Question

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

• A. $\frac{q}{4\pi {ϵ}_0{a}^4}$ along $EO$
• B. $\frac{q}{2\pi {ϵ}_0{a}^2}$ along $EO$
• C. $\frac{q}{\pi {ϵ}_0{a}^2}$ along $EO$
• D. $\frac{q}{4\pi {ϵ}_0{a}^2}$ along $EO$

Answer 1

The correct option is D $\frac{q}{4\pi {ϵ}_0{a}^2}$ along $EO$

Let the four charges q each be placed at corners A, B, C and D. If another charge q is placed at 5th corner E. The electric intensity at O is zero by symmetry. It is because a unit positive charge placed at O will have five forces equal in magnitude enclosing an angle of $\frac{360}{5}={72}^o$ between two consecutive forces. They thus form a closed polygon of vectors and hence form a null vector. The electric field at the center due to charges q at A, B,C and D is equal and opposite to that due to charge q alone which is $\frac{q}{4\pi {\epsilon }_0{a}^2}$ along EO.

Therefore, the electric field at O due to the given four charges is $\frac{q}{4\pi {\epsilon }_0{a}^2}$ along EO.