Problem 19.1

19.1. a)
The force between the two charges is an attractive force. Its magnitude is given by
$F = \frac{kq_{1}q_{2}}{r^{2}}$
$F = (9 \times 10^{9} N m^{2}/C^{2})\frac{(12 \times 10^{-9}C)(18 \times 10^{-9}C)}{(.3m)^{2}}$
$F = 2.16 \times 10^{-5} N$

b)
The net charge of -6 nC will be distributed evenly over the two spheres, leaving a charge of -3 nC on each one. Therefore, the force now will be one of repulsion, and it will have a magnitude of $F = \frac{kq_{1}q_{2}}{r^{2}}$
$F = (9 \times 10^{9} N m^{2}/C^{2})\frac{(-3 nC)(-3 nC)}{(.3m)^{2}}$
$F = 8.99 \times 10^{-7} N$