If a charge q is placed at each vertex of a regular polygon, the net electric field at its centre is zero. Show it.

Explanation

The distance of the centre of a regualr polygon from each vertex will be same. Therefore,
`|vec(E_(1))|=|vec(E_(2))|=|vec(E_(3))|` in (a)
`|vec(E_(1))|=|vec(E_(2))|=|vec(E_(3))|=|vec(E_(4))|` in (b)
and `|vec(E_(1))|=|vec(E_(2))|=|vec(E_(3))|=|vec(E_(4))|=|vec(E_(5))|` in ( c )
The angle between any two consecutive field vectors for each of a polygon is also same.
Hence, `vec(E_(1))+vec(E_(2))+vec(E_(3))+......+vec(E_(n))=vec(0)`
where n = 3, 4, 5, 6..............
Note : `vec(E)_(1)+vec(E)_(2)+............+vec(E)_(n)iff vec(E)_(1)+vec(E)_(2)+............+vec(E)_(n-1)=-vec(E)_(n)`
Hence, if a charge q is placed at each vertex except one vertex of regular polygon, then the net electric field at its centre, distant r from each vertex is `(1)/(4pi epsilon_(0))(q)/(r^(2))`, directed towards or away from the empty vertex depending on whether q is positive or negative.