Question

Four changed particles, each having a charge $q$, are placed at the four vertices of a regular pentagon. The sides of the Pentagon have equal length '$a$'. The electric field at the centre of the Pentagon is :

• A. $\frac{q}{4\pi {\epsilon }_{0}a}$
• B. $\frac{2q}{4\pi {\epsilon }_0{a}^2}$
• C. $\frac{q}{4\pi {\epsilon }_0{a}^2}$
• D. None of these

Answer 1

The correct option is C $\frac{q}{4\pi {\epsilon }_0{a}^2}$

Suppose if charge ${}^{\prime }{q}^{\prime }$ is placed at E as well as

$\overrightarrow{E}\left(atpointO\right)=0$ because of symmetry.

$\overrightarrow{E}$(center because of charge,A,B,C,D)$=\overrightarrow{E}$(center because of charge E alone)$

$\overrightarrow{E}$(at center due to charge $q$)$=\frac{q}{4\pi {ϵ}_0{a}^2}$

$\therefore$ Electric field at O will be same i.e. $\frac{q}{4\pi {ϵ}_0{a}^2}$ along OE due to given system of charge.