Add solutions and graphics to Serway and Jewett v8's chapter 26 problems.

[course.git] / latex / problems / Serway_and_Jewett_8 / problem26.57.tex

diff --git a/latex/problems/Serway_and_Jewett_8/problem26.57.tex b/latex/problems/Serway_and_Jewett_8/problem26.57.tex
index b029541b1a9d9832b45a4fc27ff7c7bcbcc11a0b..1af35e0947638d79e5624b287bdf43d17a97d5dd 100644 (file)
--- 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@@ -8,4 +8,40 @@ capacitor after the mica is withdrawn?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}