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.

Explanation

Let the charges be placed at the vertices A,B,C and D
of the pentagon ABCDE. 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
`(q)/(4piepsilon_0 alpha^2)` along EO.
Thus the field at O due to the given system of charge
is `(q)/(4piepsion_0alpha^2)` along OE.