Question

A charge of -2.99 μC is fixed at the center of a compass. Two additional charges are fixed on the circle of the compass (radius = 0.116 m). The charges on the circle are -4.08 μC at the position due north and +5.18 μC at the position due east. What is the magnitude of the net electrostatic force acting on the charge at the center?

Answer 1

Solution
The formula for force between charged particles is

F = Kq1q2/r^2
The charges are given as -2.99 C, -4.08C and 5.18 C - these are really large charges, problems usually give
charges in the order of μC or nC, so the force is going to be really large.
The charge from due north is negative, and so it the charge in the middle. Therefore the force will be repulsive
and push directly south (We can eliminate the negatives since they are used to find the direction)

F1 = (9*10^9)(2.99)(4.08)/(0.116)^2 = 8.159*10^12 N south

From the other charge, since one is negative and the other is positive, the force will pull to the east

F1 = (9*10^9)(2.99)(5.18)/(0.116)^2 = 10.359*10^12 N east

Net force is found by the Pythagorean theorem

F = √(8.159*10^12)^2+(10.359*10^12)^2 = 1.3184*10^13 N