Four particles each having a charge q are placed on the four vertices of a regular pentagon. The distance of each comer from !be centre is 'a'. Find the electric field at the centre of pentagon.

Explanation

We can calculate the electric field at centre by the superposition method i.e., by adding vectorially the electric field due to all the 4 charges at centre which will come out to be :
`vec(F)_("centre")=vec(F_1)+vec(F_2)+vec(F_3)+vec(F_4)=(Kq)/(a^(2))`

In the direction of the vector with no charge as shown in figure shown.
Alternate :
Consider pentagon with charges on all vector. Now, E.F. at centre must be zero due to symmetry

Thus E.F. due to 4 charge + E. F. due to 1 charge = 0 or E.F. due to 4 charges = `– E.F` . due to 1 charge Where – sign denotes that both the forces are in opposite direction.
Thus E.F. dut to 4 charges `= – E.F.` due to 1 charge
`=(Kq)/(a^(2))`
[Another good example of superposition theorem]