Practice!

Sample Problem

If two point charges both of charge 3.0 nC are placed 0.01 meter apart, what is the force between them? k = 9.0 × 109

Solution

We will use the formula FE = k (q1 q2) /r2 . We plug in our known values and we get FE = k (3.0 × 10-9) 2/.012 = 8.1 × 10-4 N. Also, since the charges are of the same sign, there will be a repulsive force.