(initially uncharged) capacitor, $C_2$, the final voltage on each
capacitor is $V_2 = 15\U{V}$. What is the value of $C_2$?
[\emph{Hint:} charge is conserved.]
-\end{problem*}
+\begin{center}
+\begin{asy}
+% C1 C1
+% +--||--+ +--||--+
+% | | | |
+% +--i|--+ +--||--+
+% V C2
+%
+% (a) (b)
+import Circ;
-\empaddtoprelude{
- input makecirc; % circuit drawing functions
- initlatex("");
- pair A, B, Cl, Cr, Dl, Dr, El, Er;
- numeric a;
- a := 3cm;
- A := (-3a/4,0);
- B := (3a/4, 0);
- def drawA =
- % add elements
- centreof.b(A+(1,0), A-(1,0), bat);
- battery.B(c.b, phi.b, "V", "");
- centreof.c(B.B.n, B.B.p, cap);
- capacitor.C(c.c+(0,a), normal, phi.c, "C_1", "");
- % add wiring along the bottom
- wire(B.B.n, A+(a/2,0), nsq);
- wire(B.B.p, A-(a/2,0), nsq);
- % add wiring along the sides and top
- wire(A+(a/2,0), C.C.l, udsq);
- wire(A-(a/2,0), C.C.r, udsq);
- puttext.bot("($a$)", A-(0,24pt));
- Cl = C.C.r;
- Cr = C.C.l;
- enddef;
- def drawB =
- % add elements
- centreof.d(B+(1,0), B-(1,0), cap);
- capacitor.D(c.d, normal, phi.d, "C_2", "");
- centreof.e(C.D.l, C.D.r, cap);
- capacitor.E(c.e+(0,a), normal, phi.e, "C_1", "");
- % add wiring along the bottom
- wire(C.D.l, B+(a/2,0), nsq);
- wire(C.D.r, B-(a/2,0), nsq);
- % add wiring along the sides and top
- wire(B+(a/2,0), C.E.l, udsq);
- wire(B-(a/2,0), C.E.r, udsq);
- puttext.bot("($b$)", B-(0,24pt));
- Dl = C.D.r;
- Dr = C.D.l;
- El = C.E.r;
- Er = C.E.l;
- enddef;
-}
+real u = 1cm;
+real dx = u;
+real dy = 1.5u;
-\begin{nosolution}
-\begin{center}
-\begin{empfile}[5p]
-\begin{emp}(0cm, 0cm)
- drawA;
- drawB;
-\end{emp}
-\end{empfile}
+TwoTerminal bat = battery("$V$");
+TwoTerminal c1 = capacitor("$C_1$", draw=false);
+centerto(bat, c1, dy); c1.draw();
+wire(bat.end, c1.end, rlsq, dx/2);
+wire(bat.beg, c1.beg, rlsq, -dx/2);
+label("(a)", bat.mid + (0, -dy/2));
+
+c1.shift(3*dx); c1.draw();
+TwoTerminal c2 = capacitor("$C_2$", draw=false);
+centerto(c1, c2, -dy); c2.draw();
+wire(c2.end, c1.end, rlsq, dx/2);
+wire(c2.beg, c1.beg, rlsq, -dx/2);
+label("(b)", c2.mid + (0, -dy/2));
+\end{asy}
\end{center}
-\end{nosolution}
+\end{problem*}
\begin{solution}
\begin{center}
-\begin{empfile}[5]
-\begin{emp}(0cm, 0cm)
- drawA;
- drawB;
- puttext.ulft("$Q_{1a}$", Cl);
- puttext.urt("$-Q_{1a}$", Cr);
- puttext.ulft("$Q_{2b}$", Dl);
- puttext.urt("$-Q_{2b}$", Dr);
- puttext.ulft("$Q_{1b}$", El);
- puttext.urt("$-Q_{1b}$", Er);
- labeloffset := 9pt;
- puttext.top("$V$", (Cl+Cr)/2);
- puttext.top("$V_2$", (Dl+Dr)/2);
- puttext.top("$V_2$", (El+Er)/2);
-\end{emp}
-\end{empfile}
+\begin{asy}
+import Circ;
+
+real u = 1cm;
+real dx = u;
+real dy = 1.5u;
+
+TwoTerminal bat = battery("$V$");
+TwoTerminal c1 = capacitor("$C_1$", draw=false);
+centerto(bat, c1, dy); c1.draw();
+wire(bat.end, c1.end, rlsq, dx/2);
+wire(bat.beg, c1.beg, rlsq, -dx/2);
+label("(a)", bat.mid + (0, -dy/2));
+label("$Q_{1a}$", c1.end, NE);
+label("$-Q_{1a}$", c1.beg, NW);
+
+c1.shift(3*dx); c1.draw();
+TwoTerminal c2 = capacitor("$C_2$", draw=false);
+centerto(c1, c2, -dy); c2.draw();
+wire(c2.end, c1.end, rlsq, dx/2);
+wire(c2.beg, c1.beg, rlsq, -dx/2);
+label("(b)", c2.mid + (0, -dy/2));
+label("$Q_{1b}$", c1.end, NE);
+label("$-Q_{1b}$", c1.beg, NW);
+label("$Q_{2b}$", c2.end, SE);
+label("$-Q_{2b}$", c2.beg, SW);
+\end{asy}
\end{center}
Because the voltage drop across $C_1$ in situation $a$ is the same as