Homework Answers
Electric force is given by
F = kQ1Q2/d^2
F13 = 9*10^(9)*6.42*10^(-9)*5.03*10^(-8)/(5)^2 = 1.16*10^(-7) N
horizontal componet of F13 = F13*cos(37) = 0.926*10^(-7) N
verticle componet of F13 = F13*sin(37) = 0.698*10^(-7) N
F23 = 9*10^(9)*2.38*10^(-9)*5.03*10^(-8)/(5)^2 = 0.43*10^(-7) N
horizontal componet of F23 = - 0.43*10^(-7) N
verticle componet of F23 = 0
Fx = 0.926*10^(-7) - 0.43*10^(-7) N = 0.496*10^(-7) N
Fy = 0.698*10^(-7) N
magnitude = sqrt(0.926^2 + 0.496^2)*10^(-7) = 1.05*10^(-7) N
direction
theta = arctan(0.496/0.926) = 28.17 degrees
direction 90 + 28.17 = 118.17 degrees from +x axis
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