Question

Four point charges are positioned as such: A(0,1), B(2,1), C(0,0), D(2,0). Charge A = -3 microcoulombs, Charge B = 3 microcoulombs, C = 2 microcoulombs, and D = -2 microcoulombs. What is the net electric field experienced at the origin.

Answer 1

$E=k(\frac{q_A}{r_A^2}+\frac{q_B}{r_B^2}+\frac{q_D}{r_D^2})=6~\frac{kN}C.$