1 \begin{problem*}{23.59}
2 A charged cork ball of mass $1.00\U{g}$ is suspended on a light string
3 in the presence of a uniform electric field as shown in Figure~P23.59.
4 When $\vect{E}=(3.00\ihat+5.00\jhat)\E{5}\U{N/C}$, the ball is in
5 equilibrium at $\theta=37.0\dg$. Find \Part{a} the charge on the ball
6 and \Part{b} the tension in the string.
11 Drawing a free-body diagram for the ball,
22 real x = L*Sin(theta);
23 real y = -L*Cos(theta);
24 real E_mag = length((E_x, E_y));
25 real E_dir = degrees((E_x, E_y));
26 real T_mag = E_x/Sin(theta);
27 real G_mag = E_y+E_x/Tan(theta);
30 draw((0,y)--(0,0), dashed);
33 Angle t = Angle((0,y), (0,0), (x,y), "$\theta$");
36 Charge a = pCharge((x,y));
38 Vector T = Vector(a.center(), mag=T_mag, dir=90+theta, "$T$");
40 Vector G = Vector(a.center(), mag=G_mag, dir=-90, "$mg$");
42 Vector E = Vector(a.center(), mag=E_mag, dir=E_dir, "$F_E$");
47 draw_ijhat((-0.7*x,y));
51 Balancing forces on the ball,
53 0 &= \sum F_x = qE_x - T\sin(\theta) \\
54 T &= \frac{qE_x}{\sin(\theta)} \\
55 0 &= \sum F_y = qE_y + T\cos(\theta) - mg
56 = qE_y + \frac{qE_x}{\sin(\theta)}\cos(\theta) - mg
57 = qE_y + qE_x\cot(\theta) - mg \\
58 q [E_y + E_x\cot(\theta)] &= mg \\
59 q &= \frac{mg}{E_y + E_x\cot(\theta)}
60 = \ans{1.09\E{-8}\U{C} = 10.9\U{nC}} \;.
64 Plugging back in for $T$,
66 T = \frac{qE_x}{\sin(\theta)} = \ans{5.44\U{mN}} \;.