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.]
11 input makecirc; % circuit drawing functions
13 pair A, B, Cl, Cr, Dl, Dr, El, Er;
20 centreof.b(A+(1,0), A-(1,0), bat);
21 battery.B(c.b, phi.b, "V", "");
22 centreof.c(B.B.n, B.B.p, cap);
23 capacitor.C(c.c+(0,a), normal, phi.c, "C_1", "");
24 % add wiring along the bottom
25 wire(B.B.n, A+(a/2,0), nsq);
26 wire(B.B.p, A-(a/2,0), nsq);
27 % add wiring along the sides and top
28 wire(A+(a/2,0), C.C.l, udsq);
29 wire(A-(a/2,0), C.C.r, udsq);
30 puttext.bot("($a$)", A-(0,24pt));
36 centreof.d(B+(1,0), B-(1,0), cap);
37 capacitor.D(c.d, normal, phi.d, "C_2", "");
38 centreof.e(C.D.l, C.D.r, cap);
39 capacitor.E(c.e+(0,a), normal, phi.e, "C_1", "");
40 % add wiring along the bottom
41 wire(C.D.l, B+(a/2,0), nsq);
42 wire(C.D.r, B-(a/2,0), nsq);
43 % add wiring along the sides and top
44 wire(B+(a/2,0), C.E.l, udsq);
45 wire(B-(a/2,0), C.E.r, udsq);
46 puttext.bot("($b$)", B-(0,24pt));
71 puttext.ulft("$Q_{1a}$", Cl);
72 puttext.urt("$-Q_{1a}$", Cr);
73 puttext.ulft("$Q_{2b}$", Dl);
74 puttext.urt("$-Q_{2b}$", Dr);
75 puttext.ulft("$Q_{1b}$", El);
76 puttext.urt("$-Q_{1b}$", Er);
78 puttext.top("$V$", (Cl+Cr)/2);
79 puttext.top("$V_2$", (Dl+Dr)/2);
80 puttext.top("$V_2$", (El+Er)/2);
85 Because the voltage drop across $C_1$ in situation $a$ is the same as
86 the voltage drop across the battery ($V$), we have
90 When we connect $C_2$ in situation $b$, this charge redistributes
91 between $C_1$ and $C_2$. Because charge is conserved, we know
93 Q_{1a} = Q_{1b} + Q_{2b}
95 We also know that the voltage drop across both capacitors in situation
96 $b$ must be equal ($\text{both} = V_2$), so
101 Plugging each of these formulas for charge ($Q_{1a}$, $Q_{1b}$, and
102 $Q_{2b}$) into the charge conservation formula yeilds
104 C_1 V &= C_1 V_2 + C_2 V_2 \\
105 C_1 (V - V_2) &= C_2 V_2 \\
106 V_2 C_2 &= C_1 (V - V_2) \\
107 C_2 &= C_1 \p({\frac{V}{V_2} - \frac{V_2}{V_2}}) \\
108 C_2 &= C_1 \p({\frac{V}{V_2} - 1}) \\
109 C_2 &= 7.7\U{$\mu$F} \cdot \p({\frac{125\U{V}}{15\U{V}} - 1})