Four charges on a rectangle that is 6 cm by 3 cm. All but one of the corners has a positive charge. The magnitudes of the charges are all 8.40e-12 C. What is the net electric field at the center of the rectangle?

Answer 1

The electric field contribution of the two opposite positive charges cancel out. The two charges on opposite sides that are of opposite sign will double the field due to a single charge. The center of the rectangle is at a distance R from corners that is given by the equation
R^2 = 3^2 + 1.5^2
R = 3.354 cm
Q = 8.40*10^-12 C

E = 2 * kQ^2/R^2