Question

Interactive Exercises 21.09: An Application of Superposition In these exercises, let's return to the geometry of Fig. 21.8.1 (reproduced below). Particles #1 and #2 are fixed on the x axis with L = 5.00 m. The charges of the three particles are as follows q! = 1.50 μC.q2 2.00pC, and q3 = 3.50 μ . Our task is to find the initial acceleration of particle 3 (mass m = 0.0200 kg), which s free to move. The plan of attack consists of two key steps 1) Find the net electrostatic force on particle 3 for an arbitrary initial location on the y axis, such as shown in the figure. 2) Use Newton's second law to find the acceleration of particle 3 for this arbitrary initial location on the y axis. Top View Horizontal, smooth surface Particle #3 (free to move) charge 43 Particle #2 (hxed in place) charge 92 Particle #1 (fixed in place) charge 4 Interactive Figure 21.8.1: Three particles lie in an xy coordinate system. All three particles have positive electric charge. Particles #1 and #2 are fixed in place on the x axis. Particle #3 is free to move and is initially located on the y axis with coordinate y as shown You complete step 1) in the plan above by apply the principle of superposition in the following way A. Find the electrostatic force on particle #3 due to particle #1, F31 , by ignoring the presence of particle #2 B. Find the electrostatic force on particle #3 due to particle #2, F32 , by ignoring the presence of particle #1 C. Find the net electrostatic force on particle #3 by using vector addition, Fnm3 ; F31 + F 32 [9] By the way, this process can be continued indefinitely. In other words, if there were, say, a total of four charged particles in the xy coordinate system, then you can find the net force on any one of the four particles by finding the force on that particle due to the other particles one at a time, and then adding all the forces using vector addition.

Answer 1

L = 5.0 m q_1 = 1.5 mu C q_2 = 2.0 mu c q_3 = 3.5 mu c m = 0.02 kg r = squareroot L^2 + y^2 sin theta = y/ squareroot L^2 + y^2 and cos theta = L/ squareroot L^2 + y^2 The electrostate force on particle 3 due to 1 F_31 = k q_1 q_3/(L^2 + y^2)(-cos theta i + sin theta j) F_31 = k q_1 q_3/L^2 + y^2 [-L/ squareroot L^2 + y^2 i + y/ squareroot L^2 + y^2 j] F_31 = -k q_1 q_3 L/(L^2 + y^2)^3/2 i + k q_1 q_3 y/(L^2 + y^2)^3/2 j The electrostatic force on particle 3 due to 1 F_32 = kq_2 q_3/L^2 + y^2 [cos theta i + sin theta j] F_32 = k q_2 q_3/L^2 + y^2 [1/ squareroot L^2 + y^2 i + y/ squareroot L^2 + y^2 j] F_32 = k q_2 q_3/L^2 + y^2 [L/ squareroot L^2 + y^2 i + y/ SquarerootL^2 + y^2 j]