontalcomponentcanceleachother \\ Netverticalforceis \\ 2F\cos \theta =2\times \frac{1}{4\pi \in }\frac{q^2}{r^2}\cos \theta \\ Netverticalforce=2\times \frac{1}{4\pi \in }\frac{q^2}{r^2}\cos \theta =\frac{1}{2\pi \in }\frac{q^2}{r^2}\frac{a}{r} \\ Netverticalforce=\frac{1}{2\pi \in }\frac{q^2\times a}{r^3} \\ r=\sqrt{a^2+\frac{a^2}{4}}=\sqrt{\frac{5a^2}{4}}=\frac{a}{2}\sqrt{5} \\ {r}^3=\frac{5\sqrt{5}{a}^3}{8} \\ Netverticalforce=\frac{1}{2\pi \in }\frac{q^2\times a}{\frac{5\sqrt{5}{a}^3}{8}} \\ Netverticalforce=\frac{1}{2\pi \in }\frac{8q^2\times a}{5\sqrt{5}{a}^3} \\ Netverticalforce=\frac{1}{2\pi \in }\frac{8q^2}{5\sqrt{5}{a}^2}=\frac{1}{\pi \in }\frac{4q^2}{5\sqrt{5}{a}^2} \\ Regards \end{gathered}$$