Coulomb's Law Problem- Why is this incorrect?

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SUMMARY

The discussion focuses on calculating the force on a 1.0 nC charge in an equilateral triangle configuration with a 2.0 nC charge and a -2.0 nC charge. The user initially miscalculates the net force as zero by incorrectly applying Newton's third law without considering the vector nature of the forces involved. The correct approach involves recognizing that the attractive force from the -2.0 nC charge and the repulsive force from the 2.0 nC charge do not cancel each other out completely, leading to a net force of 1.80e-4 N directed at zero degrees from the x-axis.

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charges.jpg

Homework Statement


What is the force F on the 1.0 nC charge in the figure? Give your answer as a magnitude and a direction.


Homework Equations





The Attempt at a Solution



Applying coulomb's law for the 1.0 nC charge and the 2.0 nc Charge: F= kq1q2/r^2 F=(8.99e9)(1.0e-9C)(2.0e-9C)/(.01m)2
F= 1.798e-4 N

Fx= (1.798e-4)(cos60) = 8.99e-5 N
Fy=(1.798e-4)(sin60) = 1.56 e-4 N

and If I do the action reaction pair between the 1.0 nC charge and the negative 2.0 nC charge, I will get the same magnitudes with opposite signs.

So all together, I come up with a net force of zero, but my computer says this is wrong. What am I doing incorrectly?
 
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Well I can't see the picture yet because it's still pending approval, and I'm not the best at electricity problems, but I do know that just because the action reaction pair is the same (which it always is because of Newton's third law) , that doesn't mean that the net force is zero. Sorry I can't help more until I can see the picture for myself!
 
Maybe if I describe it:

There are three charges arranged in an equilateral triangle with distance 1.0 cm between each of the charges as the "sides." The top charge is 1.0 nC, the bottom left charge is 2.0 nC, and the bottom right charge is -2.0 nC. Each of the angles in this "triangle" are 60 degrees. The question asks what is the force on the top 1.0 nC charge.
 
Try to think about what is happening. You know that the charge will be going towards the negative 2nC charge because opposites attract.

I'd suggest drawing a free body diagram :)
 
Ok, I just did that, but I'm still thinking that the +2.0 nC repels equally as much as the -2.0nC attracts, so don't those cancel out when you sum up all the forces on the 1.0 nC charge on both x &y? Don't the 2.0 nC & -2.0 nC exert the same magnitude force, just in opposite directions?
 
Some forces will cancel out, but not all of them. Look at your free body diagram. The only way the net force could be zero is if all the arrows pointed in opposite directions. However, I do believe that some arrows will be pointing in the same direction. Think about it
 
Ok, yes I see it. So the total force would be 2(8.99e-5)= 1.80 e-4 N

Part B:

Assume the x-axis is directed from the 2.0nC charge to the -2.0 nC charge. Give the direction.

So, the direction is zero degrees from the x-axis, correct? That's what it looks like on my diagram.
 
I believe so. However, I don't pretend to be an expert at using Coulomb's Law, as it's been several months since I've seen these kinds of problems. From what my diagram looks like, I can only assume that it'd be zero degrees from the x-axis.
 

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