From b9420d6d91f7cb022245b0ce235a65cb9c179ccd Mon Sep 17 00:00:00 2001 From: "W. Trevor King" Date: Fri, 8 Apr 2011 11:29:24 -0400 Subject: [PATCH] Add week 2 recitation problems for phys-102. --- .../Serway_and_Jewett_8/problem23.17.tex | 18 +++++ .../Serway_and_Jewett_8/problem23.43.tex | 24 +++++++ .../Serway_and_Jewett_8/problem23.50.tex | 19 +++++ .../Serway_and_Jewett_8/problem23.59.tex | 69 +++++++++++++++++++ .../Serway_and_Jewett_8/problem23.62.tex | 38 ++++++++++ 5 files changed, 168 insertions(+) create mode 100644 latex/problems/Serway_and_Jewett_8/problem23.17.tex create mode 100644 latex/problems/Serway_and_Jewett_8/problem23.43.tex create mode 100644 latex/problems/Serway_and_Jewett_8/problem23.50.tex create mode 100644 latex/problems/Serway_and_Jewett_8/problem23.59.tex create mode 100644 latex/problems/Serway_and_Jewett_8/problem23.62.tex diff --git a/latex/problems/Serway_and_Jewett_8/problem23.17.tex b/latex/problems/Serway_and_Jewett_8/problem23.17.tex new file mode 100644 index 0000000..aea6dc2 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem23.17.tex @@ -0,0 +1,18 @@ +\begin{problem*}{23.17} +A point charge $+2Q$ is at the origin and a point charge $-Q$ is +located along the $x$ axis at $x=d$ as in Figure P23.17. Find a +symbolic expression for the net force on a third point charge $+Q$ +located along the $y$ axis at $y=d$. +\end{problem*} + +\begin{solution} +Summing the forces from the other two charges +\begin{equation} + \vect{F} = \vect{F}_0 + \vect{F}_x + = k\frac{Q\cdot2Q}{d^2}\jhat + + k\frac{Q\cdot(-Q)}{d^2+d^2}\frac{-\ihat+\jhat}{\sqrt{2}} + = k\frac{Q^2}{d^2}\p({ 2\jhat - \frac{1}{2\sqrt{2}}(-\ihat+\jhat) }) + = \ans{k\frac{Q^2}{d^2}\p[{ \frac{\ihat}{2\sqrt{2}} + + \p({2-\frac{1}{2\sqrt{2}}})\jhat }] } +\end{equation} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem23.43.tex b/latex/problems/Serway_and_Jewett_8/problem23.43.tex new file mode 100644 index 0000000..b0f2599 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem23.43.tex @@ -0,0 +1,24 @@ +\begin{problem*}{23.43} +An electron and a proton are each placed at rest in a uniform electric +field opf magnitude $520\U{N/C}$. Calculate the speed of each +particle $48.0\U{ns}$ after being released. +\end{problem*} + +\begin{solution} +The acceleration magnitudes are +\begin{align} + a_p &= \frac{|F_p|}{m_p} = \frac{|q_e|}{m_p} \cdot E + = \frac{1.60\E{-19}\U{C}}{1.67\E{-27}\U{kg}} \cdot 520\U{N/C} + = 49.8\U{Gm/s$^2$} \\ + a_e &= \frac{|F_e|}{m_e} = \frac{|q_e|}{m_e} \cdot E + = \frac{1.60\E{-19}\U{C}}{9.11\E{-31}\U{kg}} \cdot 520\U{N/C} + = 91.3\U{Tm/s$^2$} \;. +\end{align} + +Using our constant-acceleration formula, the speed after +$t=48.0\U{ns}$ is given by +\begin{align} + v_p &= a_p t = \ans{2.39\U{km/s}} \\ + v_e &= a_e t = \ans{4.38\U{Mm/s}} \;. +\end{align} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem23.50.tex b/latex/problems/Serway_and_Jewett_8/problem23.50.tex new file mode 100644 index 0000000..f46e040 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem23.50.tex @@ -0,0 +1,19 @@ +\begin{problem*}{23.50} +A small sphere of charge $q_1=0.800\U{$\mu$C}$ hangs from the end of a +spring as in Figure P23.50a. When another small sphere of charge +$q_2=-0.600\U{$\mu$C}$ is held beneath the first sphere as in Figure +P23.50b, the spring stretches by $d=3.50\U{cm}$ from its original +length and reaches a new equilibrium position with a separation +between the charges of $r=5.00\U{cm}$. What is the force constant of +the spring? +\end{problem*} + +\begin{solution} +The addition downward force from electric attraction is balanced by +the additional upward force from spring extension, so +\begin{align} + \kappa d = |F_\text{spring}| &= |F_e| = k\frac{|q_1 q_2|}{r^2} \\ + \kappa &= k\frac{|q_1 q_2|}{dr^2} + = \ans{49.3\U{N/m}} \;. +\end{align} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem23.59.tex b/latex/problems/Serway_and_Jewett_8/problem23.59.tex new file mode 100644 index 0000000..7ee8611 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem23.59.tex @@ -0,0 +1,69 @@ +\begin{problem*}{23.59} +A charged cork ball of mass $1.00\U{g}$ is suspended on a light string +in the presence of a uniform electric field as shown in Figure P23.59. +When $\vect{E}=(3.00\ihat+5.00\jhat)\E{5}\U{N/C}$, the ball is in +equilibrium at $\theta=37.0\dg$. Find \Part{a} the charge on the ball +and \Part{b} the tension in the string. +\end{problem*} + +\begin{solution} +\Part{a} +Drawing a free-body diagram for the ball, +\begin{center} +\begin{asy} +import Mechanics; +import ElectroMag; + +real theta = 37; +real E_x = 3cm / 8; +real E_y = 5cm / 8; +real L = 2cm; + +real x = L*Sin(theta); +real y = -L*Cos(theta); +real E_mag = length((E_x, E_y)); +real E_dir = degrees((E_x, E_y)); +real T_mag = E_x/Sin(theta); +real G_mag = E_y+E_x/Tan(theta); + +draw((x,y)--(0,0)); +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)); + +Vector T = Vector(a.center, mag=T_mag, dir=90+theta, "$T$"); +T.draw(); +Vector G = Vector(a.center, mag=G_mag, dir=-90, "$mg$"); +G.draw(); +Vector E = Vector(a.center, mag=E_mag, dir=E_dir, "$F_E$"); +E.draw(); + +a.draw(); + +draw_ijhat((-0.7*x,y)); +\end{asy} +\end{center} + +Balancing forces on the ball, +\begin{align} + 0 &= \sum F_x = qE_x - T\sin(\theta) \\ + T &= \frac{qE_x}{\sin(\theta)} \\ + 0 &= \sum F_y = qE_y + T\cos(\theta) - mg + = qE_y + \frac{qE_x}{\sin(\theta)}\cos(\theta) - mg + = qE_y + qE_x\cot(\theta) - mg \\ + q [E_y + E_x\cot(\theta)] &= mg \\ + q &= \frac{mg}{E_y + E_x\cot(\theta)} + = \ans{1.09\E{-8}\U{C} = 10.9\U{nC}} \;. +\end{align} + +\Part{b} +Plugging back in for $T$, +\begin{equation} + T = \frac{qE_x}{\sin(\theta)} = \ans{5.44\U{mN}} \;. +\end{equation} + +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem23.62.tex b/latex/problems/Serway_and_Jewett_8/problem23.62.tex new file mode 100644 index 0000000..6c9dc06 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem23.62.tex @@ -0,0 +1,38 @@ +\begin{problem*}{23.62} +Four identical charged particles ($q=+10.0\U{$\mu$C}$) are located on +the corners of a rectangle as shown in Figrue P23.62. The dimensions +of the rectangle are $L=60.0\U{cm}$ and $W=15.0\U{cm}$. +Calculate \Part{a} the magnitude and \Part{b} the direction of the +total electric force exerted on the charge at the lower left corner by +the other three charges. +% L in x direction, W in y +\end{problem*} + +\begin{solution} +\Part{a} +Summing the electric force due to each source, +\begin{align} + \vect{F} &= kq^2 \p({ + \frac{-\ihat}{L^2} + \frac{-\jhat}{W^2} + + \frac{-\frac{L}{\sqrt{L^2+W^2}}\ihat - \frac{W}{\sqrt{L^2+W^2}}\jhat} + {L^2+W^2} + }) + = -kq^2 \p[{ \p({\frac{1}{L^2}+\frac{L}{(L^2+W^2)^{3/2}}})\ihat + + \p({\frac{1}{W^2}+\frac{W}{(L^2+W^2)^{3/2}}})\jhat}] \\ + &= \p({-0.478\ihat - 4.05\jhat})\U{MN} \;. +\end{align} + +The magnitude of $\vect{F}$ is therefore +\begin{equation} + |\vect{F}| = \sqrt{F_x^2 + F_y^2} + = \ans{4.08\U{MN}} \;. +\end{equation} + +\Part{b} +Using basic trig +\begin{equation} + \theta = \arctan\p({\frac{F_y}{F_x}}) + = 83.3\dg + 180\dg = \ans{263\dg} +\end{equation} +measured counter clockwise from the $x$ axis. +\end{solution} -- 2.26.2