# Homework Help: Problem related to Coloumb's force

1. Aug 31, 2015

### gracy

1. The problem statement, all variables and given/known data
One more query.Two identical small bodies each of mass m and charge q are suspended from two strings each of length l from a fixed point.This whole system is taken into an orbiting artificial satellite, then find the tension in the two strings?

2. Relevant equations

3. The attempt at a solution
I know in satellite g=0. Therefore there won't be any weight acting downwards
only forces which exists between the balls will be
Coulombs force of repulsion + mutual gravitational force attraction between bodies + gravitational force of attraction of the satellite on the bodies

However, magnitude of above said gravitational forces will be negligible as compared to coulombs force. Hence this coulombs force of repulsion can be considered the only force between the balls which will repel them .But the answer given is T= (1/4πε0)*q^2/4l^2
This suggests r=2l
I want to know how to find "r"how r came out to be 2l?
probably the following image will suit the situation in the question.

2. Aug 31, 2015

### Orodruin

Staff Emeritus
In your image, what will the angle $\theta$ be in the absence of an external gravitational field?

3. Aug 31, 2015

### gracy

No.idea.help me out.Give me a hint.

4. Aug 31, 2015

### Orodruin

Staff Emeritus
What forces are acting on the masses? What will be the requirement for the masses being in equilibrium?

5. Aug 31, 2015

### gracy

I think i have put all that I know.

6. Aug 31, 2015

### Orodruin

Staff Emeritus
So if this is the only force, how will the masses end up in relation to each other? What is the only possible equilibrium position?

7. Aug 31, 2015

### gracy

is it wrong?

8. Aug 31, 2015

### Orodruin

Staff Emeritus
There is also the tension from the strings. You need to consider that the masses should be in equilibrium.

9. Aug 31, 2015

### gracy

Yes it is.I meant only force to balance tension.

10. Aug 31, 2015

### Orodruin

Staff Emeritus
So in which direction is the tension?

11. Aug 31, 2015

### gracy

opposite to repulsive coulomb force.

12. Aug 31, 2015

### Orodruin

Staff Emeritus
Which means that the configuration must look how?

13. Aug 31, 2015

### gracy

as if bodies were attracting each other

14. Aug 31, 2015

### Staff: Mentor

Lets do a thought experiment. Imagine that the two charges are held exactly where they are in the picture in the OP. You then let go of the charges. What happens?

15. Aug 31, 2015

### Staff: Mentor

This is a bit unclear.

Do this: Name the forces that act on each mass. Hint: Only two (significant) forces act.

Draw (or describe) the configuration of the masses. What angle will the strings make?

16. Aug 31, 2015

### gracy

I think theta will start increasing because of repulsion.

17. Aug 31, 2015

### gracy

Tension and repulsion.

18. Aug 31, 2015

### Staff: Mentor

Does it increase forever, or does it stop at one point?

19. Aug 31, 2015

### gracy

I don't know whether it is correct but i think tension force will be able to be split into one horizontal force and one vertical.The horizontal one would cancel the repulsive force.But one problem with this is then vertical force would be unbalanced so net force would be vertical so motion should be in vertical direction.
They would stop when The horizontal component of tension would cancel the repulsive force.

20. Aug 31, 2015

### Staff: Mentor

Good. The tension force from the string and the Coulomb electrostatic force.

21. Aug 31, 2015

### Orodruin

Staff Emeritus
So how does the setup look then?

22. Aug 31, 2015

### gracy

Am I supposed to draw the situation.

23. Aug 31, 2015

### Orodruin

Staff Emeritus
Try to describe it, what is the angle theta in your figure going to be?

24. Aug 31, 2015

### gracy

One thing i would like to clarify is the above picture was not a part of the question in op.I copied it from google images.So as we can see nothing bout theta is mentioned in the question.

25. Aug 31, 2015

### Orodruin

Staff Emeritus
I know that, but I am asking you what it will be anyway. It is fundamental for the understanding of the problem.