2 Two identical conducting small spheres are placed with their centers
3 $r = 0.300\U{m}$ apart. One is given a charge of $q_1 = 12.0\U{nC}$
4 and the other a charge of $q_2 = -18.0\U{nC}$.
5 \Part{a} Find the electric force exerted by one sphere on the other.
6 \Part{b} Next, the spheres are connected by a conducting wire.
7 Find the electric force between the two after they have come to equilibrium.
8 \end{problem*} % problem 19.7
17 real u = 200cm; // Length of 1 m on the page
18 real Fscale = 5e9cm; // Length of 1 N on the page
22 Charge a = aCharge((0,0)*u, q=qa, L="$q_1$");
23 Charge b = aCharge((0.03,0)*u, q=qb, L="$q_2$");
24 Distance r = Distance(a.center, b.center, L="$r$");
25 Vector Fab = CoulombForce(a, b, scale=Fscale, L="$F$");
26 Vector Fba = CoulombForce(b, a, scale=Fscale, L="$F$");
27 Fab.mag /= 10.0; // Not part of Fscale b/c of rounding
38 F = k_e \frac{q_1 q_2}{r^2}
39 = 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}
40 = \ans{-2.16\E{-5}\U{N}}
42 And the force is towards the other sphere for each sphere because
51 real u = 200cm; // Length of 1 m on the page
52 real Fscale = 5e9cm; // Length of 1 N on the page
56 Charge a = aCharge((0,0)*u, q=(qa+qb)/2, L="$Q/2$");
57 Charge b = aCharge((0.03,0)*u, q=(qa+qb)/2, L="$Q/2$");
58 Distance r = Distance(a.center, b.center, L="$r$");
59 Vector Fab = CoulombForce(a, b, scale=Fscale, L="$F$");
60 Vector Fba = CoulombForce(b, a, scale=Fscale, L="$F$");
69 The total charge on the both spheres is $Q = q_1 + q_2 = -6.0\U{nC}$.
70 The spheres are identical, so at equilibrium, there will be $Q/2 =
71 -3.0\U{nC}$ on each sphere. The repulsive (since now they have the
72 same charge sign) force between them is given by
74 F = k_e \frac{(Q/2)^2}{r^2}
75 = 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
76 = \ans{8.99\E{-7}\U{N}}