Question

Please help me answer this physics problem.

Question 6: Three Plastic Spheres in a Triangle +i Three plastic spheres, each holding a charge with magnitude 6.00 x 10 C, are arranged as shown in the figure to the right. Charges q, and q, are positively charged while qs is negatively charged. +i (0, 0.004)m$93 (a) What is the Net Electric Force on charge qs? (-0.002, 0)m (0.002, 0)m 9i 42 Model(s) / Definition(s)/Theorem(s) /Law(s): Assumptions:

Answer 1

Here, given, q_1 = q_2 = q_3 = 6 times 10^- 9 C so, F_1 = k q_1 q_3/r^2 where, r^2 = (0.002)^2 + (0.004)^2 = 0.0002 since, magnitude of q_1, q_2 and q_3 are equal therefore, therefore Vertical-bar F_1 Vertical-bar = Vertical-bar F_2 Vertical-bar = kq^2/r^2 therefore F_1 = F_2 = (9 times 10^9)(6 times 10^- 9)^2/(0.00002) = 0.0162N Here, theta = p tan^- 1 [0.002/0.004] = 26.56^0 Due to symmetry: Horizontal component of et force on charge will cancell each other. And, therefore: F_net = F_1 cos theta + F_2.cos theta = 2F.cos theta therefore F_net = 2(0.0162)cos(26.56) therefore F_net = 0.029N and, in vector form: F_net = -(0.029N)j

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