2 A small sphere with mass $1.50\U{g}$ hangs by a thread between two
3 parallel vertical plates $5.00\U{cm}$ apart (Fig.~23.36). The plates
4 are insulateing and have uniform surface charge densitied $+\sigma$
5 and $-\sigma$. The charge on the sphere is $q=8.90\E{-6}\U{C}$. What
6 potential difference between the plates will cause the thread to
7 assume an angle of $30.0\dg$ with the vertical?
17 real dy = 1.8*L*Cos(phi);
20 Charge q = pCharge(dir(-90+phi)*L*u, q=1, L="$q$");
21 Wire wire = Wire((0,0), q.center());
22 Angle theta = Angle(q.center(), (0,0), (0,-1), L="$\theta$");
23 Surface s = Surface((ds/2,0)*u, (-ds/2,0)*u);
24 Wire right_plate = Wire((ds,0)*u, (ds,-dy)*u, outline=green);
25 Wire left_plate = Wire((-ds,0)*u, (-ds,-dy)*u, outline=green);
26 Distance Ds = Distance(left_plate.pTo, right_plate.pTo,
27 L=Label("$5.00\U{cm}$", align=S));
34 draw((0,0)--(0,q.center().y), dashed);
42 The sphere is stationary, so we know the forces must balance.
44 \sum F_y &= T\cos\theta - mg = 0 \\
45 T &= \frac{mg}{\cos\theta} \\
46 \sum F_x &= qE - T\sin\theta = 0 \\
47 qE &= T\sin\theta = mg\tan\theta \\
48 E &= \frac{mg}{q}\tan\theta \;.
51 The $E$ field is constant, so the potential difference between the two
54 \Delta V = \int_\text{left}^\text{right} \vect{E}\cdot\vect{dx}
55 = E\int_\text{left}^\text{right} dx = Ed \;,
57 and the potential difference between the plates is
59 \Delta V = Ed = \frac{mgd}{q}\tan\theta
60 = \frac{1.50\E{-3}\U{kg}\cdot9.80\U{m/s$^2$}\cdot5.00\E{-2}\U{m}}
61 {8.90\E{-6}\U{C}} \tan(30.0\dg)