In Fig. 21-22, the particles have charges 91 = -92 = 137 nC and 93 = -94 = 148 nC, and distance a = 5.2 cm. (a) Find the net electrostatic force on particle 3: F net,3 = < > N (b) What is the vector sum of net electric forces acting on particle 1, 2, 3 and 4? Fnet, 1+ Fnet, 2 + Fnet,3 + Fnet,4 = < | > N 3 a Fig. 21-22

Your answer is partially correct. The charges and coordinates of two charged particles held fixed in an xy plane are q1 = 2.13 µC, x1 = 4.35 cm, Y1 = 0.304 cm and 92 = -5.93 µC, x2 = -2.30 cm, y2 = 1.07 cm. Find the (a) magnitude and (b) direction (with respect to +x-axis in the range (-180°;180°]) of the electrostatic force on particle 2 due to particle 1. At what (c) x and (d) y coordinates should a third particle of charge q3 = 6.16 µC be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero? (a) Number 25.3 Units N (b) Number i 173.44 Units ° (degrees) (c) Number -3.6 Units m Number i 0.42 Units m

Particle A of charge 3.00 X Yo-C is at the origin, par- ticle B of charge -6.00 X 10-4 C is at (4.00 m, 0), and particle C of charge 1.00 x 10-4C is at (0, 3.00 m). We wish to find the net electric force on C. (a) What is the x component of the electric force exerted by A on C? (b) What is the y component of the force exerted by A on C? (c) Find the magnitude of the force exerted by B oon C. (d) Calculate the x component of the force exerted by B on C. (e) Calculate the y component of the force exerted by B on C. (f) Sum the two x compo- nents from parts (a) and (d) to obtain the resultant x component of the electric force acting on C. (g) Simi- larly, find the y component of the resultant force vector acting on C. (h) Find the magnitude and direction of the resultant electric force acting on C.

18 In Fig. 21-29a, three positively charged particles are fixed on an x BC (a) axis. Particles B and C are so close to each other that they can be con- sidered to be at the same distance B. A from particle A. The net force on particle A due to particles B and C is 2.014 x 10-23 N in the negative direction of the x axis. In Fig. 21- 29b, particle B has been moved to the opposite side of A but is still at the same distance from it. The net force on A is now 2.877 x 10-24 N in the negative direction of the x axis. What is the ratio Id9n? (b) Figure 21-29 Problem 18.