Five charges, q each are placed at the corners of a regular pentagon of side a. (Refer the adjoining figure)
(a) (i) What will be the electric field at O, if 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 magnitude of the electric field intensity at the centre of the pentagon is the same for all the charges, and all these five electric field vectors are directed away from each corner of the pentagon towards the centre. These electric field vectors will make a `72^@` angle with the electric field of its adjacent particle. If we arrange all the five vectors in an order such that the tail of one is placed on the head of the other, then we can see that the polygon will get completed and the resultant will become zero.
(ii)The resultant of all five vectors is zero, hence we can say that the resultant of any four of them will be equal and opposite to the electric field intensity of the fifth particle. If the charged particle at A is removed, then the net electric field intensity at point O will be directed from point O towards point A with the same magnitude as due to the particle at point A. Hence the electric field intensity at the centre becomes `q/(4piepsilon_0r^2)` from point O towards point A.
(iii) If the charge at point A is replaced by another -q charge, then the electric field of-q charge will also be added in the same direction along OA to make the resultant two times the value mentioned above in (ii), hence the electric field will be `(2q)/(4piepsilon_0r^2)` from point O towards point A.
(b)The answers will not be affected if it were any other regular polygon instead of a pentagon.