\begin{problem*}{5.34}
In a thundercloud, there may be electric charges of $q_t = +40.0\U{C}$
near the top of the cloud and $q_b = -40.0\U{C}$ near the bottom of
the cloud.  These charges are separated by $r = 2.00\U{km}$.  What is
the electric force on the top charge?
\end{problem*} % problem 5.34

\begin{solution}
We'll be using Coulomb's law for the electro-magnetic force (text p. 144):
\begin{equation}
 F_e = k_e \frac{q_1 q_2}{r^2}
\end{equation}
with $k_e = 8.99\E{9}\U{Nm$^2$/C$^2$}$.

So
\begin{equation}
 F_e = 8.99\E{9}\U{Nm$^2$/C$^2$}
	\frac{-40.0\U{C} \cdot 40.0\U{C}}{(2000\U{m})^2}
	= -8.99\E{9}\U{Nm$^2$/C$^2$}
		\left( \frac{40\U{C}}{2000\U{m}} \right)^2
	= \ans{-3.596\E{6}\U{N}}
\end{equation}
where the $-$ sign indicates an attractive force.
\begin{solution}
