From c912647ec9c1fbc4a8c41db7f745a908b417db55 Mon Sep 17 00:00:00 2001 From: "W. Trevor King" Date: Fri, 1 Apr 2011 14:51:01 -0400 Subject: [PATCH] Add week 1 recitation problems for phys-102. --- .../Serway_and_Jewett_8/problem23.09.tex | 41 +++++++++ .../Serway_and_Jewett_8/problem23.10.tex | 68 +++++++++++++++ .../Serway_and_Jewett_8/problem23.11.tex | 85 +++++++++++++++++++ .../Serway_and_Jewett_8/problem23.15.tex | 28 ++++++ 4 files changed, 222 insertions(+) create mode 100644 latex/problems/Serway_and_Jewett_8/problem23.09.tex create mode 100644 latex/problems/Serway_and_Jewett_8/problem23.10.tex create mode 100644 latex/problems/Serway_and_Jewett_8/problem23.11.tex create mode 100644 latex/problems/Serway_and_Jewett_8/problem23.15.tex diff --git a/latex/problems/Serway_and_Jewett_8/problem23.09.tex b/latex/problems/Serway_and_Jewett_8/problem23.09.tex new file mode 100644 index 0000000..3f63004 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem23.09.tex @@ -0,0 +1,41 @@ +\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} diff --git a/latex/problems/Serway_and_Jewett_8/problem23.10.tex b/latex/problems/Serway_and_Jewett_8/problem23.10.tex new file mode 100644 index 0000000..dc6e5ed --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem23.10.tex @@ -0,0 +1,68 @@ +\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} diff --git a/latex/problems/Serway_and_Jewett_8/problem23.11.tex b/latex/problems/Serway_and_Jewett_8/problem23.11.tex new file mode 100644 index 0000000..fa42222 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem23.11.tex @@ -0,0 +1,85 @@ +\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 $x0$) 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} diff --git a/latex/problems/Serway_and_Jewett_8/problem23.15.tex b/latex/problems/Serway_and_Jewett_8/problem23.15.tex new file mode 100644 index 0000000..1f32617 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem23.15.tex @@ -0,0 +1,28 @@ +\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} -- 2.26.2