Five charges, q each are placed at the corners of a regular pentagon of side 'a' as in figure:

(a) (i) What will be the electric field at O, the centre of the pentagon ?
(ii) What will be the electric field at O if the charge from one of the corners (say A) is removed ?
(iii) What will be the electric field at O if the charge q at A is replaced by -q ?
(b) How would your answer to (a) be affected if pentagon is replaced by n-sided regular polygon with charge q at each of its corners ?

Explanation

(a) (i) The point O, the centre of the pentago, is equidistant from all the charges at the end point of pentagon. Thus, due to symmetry the electric field due to all th charges are cancelled out. As a resul electric field at O is zero.
(ii) When charge q is removed from A ne electric field at the centre due to remaining charges `E=(kq)/r^(2)` along OA.
(iii) If charge q at A is replaced by - q then electric field due to this negative charge `vecE_(-q) =(kq)/r^(2)` along OA.

(b) If pentagon is replaced by n-sided regulai polygon with charge q at each of its corners Here, again charges are symmetrical about the centre. The net electric field at O would continue to be zero, it doesn.t depend on the number of sides or the number oi charges. Hence, the answer of (a) would not be affected.