--- /dev/null
+\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}
--- /dev/null
+\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 $0<Q_{1L}=Q_1=-Q_{1R}$ and $0<Q_{2L}=Q_2=-Q_{2R}$.
+Balancing the charge on the wire $d$
+\begin{align}
+ 0 &= Q_{1R} + Q_{2R} + Q_{3R} \\
+ Q_{3R} &= -Q_{1R} - Q_{2R} = Q_1+Q_2 = \ans{120\U{$\mu$C}} > 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}
--- /dev/null
+\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}
--- /dev/null
+\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}
projects / course.git / commitdiff