# Conductors and co-axial cables?

1. Mar 6, 2007

### Swapnil

Conductors and co-axial cables!??

Say you have a long conducting hollow cylinder with some surface charge density $\rho$ which is inside another hollow (uncharged) cylinder. What would the electric field be outside the outer cylinder?

I am guessing that the e-field outside the outer cylinder should be zero because the inner cylinder would produce induced charges on the surface of the outer cylinder equal but opposite in magnitude and therefore the total charge enclosed in both of the cylinders would be zero.

Is this correct?

2. Mar 6, 2007

### marcusl

What does Gauss' law (in integral form), applied outside the outer cylinder, tell you? Can you spot the mistake in your reasoning?

3. Mar 6, 2007

### Swapnil

If we choose a cylindrical gaussian surface $S$ outside the outer cylinder then

$$\oint_{S} \vec{E}\cdot d\vec{S} = Q_\text{enc}$$

which is equal to zero since the total charge enclosed inside the surface is 0 (Q on the inner cylinder and -Q on the outer cylinder).

4. Mar 6, 2007

### marcusl

Your problem specifies only charge on the inner cylinder so Q enclosed is nonzero.

5. Mar 6, 2007

### Dick

If the outer cylinder is conducting, put the gaussian surface inside the outer cylinder.

6. Mar 6, 2007

### Swapnil

But wouldn't there be an induced charge -Q on the outer cylinder due to the inner cylinder?

7. Mar 6, 2007

### Swapnil

But the outer cylinder/conductor is not a shell -- it doesn't have any thickness.

8. Mar 6, 2007

### Dick

That's what a gaussian surface inside the outer conductor (is it a conductor) would tell you. That there is surface charge inside it.

Last edited: Mar 6, 2007
9. Mar 6, 2007

### Dick

That's an interesting problem in metaphysics.

10. Mar 6, 2007

### robb_

lol
Remember what must be true for the outer conductor in a statics situation. Hint: think about the E field.

11. Mar 6, 2007

### marcusl

Yes, but since the outer conductor must remain overall neutral there will be an opposite charge on its outer surface.

12. Mar 7, 2007

### Swapnil

Oh I see. So the net charge outside would then be Q (+Q on the inner conductor, -Q on the inner suface of the outer conductor, and +Q on the outer suface of the outer conductor).

13. Mar 7, 2007

### Dick

Exactly as you say.