# Point Charges on Equilateral Triangle - Need Geometry Help

## Homework Statement

http://www.ridemtl.com [Broken]

F = k(q1q2/r^2)

## The Attempt at a Solution

I know pretty much what to do for the problem, but I cannot get the right answer. I think my issue lies within what angles I am using for the components. Because it is an equilateral triangle, all angles within that triangle are 60 deg. But when I use 60 deg to calculate the components in the x & y directions, I can't come up with the right answer.

F of 2 on 1 = 0.0579 N
F of 3 on 1 = 0.0358 N

x: 0.0579 cos 60 -i, 0.0358 cos 60 +i
y: 0.0579 sin 60 -j, 0.0358 sin 60 -j

Sum these together and use Pythagoras for overall magnitude
The magnitude the above produces is 0.0818 N

I am not getting the right answer. Thus I think my problem lies in the angles I am using. Can anyone shed some light?

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PeterO
Homework Helper

## Homework Statement

http://www.ridemtl.com [Broken]

F = k(q1q2/r^2)

## The Attempt at a Solution

I know pretty much what to do for the problem, but I cannot get the right answer. I think my issue lies within what angles I am using for the components. Because it is an equilateral triangle, all angles within that triangle are 60 deg. But when I use 60 deg to calculate the components in the x & y directions, I can't come up with the right answer.

F of 2 on 1 = 0.0579 N
F of 3 on 1 = 0.0358 N

x: 0.0579 cos 60 -i, 0.0358 cos 60 +i
y: 0.0579 sin 60 -j, 0.0358 sin 60 -j

Sum these together and use Pythagoras for overall magnitude
The magnitude the above produces is 0.0818 N

I am not getting the right answer. Thus I think my problem lies in the angles I am using. Can anyone shed some light?
Nothing wrong with what you have written so far. Could be a problem with what you haven't written.

I would have liked to see what you thought 0.0579 sin 60 was equal to, for example.

I would then like to see how you thought the 0.0818 came about???

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