Question

2. Three charges are fixed at positions along the a-axis as positions -d, o, and +d. The charges at -d and d are both and the charge at O is positive. (a) A positively charged object of mass m is placed on the x-axis between O and d, close to the positlicn the three charges described above do not move as a result of this new charged object, describe the mois object after it is released as it moves in the region o<x<d. The charge at O has a magnitude of 2 Q, while the other two charges have a magnitude of o. (b) On the axes below, sketch the electric field along the x-axis in the vicinity of the charges. An elect the right should be graphed as positive and a field pointing left should be graphed as negative. e(Jeot iS uwouh4.to Hy hオ, because t repeal the ou theoret cts (c) Write a mathematical function, E(x), that gives the electric field at any position along the x-axis for 0 << d. Give your answer in terms of Q, d, and fundamental constants. (d) In order to originally assemble the three charges on the x-axis, some w ork had to be done., Consider arrang. es along the x-axis in the following manner: first, bring the +2Q charge to position x = 0, then bringing the-Q charge tox required no work. Bringing in the second charge required an amount of the third charge will require more work, less work, or an amount of work equal to W +d, and finally, bring in the last charge. Bringing the +2Q charge to position。 work W. Explain whether bringing in

Answer 1

Initial situation negative charge will attract the object while (at + d) positive (at origin) will repel. So object will move along + x axis. After paring + d it will feel net force in - x direction so it will record and starts motion along - x axis as a result of this it will execute periodic motion about (+ d) \r\n Electric field at x due to - emptyset at - d E_1 = -k theta/(d + x)^2 due to + 2 theta at 0 E_2 = 2 k theta/x^2 due to - theta at + d E_3 = + k theta/ (d - x)^2 E_net = E_1 + E_2 + E_3 E_net = k theta [- 1/(x + d)^2 + 2/x^2 + 1/(d - x)^2] k = 1/ 4 pi E_0 total work done = + (change in potential energy) = u_f - u_i \r\n w = k (- theta) 2 (theta)/d + k (- theta) (2 theta)/d + k(- theta) (- theta)/2d