From f5cd4b106eedf9240af3af1359f6d23a63d51fa1 Mon Sep 17 00:00:00 2001
From: "W. Trevor King" <wking@drexel.edu>
Date: Tue, 7 Jul 2009 21:52:57 -0400
Subject: [PATCH] Added recitation week 3 recitation solutions

---
 .../Young_and_Freedman_12/problem23.21.tex    | 33 ++++++++++
 .../Young_and_Freedman_12/problem23.22.tex    | 60 +++++++++++++++++
 .../Young_and_Freedman_12/problem23.25.tex    | 65 +++++++++++++++++++
 .../Young_and_Freedman_12/problem23.27.tex    | 18 ++++-
 .../Young_and_Freedman_12/problem23.58.tex    | 28 ++++++++
 .../Young_and_Freedman_12/problem23.60.tex    | 22 +++++++
 .../Young_and_Freedman_12/problem23.63.tex    | 54 +++++++++++++++
 7 files changed, 279 insertions(+), 1 deletion(-)

diff --git a/latex/problems/Young_and_Freedman_12/problem23.21.tex b/latex/problems/Young_and_Freedman_12/problem23.21.tex
index ca5cb0d..63b5dd5 100644
--- a/latex/problems/Young_and_Freedman_12/problem23.21.tex
+++ b/latex/problems/Young_and_Freedman_12/problem23.21.tex
@@ -42,4 +42,37 @@ qO.draw(); qT.draw();
 \end{problem*}
 
 \begin{solution}
+\Part{a}
+One of the nice features of electric potential is that it is a scalar,
+so it's a lot easier to sum up then the vector electric field.
+\begin{equation}
+  V_A = V_{1A} + V_{2A}
+      = \frac{kq_1}{r_{1A}} + \frac{kq_2}{r_{2A}}
+      = 8.99\E{9}\U{Nm$^2$/C$^2$}\cdot\p({
+        \frac{2.40\E{-9}\U{C}}{0.050\U{m}}+\frac{-6.5\E{-9}\U{C}}{0.050\U{m}}})
+      = \ans{-737\U{J/C}} \;.
+\end{equation}
+
+\Part{b}
+\begin{equation}
+  V_B = V_{1B} + V_{2B}
+      = \frac{kq_1}{r_{1B}} + \frac{kq_2}{r_{2B}}
+      = 8.99\E{9}\U{Nm$^2$/C$^2$}\cdot\p({
+        \frac{2.40\E{-9}\U{C}}{0.080\U{m}}+\frac{-6.5\E{-9}\U{C}}{0.060\U{m}}})
+      = \ans{-704\U{J/C}} \;.
+\end{equation}
+
+\Part{c}
+The electric potential energy change of the moving charge is given by
+\begin{equation}
+  \Delta U_{BA} = q\Delta V_{BA} = q(V_A-V_B)
+    = 2.50\E{-9}\U{C}\cdot (-737 + 704)\U{J/C}
+    = -82\U{nJ} \;.
+\end{equation}
+It makes sense that the charge lost electric potential energy, since
+it is a positive charge moving to the lower potential point $A$.  The
+electric potential energy lost went into some other form of energy
+(kinetic, heat, work, mechanical), but the electric field was giving
+energy to that other form, so it does \emph{positive} work:
+$\ans{82\U{nJ}}$.
 \end{solution}
diff --git a/latex/problems/Young_and_Freedman_12/problem23.22.tex b/latex/problems/Young_and_Freedman_12/problem23.22.tex
index c9dd3ac..6861f01 100644
--- a/latex/problems/Young_and_Freedman_12/problem23.22.tex
+++ b/latex/problems/Young_and_Freedman_12/problem23.22.tex
@@ -14,4 +14,64 @@ $x\gg a$?  Explain why this result is obtained.
 \end{problem*}
 
 \begin{solution}
+\Part{a}
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 1cm;
+real a = 1;
+
+Charge qt = pCharge((0,a)*u);
+Charge qb = pCharge((0,-a)*u);
+
+qt.draw(); qb.draw();
+draw_ijhat((0,0));
+\end{asy}
+\end{center}
+
+\Part{b}
+This is particularly easy, since both scalar electric potentials are
+the same (same $q$, same $r=a$, direction doesn't matter).
+\begin{equation}
+  V_0 = V_{0t} + V_{0b} = 2V_{0t} = \ans{\frac{2kq}{a}} \;.
+\end{equation}
+
+\Part{c}
+The $x$-axis is still symmetric (same distance to each charge), so
+$V=2V_t$, and we can use the Pythagorean theorem for $r$.
+\begin{equation}
+  V = 2V_t = \frac{2kq}{r} = \frac{1}{4\pi\varepsilon_0}\frac{2q}{r}
+    = \ans{\frac{1}{4\pi\varepsilon_0}\frac{2q}{\sqrt{a^2+x^2}}} \;.
+\end{equation}
+
+\Part{c}
+\begin{center}
+\begin{asy}
+import graph;
+size(5cm, 2cm, IgnoreAspect);
+
+real a=1;
+real V(real x){
+  return 1.0/(a**2 + x**2)**0.5;
+}
+
+xaxis("$x$");
+yaxis("$V$");
+draw(graph(V, -2*a, 2*a, n=600), red);
+label("$a$", align=S, Scale((a,0)));
+label("$-a$", align=S, Scale((-a,0)));
+\end{asy}
+\end{center}
+
+\Part{e}
+As $x$ gets much bigger than $a$, $a^2+x^2\rightarrow x^2$, so
+\begin{equation}
+  V = \frac{2kq}{r^2} \approx \ans{\frac{k\cdot2q}{x^2}} \;.
+\end{equation}
+In other words, the electric field is pretty much the same as it would
+be if both charges were sitting at the origin.  This makes sense since
+the distinction that the charge is actually some small distance $a$
+off the origin isn't very important for $x\gg a$.
 \end{solution}
diff --git a/latex/problems/Young_and_Freedman_12/problem23.25.tex b/latex/problems/Young_and_Freedman_12/problem23.25.tex
index d98b091..c97757b 100644
--- a/latex/problems/Young_and_Freedman_12/problem23.25.tex
+++ b/latex/problems/Young_and_Freedman_12/problem23.25.tex
@@ -12,4 +12,69 @@ result is obtained.
 \end{problem*}
 
 \begin{solution}
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 1cm;
+real a = 1;
+
+Charge p = pCharge((0,0), L="$q$");
+Charge n = pCharge((a,0)*u, L="$-2q$");
+
+draw_ijhat((0,0));
+p.draw(); n.draw();
+\end{asy}
+\end{center}
+
+\Part{b}
+\begin{equation}
+  V = V_+ + V_- = \frac{kq}{|x|} + \frac{-2kq}{|x-a|}
+    = kq \p({\frac{1}{|x|} - \frac{2}{|x-a|}}) \;.
+\end{equation}
+
+\Part{c}
+$kq > 0$ so $V=0$ only when
+\begin{align}
+    &&   0 &= \frac{1}{|x|} - \frac{2}{|x-a|} \\
+    &&   2|x| &= |x-a| \\
+  -2x &= a-x       &   2x &= a-x              &  2x &= x-a  \\
+    x &= -a        &    x &= \frac{a}{3}      &   x &= -a < 0 \\
+    x &= \ans{-a}  &  x &= \ans{\frac{a}{3}}  &     &\text{contradiction} \\
+\end{align}
+Where the split to three bands comes from taking all possible absolute
+value conditions ($x<0<a$; $0<x<a$; $0<a<x$and respectively).
+
+\Part{d}
+\begin{center}
+\begin{asy}
+import graph;
+size(5cm, 2cm, IgnoreAspect);
+
+real a=1;
+real V(real x){
+  return 1.0/abs(x) - 2/abs(x-a);
+}
+
+xaxis("$x$");
+yaxis("$V$");
+draw(graph(V, -2*a, 2*a, n=333), red);
+label("$a$", align=S, Scale((a,0)));
+label("$-a$", align=S, Scale((-a,0)));
+ylimits(-6,6,Crop);
+\end{asy}
+\end{center}
+Note that I've cropped the graph, since the potential explodes to
+$\pm\infty$ at $x=0$ and $a$ respectively.
+
+\Part{e}
+As $x$ gets much larger than $a$, $|x-a|\rightarrow |x|$, so
+\begin{equation}
+  V \approx kq \p({\frac{1}{|x|} - \frac{2}{|x|}}) = \ans{\frac{-kq}{|x|}} \;.
+\end{equation}
+In other words, the electric field is pretty much the same as it would
+be if both charges were sitting at the origin.  This makes sense since
+the distinction that one charge is actually some small distance $a$
+off the origin isn't very important for $x\gg a$.
 \end{solution}
diff --git a/latex/problems/Young_and_Freedman_12/problem23.27.tex b/latex/problems/Young_and_Freedman_12/problem23.27.tex
index dfa2209..2f8c5df 100644
--- a/latex/problems/Young_and_Freedman_12/problem23.27.tex
+++ b/latex/problems/Young_and_Freedman_12/problem23.27.tex
@@ -1,6 +1,6 @@
 \begin{problem*}{27}
 Before the advent of solid-state electronics, vacuum tubes were widely
-used in radious and other devices.  A simple type of vacuum tube known
+used in radios and other devices.  A simple type of vacuum tube known
 as a \emph{diode} consists of essentially two electrodes within a
 highly evacuated enclosure.  One electrode, the \emph{cathode}, is
 maintained at a high temperature and emits electrons from its surface.
@@ -13,4 +13,20 @@ zero initial speed.  Find its speed when it strikes the anode.
 \end{problem*}
 
 \begin{solution}
+The electric potential is higher at the anode, so the electric
+potential energy for an electron will be lower there (because $q<0$
+and $U=qV$).  This means that the electron is \emph{losing} electrical
+potential energy as it moves from the cathode towards the anode.  This
+energy has to go somewhere, and in this case it ends up as kinetic
+energy, since there are no other forms of energy in the problem
+besides kinetic and electric potential energy.  Conserving energy, we
+have
+\begin{align}
+  U_i = U_{Ei} &= U_f = K+U_{Ef} \\
+  K &= U_{Ei}-U_{Ef} = -\Delta U_E = q(V_i-V_f) = \frac{1}{2} m v^2 \\
+  v &= \sqrt{\frac{2q(V_i-V_f)}{m}}
+    = \sqrt{\frac{2(-1.60\E{-19}\U{C})(0-295\U{J/C})}{9.11\E{-31}\U{kg}}}
+    = \ans{10.2\U{Mm/s}} \;.
+\end{align}
+Which is pretty fast (but not relativistic).
 \end{solution}
diff --git a/latex/problems/Young_and_Freedman_12/problem23.58.tex b/latex/problems/Young_and_Freedman_12/problem23.58.tex
index 0be5b32..d526960 100644
--- a/latex/problems/Young_and_Freedman_12/problem23.58.tex
+++ b/latex/problems/Young_and_Freedman_12/problem23.58.tex
@@ -13,4 +13,32 @@ the fixed sphere.)
 \end{problem*}
 
 \begin{solution}
+\Part{a}
+Imagine assembling this system from scratch (when $V=0$ and the
+particles are all infinitely seperated).  Bringing the first particle
+in from infinitey takes no energy, since the other particles are still
+infinitely far away.  Once the first charge is in place, the second
+charge is attracted to the first, and we have to resist this
+attraction as we bring the charge in slowly, so the system is dropping
+into a lower potential energy state.
+\begin{equation}
+  U = q_- V_+ = q_- \frac{kq_+}{r}
+    = -3.50\E{-6}\U{C} \cdot
+      \frac{8.99\E{9}\U{Nm$^2$/C$^2$}\cdot2.00\E{-6}\U{C}}{0.250\U{m}}
+    = -0.252\U{J}
+\end{equation}
+
+\Part{b}
+The sphere will need $K=-U=0.252\U{J}$ of energy to escape, since
+stopping at infinity will yield (conserving energy)
+\begin{equation}
+  E_i = K + U = E_f = (K_f=0) + (U_f=0) = 0 \;.
+\end{equation}
+This corresponds to a velocity of
+\begin{align}
+  K &= \frac{1}{2} m v^2 \\
+  v &= \sqrt{\frac{2K}{m}}
+    = \sqrt{\frac{2\cdot0.252\U{J}}{1.50\E{-3}\U{kg}}}
+    = \ans{18.3\U{m/s}} \;.
+\end{align}
 \end{solution}
diff --git a/latex/problems/Young_and_Freedman_12/problem23.60.tex b/latex/problems/Young_and_Freedman_12/problem23.60.tex
index c131511..2845875 100644
--- a/latex/problems/Young_and_Freedman_12/problem23.60.tex
+++ b/latex/problems/Young_and_Freedman_12/problem23.60.tex
@@ -39,4 +39,26 @@ theta.draw();
 \end{problem*}
 
 \begin{solution}
+The sphere is stationary, so we know the forces must balance.
+\begin{align}
+  \sum F_y &= T\cos\theta - mg = 0 \\
+  T &= \frac{mg}{\cos\theta} \\
+  \sum F_x &= qE - T\sin\theta = 0 \\
+  qE &= T\sin\theta = mg\tan\theta \\
+  E &= \frac{mg}{q}\tan\theta \;.
+\end{align}
+
+The $E$ field is constant, so the potential difference between the two
+plates is given by
+\begin{equation}
+  \Delta V = \int_\text{left}^\text{right} \vect{E}\cdot\vect{dx}
+    = E\int_\text{left}^\text{right} dx = Ed \;,
+\end{equation}
+and the potential difference between the plates is
+\begin{equation}
+  \Delta V = Ed = \frac{mgd}{q}\tan\theta
+    = \frac{1.50\E{-3}\U{kg}\cdot9.80\U{m/s$^2$}\cdot5.00\E{-2}\U{m}}
+           {8.90\E{-6}\U{C}} \tan(30.0\dg)
+    = \ans{47.7\U{V}} \;.
+\end{equation}
 \end{solution}
diff --git a/latex/problems/Young_and_Freedman_12/problem23.63.tex b/latex/problems/Young_and_Freedman_12/problem23.63.tex
index 264f5e9..38be8d0 100644
--- a/latex/problems/Young_and_Freedman_12/problem23.63.tex
+++ b/latex/problems/Young_and_Freedman_12/problem23.63.tex
@@ -55,4 +55,58 @@ q.draw();
 \end{problem*}
 
 \begin{solution}
+\Part{a}
+\begin{equation}
+  \vect{F} = q\vect{E} = -1.60\E{-19}\U{C} \cdot 1.10\E{3}\U{N/C} \; \jhat
+    = \ans{-1.76\E{-16}\U{N} \; \jhat} \;.
+\end{equation}
+
+\Part{b}
+\begin{align}
+  \vect{F}&=m\vect{a} &
+  \vect{a}&=\frac{\vect{F}}{m}
+    = \frac{-1.76\E{-16}\U{N}}{9.11\E{-31}\U{kg}} \jhat
+    = \ans{-1.93\E{14}\U{m/s$^2$} \; \jhat} \;.
+\end{align}
+
+\Part{c}
+This part is just like the inkjet printer problem from first week
+(Ch.~21, Prob.~86).  Dusting off the constant acceleration
+equations\ldots
+
+No acceleration in the $x$ direction means
+$v_x$ is constant, so the time-of-flight is given by.
+\begin{align}
+  \Delta x &= v_x \Delta t  &
+  \Delta t &= \frac{\Delta x}{v_x} = \frac{2.00\E{-2}\U{m}}{6.50\E{6}\U{m/s}}
+    = 9.23\U{ns} \;.
+\end{align}
+We can plug this time-of-flight into our constant-acceleration
+equation for $y(t)$,
+\begin{align}
+  y(t) &= \frac{a_y}{2}t^2 + v_{y0} t + y_0 \\
+  \Delta y &= \frac{a_y}{2} \Delta t^2 = -8.24\U{mm} \;,
+\end{align}
+or \ans{8.24\U{mm}} below the axis.
+
+\Part{d}
+Rembering our constant acceleration equations the $y$ velocity upon
+exitting the plates will be
+\begin{equation}
+  v_y = a_y t + v_{y0} = a_y \Delta t = -1.79\U{Mm/s} \;,
+\end{equation}
+so the direction of departure is given by
+\begin{align}
+  \theta = \arctan\p({\frac{v_y}{v_x}})
+    = \arctan\p({\frac{-1.79}{6.50}})
+    = \ans{-15.4\dg}
+\end{align}
+
+\Part{e}
+The total drop is a combination of the drop from the plates
+(calculated in \Part{c}) and the drop afterward (from simple trig)
+\begin{equation}
+  \Delta y = \Delta y_c + 12.0\E{-2}\U{m}\cdot\sin(15.4\dg) \\
+    = \ans{4.01\U{cm}}
+\end{equation}
 \end{solution}
-- 
2.26.2