Question

The drawings show three charges that have the same magnitude but may have different signs. In all cases the distance d between the charges is the same. The magnitude of the charges is |q|=6.6 μC, and the distance between them is d=6.8 mm. Determine the magnitude of the net force on charge 2 for each of the three drawings.

Answer 1

q1 = q3 = +6.6*10^-6 C

d = 6.8 mm = 6.8*10^-3 m

q2 = -6.6*10^-6 C

Force on q2 due to q1 and q3 are same in magnitude but they are at right angle(θ = 90) to each other.

F21 = F23 = Kq^2/d^2

                 = [9*10^9 * (6.6*10^-6)^2]/(6.8*10^-3)^2

                 = 8478.37 N

Resultant force F = Sqrt[(F21)^2 + (F23)^2 + 2 F21 F23 cos90]

                           = (√2)(F21)

                           = 11988.41 N