Question

Practice EX 1 physics 205 Spring 2018 Please show all your work to receive full credit for the correct answers. Particles of charge +75 με 48 με and-8sc are placed in a line. The 48 μC ís 0.35 m from each of the other charged particles. a) Calculate the net force on the charge +75 μC due to other charges. (magnitude and direction)

Answer 1

Three charges are at A, B & C AB = BC = 0.35m Force on charge A, due to charge at B = F_1 Vertical-bar F^vector_1 Vertical-bar = K q_1 q_2/AB^2 = K (75 times 48)/(0.35)^2 times 10^-12 N Direction of F^vector_1 is along BA^vector Force on charge A, due to charge at C = F_2 Vertical-bar F^vector_2 Vertical-bar = K q_1 q_3/AC^2 = K (75 times 85)/(0.7)^2 times 10^-12 N Direction of F^vector_2 is along AC^vector Now Vertical-bar F^vector_1 Vertical-bar = K (29387.7551) times 10^-12 N, along BA^vector Vertical-bar F^vector_2 Vertical-bar = K (13010.2041) times 10^-12 N, along AC^vector Since Vertical-bar F^vector_2 Vertical-bar < Vertical-bar F^vector_2 Vertical-bar , direction of is along BA^vector Resultant force [magnitude] = Vertical-bar F^vector Vertical-bar Vertical-bar F^vector Vertical-bar = Vertical-bar 29387.7551 - 13010.2041) times (9.1 times 10^9) times 10^-12 N = 149035.714 times 10^-3 N tildeequal 1.490 times 10^2 N, along BA^vector