Why Does SQRT(3) Factor into the Coulomb Force in an Equilateral Triangle?

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Homework Help Overview

The problem involves three identically charged particles positioned at the corners of an equilateral triangle, and the original poster is attempting to calculate the force experienced by each charge. The poster has calculated a force expression but notes a discrepancy involving a factor of SQRT(3).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to consider vector addition of forces rather than simple arithmetic addition due to the directional nature of the forces. There is a suggestion to draw a vector diagram to visualize the forces acting on one of the charges. One participant questions the angle used in the vector analysis, suggesting a misunderstanding of the geometry involved.

Discussion Status

The discussion is ongoing, with some participants providing guidance on how to approach the vector addition of forces. There is a focus on clarifying the angles involved in the force vectors, indicating that multiple interpretations of the geometry are being explored.

Contextual Notes

There may be assumptions regarding the symmetry of the problem and the nature of the forces acting on the charges that are under discussion. The original poster's calculations and the resulting factor of SQRT(3) are points of contention that have not yet been resolved.

Chronos000
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Homework Statement



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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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.
 
Could you please explain how I get an angle of 120. I can only see that it would be 60.
 
Force Pic.png


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

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