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 × 10^{9}

We will use the formula F_{E} = k (q_{1} q_{2})
/r^{2} . We plug in our known values and we get F_{E} = k (3.0
× 10^{-9}) ^{2}/.01^{2} = 8.1 × 10^{-4} N.
Also, since the charges are of the same sign, there will be a repulsive force.