Question

Two point charges each of 1.0 μC are placed at the positions (–1 m, 0) and (1 m, 0) respectively. What is the electric field at the point (0, 1 m)?

Full explanation and solution please! I dont understand how symmetry works along the x-axis and what is cancelled out.

Answer 1

charge 1 uC is placed at A(-1, 0) and B(1,0) .

and field due charge Q at distance r , E = kQ/r^2

and away from charge if Q is +ve.

at point P:

due to charge at A, AP = sqrt(1^2 + 1^2) = 1.414 m

and direction of AP = tan^-1((1-0)/(0-(1))) = 45 deg   

E1 = [(9 x 10^9 x 1. x 10^-6)/(1.414)^2] (cos45i + sin45j)

E1 = 3181.98i + 3181.98j N/C

due to charge at B, BP = sqrt(1^2 + 1^2) = 1.414 m

and direction of BP = tan^-1((1-0)/(0-(-1))) = 180-45 = 135 deg   

E2 = [(9 x 10^9 x 1. x 10^-6)/(1.414)^2] (cos(135)i + sin(135)j)

E2 = - 3181.98i + 3181.98j N/C

Enet = E1 + E2 = 6363.96 N/C j

field is 6363.96 N/C along +ve y axis.