Electrostatics problem with pith balls hung using threads

[Sunil Simha]

Homework Statement

Please open the attachment for the question

Homework Equations

1)coulomb's law

The Attempt at a Solution

I'm at a loss regarding what to do. Upon reading the question (ignoring the figure), I assumed that the charges will form a square and thus the angle between two adjacent threads will depend on the length of the thread. But the solution key given to me said option (d) was correct. Please help.

 

[mfb]

That problem statement is ill. The angle depends on the charges and masses - for negligible charges, it will be nearly zero, which is not covered in any answer.
Without or with negligible gravity, the masses will arrange like a tetrahedron, and d is correct. But that requires at least one charge to point "upwards" relative to the point where they are attached.

 

[Sunil Simha]

Without or with negligible gravity, the masses will arrange like a tetrahedron, and d is correct.

Without gravity, won't they get arranged as a square in the plane of the ceiling to which the threads are attached?

Thanks in advance mfb

 

[mfb]

As far as I know, a square is not the ideal arrangement, a tetrahedron is better.

 

[Sunil Simha]

But with a tetrahedral arrangement, won't one of the threads slacken? Also, could you please explain how the tetrahedral arrangement is favored to the square one where the balls are far apart from each other and thus the system has the lowest potential energy.

 

[mfb]

won't one of the threads slacken?

Electrostatic repulsion keeps it there.

With a tetrahedral arrangement, the angle is somewhere at ~110°, compared to ~90° for the square. On the other hand, you have 3 nearest neighbors instead of 2, so it would need a calculation to compare both arrangements.