Question

Three equal charges are placed at the corners of an equilateral triangle 0.50 m on a side. What are the magnitude of the force on each charge if the charges are each -2.3×10^-9 C?

Answer 1

The charges push away. The big components of the repulsion add. The small components (horizontal red lines) cancel.

The triangles formed by the components are 30, 60, 90 right triangles.

So to add the components we want 2F cos 30. This would be the long legs of both force component triangles.

No we need to find what F is. We use Coulomb's law

F = (1/4Pi eo) q1 q2/r^2 = 9E9Nm^2/C^2*(-2.3E-9C)^2/(0.50m)^2 = 1.9E-7N

So, the magnitude of the force on each charge is

F net = 2*1.9E-7N cos 30 = 3.3 x 10 ^-7 N

If you weren't sure whether to use 30 or 60, remember that you wanted the big side of the triangle (the vertical red line in the sketch). Cos(60) = 1/2 and cos(30) = SQRT(3)/2, so cos(30) is what we want