Question

Four particles each having a charge $q,$ are placed on the four vertices of a regular pentagon. The distance of each corner from the center is $a.$ Find the electric field at the center of the pentagon.

Answer 1

Field at center due to charge a B = $\frac{1}{4\pi ϵ}\frac{q}{a^2}\overrightarrow{BO}$
Field at center due to charge a C = $\frac{1}{4\pi ϵ}\frac{q}{a^2}\overrightarrow{CO}$
Field at center due to charge a D = $\frac{1}{4\pi ϵ}\frac{q}{a^2}\overrightarrow{DO}$
Field at center due to charge a E = $\frac{1}{4\pi ϵ}\frac{q}{a^2}\overrightarrow{EO}$
Total field at center= $\left[2\frac{1}{4\pi ϵ}\frac{q}{a^2}cos\right(\frac{{72}^{\circ }}{2})-2\frac{1}{4\pi ϵ}\frac{q}{a^2}cos({72}^{\circ }\left)\right]\overrightarrow{OA}$
= $2\frac{1}{4\pi ϵ}\frac{q}{a^2}\left[cos\right({36}^{\circ }-cos\left({72}^{\circ }\right)]\overrightarrow{OA}$
= $0.93399\frac{1}{4\pi ϵ}\frac{q}{a^2}\overrightarrow{OA}$