1 \begin{problem*}{19.13}
2 Three point charges are arranged as shown in Figure P19.13.\\
4 % \begin{tabular}{|l|r|r|r|}
5 % Name & Charge (nC) & x (m) & y (m) \\
7 % $q_1$ & $5.00$ & $0$ & $0$ \\
8 % $q_2$ & $6.00$ & $0.300$ & $0$ \\
9 % $q_3$ & $-3.00$ & $0$ & $-0.100$ \\
13 \Part{a} Find the vector electric field \vect{E} that $q_2$ and $q_3$
14 together create at the origin.
15 \Part{b} Find the vector force \vect{F} on $q_1$.
16 \end{problem*} % problem 19.13
26 label.top(btex 0.300\mbox{ m} etex, draw_length(B, A, 8pt));
27 label.lft(btex 0.100\mbox{ m} etex, draw_length(A, C, 8pt));
30 label.rt(btex $q_1 = 5\mbox{ nC}$ etex, A);
31 draw_pcharge(B, 4.2pt);
32 label.rt(btex $q_2 = 6\mbox{ nC}$ etex, B);
34 label.rt(btex $q_3 = -3\mbox{ nC}$ etex, C);
52 label.lft(btex $E_{21}$ etex, draw_Efield(B, A, a/8));
53 label.rt(btex $E_{31}$ etex, draw_Efield(C, A, -a/5));
54 draw_ijhats(-(a, a/3), 0, a/6);
62 \vect{E} = k_e \sum_i \frac{q_i}{r_i^2}\rhat_i
63 = k_e \left[\frac{q_2}{x_2^2}(-\ihat) + \frac{q_3}{y_3^2}\jhat\right]
64 = 8.99\E{9}\U{N$\cdot$m$^2$/C$^2$} \left(\frac{-6.00\ihat}{0.300^2} - \frac{3.00\jhat}{0.100^2}\right)\E{-9}{C/m^2}
65 = \ans{\left( -0.599\ihat - 2.70\jhat \right)\U{kN/C}}
70 \vect{F} = q_1 \vect{E}
71 = \ans{\left( -3.00\ihat -13.5\jhat\right)\U{$\mu$N}}