--- /dev/null
+\begin{problem*}{23.9}
+Thre point charges are arranged as shown in Figure~P23.9.
+Find \Part{a} the magnitude and \Part{b} the direction of the electric
+force on the particle at the origin.
+% y5.00nC 6.00nC
+% o-----------o----x
+% |0.100m 0.300m
+% o-3.00nC
+\end{problem*}
+
+\begin{solution}
+The electric field at the origin due to the other two charges is
+\begin{equation}
+ \vect{E}_0 = k\frac{6.00\U{nC}}{0.300\U{m}^2}(-\ihat)
+ + k\frac{-3.00\U{nC}}{0.100\U{m}^2}\jhat
+ = (-599.3\ihat - 2697\jhat)\U{N/C}
+\end{equation}
+The electric force on the particle at the origin can be found from the
+electric field
+\begin{equation}
+ \vect{F}_0 = q_0\vect{E}_0
+ = 5.00\U{nC}\cdot(-599.3\ihat - 2697\jhat)\U{N/C}
+ = (-2.997\E{-6}\ihat - 1.348\E{-5}\jhat)\U{N}
+\end{equation}
+
+\Part{a}
+Using the Pythagorean theorem, the magnitude is
+\begin{equation}
+ |\vect{F}_0| = \sqrt{F_{0,x}^2 + F_{0,y}^2}
+ = \ans{1.38\E{-5}\U{N}}
+\end{equation}
+
+\Part{b}
+Using basic trig, the angle is
+\begin{equation}
+ \theta = 180\dg + \arctan\p({\frac{F_{0,y}}{F_{0,x}}})
+ = 180\dg + 77.47\dg = \ans{257\dg}
+\end{equation}
+counter clockwise from the $x$ axis, where the $180\dg$ adjusts for
+the back-side ($x<0$) arctangent.
+\end{solution}
--- /dev/null
+\begin{problem*}{23.10}
+Two small metallic spheres, each of mass $m=0.200\U{g}$, are suspended
+as pendulums by light strings of length $L$ as shown if Figure~P23.10.
+The spheres are given the same electric charge of $7.20\U{nC}$, and
+they come to equilibrium when each string is at an angle of
+$\theta=5.00\dg$ with the vertical. How long are the strings?
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real theta = 20; // 5; Exaggerated for clarity
+real L = 2cm;
+
+real x = L*Sin(theta);
+real y = -L*Cos(theta);
+
+draw((x,y)--(0,0)--(-x,y));
+draw((0,y)--(0,0), dashed);
+dot((0,0));
+
+Angle t = Angle((0,y), (0,0), (x,y), "$\theta$");
+t.draw();
+
+Charge a = pCharge((-x,y));
+Charge b = pCharge((x,y));
+a.draw(); b.draw();
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+By symmetry the two spheres will be at the same height, so the
+distance between them is $2L\sin(\theta)$ and we can draw a free body
+diagram for the right-hand sphere:
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real theta = 5;
+real E_mag = 0.1cm;
+
+Vector T = Vector((0,0), mag=E_mag/Sin(theta), dir=90+theta, "$T$");
+Vector G = Vector((0,0), mag=E_mag/Tan(theta), dir=-90, "$mg$");
+Vector E = Vector((0,0), mag=3*E_mag, dir=0, "$F_E$");
+
+T.draw();
+G.draw();
+E.draw();
+dot((0,0));
+\end{asy}
+\end{center}
+
+Because the system is in equilibrium, the net force on the sphere must
+be zero.
+\begin{align}
+ 0 &= \sum F_y = T\cos(\theta) - mg \\
+ T &= \frac{mg}{\cos(\theta)} \\
+ 0 &= \sum F_x = \frac{kq^2}{[2L\sin(\theta)]^2} - T\sin(\theta) \\
+ \frac{kq^2}{[2L\sin(\theta)]^2} &= T\sin(\theta)
+ = \frac{mg}{\cos(\theta)}\sin(\theta) = mg\tan(\theta) \\
+ [2L\sin(\theta)]^2 &= \frac{kq^2}{mg\tan(\theta)} \\
+ 2L\sin(\theta) &= \sqrt{\frac{kq^2}{mg\tan(\theta)}}
+ = q\sqrt{\frac{k}{mg\tan(\theta)}} \\
+ L &= \frac{q}{2\sin(\theta)}\sqrt{\frac{k}{mg\tan(\theta)}}
+ = \ans{29.9\U{cm}}
+\end{align}
+\end{solution}
--- /dev/null
+\begin{problem*}{23.11}
+Two small beads having positive charges $q_1=3q$ and $q_2=q$ are fixed
+at the opposite ends of a horizontal insulating rod of length
+$d=1.50\U{m}$. The bead with charge $q_1$ is at the origin. As shown
+in Figure~P23.11, a third small, charged bead is free to slide on the
+rod. \Part{a} At what position $x$ is the third bead in
+equilibrium? \Part{b} Can the equilibrium be stable?
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real d = 2cm;
+real q2oq1 = 1/3.;
+real x = (1-sqrt(q2oq1))/(1-q2oq1) * d;
+
+Charge q1 = aCharge((0,0), q=3, "$q_1$");
+Charge q2 = aCharge((d,0), q=1, "$q_2$");
+Charge q3 = aCharge((x,0), q=0, "$q_3$");
+
+Distance dx = Distance((0,0), (x,0), offset=-30pt, "$x$");
+Distance dd = Distance((0,0), (d,0), offset=-12pt, "$d$");
+
+dx.draw();
+dd.draw();
+q1.draw();
+q2.draw();
+q3.draw();
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+The electric field along the rod due to the two end charges has a $y$
+component of $0$ and an $x$ component of
+\begin{equation}
+ E = \frac{kq_1}{r_1^2} - \frac{kq_2}{r_2^2}
+ = k\p({\frac{3q}{x^2} - \frac{q}{(d-x)^2}})
+ = kq\p({\frac{3}{x^2} - \frac{1}{(d-x)^2}})
+\end{equation}
+\begin{center}
+\begin{asy}
+import graph;
+
+size(6cm,4cm,IgnoreAspect);
+
+real q2oq1 = 1/3.;
+real X = (1-sqrt(q2oq1))/(1-q2oq1);
+
+real E(real x) { return 3./x^2 - 1./(1.-x)^2; }
+real Z(real x) { return 0; }
+
+real xmin = 0.15;
+real xmax = 0.9;
+
+draw((xmin,0)--(xmax,0), grey+dashed);
+draw((X,0)--(X,E(xmax)), grey+dashed);
+draw(graph(E, xmin, xmax), red);
+
+xaxis("$x/d$", YEquals(E(xmax)), xmin=xmin, xmax=xmax, LeftTicks);
+yaxis("$\frac{E}{kq}$", XEquals(xmin), ymin=E(xmax), ymax=E(xmin), RightTicks);
+\end{asy}
+\end{center}
+
+The third bead will be in equilibrium when this electric field is
+zero, which occurs when
+\begin{align}
+ 0 &= \frac{3}{x_0^2} - \frac{1}{(d-x_0)^2} \\
+ \frac{1}{(d-x_0)^2} &= \frac{3}{x_0^2} \\
+ x_0^2 &= 3(d^2 - 2dx_0 + x_0^2) \\
+ 0 &= 2x_0^2 - 6 dx_0 + 3d^2 \\
+ 0 &= 2\p({\frac{x_0}{d}})^2 - 6 \frac{x_0}{d} + 3 \\
+ \frac{x_0}{d} &= \frac{6 \pm \sqrt{(-6)^2 - 4\cdot2\cdot3}}{2\cdot2}
+ = 2.366 \text{ or } 0.6340 \\
+ x_0 &= \ans{0.951\U{m}} \;.
+\end{align}
+The second $x_0$ is greater than $d$, so it is non-physical.
+
+\Part{b}
+If the third bead has a positive charge, it will be pushed to the
+right for $x<x_0$ ($E(x<x_0)>0$) and to the left for $x>x_0$
+($E(x>x_0)<0$), so it will be stable. If it has a negative charge it
+will not be stable.
+\end{solution}
--- /dev/null
+\begin{problem*}{23.15}
+In the Bohr theory of the hydrogen atom, an electron moves in a
+circular orbit about a proton, where the radius of the orbit is
+$5.29\E{-11}\U{m}$. \Part{a} Find the magnitude of the electric force
+exerted on each particle. \Part{b} If this force causes the
+centripetal acceleration of the electron, what is the speed of the
+electron?
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+\begin{equation}
+ F = k\frac{q_1q_2}{r^2}
+ = 8.99\E{9}\U{Nm$^2$/C$^2$}\frac{(1.60e-19\U{C})^2}{(5.29\E{-11}\U{m})^2}
+ = \ans{8.22\E{-8}\U{N}}
+\end{equation}
+
+\Part{b}
+Because the mass of the electron is much less than the mass of the
+proton, the proton is relatively stationary and the rotating mass is
+nearly the mass of the electron. Using our circular motion formula
+\begin{align}
+ F &= \frac{mv^2}{r} \\
+ v &= \sqrt{\frac{Fr}{m}}
+ = \sqrt{\frac{8.22\E{-8}\U{N}\cdot5.29\E{-11}\U{m}}{9.11\E{-31}\U{kg}}}
+ = \ans{2.19\E{6}\U{m/s}}
+\end{align}
+\end{solution}
projects / course.git / commitdiff