Answer to Question #110878 in Electricity and Magnetism for Pace

Question

Four charges are arranged in a square with one side of the square measuring 0.2m. From upper left going clockwise, q1 = 2µC; q2 = -4µC; q3 = 6µC; and q4 = -8µC. Solve for the amount of electric force experienced by q4 due to the other charges.

Answer 1

$$F_1=\frac{kq_1q_4}{r_1^2}=\frac{(9\cdot10^{9})(2\cdot10^{-6})(8\cdot10^{-6})}{0.2^2}=3.6\ N$$

$$F_3=\frac{kq_3q_4}{r_3^2}=\frac{(9\cdot10^{9})(6\cdot10^{-6})(8\cdot10^{-6})}{0.2^2}=10.8\ N$$

$$F_2=\frac{kq_2q_4}{r_2^2}=\frac{(9\cdot10^{9})(4\cdot10^{-6})(8\cdot10^{-6})}{(\sqrt{2}\cdot 0.2)^2}=3.6\ N$$

$$F=\sqrt{\left(3.6-\frac{10.8}{\sqrt{2}}\right)^2+\left(3.6-\frac{10.8}{\sqrt{2}}\right)^2}=5.7\ N$$