Question

Use the worked example above to help you solve this problem. Consider three point charges at the corners of a triangle, as shown in the figure, where q1 = 6.42 x 10-9 C, q2 =-2.38 x 10.9 C, and 93 = 5.03 x 10-8 C.

Answer 1

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