Net force on a charge - Coulomb's law

[rms5643]

Homework Statement

What is the force F on the 1 nC charge at the bottom?

Figure: http://session.masteringphysics.com/problemAsset/1384340/5/pic1.jpg

Homework Equations

Coulomb's law - k*qq/r^2

The Attempt at a Solution

1) After splitting up the net force into their respective vectors, I assume that the X vectors from the two 2.0 nC charges cancel out. As a result, I believe this problem only involves computation of the Y direction.

2) -6 nC on the 1.0 nC:

8.99*10^-9 * (6.6*10^-9) * (1.0^-9) / (5*10^-2)^2

* Pulls upwards, thus positive force

3) 2.0 nc on the 1.0 nC (Y component):

(8.99*106-9 * (2.0*10^-9) * (1.0*10^-9) / (5.10^-2)^2 ) * cos(45) * 2 (<-- Since there are two 2.0 nC charges contributing to the net Y force)

* Pulls downwards, thus negative force

4) Net X = 0
Net Y = (8.99*10^-9 * (6.6*10^-9) * (1.0^-9) / (5*10^-2)^2) - ((8.99*106-9 * (2.0*10^-9) * (1.0*10^-9) / (5.10^-2)^2 ) * cos(45) * 2)

Net Y = -0.000029 (Rounded to two sig figs)

This answer is incorrect. Where did I go wrong?

 

[Redbelly98]

Welcome to Physics Forums.

Your method is correct, so it looks like there is an arithmetic error. A negative final answer doesn't make sense: each +2 nC charge exerts less then half of the force exerted by the -6 nC charge -- plus there is the cos(45) factor for the +2 nC charges. So the net force must be in the same direction as that due to the -6 nC charge, hence upward.

Is the "6.6" just a typo in your post, or did you really calculate using that number?

Net Y = (8.99*10^-9 * (6.6*10^-9) * (1.0^-9) / (5*10^-2)^2) - ((8.99*106-9 * (2.0*10^-9) * (1.0*10^-9) / (5.10^-2)^2 ) * cos(45) * 2)

I would take out the common factor of [itex]8.99 \cdot 10^{+9} \cdot \frac{10^{-9} \cdot 10^{-9}}{(5 \cdot 10^{-2})^2}[/itex], then calculate the [itex]6\cdot 1 - 2\cdot 1 \cdot \cos(45) \cdot 2[/itex] part next.