\begin{problem*}{19.7}
Two identical conducting small spheres are placed with their centers
$r = 0.300\U{m}$ apart.  One is given a charge of $q_1 = 12.0\U{nC}$
and the other a charge of $q_2 = -18.0\U{nC}$.
\Part{a} Find the electric force exerted by one sphere on the other.
\Part{b} Next, the spheres are connected by a conducting wire.
Find the electric force between the two after they have come to equilibrium.
\end{problem*} % problem 19.7

\begin{solution}
\Part{a}
\begin{center}
\begin{asy}
import Mechanics;
import ElectroMag;

real u = 200cm; // Length of 1 m on the page
real Fscale = 5e9cm; // Length of 1 N on the page
real qa = 12e-9;
real qb = -18e-9;

Charge a = aCharge((0,0)*u, q=qa, L="$q_1$");
Charge b = aCharge((0.03,0)*u, q=qb, L="$q_2$");
Distance r = Distance(a.center, b.center, L="$r$");
Vector Fab = CoulombForce(a, b, scale=Fscale, L="$F$");
Vector Fba = CoulombForce(b, a, scale=Fscale, L="$F$");
Fab.mag /= 10.0; // Not part of Fscale b/c of rounding
Fba.mag /= 10.0;

r.draw();
Fab.draw();
Fba.draw();
a.draw();
b.draw();
\end{asy}
\end{center}
\begin{equation}
 F = k_e \frac{q_1 q_2}{r^2}
   = 8.99\E{9}\U{N$\cdot$m$^2$/C$^2$} \frac{12.0\E{-9}\U{C}\cdot(-18.0)\E{-9}\U{C}}{(0.300\U{m})^2}
   = \ans{-2.16\E{-5}\U{N}}
\end{equation}
And the force is towards the other sphere for each sphere because
opposites attract.

\Part{b}
\begin{center}
\begin{asy}
import Mechanics;
import ElectroMag;

real u = 200cm; // Length of 1 m on the page
real Fscale = 5e9cm; // Length of 1 N on the page
real qa = 12e-9;
real qb = -18e-9;

Charge a = aCharge((0,0)*u, q=(qa+qb)/2, L="$Q/2$");
Charge b = aCharge((0.03,0)*u, q=(qa+qb)/2, L="$Q/2$");
Distance r = Distance(a.center, b.center, L="$r$");
Vector Fab = CoulombForce(a, b, scale=Fscale, L="$F$");
Vector Fba = CoulombForce(b, a, scale=Fscale, L="$F$");

r.draw();
Fab.draw();
Fba.draw();
a.draw();
b.draw();
\end{asy}
\end{center}
The total charge on the both spheres is $Q = q_1 + q_2 = -6.0\U{nC}$.
The spheres are identical, so at equilibrium, there will be $Q/2 =
-3.0\U{nC}$ on each sphere.  The repulsive (since now they have the
same charge sign) force between them is given by
\begin{equation}
 F = k_e \frac{(Q/2)^2}{r^2}
   = 8.99\E{9}\U{N$\cdot$m$^2$/C$^2$} \left(\frac{-3.0\E{-9}\U{C}}{0.300\U{m}}\right)^2
   = \ans{8.99\E{-7}\U{N}}
\end{equation}
\end{solution}
