Question

Five point charges, each of value q are placed on 5 vertices of a regular hexagon of side l. Find the magnitude of the force on a point charge of value −q placed at the centre of the hexagon.

Answer 1

$$\begin{gathered} \mathrm{Among}\mathrm{the}\mathrm{five}\mathrm{forces}\mathrm{on}\mathrm{five}\mathrm{charges},2\mathrm{pairs}\left(\mathrm{four}\right)\mathrm{of}\mathrm{forces}\mathrm{are}\mathrm{cancelled}\mathrm{with}\mathrm{each}\mathrm{other}\mathrm{since}\mathrm{they}\mathrm{are}\mathrm{equal}\mathrm{in}\mathrm{magnitude}\mathrm{and}\mathrm{opposite}\mathrm{in}\mathrm{direction}. \\ \mathrm{The}\mathrm{resultant}\mathrm{force}\mathrm{is}\mathrm{due}\mathrm{to}\mathrm{a}\mathrm{single}\mathrm{force}\mathrm{given}\mathrm{below}. \\ F=\frac{1}{4\pi {\epsilon }_0}\times \frac{\left(q\right)\left(q\right)}{l^2} \\ F=\frac{1}{4\pi {\epsilon }_0}\times \frac{q^2}{l^2}(\mathrm{attractive}\mathrm{in}\mathrm{nature}) \end{gathered}$$