Three point charges,each +q ,are placed at the corners of an equilateral triangle of side r .the electric field at the circa centre will be,where k is 1/4*22/7£

Let k = 1/4π€°

The electric field intensity (E) at O due to A is |E1 |= k (q/r^2) [directed towards midpoint of BC]——-(1)

And electric field intensity (E) at O due to B is |E2| = k (q/r^2) [directed towards midpoint of AC] ———(2)

The electric field intensity (E) at O due to C is |E3| = k (q/r^2) [directed towards midpoint of AB] ——-(3)

From (1),(2),(3) we get the net electric intensity at O due to A,B,C combined is given by ,

E = E1 + E2 + E3

Note that electric field intensity E1,E2,E3 are vector quantities and they form a triad with angle between them = 120°, and since they are equal in magnitude they cancel out each other

Thus ,net electric intensity

E = 0