2 Thre point charges are arranged as shown in Figure~P23.9.
3 Find \Part{a} the magnitude and \Part{b} the direction of the electric
4 force on the particle at the origin.
12 The electric field at the origin due to the other two charges is
14 \vect{E}_0 = k\frac{6.00\U{nC}}{0.300\U{m}^2}(-\ihat)
15 + k\frac{-3.00\U{nC}}{0.100\U{m}^2}\jhat
16 = (-599.3\ihat - 2697\jhat)\U{N/C}
18 The electric force on the particle at the origin can be found from the
21 \vect{F}_0 = q_0\vect{E}_0
22 = 5.00\U{nC}\cdot(-599.3\ihat - 2697\jhat)\U{N/C}
23 = (-2.997\E{-6}\ihat - 1.348\E{-5}\jhat)\U{N}
27 Using the Pythagorean theorem, the magnitude is
29 |\vect{F}_0| = \sqrt{F_{0,x}^2 + F_{0,y}^2}
30 = \ans{1.38\E{-5}\U{N}}
34 Using basic trig, the angle is
36 \theta = 180\dg + \arctan\p({\frac{F_{0,y}}{F_{0,x}}})
37 = 180\dg + 77.47\dg = \ans{257\dg}
39 counter clockwise from the $x$ axis, where the $180\dg$ adjusts for
40 the back-side ($x<0$) arctangent.