Five charges , q each are placed at the corners of a regular pentagon of side.

(a) (i) What will be the electric field at 0, the centre of the pentagon?
(ii) What will be the electric field at 0 if the charge from one of the corners (say A) is removed?
(iii). What will be the electric field at 0 if the pentagon is replaced by n-sided regular polygon with charge q at each of its corner?

Explanation

(a) (i) The point O is equidistant from all the charges at the end point of pentagon. Thus, due to symmetry, the forces due to all the charges are cancelled out. As a result electric field at O is zero.
(ii). When charge q is removed a negative charge will develop at A giving electric field `E=(qxx1)/(4piepsi_(0)r^(2))` along OA.
(iii) If charge q at A is replaced by -q, then two negative charges -2q will develop there. thus, the value of electric field `E=(2q)/(4piepsi_(0)r^(2))` along OA.
(b). When pentagon is replaced by n sided regular polygon with charge q at each of its corners, the electric field at O would continue to be zero as symmetricity of the charges is due to the regularity of the polygon it doesn't depend on the number of sided or the number of charges.