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Superposition of charges

by Chronos000
Tags: charges, superposition, triangle
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Chronos000
#1
Mar10-11, 06:48 AM
P: 80
1. The problem statement, all variables and given/known data

3 identically charged charges are placed at corners of an equilateral triangle of sides a. I need to find the force felt by each charge.

To do this I have calculated the force from charges - "2 on 1" and then "3 on 1". The answer i get is (2*q^2)/4*Pi*e*a^2. This is not correct however. A factor of SQRT(3) times the answer i got is correct. I don't see how to get to this
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Mr.A.Gibson
#2
Mar10-11, 07:16 AM
P: 41
The charges are not pushing in the same direction so they cannot be added together, you'll have to add vectors and do some trig.

If you draw out the diagram then draw the direction of force on each particle, i.e. two forces on each, all forces the same size, this'll help you see they are pushing in different directions and hence cannot be added.

You then draw a vector diagram for the force on one of the particles. Draw the arrow for one force, where that arrow ends start the arrow for the other force. This should form an isosceles triangle with an angel of 120 between the two force vectors.

Since you know the size of one of the force vectors, and hence two sides of the triangle and the angle between them you can now calculate the length of the other side of the triangle. This will be the size of the resultant force on each charge.

Hope this is clear it's much easier to draw it than explain it words.
Chronos000
#3
Mar10-11, 07:32 AM
P: 80
Could you please explain how I get an angle of 120. I can only see that it would be 60.

Mr.A.Gibson
#4
Mar10-11, 08:05 AM
P: 41
Superposition of charges

Click image for larger version

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The green line is the resultant force, the angle in black is the 120 degree angle.


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