Question

A charge 5.01 nC is placed at the origin of an xy-coordinate system, and a charge -2.04 nC is placed on the positive x-axis at x = 3.98 cm . A third particle, of charge 5.99 nC is now placed at the point x = 3.98 cm , y = 2.98 cm find the x and y component for the force acted on the third charge by the other two as well as the total force.

Answer 1

force due to 5.01 nC

F 1 = 9* 10^9* 5.01* 10^-18* 5.99 / (0.0398^2 + 0.0298^2)

F1 = 1.09* 10^-4 N

angle of F1 with horizontal

x = arctan (2.98 / 3.98) = 36.82

force due to - 2.04 nC

F2 = 9* 10^9*2.04* 5.01* 10^-18 / 0.0298^2

F2 = 1.036* 10^-4 N

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net horizontal force

Fx = F1 cos x = 8.726* 10^-5 N

net vertical force

Fy = F2 - F1 sin x = - 3.95* 10^-5 N

net force,

F^2 = Fx^2 + Fy^2

F = 9.58* 10^-5 N

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