Question

Three particles with charges of 1.0 uc.-1.0 uc, and 0.50 pC are placed at the corners A, a and Cof an equ lateral triangle with side length S-O72 m as shown in the figure below. Find the net force on the charge at point C. (Express your answer in vector form. Assume the *x direction is to the right and the ty direction is upward.) F-1-0.0085-0.01 47j | X N 1.0m -10 pC 0.50 C Noed Help?

Answer 1

Equilateral delta^le of side 0.72 m theta = 60 degree sin 60 degree = squareroot 3/2 cos 60 degree = 1/2 force on 'c' due to B charge Vertical-bar F_BC Vertical-bar = 1/4 pi epsilon_0 middot (-1.0 times 10^-6) times 0.5 times 10^-6/(0.72)^2 Vertical-bar F_BC Vertical-bar = 8.65 times 10^-3 (-i) N force OA 'C' due to charge A Vertical-bar F_AC Vertical-bar = 1/4 pi epsilon_0 times 1 times 10^-6 times 0.5 times 10^-6/(0.7)^2 Vertical-bar F_AC Vertical-bar = 9 times 10^-9 times 0.5 times 10^-2/0.52 = > Vertical-bar F_AC Vertical-bar - 8.65 times 10^-3 N Net force on charge 'c' is F = (F_AC cos theta + F_BC) i^^+ F_A sin theta (-j^^) F = (8.65 times 10^-3 times cos 60 degree - 8.65 times 10^-3) i^^N - 8.65 times 10^-3 sin 60 degree (j^^) N F = -4.33 times 10^-3 (i^^) N - 7.49 times 10^-3 (j^^) N (or) F = 4.33 times 10^-3 (-i^^) N + 7.49 times 10^-3 (-j^^) N