Two identical tiny metal balls in air have charges $q_{1}$ and $q_{2}$. The repulsive force one exerts on the other when they are $20 \mathrm{~cm}$ apart is $1.35 \times 10^{-4} \mathrm{~N}$. After the balls are touched together and then separated once again to $20 \mathrm{~cm}$, the repulsive force is found to be $1.406 \times 10^{-4} \mathrm{~N}$. Find $q_{1}$ and $q_{2}$.
Because the force is one of repulsion, $q_{1}$ and $q_{2}$ have the same sign. After the balls are touched, they share charge equally, so each has a charge $\frac{1}{2}\left(q_{1}+q_{2}\right)$. Writing Coulomb's Law for the two situations described, we have
and
$$\begin{array}{c}
0.000135 \mathrm{~N}=k_{0} \frac{q_{1} q_{2}}{0.040 \mathrm{~m}^{2}} \\
0.0001406 \mathrm{~N}=k_{0} \frac{\left[\frac{1}{2}\left(q_{1}+q_{2}\right)\right]^{2}}{0.040 \mathrm{~m}^{2}}
\end{array}$$
After substitution for $k_{0}$, these equations reduce to
$$q_{1} q_{2}=6.00 \times 10^{-16} \mathrm{C}^{2} \quad \text { and } \quad q_{1}+q_{2}=5.00 \times 10^{-8} \mathrm{C}$$
Solving these equations simultaneously leads to $q_{1}=20 \mathrm{n} \mathrm{C}$ and $q_{2}=30 \mathrm{nC}$ (or vice versa). Alternatively, both charges could have been negative.