Point charges in a equilateral triangle - typo in solution?

AI Thread Summary
The discussion revolves around calculating the forces on three equal point charges positioned at the corners of an equilateral triangle. A participant questions a potential typo in the solution regarding the angle calculations, specifically the use of arctan for determining the angle between forces. There is a consensus that symmetry simplifies the problem, suggesting that focusing on one charge first can clarify the forces acting on the others. Participants also emphasize the importance of correctly visualizing vector placements to accurately calculate angles. Overall, the conversation highlights common challenges in understanding force interactions in electrostatics.
chopnhack
Messages
53
Reaction score
3

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?
hw1q5.jpg
Lesson 2.1 Homework Solutions_Page_2.jpg

Lesson 2.1 Homework Solutions_Page_3.jpg
 
Physics news on Phys.org
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?
 
  • Like
Likes chopnhack
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...
 
  • Like
Likes chopnhack
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.
 
Back
Top