A charge, q1, of +17 micro-coulombs is located at x = 0, y = 9.2 cm, a charge q2 of -42 micro-coulombs is located at x = 0, y = -3.9 cm, an unknown charge of q3, is located at x =- 3.9 cm, y = 0 cm, and a charge, q4, -6 micro-coulombs is located at x = +10.7 cm, y = 0 cm. The direction of the total electric force on q4 is 135.4 degrees from the +x axis, counter-clockwise. What is the unknown charge, q3, in micro-coulombs? If negative, include a negative sign in your answer.
FA = force by charge at A = k q1 q4/AD2 = (9 x 109) (17 x 10-6) (6 x 10-6)/(0.141)2 = 46.2 N
FAy = Y-component of force FA = FA Sin = 46.2 Sin40.7 = 30.13 N
FAx = X-component of force FA = - FA Cos = - 46.2 Cos40.7 = - 35.03 N
FB = force by charge at B = k q2 q4/BD2 = (9 x 109) (42 x 10-6) (6 x 10-6)/(0.114)2 = 174.5 N
FBy = Y-component of force FB = FB Sin' = 174.5 Sin20 = 59.7 N
FBx = X-component of force FB = FB Cos' = 174.5 Cos20 = 164 N
Fc = force by charge at C = k q3 q4/CD2 = (9 x 109) q3 (6 x 10-6)/(0.146)2
FCy = Y-component of force FC = 0
FCx = X-component of force FC = (9 x 109) q3 (6 x 10-6)/(0.146)2
Fy = Y-component of the net force = Fnet Sin135.4
Along the Y-direction , force equation is given as
FAy + FBy = Fy
30.13 + 59.7 = Fnet Sin135.4
Fnet = 128 N
Along the X-direction , force equation is given as
FCx + FBx + FAx = Fx
FCx + (164) + ( - 35.03) = (128) Cos135.4
FCx = - 220.11 N
magnitude of force Fc = 220.11 N
(9 x 109) q3 (6 x 10-6)/(0.146)2 = 220.11
q3 = 86.88 x 10-6 C
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