\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}
\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}
\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}
\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.
\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}
\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}
\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}
\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}
projects / course.git / commitdiff