Point charges in a equilateral triangle - typo in solution?

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chopnhack
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Homework Statement


Three point charges each carrying a charge of 11.0 µC are located at the corners of an equilateral triangle of side 15.0 cm. Calculate the magnitude and direction of the force on each charge.

Homework Equations


k = 9.0x109NM2C-2
F = k⋅(Q1⋅Q2)/r2

The Attempt at a Solution


See attached. My main problem lies with the solution provided which shows on ii) the solution for the left vertex as arctan of 41.9/72.6 whereas I got 72.6/41.9 as in tan = opp/adj. Is it a typo or have I totally missed the mark?
Handwritten worked solution for three equal point charges at the corners of an equilateral triangle, with a diagram of the charges and force vectors on the left and calculations of forces F21 and F31 and their resultant of 183.8 N at 240 degrees.
Solution diagram of an equilateral triangle with charges q1, q2, q3 at the corners, side 0.15 m, showing forces F21, F31, F12, F32, F13, F23 and a resultant of 83.83 N, with 60 degree angles marked, followed by worked equations for F21 and F31 each equal to 48.4 N.

Handwritten worked solution on a printed page: part (ii) gives the force on q3 as components F = -72.6 i - 41.9 j N, magnitude 83.8 N, and angle arctan(-41.9/-72.6) = 30 degrees, third quadrant 210 degrees, followed by part 6 solving for distance r from a Coulomb force equation.
 
Isn't it easier to find the force on charge 2 first, and then use that to get the magnitude and direction of the forces on charges 1 and 3?
 
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Hard to read, your writing is...
However, I think you could see from symmetry alone that the 30 degree angle (or 210, if you want), is correct. Not 240 !

Your division finds the angle between the y-axis and the force; we are used to reporting the angle between the force and the postiive x-axis...
 
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Chestermiller said:
Isn't it easier to find the force on charge 2 first, and then use that to get the magnitude and direction of the forces on charges 1 and 3?
Hi Chester! I honestly picked the first vertex and went to work. Being that they are all equal charges, I didn't see the difference in locations.
Can you comment on my tan theta question?

Thanks as always.
 
BvU said:
Hard to read, your writing is...
However, I think you could see from symmetry alone that the 30 degree angle (or 210, if you want), is correct. Not 240 !

Your division finds the angle between the y-axis and the force; we are used to reporting the angle between the force and the postiive x-axis...

Sorry about that! The work was to only be for internal use and discarded, practice to understand the application - I literally have a dozen more of these to do before I will feel comfortable moving on to the next chapter and they each take quite a bit of time... so fast and sloppy it is!

I think I see what you mean, I believe that my force diagram is misdrawn... what I have labelled resultant in yellow is actually the extension of the force between 2 and 1... and the erased green line was correct - the resultant would have been drawn somewhere between them!

I am still having trouble visualizing the proper placement of the vectors to give me tan 41/72 though...

Thanks
 
Thanks all - I had to redraw to see it. Thank you again.
 
chopnhack said:
Hi Chester! I honestly picked the first vertex and went to work. Being that they are all equal charges, I didn't see the difference in locations.
Can you comment on my tan theta question?

Thanks as always.
I'm too lazy to look at that, since the problem is so much easier if the focus is on charge 2. For the other two charges, the direction will be the bisector of the included angle; that will be at an angle of 30 degrees to each of the two sides.