# Point charges in a equilateral triangle - typo in solution?

1. Jul 4, 2017

### chopnhack

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
k = 9.0x109NM2C-2
F = k⋅(Q1⋅Q2)/r2
3. 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???

2. Jul 4, 2017

### Staff: Mentor

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?

3. Jul 4, 2017

### BvU

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...

4. Jul 4, 2017

### chopnhack

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.

5. Jul 4, 2017

### chopnhack

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

6. Jul 4, 2017

### chopnhack

Thanks all - I had to redraw to see it. Thank you again.

7. Jul 4, 2017

### Staff: Mentor

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.