1 \begin{problem*}{17.40} % capacitor circuits
2 A $C_1 = 7.7\U{$\mu$F}$ capacitor is charged by a $V = 125\U{V}$
3 battery (Fig. 17-29a) and then is disconnected from the battery. When
4 this capacitor ($C_1$) is then connected (Fig. 17-29b) to a second
5 (initially uncharged) capacitor, $C_2$, the final voltage on each
6 capacitor is $V_2 = 15\U{V}$. What is the value of $C_2$?
7 [\emph{Hint:} charge is conserved.]
23 MultiTerminal bat = battery("$V$");
24 MultiTerminal c1 = capacitor("$C_1$", draw=false);
25 two_terminal_centerto(bat, c1, dy); c1.draw();
26 wire(bat.terminal[1], c1.terminal[1], rlsq, dx/2);
27 wire(bat.terminal[0], c1.terminal[0], rlsq, -dx/2);
28 label("(a)", bat.center + (0, -dy/2));
30 c1.shift(3*dx); c1.draw();
31 MultiTerminal c2 = capacitor("$C_2$", draw=false);
32 two_terminal_centerto(c1, c2, -dy); c2.draw();
33 wire(c2.terminal[1], c1.terminal[1], rlsq, dx/2);
34 wire(c2.terminal[0], c1.terminal[0], rlsq, -dx/2);
35 label("(b)", c2.center + (0, -dy/2));
49 MultiTerminal bat = battery("$V$");
50 MultiTerminal c1 = capacitor("$C_1$", draw=false);
51 two_terminal_centerto(bat, c1, dy); c1.draw();
52 wire(bat.terminal[1], c1.terminal[1], rlsq, dx/2);
53 wire(bat.terminal[0], c1.terminal[0], rlsq, -dx/2);
54 label("(a)", bat.center + (0, -dy/2));
55 label("$Q_{1a}$", c1.terminal[1], NE);
56 label("$-Q_{1a}$", c1.terminal[0], NW);
58 c1.shift(3*dx); c1.draw();
59 MultiTerminal c2 = capacitor("$C_2$", draw=false);
60 two_terminal_centerto(c1, c2, -dy); c2.draw();
61 wire(c2.terminal[1], c1.terminal[1], rlsq, dx/2);
62 wire(c2.terminal[0], c1.terminal[0], rlsq, -dx/2);
63 label("(b)", c2.center + (0, -dy/2));
64 label("$Q_{1b}$", c1.terminal[1], NE);
65 label("$-Q_{1b}$", c1.terminal[0], NW);
66 label("$Q_{2b}$", c2.terminal[1], SE);
67 label("$-Q_{2b}$", c2.terminal[0], SW);
71 Because the voltage drop across $C_1$ in situation $a$ is the same as
72 the voltage drop across the battery ($V$), we have
76 When we connect $C_2$ in situation $b$, this charge redistributes
77 between $C_1$ and $C_2$. Because charge is conserved, we know
79 Q_{1a} = Q_{1b} + Q_{2b}
81 We also know that the voltage drop across both capacitors in situation
82 $b$ must be equal ($\text{both} = V_2$), so
87 Plugging each of these formulas for charge ($Q_{1a}$, $Q_{1b}$, and
88 $Q_{2b}$) into the charge conservation formula yeilds
90 C_1 V &= C_1 V_2 + C_2 V_2 \\
91 C_1 (V - V_2) &= C_2 V_2 \\
92 V_2 C_2 &= C_1 (V - V_2) \\
93 C_2 &= C_1 \p({\frac{V}{V_2} - \frac{V_2}{V_2}}) \\
94 C_2 &= C_1 \p({\frac{V}{V_2} - 1}) \\
95 C_2 &= 7.7\U{$\mu$F} \cdot \p({\frac{125\U{V}}{15\U{V}} - 1})