Question

IP The point charges in the figure below have the following values: q1 = 1.8 uC, q2 = 6.1 uC, q3 = -0.88 uC. 91 + ----- --------93 Assume that charge qı is at the origin and q3 lies on the positive x-axis.

A) Given that the distance d in the figure is 4.31 cm, find the magnitude of the net electrostatic force exerted on the point charge q1.

B) Find the direction of the net electrostatic force exerted on the point charge q1q1 in Part A.

C) How would your answer to Part A change if the distance dd were doubled?

Answer 1

solution arge q = 1.8ML = 1.8x106c %= 6.14C = 61X166C = 0.88 = -0.88X1060 d=4.311m = 431X15?m 12. Er 4E12 E3 Ez = I NEO 23 - ? Electric field at point q, due to charge q, is given by 6.18166 4x3.14 X8.85 x16/2 (4:31x16 22 E = 2.95x10 NIC E = 2.958107 N/ Electric field at point q, due to charge 2 is given by (-0.88x166) 4*3.14-x8-85X16/2 4.37 X1o2j7 E = 0.43410 N/ Now Net Electric field E = Ezt = a + 6 + 2 E2 Ez las E = 2.95)2+ (0.431+ 2x(2.95JX(043) C0g/20 X107 E = 10.7025 +0.1849 –1-2005X/04 = 77.6109x77 E = 2.76X10 NIC a) Force On & is given by F = 2, E = 1.8 X10 X 2.76X107 N - 49.68N

b) tana - 2.735 Direction of field of Direction of Force E = E sino – 0.438107 Sinn Er + Ez Cose 2.95x10+ 0.43x10 6x120 tana – 0.372 = 0.372 2.735 a = fan (0.136) =(7.74° 2 = (1.740° II distance d=2d or doubled then Force is reduced to Fr. F = F = 49.68 = 12.42N c) 4