The vector sum of the electric forces exerted on a particle

[msw1]

Homework Statement

Two particles are located on the x axis of a Cartesian coordinate system. Particle 1 carries a charge of +3.0 nC and is at x = -30 mm and particle 2 carries a charge of -3.0 nC and is at x = 30 mm . Particle 3, which carries a charge of +5.0 μC , is located on the positive y axis 90 mm from the origin. What is the magnitude of the vector sum of the electric forces exerted on particle 3?

Relevant Equations

F = k(Q1Q2/r^2)
c = sqrt(a^2+b^2)

$$r_{13}=r_{23}=\sqrt{(30*10^{-3})^2+(90*10^{-3})^2}=\sqrt{9*10^{-3}}\\<br /> F^E_{13}=F^E_{23}=9E9\cdot\frac{5*10^{-9}\cdot3*10^{-9}}{9*10^{-3}}=1.5*10^{-5}\\<br /> \theta=tan^{-1}(\frac{90*10^{-3}}{30*10^{-3}})=71.565\,degrees\\<br /> \vec{F}^E_{13}=<F^E_{13}cos\theta, F^E_{13}sin\theta> = <4.743*10^{-6}, 1.423*10^{-5}>\\<br /> \vec{F}^E_{23}=<F^E_{23}cos(-\theta), F^E_{23}sin(-\theta)> = <4.743*10^{-6}, -1.423*10^{-5}>\\$$
Because particle 3 is positively charged, particle 1 is positively charged, and particle 2 is negatively charged, and the magnitude of the force by 1 on 3 is the same as magnitude of the force by 2 on 3, the vertical components of the force should cancel out, while the horizontal components should be additive, so my final answer was
$$F=2*(4.743*10^{-6})=9.487*10^{-6}$$
But this is incorrect. Any help would be appreciated.

 

[haruspex]

Homework Statement:: Two particles are located on the x-axis of a Cartesian coordinate system. Particle 1 carries a charge of +3.0 nC and is at x = -30 mm and particle 2 carries a charge of -3.0 nC and is at x = 30 mm . Particle 3, which carries a charge of +5.0 μC , is located on the positive y-axis 90 mm from the origin. What is the magnitude of the vector sum of the electric forces exerted on particle 3?
Relevant Equations:: F = k(Q1Q2/r^2)
c = sqrt(a^2+b^2)

$$r_{13}=r_{23}=\sqrt{(30*10^{-3})^2+(90*10^{-3})^2}=\sqrt{9*10^{-3}}\\<br /> F^E_{13}=F^E_{23}=9E9\cdot\frac{5*10^{-9}\cdot3*10^{-9}}{9*10^{-3}}=1.5*10^{-5}\\<br /> \theta=tan^{-1}(\frac{90*10^{-3}}{30*10^{-3}})=71.565\,degrees\\<br /> \vec{F}^E_{13}=<F^E_{13}cos\theta, F^E_{13}sin\theta> = <4.743*10^{-6}, 1.423*10^{-5}>\\<br /> \vec{F}^E_{23}=<F^E_{23}cos(-\theta), F^E_{23}sin(-\theta)> = <4.743*10^{-6}, -1.423*10^{-5}>\\$$
Because particle 3 is positively charged, particle 1 is positively charged, and particle 2 is negatively charged, and the magnitude of the force by 1 on 3 is the same as magnitude of the force by 2 on 3, the vertical components of the force should cancel out, while the horizontal components should be additive, so my final answer was
$$F=2*(4.743*10^{-6})=9.487*10^{-6}$$
But this is incorrect. Any help would be appreciated.

I get the same as you do. Do you know what the answer is claimed to be?

But it is much better to keep everything algebraic and avoid plugging in numbers until the final step. Many advantages, one of which is a better strike rate on getting help on this forum!

You can avoid finding the angle. You had to find the distance anyway, so the cos and sin values can be had by simple division.

 

[msw1]

I get the same as you do. Do you know what the answer is claimed to be?

But it is much better to keep everything algebraic and avoid plugging in numbers until the final step. Many advantages, one of which is a better strike rate on getting help on this forum!

You can avoid finding the angle. You had to find the distance anyway, so the cos and sin values can be had by simple division.

Thanks for the feedback! I found my mistake. It was very subtle. The charge of particle 3 is given in microcoulombs, not nanocoulombs. Very annoying!

 

[haruspex]

Thanks for the feedback! I found my mistake. It was very subtle. The charge of particle 3 is given in microcoulombs, not nanocoulombs. Very annoying!

Well spotted.