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 $a$. Find the electric field at the centre of the pentagon.

• A. $\frac{kq}{a^2}$
• B. $\frac{kq}{\sqrt{2}{a}^2}$
• C. $\frac{kq}{2a^2}$
• D. $\frac{kq}{\sqrt{3}{a}^2}$

Answer 1

The correct option is A $\frac{kq}{a^2}$



Let the electric field at the centre due to four charges be $E_o$
Let us assume a fifth particle having charge $q$ at $E$
By Symmetry, ${\overrightarrow{E}}_{\text{net}}={\overrightarrow{E}}_A+{\overrightarrow{E}}_B+{\overrightarrow{E}}_C+{\overrightarrow{E}}_D+{\overrightarrow{E}}_E=0$
$\Rightarrow {\overrightarrow{E}}_A+{\overrightarrow{E}}_B+{\overrightarrow{E}}_C+{\overrightarrow{E}}_D=-{\overrightarrow{E}}_E$
$\Rightarrow E_0=|-{\overrightarrow{E}}_E|$
$\Rightarrow E_0=\frac{kq}{a^2}$