Five charges, q each are placed at the corners of regular pentagon of side a as shown in the figure.
(a) What will be the electrric field at O, the centre of the pentagon ?

(b) What will be the electric field at O, if the charge from one of the corners (say A) is removed ?
(c ) What will be the electric field at O, if the charge q at A is replaced by `-q` ?

Explanation

(a) The point O is equidistant from all the charges at the end points 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.
(b) When charge q is removed from a, electric field at O would become
`E=(qxx1)/(4piepsilon_(0)r^(2))" (along OA)"`
(c) If charge q at A is replaced by `-q`, then it is equivalent to adding charge `-2q`. Thus, the electric field at O would become
`E=(2q)/(4piepsilon_(0)r^(2))" (along OA)"`