1 \begin{problem*}{23.11}
2 Two small beads having positive charges $q_1=3q$ and $q_2=q$ are fixed
3 at the opposite ends of a horizontal insulating rod of length
4 $d=1.50\U{m}$. The bead with charge $q_1$ is at the origin. As shown
5 in Figure~P23.11, a third small, charged bead is free to slide on the
6 rod. \Part{a} At what position $x$ is the third bead in
7 equilibrium? \Part{b} Can the equilibrium be stable?
15 real x = (1-sqrt(q2oq1))/(1-q2oq1) * d;
17 Charge q1 = aCharge((0,0), q=3, "$q_1$");
18 Charge q2 = aCharge((d,0), q=1, "$q_2$");
19 Charge q3 = aCharge((x,0), q=0, "$q_3$");
21 Distance dx = Distance((0,0), (x,0), offset=-30pt, "$x$");
22 Distance dd = Distance((0,0), (d,0), offset=-12pt, "$d$");
35 The electric field along the rod due to the two end charges has a $y$
36 component of $0$ and an $x$ component of
38 E = \frac{kq_1}{r_1^2} - \frac{kq_2}{r_2^2}
39 = k\p({\frac{3q}{x^2} - \frac{q}{(d-x)^2}})
40 = kq\p({\frac{3}{x^2} - \frac{1}{(d-x)^2}})
46 size(6cm,4cm,IgnoreAspect);
49 real X = (1-sqrt(q2oq1))/(1-q2oq1);
51 real E(real x) { return 3./x^2 - 1./(1.-x)^2; }
52 real Z(real x) { return 0; }
57 draw((xmin,0)--(xmax,0), grey+dashed);
58 draw((X,0)--(X,E(xmax)), grey+dashed);
59 draw(graph(E, xmin, xmax), red);
61 xaxis("$x/d$", YEquals(E(xmax)), xmin=xmin, xmax=xmax, LeftTicks);
62 yaxis("$\frac{E}{kq}$", XEquals(xmin), ymin=E(xmax), ymax=E(xmin), RightTicks);
66 The third bead will be in equilibrium when this electric field is
67 zero, which occurs when
69 0 &= \frac{3}{x_0^2} - \frac{1}{(d-x_0)^2} \\
70 \frac{1}{(d-x_0)^2} &= \frac{3}{x_0^2} \\
71 x_0^2 &= 3(d^2 - 2dx_0 + x_0^2) \\
72 0 &= 2x_0^2 - 6 dx_0 + 3d^2 \\
73 0 &= 2\p({\frac{x_0}{d}})^2 - 6 \frac{x_0}{d} + 3 \\
74 \frac{x_0}{d} &= \frac{6 \pm \sqrt{(-6)^2 - 4\cdot2\cdot3}}{2\cdot2}
75 = 2.366 \text{ or } 0.6340 \\
76 x_0 &= \ans{0.951\U{m}} \;.
78 The second $x_0$ is greater than $d$, so it is non-physical.
81 If the third bead has a positive charge, it will be pushed to the
82 right for $x<x_0$ ($E(x<x_0)>0$) and to the left for $x>x_0$
83 ($E(x>x_0)<0$), so it will be stable. If it has a negative charge it