There is a pentagon ABCDE, whose centre is O. Five identical charges Q are placed at the corners of pentagon.
(a)What will be the net electric field intensity at the centre of pentagon ?
(b)If charge at corner A is moved to centre O of pentagon, then what will be the net force experienced by it ?
Distance of any vertex of pentagon from its centre is d.

Explanation

(i)Electric field at centre , due to all five charges will be equal in magnitude and inclined at an angle `72^@` to each other . If we join the vectors in a sequence , such that the tail of one lies on the head of the other, then we will get a closed polygon. So, the net electric field at the centre will be zero.
(ii)Electric field due to all five charges is zero at the centre, so if we remove one charge from A, then the electric field at O due to rest four must be equal and opposite to electric field created by charge on A. The net field due to four charges at the centre will be `1/(4piepsilon_0)Q/d^2` towards A.
Magnitude of electric force experienced by charge at O :
F=QE
`F=1/(4piepsilon_0) Q^2/d^2`
Since the charges are positive , so force will also be directed toward A. .