Question #222839

A +100μC charged is place at the origin. A +200μC charge is place 2m above the origin and a -300μC charged is placed on 3m to the right of the origin. What is the magnitude and direction of the net force acting on the charge placed at the origin?

Expert's answer


The charges and forces are shown in the figure. Let q1=100μC=104Cq_1 = 100\mu C = 10^{-4}C, q2=2×104Cq_2 = 2\times 10^{-4} C, q3=3×104Cq_3 = 3\times 10^{-4} C.

Then the forces according to the Coulom's law are:


F12=9×1091042×10422=45NF_{12} = \dfrac{9\times 10^9\cdot 10^{-4}\cdot 2\times 10^{-4}}{2^2} = 45N

F13=9×1091043×10432=3NF_{13} = \dfrac{9\times 10^9\cdot 10^{-4}\cdot 3\times 10^{-4}}{3^2} = 3N

Since the angle betwen them is right one, the magnitude of the net force can be found from Pyphagorean's theorem:


F=452+3245.1NF = \sqrt{45^2 + 3^2} \approx 45.1N

The direction is:


θ=360°arctan(F12F13)=360°arctan(453)274°\theta = 360\degree - \arctan\left( \dfrac{F_{12}}{F_{13}} \right)=360\degree - \arctan\left( \dfrac{45}{3} \right) \approx 274\degree

Answer. 45.1 N, 274 degrees with positive x-axis.


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