From c1f2a55d2d987f726319e3155009efc4fd8c450f Mon Sep 17 00:00:00 2001 From: "W. Trevor King" Date: Fri, 27 Apr 2012 15:47:27 -0400 Subject: [PATCH] Add solutions and graphics to Serway and Jewett v8's chapter 26 problems. --- .../Serway_and_Jewett_8/problem26.04.tex | 48 ++++++++ .../Serway_and_Jewett_8/problem26.16.tex | 53 ++++++++ .../Serway_and_Jewett_8/problem26.18.tex | 113 ++++++++++++++++++ .../Serway_and_Jewett_8/problem26.31.tex | 7 ++ .../Serway_and_Jewett_8/problem26.32.tex | 7 ++ .../Serway_and_Jewett_8/problem26.57.tex | 36 ++++++ .../Serway_and_Jewett_8/problem26.67.tex | 45 +++++++ 7 files changed, 309 insertions(+) diff --git a/latex/problems/Serway_and_Jewett_8/problem26.04.tex b/latex/problems/Serway_and_Jewett_8/problem26.04.tex index 1a7f8fd..01caccc 100644 --- a/latex/problems/Serway_and_Jewett_8/problem26.04.tex +++ b/latex/problems/Serway_and_Jewett_8/problem26.04.tex @@ -7,4 +7,52 @@ is the magnitude of the uniform electric field between the plates? \end{problem*} \begin{solution} +\Part{a} +Using Gauss' law, we can find the electric field from an infinite +plate with charge density $\sigma$. +\begin{align} + \Phi_E &= \oint_S \vect{E}\cdot\vect{dA} = 2EA \\ + &= \frac{q_\text{in}}{\kappa\varepsilon_0} + = \frac{\sigma A}{\kappa\varepsilon_0} \\ + E &= \frac{\sigma}{2\kappa\varepsilon_0} \;. +\end{align} + +The capacitance of a ciruit element is defined by $C\equiv q/V$, where +the potential difference $V$ is measured with $+q$ on one side of the +element and $-q$ on the other. The electric field between the plates +will be +\begin{equation} + E = 2\frac{\sigma}{2\kappa\varepsilon_0} + = \frac{q}{\kappa\varepsilon_0 A} \;, +\end{equation} +and the potential difference is +\begin{equation} + V = Ed = \frac{qd}{\kappa\varepsilon_0 A} \;. +\end{equation} + +Now that we have a formula for $V(q)$, we can plug into the capacitor +equation to find the capacitance. +\begin{equation} + C \equiv \frac{q}{V} = \frac{\kappa\varepsilon_0 A}{d} + = \frac{1.00 \cdot 8.85\E{-12}\U{C$^2$/N$\cdot$m$^2$} + \cdot 2.30\E{-4}\U{m$^2$}} + {1.50\E{-3}\U{m}} + = \ans{1.36\E{-12}\U{F}} = \ans{1.36\U{pF}} \;. +\end{equation} +As you'd expect, $C$ increases with increasing $A$, and decreases with +increasing $d$. + +\Part{b} +Just plugging into the capacitance equation, +\begin{equation} + q = CV = 1.36\E{-12}\U{F} \cdot 12.0\U{V} + = \ans{1.63\E{-11}\U{C}} = \ans{16.3\U{pC}} \;. +\end{equation} + +\Part{c} +Plugging into my earlier equation for electric field, +\begin{equation} + E = \frac{V}{d} = \frac{12.0\U{V}}{1.50\E{-3}\U{m}} + = \ans{8.00\U{kV/m}} \;. +\end{equation} \end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem26.16.tex b/latex/problems/Serway_and_Jewett_8/problem26.16.tex index 176b304..84d5e50 100644 --- a/latex/problems/Serway_and_Jewett_8/problem26.16.tex +++ b/latex/problems/Serway_and_Jewett_8/problem26.16.tex @@ -6,4 +6,57 @@ parallel across the battery. \end{problem*} \begin{solution} +\begin{center} +\begin{asy} +import Circ; + +real u = 1cm; + +TwoTerminal C1 = capacitor((0,0), ang=0, "$C_1$", "2.50\U{$\mu$F}"); +TwoTerminal C2 = capacitor(C1.end + (u,0), ang=0, "$C_2$", "6.25\U{$\mu$F}"); +TwoTerminal B = battery("$V$", "6.00\U{V}", draw=false); +B.centerto(C1.beg, C2.end, offset=2u); +B.draw(); +wire(B.beg, C1.beg, rlsq); +wire(B.end, C2.end, rlsq); +wire(C1.end, C2.beg); + +label("\Part{a}", B.mid + (0, -5u)); + +B.shift((4u, 0)); +B.draw(); +centerto(B, C1, offset=-2u); +C1.draw(); +centerto(B, C2, offset=-4u); +C2.draw(); +wire(B.beg, C1.beg, rlsq, dist=-u/2); +wire(B.beg, C2.beg, rlsq, dist=-u/2); +wire(B.end, C1.end, rlsq, dist=u/2); +wire(B.end, C2.end, rlsq, dist=u/2); + +label("\Part{b}", B.mid + (0, -5u)); +\end{asy} +\end{center} + +\Part{a} +The net capacitance of the two capacitors in series is +\begin{equation} + C = \p({\frac{1}{C_1} + \frac{1}{C_2}})^{-1} + = 1.79\U{$\mu$F} \;. +\end{equation} +The equivalent capacitor has the total battery voltage across itself, +so it carries a charge of +\begin{equation} + Q = CV = 1.79\U{$\mu$F} \cdot 6.00\U{V} = \ans{10.7\U{$\mu$C}} \;. +\end{equation} +Because $C_1$ and $C_2$ are in series, they each have the same $Q$ as +the equivalent capacitor. + +\Part{b} +When the capacitors are in parallel, they each have the total battery +voltage, so +\begin{align} + Q_1 &= C_1 V = 2.50\U{$\mu$F} \cdot 6.00\U{V} = \ans{15.0\U{$\mu$C}} \\ + Q_2 &= C_2 V = 6.25\U{$\mu$F} \cdot 6.00\U{V} = \ans{37.5\U{$\mu$C}} \;. +\end{align} \end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem26.18.tex b/latex/problems/Serway_and_Jewett_8/problem26.18.tex index 354838f..d45c384 100644 --- a/latex/problems/Serway_and_Jewett_8/problem26.18.tex +++ b/latex/problems/Serway_and_Jewett_8/problem26.18.tex @@ -2,13 +2,126 @@ Find \Part{a} the equivalent capacitance of the capacitors in Figure~P26.18, \Part{b} the charge on each capacitor, and \Part{c} the potential difference across each capacitor. +\begin{center} % +-6uF-+ % +-+ +-+ % | +-2uF-+ | % 8uF 8uF % | 9V | % +---|i----+ +\begin{asy} +import Circ; + +real u = 1cm; + +TwoTerminal B = battery((0,0), ang=180, "$V$", "9.00\U{V}"); +TwoTerminal C1 = capacitor("$C_1$", "8.00\U{$\mu$F}", draw=false); +TwoTerminal C2 = capacitor("$C_2$", "6.00\U{$\mu$F}", draw=false); +TwoTerminal C3 = capacitor("$C_3$", "2.00\U{$\mu$F}", draw=false); +TwoTerminal C4 = capacitor("$C_4$", "8.00\U{$\mu$F}", draw=false); + +centerto(B, C3, offset=3u, reverse=true); C3.draw(); +centerto(C3, C2, offset=2u); C2.draw(); +pair a = (B.end.x, (C2.beg.y + C3.beg.y)/2); +pair b = (B.beg.x, a.y); +C1.centerto(B.end, a, offset=2u); C1.draw(); +C4.centerto(B.beg, b, offset=-2u); C4.draw(); + +a = (a.x - u, a.y); // shift to split x difference between C1.end and B.end +b = (b.x + u, b.y); + +wire(B.end, C1.beg, rlsq); +wire(C1.end, a, udsq); +wire(a, C2.beg, udsq); +wire(a, C3.beg, udsq); +dot(a); +wire(C2.end, b, rlsq); +wire(C3.beg, b, rlsq); +wire(b, C4.end, rlsq); +dot(b); +wire(C4.beg, B.beg, udsq); +\end{asy} +\end{center} \end{problem*} \begin{solution} +\Part{a} +Bundling down equivalent capacitances, +\begin{center} +\begin{asy} +import Circ; + +real u = 1cm; + +TwoTerminal B = battery((0,0), ang=180, "$V$"); +TwoTerminal C1 = capacitor("$C_1$", draw=false); +TwoTerminal C23 = capacitor("$C_{23}$", draw=false); +TwoTerminal C4 = capacitor("$C_4$", draw=false); +centerto(B, C23, offset=2u, reverse=true); C23.draw(); +pair a = (B.end.x, C23.beg.y); +pair b = (B.beg.x, a.y); +C1.centerto(B.end, a, offset=u); C1.draw(); +C4.centerto(B.beg, b, offset=-u); C4.draw(); +wire(B.end, C1.beg, rlsq); +wire(C1.end, C23.beg, udsq); +wire(C23.end, C4.end, rlsq); +wire(C4.beg, B.beg, udsq); + +B.shift((6u, 0)); +B.draw(); +TwoTerminal C1234 = capacitor("$C_{1234}$", draw=false); +centerto(B, C1234, offset=2u, reverse=true); C1234.draw(); +wire(B.end, C1234.beg, rlsq, dist=-u/2); +wire(B.beg, C1234.end, rlsq, dist=+u/2); + +pair c = ((C4.mid.x + B.end.x - u/2)/2, C4.mid.y); +draw((c-(u,0)) -- (c+(u,0)), kirchhoff_pen, Arrows); +\end{asy} +\end{center} +$C_{23}$ comes from $C_2$ and $C_3$ in parallel, so the equivalent +capacitance is +\begin{equation} + C_{23} = C_2 + C_3 = (6.00 + 2.00)\U{$\mu$F} = 8.00\U{$\mu$F} \;. +\end{equation} +$C_{1234}$ comes from $C_1$, $C_{23}$, and $C_4$ in series, so the +equivalent capacitance is +\begin{equation} + C_{1234} = \p({\frac{1}{C_1} + \frac{1}{C_{23}} + \frac{1}{C_4}})^{-1} + = \frac{8.00\U{$\mu$F}}{3} = \ans{2.67\U{$\mu$F}} \;. +\end{equation} + +\Part{b} +The equivalent capacitor has the total battery voltage across itself, +so it carries a charge of +\begin{equation} + Q_{1234} = C_{1234} V = (2.67\U{$\mu$F}) \cdot (9.00\U{V}) + = 24.0\U{$\mu$C} \;. +\end{equation} +Because $C_1$, $C_{23}$, and $C_3$ are in series, they each have the +same $Q$ as the equivalent capacitor. +\begin{equation} + Q_1 = Q_{23} = Q_3 = Q_{1234} = \ans{24.0\U{$\mu$C}} \;. +\end{equation} +The voltage across $C_{23}$ is therefore +\begin{equation} + V_{23} = \frac{Q_{23}}{C_{23}} = \frac{24.0\U{$\mu$C}}{8.00\U{$\mu$F}} + = 3.00\U{V} \;. +\end{equation} +$C_2$ and $C_3$ are in parallel, so they each have the total $V_{23}$ +voltage. +\begin{align} + Q_2 &= C_2 V_{23} = 6.00\U{$\mu$F} \cdot 3.00\U{V} = \ans{18.0\U{$\mu$C}} \\ + Q_3 &= C_3 V_{23} = 2.00\U{$\mu$F} \cdot 3.00\U{V} = \ans{6.00\U{$\mu$C}} \;. +\end{align} +Note that $Q_2+Q_3=Q_{23}$, which must be true because charge is conserved. + +\Part{c} +During \Part{b}, we already found $\ans{V_{23}=V_2=V_3=3.00\U{V}}$. +Using the charges on $C_1$ and $C_4$, we find +\begin{align} + V_1 &= \frac{Q_1}{C_1} = \frac{24.0\U{$\mu$C}}{8.00\U{$\mu$F}} + = \ans{3.00\U{V}} \\ + V_4 &= \frac{Q_4}{C_4} = \frac{24.0\U{$\mu$C}}{8.00\U{$\mu$F}} + = \ans{3.00\U{V}} \;. +\end{align} \end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem26.31.tex b/latex/problems/Serway_and_Jewett_8/problem26.31.tex index c7ff687..fe10256 100644 --- a/latex/problems/Serway_and_Jewett_8/problem26.31.tex +++ b/latex/problems/Serway_and_Jewett_8/problem26.31.tex @@ -5,4 +5,11 @@ is stored in the capacitor? \end{problem*} \begin{solution} +Just plug into the capacitor energy formula, +\begin{equation} + U = \frac{1}{2}CV^2 + = \frac{1}{2}QV + = \frac{1}{2} \cdot (54.0\E{-6}\U{C}) \cdot (12.0\U{V}) + = \ans{324\U{$\mu$J}} \;. +\end{equation} \end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem26.32.tex b/latex/problems/Serway_and_Jewett_8/problem26.32.tex index 9e27976..0009032 100644 --- a/latex/problems/Serway_and_Jewett_8/problem26.32.tex +++ b/latex/problems/Serway_and_Jewett_8/problem26.32.tex @@ -14,4 +14,11 @@ potential difference must it be charged? \end{problem*} \begin{solution} +Using the capacitor energy formula, +\begin{align} + U &= \frac{1}{2}CV^2 \\ + V &= \sqrt{\frac{2U}{C}} + = \sqrt{\frac{2 \cdot 300\U{J}}{30.0\E{-6}\U{F}}} + = \ans{4.47\U{kV}} \;. +\end{align} \end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem26.57.tex b/latex/problems/Serway_and_Jewett_8/problem26.57.tex index b029541..1af35e0 100644 --- a/latex/problems/Serway_and_Jewett_8/problem26.57.tex +++ b/latex/problems/Serway_and_Jewett_8/problem26.57.tex @@ -8,4 +8,40 @@ capacitor after the mica is withdrawn? \end{problem*} \begin{solution} +\Part{a} +The capacitance of a parallel plate capacitor is +\begin{equation} + C = \frac{\kappa\varepsilon_0 A}{d} \;. +\end{equation} +For other geometries, the constants change, but the capacitance is +always proportional to $\kappa$, the dielectric constant of the gap. +After removing the mica sheet, the capacitor will have a new +capacitance of +\begin{equation} + C' = \frac{\kappa'}{\kappa} C = \frac{1.00}{5.00} \cdot 2.00\U{nF} + = 400\U{pF} \;, +\end{equation} +where $\kappa' \approx 1.00$ is the dielectric constant for air (which +replaces the mica). + +During the withdrawing process, the capacitance changes, and the +voltage between the plates changes, but because the capacitor is +isolated, the charge does not change. We can use the initial voltage +to find that charge. +\begin{equation} + Q = C \Delta V_i = (2.00\E{-9}\U{F}) \cdot (100\U{V}) = 200\U{nC} +\end{equation} + +Conserving energy during the withdrawing process, +\begin{align} + U_i + W &= U_f \\ + \frac{Q^2}{2C} + W &= \frac{Q^2}{2C'} \\ + W &= \frac{Q^2}{2} \cdot \p({\frac{1}{C'} - \frac{1}{C}}) + = \frac{Q^2}{2} \cdot \p({\frac{\kappa}{\kappa' C} - \frac{1}{C}}) + = \frac{Q^2}{2\kappa' C} \cdot (\kappa - \kappa') + = \frac{C\Delta V_i^2}{2\kappa'} \cdot (\kappa - \kappa') \\ + &= \frac{(2.00\E{-9}\U{F}) \cdot (100\U{V})^2}{2 \cdot 1.00} + \cdot(5.00 - 1.00) + = \ans{40.0\U{$\mu$J}} \;. +\end{align} \end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem26.67.tex b/latex/problems/Serway_and_Jewett_8/problem26.67.tex index df197ec..8437dee 100644 --- a/latex/problems/Serway_and_Jewett_8/problem26.67.tex +++ b/latex/problems/Serway_and_Jewett_8/problem26.67.tex @@ -7,6 +7,51 @@ positive plate. Calculate the resulting charge on each capacitor. \end{problem*} \begin{solution} +\begin{center} +\begin{asy} +import Circ; + +real u = 1cm; +real s = 6u; + +TwoTerminal B = battery("$V$"); +TwoTerminal C1 = capacitor("$C_1$", draw=false); +TwoTerminal C2 = capacitor("$C_2$", draw=false); +centerto(B, C1, offset=2u); C1.draw(); +centerto(C1, C2, offset=2u); C2.draw(); +wire(B.end, C1.end, rlsq, dist=u/2); +wire(B.end, C2.end, rlsq, dist=u/2); +wire(B.beg, C1.beg, rlsq, dist=-u/2); +wire(B.beg, C2.beg, rlsq, dist=-u/2); +label("$+Q_1$", C1.end, align=dir(70)); +label("$-Q_1$", C1.beg, align=dir(110)); +label("$+Q_2$", C2.end, align=dir(70)); +label("$-Q_2$", C2.beg, align=dir(110)); + +pair c = C1.mid + (s/2, 0); +draw((c-(u,0)) -- (c+(u,0)), kirchhoff_pen, Arrow); + +C1.centerto(B.mid, C2.mid, -s); C1.draw(); +centerto(C1, C2, offset=u, reverse=true); C2.draw(); +label("$+Q_1$", C1.end, align=NW); +label("$-Q_1$", C1.beg, align=SW); +label("$+Q_2$", C2.end, align=SE); +label("$-Q_2$", C2.beg, align=NE); + +c = C2.mid + (s/2, 0); +draw((c-(u,0)) -- (c+(u,0)), kirchhoff_pen, Arrow); + +C1.shift(s+u); C1.draw(); +C2.shift(s+u); C2.draw(); +wire(C1.end, C2.beg); +wire(C2.end, C1.beg); +label("$+Q_1'$", C1.end, align=NW); +label("$-Q_1'$", C1.beg, align=SW); +label("$-Q_2'$", C2.end, align=SE); +label("$+Q_2'$", C2.beg, align=NE); +\end{asy} +\end{center} + After charging, the charges on the capacitors are \begin{align} Q_1 &= C_1 V = 1.50\U{mC} \\ -- 2.26.2