1 \begin{problem*}{19.55}
2 Four identical point charges ($q = +10.0\U{$\mu$C}$) are located on
3 the corners of a rectangle as shown in Figure P19.55. The dimensions
4 of the rectangle are $L = 60.0\U{cm}$ and $W = 15.0\U{cm}$. Calculate
5 the magnitude and direction of the resultant electric force exerted on
6 the charge at the lower left corner by the other three charges.
7 \end{problem*} % problem 19.55
13 A := (0, a); % q1, labeled CCW from upper left
18 label.rt("W", draw_length(C, D, 8pt));
19 label.bot("L", draw_length(B, C, 8pt));
22 label.bot(btex $q_1$ etex, A);
24 label.top(btex $q_2$ etex, B);
26 label.top(btex $q_3$ etex, C);
28 label.bot(btex $q_4$ etex, D);
45 label.rt(btex $F_{12}$ etex, draw_force(A, B, 2cm));
46 label.lft(btex $F_{32}$ etex, draw_force(C, B, .5cm));
47 label.bot(btex $F_{42}$ etex, draw_force(D, B, .48cm));
48 draw A--(D+(a/2,0)) dashed evenly;
49 draw A--B dashed evenly;
50 draw B--D dashed evenly;
51 label.top(btex $\theta$ etex, draw_lout_angle(D+(1,0),D,B, a/3));
52 draw_right_angle(B, A, D, a/3);
53 draw_ijhats((5.5*a, a/3), 0, a/3);
60 label.rt(btex $F_{12}$ etex, draw_force(A, B, 2cm));
61 label.lft(btex $F_{32}$ etex, draw_force(C, B, .5cm));
62 label.bot(btex $F_{42}$ etex, draw_force(D, B, .48cm));
63 draw A--(D+(a/2,0)) dashed evenly;
64 draw A--B dashed evenly;
65 draw B--D dashed evenly;
66 label.top(btex $\theta$ etex, draw_lout_angle(D+(1,0),D,B, a/3));
67 draw_right_angle(B, A, D, a/3);
68 draw_ijhats((5.5*a, a/3), 0, a/3);
72 The unit vector \rhat\ diagonally across from the upper right is given by
74 \rhat &= \cos\theta \ihat + \sin\theta \jhat \\
75 \theta &= \arctan{W/L} + 180\dg = 194\dg \\
76 \cos\theta &= -0.970 \\
79 So the electric field in the lower left corner is given by
81 \vect{E} &= k_e \sum_i \frac{q_i}{r_i^2}\rhat_i
82 = k_e \left(\frac{q}{L^2}(-\ihat)
83 + \frac{q}{(L^2 + W^2)}(\cos\theta\ihat + \sin\theta\jhat)
84 + \frac{q}{W^2}(-\jhat) \right) \\
86 \left(\frac{1}{L^2} - \frac{\cos\theta}{L^2+W^2}\right)\ihat
87 + \left(\frac{1}{W^2} - \frac{\sin\theta}{L^2+W^2}\right)\jhat
91 So the magnitude of \vect{E} is given by
93 E = k_e q \sqrt{ \left(L^{-2} - \frac{\cos\theta}{L^2+W^2}\right)^2
94 + \left(W^{-2} - \frac{\sin\theta}{L^2+W^2}\right)^2 }
97 (Remembering to convert $L$ and $W$ to meters.) And the direction
98 $\phi$ (measured counter clockwise from \ihat) of \vect{E} is given by
100 \phi = \arctan\left(\frac{-W^{-2}+\frac{\sin\theta}{L^2+W^2}}{-L^{-2}+\frac{\cos\theta}{L^2+W^2}}\right) + 180\dg
103 Where the $+180\dg$ is because the tangent has a period of $180\dg$,
104 and the angle we want is in the backside $180\dg$.
106 $\vect{F} = q \vect{E}$ so the direction of \vect{F} is the same as
107 the direction of \vect{E}. The magnitude of \vect{F} is given by
109 F = 10.0\E{-6}\U{C} \cdot 4.08\E{6}\U{N/C} = \ans{40.8\U{N}}