From: W. Trevor King Date: Wed, 15 Jul 2009 13:29:32 +0000 (-0400) Subject: Rec4 solutions added with ascii drawings. X-Git-Url: http://git.tremily.us/?a=commitdiff_plain;h=d2685e457a47c6552743afda298d77e8d1919846;p=course.git Rec4 solutions added with ascii drawings. --- diff --git a/latex/problems/Young_and_Freedman_12/problem24.14.tex b/latex/problems/Young_and_Freedman_12/problem24.14.tex new file mode 100644 index 0000000..6d490d6 --- /dev/null +++ b/latex/problems/Young_and_Freedman_12/problem24.14.tex @@ -0,0 +1,42 @@ +\begin{problem*}{14} +For the system of capacitors shown in Fig.~24.24, find the equivalent +capacitance \Part{a} between $b$ and $c$, and \Part{b} between $a$ and +$c$. +\begin{center} +\begin{verbatim} + a + | + === 15pF + | + b + | + +--+--+ + |9.0pF| + === === 11pF + | | + +--+--+ + | + c +\end{verbatim} +\end{center} +\end{problem*} + +\begin{solution} +\Part{a} +The $9$ and $11\U{pF}$ capacitors are in parallel, so the equivalent +capacitance between $b$ and $c$ is +\begin{equation} + C_{bc} = C_L + C_R = (9.0+11)\U{pF} = \ans{20\U{pF}} +\end{equation} + +\Part{b} +The $15\U{pF}$ capacitor is in series with the equivalent capacitance +$C_{bc}$, so the equivalent capacitance between $a$ and $c$ is +\begin{equation} + C_{ac} = \p({\frac{1}{C_T}+\frac{1}{C_{bc}}})^{-1} + = \p({\frac{1}{15}+\frac{1}{20}})^{-1}\U{pF} + = \p({\frac{4}{12}+\frac{3}{12}})^{-1}\cdot5\U{pF} + = \frac{60}{7}\U{pF} + = \ans{8.6\U{pF}} +\end{equation} +\end{solution} diff --git a/latex/problems/Young_and_Freedman_12/problem24.18.tex b/latex/problems/Young_and_Freedman_12/problem24.18.tex new file mode 100644 index 0000000..0731fd7 --- /dev/null +++ b/latex/problems/Young_and_Freedman_12/problem24.18.tex @@ -0,0 +1,59 @@ +\begin{problem*}{18} +In Fig.~24.26, $C_1=6.00\U{$\mu$F}$, $C_2=3.00\U{$\mu$F}$, and +$C_3=5.00\U{$\mu$F}$. The capacitor network is connected to an +applied potential $V_{ab}$. After the charges on the capacitors have +reached their final values, the charge on $C_2$ is +$40.0\U{$\mu$C}$. \Part{a} What are the charges on capacitors $C_1$ +and $C_3$? \Part{b} What is the applied voltage $V_{ab}$? +\begin{center} +\begin{verbatim} + 1 + +-||-+ + | | + a--+ +-+ + | 2 | | + +-||-+ | + d + 3 | + b----||---+ +\end{verbatim} +\end{center} +\end{problem*} + +\begin{solution} +\Part{a} +Because $C_1$ and $C_2$ are in parallel, they must have the same +voltage differences. Using this and the definition of capacitance +$Q=CV$, we have +\begin{align} + \frac{Q_1}{C_1} &= V_1 = V_2 = \frac{Q_2}{C_2} \\ + Q_1 &= \frac{C_1}{C_2}\;Q_2 + = \frac{6.00}{3.00}\;40.0\U{$\mu$C} + = \ans{80.0\U{$\mu$C}} \;. +\end{align} + +Since the wire $d$ was initially uncharged, all the charge on $C_1$ +and $C_2$ had to come from $C_3$. Assuming $V_a>V_b$ (if $V_b>V_a$, +just flip all the charge signs in the following solution), the left +hand sides of $C_1$ and $C_2$ will be at a higher potential than the +right, so $0 0 \\ + Q_{3L} &= -Q_{3R} = -120\U{$\mu$C} < 0 \;. +\end{align} + +\Part{b} +\begin{equation} + V_{ab} = V_{ad} + V_{db} +\end{equation} +We can find $V_{db}$ easily enough from $Q_3=C_3V_3$. Because $C_1$ +and $C_2$ are in parallel, $V_1=V_2=V_{ad}$ (as we pointed out +in \Part{a}), so +\begin{equation} + V_{ab} = \frac{Q_3}{C_3} + \frac{Q_2}{C_2} + = \frac{120\U{$\mu$C}}{5.00\U{$\mu$F}} + \frac{40.0\U{$\mu$C}}{3.00\U{$\mu$F}} + = 24.0\U{V} + 13.3\U{V} = \ans{37.3\U{V}} +\end{equation} +\end{solution} diff --git a/latex/problems/Young_and_Freedman_12/problem24.32.tex b/latex/problems/Young_and_Freedman_12/problem24.32.tex new file mode 100644 index 0000000..3629af5 --- /dev/null +++ b/latex/problems/Young_and_Freedman_12/problem24.32.tex @@ -0,0 +1,73 @@ +\begin{problem*}{32} +For the capacitor network shown in Fig.~24.28, the potential +difference across $ab$ is $36\U{V}$. Find \Part{a} the total charge +stored in this network; \Part{b} the charge on each +capacitor; \Part{c} the total energy stored in the network; \Part{d} +the energy stored in each capacitor; \Part{e} the potential +differences across each capacitor. +\begin{center} +\begin{verbatim} +a--||--||--b + 150 120 + nF nF +\end{verbatim} +\end{center} +\end{problem*} + +\begin{solution} +\Part{a} +The equivalent capacitance is +\begin{equation} + C_{ab} = \p({\frac{1}{150}+\frac{1}{120}})^{-1}\U{nF} = 66.7\U{nF} \;. +\end{equation} +The total charge on the capacitors is then +\begin{equation} + Q = CV = 66.7\U{nF}\cdot36\U{V} = \ans{2.40\U{$\mu$C}} \;. +\end{equation} + +\Part{b} +Each capacitor must also have $Q=\ans{2.40\U{$\mu$C}}$, since the wire +connecting the capacitors must remain uncharged as a whole. + +\Part{c} +The total capacitative energy is +\begin{equation} + E_C = \frac{1}{2} C V^2 + = \frac{1}{2} \cdot 66.7\U{nF} \cdot (36\U{V})^2 + = \ans{43.2\U{$\mu$J}} +\end{equation} + +\Part{d} +We don't know the voltage of each capacitor seperately, but we could +find it using $Q=CV$, since we know $Q$ and $C$. We could also just +plug the formula in directly for $V$ in our energy formula +\begin{equation} + E_C = \frac{1}{2} C V^2 = \frac{1}{2} C \p({\frac{Q}{C}})^2 + = \frac{1}{2} \frac{Q^2}{C} +\end{equation} +The energy stored in each capacitor is then +\begin{align} + E_{CL} &= \frac{1}{2} \frac{Q_L^2}{C_L} + = \frac{1}{2} \frac{(2.40\U{$\mu$C})^2}{150\U{nF}} + = \ans{19.2\U{$\mu$J}} \\ + E_{RL} &= \frac{1}{2} \frac{Q_R^2}{C_R} + = \frac{1}{2} \frac{(2.40\U{$\mu$C})^2}{120\U{nF}} + = \ans{24.0\U{$\mu$J}} \;. +\end{align} +Note that $E_{CL}+E_{CR}=43.2\U{$\mu$J}=E_C$, as it should be, since +the equivalent capacitor situation is \emph{equivalent} +(i.e. interchangable) with the two-capacitor situation. + +\Part{e} +Here we find the potential differences the way we mentioned +in \Part{d}. +\begin{align} + V_L &= \frac{Q_L}{C_L} + = \frac{2.40\U{$\mu$C}}{150\U{nF}} + = \ans{16.0\U{V}} \\ + V_R &= \frac{Q_R}{C_R} + = \frac{2.40\U{$\mu$C}}{120\U{nF}} + = \ans{20.0\U{V}} \;. +\end{align} +Note that $V_L+V_R=36.0\U{V}=V_{ab}$. +\end{solution} diff --git a/latex/problems/Young_and_Freedman_12/problem24.52.tex b/latex/problems/Young_and_Freedman_12/problem24.52.tex new file mode 100644 index 0000000..1ab8ac0 --- /dev/null +++ b/latex/problems/Young_and_Freedman_12/problem24.52.tex @@ -0,0 +1,57 @@ +\begin{problem*}{52} +Cell membranes (the walled enclosure around a cell) are typically +about $7.5\U{nm}$ thick. They are partially permeable to allow +charged material to pass in and out, as needed. Equal but opposite +charge densities build up on the inside and outside faces of such a +membrane,and these charges prevent additional charges from passing +throughthe cell wall. We can model a cell membrane as a +parallel-plate capacitor, with the membrane itself containing proteins +embedded in an organic material to give the membrane a dielectric +constant of about 10. (See Fig.~24.30.) \Part{a} What is the +capacitance per square centimeter of such a cell wall? \Part{b} In +its normal resting state, a cell has a potential difference of +$85\U{mV}$ across its membrane. What is the electric field inside +this membrane? +\begin{center} +\begin{verbatim} + Outside axon +++++++++++++++++++++++++ +________________________ + Axon membrane |- 7.5nm +________________________ +------------------------ + Inside Axon +\end{verbatim} +\end{center} +\end{problem*} + +\begin{solution} +\Part{a} +For parallel-plate capacitors (from Gauss' law) +\begin{equation} + C = \frac{Q}{V} = \varepsilon \frac{A}{d} = k\varepsilon_0 \frac{A}{d} \;. +\end{equation} +So the capacitance per square centimeter will be +\begin{equation} + \frac{C}{A} = \frac{k\varepsilon}{d} + = \frac{10\cdot8.85\E{-12}\U{F/m}}{7.5\E{-9}\U{m}} + = 1.2\E{-2}\U{F/m$^2$} \cdot \p({\frac{1\U{m}}{100\U{cm}}})^2 + = \ans{1.2\U{$\mu$F/cm$^2$}} +\end{equation} + +\Part{b} +Since the electric field generated by parallel plates is constant and +perpendicular to the plates, we can easily integrate across the +membrane from the bottom to the top +\begin{equation} + \Delta V = \int_b^t \vect{E}\cdot\vect{dx} + = \vect{E}\cdot \int_b^t \vect{dx} = \vect{E}\cdot\vect{d} \;. +\end{equation} +Notice that for a perpendicular path (cutting straight across the +membrane), \vect{E} and \vect{d} are in the same direction, so $\Delta +V=\vect{E}\cdot\vect{d}=Ed$. Flipping this around to give $E$, we +have +\begin{equation} + E = \frac{V}{d} = \frac{85\U{mV}}{7.5\U{nm}} = \ans{11\U{MV/m}} +\end{equation} +\end{solution}