--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
projects / course.git / commitdiff