# Gauss's Law: Cylindrical Symmetry

1. Sep 30, 2004

### cepheid

Staff Emeritus
Problem : A long coaxial cable carries a uniform volume charge density $\rho$ on the inner cylinder (radius $a$), and a uniform surface charge density on the outer cylindrical shell (radius $b$). The surface charge is negative and of just the right magnitude so that the cable as a whole is electrically neutral. Find the electric field in each of the three regions: (i) inside the inner cylinder (s < a), (ii) in between the cylinders (a < s < b), (iii) outside the cable (s > b).

My thoughts: First of all, from the statement that the volume charge distribution on the inner cylinder was equal and opposite to the surface charge distribution on the outer one, I extracted this relation, hoping it would come in useful:

$$\iiint_{V_{\text{inner}}} {(\rho dV)} = - \iint_{S_\text{outer}}{(\sigma dS)}$$

Correct?

My next thought was that the electric field in region iii (outside the cable) should be zero everywhere. My reasoning was that any Gaussian surface enclosing a section of the cable encloses zero net charge. So it would not make any sense for there to be an electric field at any point on said surface.

My instinct tells me that by symmetry, the contribution to the electric field due to the outer cylinder in region i (inside the inner one) cancels itself out completely, but I'm not sure how to prove it.

I'm a bit new to this type of problem, so I appreciate hearing whether I'm right so far, and how to go about solving the rest of the problem.

2. Sep 30, 2004

### Dr Transport

Integrate each region separatly, you will see that the electric field increases linearly from the center to the edge of the inner cylinder, damps of as negative power of r between the cylinders then is zero outside because the net charge is zero. Only takes a few minutes to do the problem completely.

3. Sep 30, 2004

### Staff: Mentor

There is no need for any integration. Realize that the linear charge density (charge per unit length) must be the same for the volume and surface charge distributions.
Right.

Correct. To prove this, realize that symmetry allows you to apply Gauss's law for any radius, which shows that the field only depends on the charge within that radius.

Use Gauss's law to find the field in each region with no need to integrate.

4. Sep 30, 2004

### Dr Transport

exactly........

5. Sep 30, 2004

### cepheid

Staff Emeritus
I'm still confused. In region (i), I have decided to take your advice and use a cylindrical Gaussian surface of radius a . It encloses the region. In fact, it is exactly the same as the inner cylinder, but let's give it finite length l . First of all, does the electric field |E|outer due to the outer (surface) charge contribute to |E| in this region? I thought not, since the surface I just defined doesn't enclose any of the charge that is the source of this field. Another way to look at it is that inside the cylinder, every electric field vector coming at you from the outer cylinder's surface has an equal and opposite counterpart 180 degrees away. Fine, so Gauss's law reduces to:

$$\oint{\vec{E}_{\text{inner}} \cdot d\vec{a}} = \frac{Q_{\text{inner}}}{\epsilon_0} = \frac{\rho V}{\epsilon_0} = \frac{\rho}{\epsilon_0}\pi a^2 l$$

Normally, I would solve for |E| by taking it out of the integral sign, using some "knowledge of symmetry" to be fairly confident that the electric field is uniform over the entire Gaussian surface I just defined. Can anyone explain to me why this is so in this case? For instance, you're sitting on this Gaussian surface enclosing the volume charge. Who says all the electric field vectors are normal to the surface everywhere? Why can't you have some wacky diagonal one coming at you from some unassuming part of the charge distribution within? And how do you prove that |E| is also of the same strength everywhere on that surface? I'm sorry, but my prof didn't really justify these "symmetry arguments" in my mind. Maybe I'm just being stupid, and she felt it was obvious...

Another point that I'm not sure about. I assumed that the Gaussian surface I defined was closed. So the "can" must have a top and bottom. Is it also by "symmetry arguments" that we can assume the flux through the lid/base of the can is zero? I can see that being so for E_outer, but not for E_inner!

Last edited: Sep 30, 2004
6. Oct 1, 2004

### cepheid

Staff Emeritus

7. Oct 1, 2004

### ehild

This is a very general law ... If something has no reason to be different than it is not different.

You are right, you should include both the top and the bottom of the cylinder. But the flux means field lines perpendicular to the surface.The dot product in the Gaussian integrand is zero on the top and bottom surfaces because the field is radial everywhere, tangent to these surfaces but the very ends of the cable. This is a long coaxial cable, and you were asked to determine the field somewhere. In this case you can say "I determine the field far from the ends".

ehild

8. Oct 1, 2004

### Dr Transport

If the cylinder is long, ignore the ends, the electric field inside the inner cylinder is given by
$$\vec{E} = \frac{\rho \vec{r}}{8\pi}$$

between the cylinders

$$\vec{E} = \frac{\rho a^{2} \vec{r}}{8\pi r^{2}}$$

and zero outside both cylinders because of zero net charge.

Units may not be completely correct, but the dep[endencies are..........