struct Distance {
pair pFrom;
pair pTo;
+ real offset;
real scale;
pen outline;
Label L;
- void operator init(pair pFrom=(0,0), pair pTo=(5mm,0), real scale=5mm, pen outline=currentpen, Label L="") {
+ void operator init(pair pFrom=(0,0), pair pTo=(5mm,0), real offset=0, real scale=5mm, pen outline=currentpen, Label L="") {
this.pFrom = pFrom;
this.pTo = pTo;
+ this.offset = offset;
this.scale = scale;
this.outline = outline;
this.L = L;
position = pDiff/2
+ unit(rotate(90)*pDiff) * pLabelSize.y / 2);
//label(pic=picF, L = rotate(degrees(label_rotate)) format("%g", pDiff/scale), position = TODO);
- add(pic, picF, pFrom);
+ add(pic, picF, pFrom+offset*unit(rotate(-90)*pDiff));
}
}
--- /dev/null
+\begin{problem*}{21}
+Two point charges $q_1=+2.40\U{nC}$ and $q_2=-6.50\U{nC}$ are
+$0.100\U{m}$ apart. Point $A$ is midway between them; point $B$ is
+$0.080\U{m}$ from $q_1$ and $0.060\U{m}$ from $q_2$ (Fig.~23.31).
+Take the electric potential to be zero at infinity. Find \Part{a} the
+potential at point $A$; \Part{b} the potential at point $B$; \Part{c}
+the work done by the electric field on a charge of $2.50\U{nC}$ that
+travels from point $B$ to point $A$.
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 50cm;
+real dOT = 0.1;
+real dOB = 0.08;
+real dTB = 0.06;
+
+Charge qO = pCharge((0,0), L="$q_1$");
+Charge qT = nCharge((dOT,0)*u, L="$q_2$");
+pair A = (qO.center+qT.center)/2.0;
+// x^2 + y^2 = dOB^2
+// (x-a)^2 + y^2 = dTB^2 where a=qT.center.x
+// so
+// 2xa - a^2 = dOB^2 - dTB^2
+// x = (dOB^2 - dTB^2 + a^2)/(2a)
+// y = (dOB^2-x^2)^0.5
+real Bx = (dOB**2 - dTB**2 + dOT**2)/(2*dOT);
+pair B = (Bx,(dOB**2-Bx**2)**0.5)*u;
+Distance DOA = Distance(qO.center, A, L=Label("$0.050\U{m}$", align=S));
+Distance DTA = Distance(A, qT.center, L=Label("$0.050\U{m}$", align=S));
+Distance DOB = Distance(qO.center, B, L=Label("$0.080\U{m}$", align=NW));
+Distance DTB = Distance(B, qT.center, L=Label("$0.060\U{m}$", align=NE));
+
+DOA.draw(); DTA.draw(); DOB.draw(); DTB.draw();
+dot(A, L="$A$", align=S);
+dot(B, L="$B$", align=N);
+
+qO.draw(); qT.draw();
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{22}
+Two positive point charges, each of magnitude $q$, are fixed on the
+$y$-axis at the points $y=+a$ and $y=-a$. Take the potential to be
+zero at an infinite distance from the charges. \Part{a} Show the
+positions of the charges in a diagram. \Part{b} What is the potential
+$V_0$ at the origin? \Part{c} Show that the potential at any point on
+the $x$-axis is
+\begin{equation}
+ V = \frac{1}{4\pi\varepsilon_0}\frac{2q}{\sqrt{a^2+x^2}}
+\end{equation}
+\Part{d} Graph the potentialon the $x$-axis as a function of $x$ over
+the range from $x=-4a$ to $x=4a$. \Part{e} What is the potential when
+$x\gg a$? Explain why this result is obtained.
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{25}
+A positive charge $q$ is fixed at the point $x=0$, $y=0$, and a
+negative charge $-2q$ is fixed at the point $x=a$, $y=0$. \Part{a}
+Show the positions of the charges in a diagram. \Part{b} Derive an
+expression for the potential $V$ at points on the $x$-axis as a
+function of the coordinate $x$. Take $V$ to be zero at an infinite
+distance from the charges. \Part{c} At which positions on the
+$x$-axis is $V=0$? \Part{d} Graph $V$ at points on the $x$-axis as a
+function of $x$ in the range from $x=-2a$ to $x=+2a$. \Part{e} What
+does the answer to \Part{b} become when $x\gg a$? Explain why this
+result is obtained.
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\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
+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.
+A potential difference of a few hundred volts is maintained between
+the cathode and the other electrode, known as the \emph{anode}, with
+the anode at the higher potential. Suppose that in a particular
+vacuum tube the potential of the anode is $295\U{V}$ higher than that
+of the cathode. An electron leaves the surface of the cathode with
+zero initial speed. Find its speed when it strikes the anode.
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{58}
+\Part{a} Calculate the potential energy of a system of two small
+spheres, one carrying a charge of $2.00\U{$\mu$C}$ and the other a
+charge of $-3.50\U{$\mu$C}$, with their centers separated by a
+distance of $0.250\U{m}$. Assume zero potential energy when the
+charges are infinitely separated. \Part{b} Suppose that one of the
+spheres is held in place and the other sphere, which has a mass of
+$1.50\U{g}$, is shot away from it. What minimum initial speed would
+the moving sphere need in order to escape completely from the
+attraction of the fixed sphere? (To escape, the moving sphere would
+have to reach a velocity of zero when it was infinitely distant from
+the fixed sphere.)
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{60}
+A small sphere with mass $1.50\U{g}$ hangs by a thread between two
+parallel vertical plates $5.00\U{cm}$ apart (Fig.~23.36). The plates
+are insulateing and have uniform surface charge densitied $+\sigma$
+and $-\sigma$. The charge on the sphere is $q=8.90\E{-6}\U{C}$. What
+potential difference between the plates will cause the thread to
+assume an angle of $30.0\dg$ with the vertical?
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 0.3cm;
+real L = 4;
+real ds = 5;
+real phi = 30;
+real dy = 1.8*L*Cos(phi);
+int n=3;
+
+Charge q = pCharge(dir(-90+phi)*L*u, q=1, L="$q$");
+Wire wire = Wire((0,0), q.center);
+Angle theta = Angle(q.center, (0,0), (0,-1), L="$\theta$");
+Surface s = Surface((ds/2,0)*u, (-ds/2,0)*u);
+Wire right_plate = Wire((ds,0)*u, (ds,-dy)*u, outline=green);
+Wire left_plate = Wire((-ds,0)*u, (-ds,-dy)*u, outline=green);
+Distance Ds = Distance(left_plate.pTo, right_plate.pTo,
+ L=Label("$5.00\U{cm}$", align=S));
+
+left_plate.draw();
+right_plate.draw();
+s.draw();
+Ds.draw();
+wire.draw();
+draw((0,0)--(0,q.center.y), dashed);
+q.draw();
+theta.draw();
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{63}
+Cathode-ray tubes (CRTs) are often found in oscilloscopes and computer
+monitors. In Fig.~23.38 an electron with an initial speed of
+$6.50\E{6}\U{m/s}$ is projected along the axis midway between the
+deflection plates of a cathode-ray tube. The uniform electric field
+between the plates has a magnitude of $1.10\E{3}\U{V/m}$ and is
+upward. \Part{a} What is the force (magnitude and direction) on the
+electron when it is between the plates? \Part{b} What is the
+acceleration of the electron (magnitude and direction) when acted on
+by the force in \Part{a}? \Part{c} How far below the axis has the
+electron moved when it reaches the end of the plates? \Part{d} At
+what angle with the axis is it moving as it leaves the
+plates? \Part{e} How far below the axis will it strike the
+fluorescent screen $S$?
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 0.5cm;
+real L = 6;
+real s = 12;
+real dy = 2;
+int n=3;
+
+Charge q = nCharge((-1,0)*u, q=1);
+Vector v = Velocity(q.center, L="$v$");
+Wire top_plate = Wire((0,0.5*dy)*u, (L,.5*dy)*u);
+Wire bottom_plate = Wire((0,-.5*dy)*u, (L,-.5*dy)*u);
+Wire screen = Wire((L+s,-.7*dy)*u, (L+s,+.7*dy)*u, L="screen");
+
+Vector e;
+for (int i=0; i<n; i+=1) {
+ e = EField((L*(i+.5)/n, -.35*dy)*u, mag=.7*dy*u, dir=90);
+ e.draw();
+}
+
+Distance Ddy = Distance((L,-.5*dy)*u, (L,.5*dy)*u, offset=2mm,
+ L=Label("$2.0\U{cm}$", align=E));
+Distance DL = Distance((0,-.5*dy)*u, (L,-.5*dy)*u, offset=2mm,
+ L=Label("$6.0\U{cm}$", align=S));
+Distance Ds = Distance((L,-.5*dy)*u, (L+s,-.5*dy)*u, offset=2mm,
+ L=Label("$12.0\U{cm}$", align=S));
+
+top_plate.draw();
+bottom_plate.draw();
+screen.draw();
+Ddy.draw();
+DL.draw();
+Ds.draw();
+v.draw();
+q.draw();
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
projects / course.git / commitdiff