Four particles, each having a charge $q$, are placed on the four vertices of a regular pentagon. The

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 centre is a. The electric field at the centre of the pentagon is

Professor PC Thomas Classes · Answer

Let the charges be placed at the vertices $A, B, C$ and $D$ of the pentagon $A B C D E$. If we put a charge $q$ at the corner $E$ also, the field at $O$ will be zero by symmetry. Thus, the field at the centre due to the charges at $A, B$,$C$ and $D$ is equal and opposite to the field due to the charge $q$ at $E$ alone.
The field at $O$ due to the charge $q$ at $E$ is
$
\frac{q}{4 \pi \varepsilon_0 a^2} 	ext { along } E O 	ext {. }
$
Thus, the field at $O$ due to the given system of charges is $\frac{q}{4 \pi \varepsilon_0 a^2}$ along $O E$.