# Gauss' Theorum Applications

1. Sep 14, 2007

### mitleid

Two identical conducting spheres each having a radius of 0.460 cm are connected by a light 1.20 m long conducting wire. Determine the tension in the wire if 69.2 (micro)C is placed on one of the conductors. (Hint: Assume that the surface distribution of charge on each sphere is uniform.)

Round your answer to three significant figures. Take the Coulomb constant to be K_e = 8.99 *10^9 N*m^2/C^2

E = K_e (q/r^2) Field on conducting sphere (needed to produce force on string)
E = (1/2*$$\Pi$$*$$\epsilon$$) q/r^2 Field on -infinite- charged wire (where Epsilon = 8.854*10^-12 C^2/N*m^2)

Force = qE

So far I'm fairly stumped by this problem, though I've been applying Gauss' Theorum most of this afternoon... The charge must be distributed symmetrically, first throughout the sphere in which it is placed. I can solve for the charge at the surface of the first sphere to get something like 2.94 * 10^10 N-C.

From here, I'm thinking the positive charge will be distributed through the conducting wire to the other sphere equally... So each sphere will end up having the same surface charge density and field, which produces a tension on the wire.

Even some conceptual help here would be thoroughly appreciated. Thanks.

2. Sep 14, 2007

### learningphysics

Hint: once the charges are distributed evenly... think of the wire just like a rope, or a string... that is preventing the spheres from moving away from each other... what is the force one sphere exerts on the other?

3. Sep 15, 2007

### dynamicsolo

How did you arrive at this value? Note that the two spheres are conductors: where on the spheres is the net charge going to end up? What is the effective behavior of the charge found on each sphere?

This part is fine: the mutual repulsion of the equally-charged spheres leads to the tension in the connecting wire. So the critical question at this point is: how do we treat the fields around each sphere? There is an important theorem they should have told you about that will be valuable here... (As you said, the charge on each sphere is [spherically] symmetrically distributed.)

4. Sep 15, 2007

### mitleid

The net charge ends up distributed symmetrically over both of their surfaces. Will some of the charge be distributed in the wire as well?

I'm using the equation E = (q/r^2)Ke... but do I divide the charge by two first? Unfortunately I'm away from my notes and can't isolate the proper theorum... Also, I have used the radius of .6 cm as well as .25/2 + .6 as the radius in my calculations, but it hasn't worked as of yet. Still missing something...

5. Sep 15, 2007

### learningphysics

No, you can ignore the charge distributed on the wire...

Think of the spheres as point charges (where the distance between them is the distance between the centers of the spheres)... what is the force one sphere exerts on the other?

Where are you getting the .25/2 and .6?

6. Sep 15, 2007

### mitleid

Incredible! Thanks very much for the help, this is how I got the solution :

Based on Coulomb's Law, Force(of E field) = Ke*[(q1*q2)/r^2)], where q1 and q2 are equal to one another here (just divide the charge applied to one sphere by two, and square it).

The tricky part was, as usual, using the correct radius. The important distance is from the center of each sphere, which really treats them both as point charges. Plugging in, I got an answer of 1.21 N.

Thanks again for the help. :)

7. Sep 15, 2007

### dynamicsolo

The theorem I was referring to concerns a finite spherically symmetrical charge distribution. Your conducting spheres will each have (effectively) half the total charge "uniformly" arranged on their surfaces, thus each is such a symmetrical distribution. The theorem states that the total charge can then be treated as if it were a point at the center of the sphere. (IIRC, this was originally proved by Newton for mass and gravity using *geometry* -- the proof became simpler with his calculus methods.) So the radii of the conducting spheres is an irrelevant detail.