# Static electric calculation help

1. Jan 24, 2005

### Gogsey

Two 5.0 gram conducting spheres are hung side by side from insulating threads. The threads are each 0.4 m long. Both spheres are given the same negative charge. Due to mutual static electric repulsion the spheres now hang with a 40 degree angle between the threads(20 degrees each side of the vertical).

a) What is the magnitude of the electric force each sphere experiences?

b) What is the charge (in Coloumbs) on each sphere?

c) What is the electric field strength experienced by the sphere on the left?

Thank-you very much if you can help.

2. Jan 27, 2005

### maverick280857

What have you done so far?

3. Jan 27, 2005

### Gogsey

Well first of all, the force of the repelling charges created a distance between the spheres. That distance is (X).

X = 2Length sin theta

There is also a force keeping them in equilibrium opposite in direction from the repelling force, This is (F)

F = Kq1q2
x squared

The thirg and inal force is FG =mg sin theta

together this is

X cubed = 2LKq squared
mg

So, using this formula, i found the charge, which should be equal in magnitude and be like charges because they repel each other.

Then i used F = k q squared
d squared

But, part (b) asks you to calculate the charge, but i have already done this in part (a), thats what makes me think its incorrect. If this is correct, why would they ask for the charge in part (b) aswll?

Really its only part (a), i'mconfused about.
Part (c) is, ok, and part (b) is probably ok aswell, but i need part (a) before i can do the other parts.

Last edited: Jan 27, 2005
4. Jan 29, 2005

### maverick280857

You can express the charge in terms of the force on each sphere. I have not seen your mathematics in detail; essentially though this problem requires setting up the condition for equilibrium of either sphere when subjected to the tension in the string, its own weight and the electrostatic force (of repulsion). If you have done this, you can express either quantity in terms of the others (this however does not necessarily have a physical significance unless some process relates the left and right hand sides).

This reminds me of a problem from I.E. Irodov General Physics (to be precise, 3.10) which might interest you as well.

Now for part (a) draw a freebody diagram of one sphere. Since you are concerned only with the electric force, compute first the distance between the charges at this given equilibrium position. This turns out to be 2Lsin(20). The charges -q and -q give rise to the mutually repulsive electric force, which is given by

$$F_{elec} = \frac{1}{4\pi\epsilon_{0}}\frac{(-q)(-q)}{(2L\sin(20))^2}$$

This is directed to the left. Now clearly q is an unknown here but don't let this trouble you. The other forces on the sphere are

(a) its weight acting vertically downwards
(b) tension (resolve this yourself)

Resolving forces and applying the condition for translational equilibrium will give you a relationship between the forces. If you now eliminate the tension T between them, you can get a single equation from which you can compute q (please try this...). Now this q can be substituted back in the Coulomb's law equation to get an answer to part (a). Part(b) and part(c) are then trivial.

Hope this helps.

Cheers
Vivek

(My 500th post!! YAY!!! )

5. Jan 29, 2005

### heman

Congrats..Turn it into 5000.

6. Feb 3, 2005

### maverick280857

Mate, with your good wishes...I hope to do just that!! :tongue: