# Applicability of Gauss law

1. May 29, 2007

### neelakash

1. The problem statement, all variables and given/known data

Is it possible to use Gauss' law to find intensity of a spherical charge distribution where the charge distribution
(i)is non-homogeneous in general...i.e. dependent on (r,theta,phi)
(ii)can be thought of as a made up of concentric spherical shells of different densities?

2. Relevant equations
3. The attempt at a solution

(i) I think -No.As the distribution varies with (r,theta,phi) the spherical symmetry is destroyed---the spherical layers should encounter the field lines symmetrically.But that is not allowed here.

(ii)the spherical layers should encounter the field lines symmetrically.So,here it is allowed.

Please confirm if I am correct.

2. May 29, 2007

### stunner5000pt

Gauss law is
$$\int \vec{E}\cdot d\vec{a} = \frac{1}{\epsilon_{0}} Q_{enc}$$

since the law only cares about the total charge enclosed, the manner in which it is distributed should not matter

do u mean electric field or intensity... by the way?

3. May 29, 2007

### Staff: Mentor

That looks correct to me.

While stunner5000pt is correct that Gauss's law holds true for both cases, it won't help you find the electric field unless there's a certain degree of symmetry, which is lacking in case (i).

4. May 29, 2007

### Niishi

If it is a spherical distribution and charge density depends on radial distance the field lines would be symetrical so you can use Gauss law.

5. May 29, 2007

### Niishi

If you had to find the electric field inside the spherical charge distribution and charge density depended only on r then you would consider a gaussian surface inside the spherical volume. To find the charge enclosed you would need to integrate change density rho(r) from r=0 to r=radius of Gaussian surface and multiply it 2 pi. Or maybe I am wrong?

6. May 29, 2007

### Niishi

Actually by 2 pi^2 i think.

7. May 29, 2007

### Staff: Mentor

Why multiply by anything? The integral of the charge density is the charge.

8. May 29, 2007

### neelakash

Nishi,the factor 2*pi comes to your mind possibly because of the phi integration.For systems with Azemuthal symmetry,the integration is done from phi=0 to phi=2*pi.

Thank you.

9. May 30, 2007

### Niishi

I thought if charge density was a function of r it would be given as such.

Yes I am wrong. I should not multiply by the angle because the angle is not a length.... the length is r dpheta. thank you.

10. Jun 7, 2007

### esalihm

answering the first question using gauss' law gives you the average intensity of the spherical charge distribution.
because you are using the enclosed charge and integrating the total are (which gives you the total volume) you get an average value for intensity.

and I think, no it cannot be thought of thin concentric spherical shells
because the charge density could be something different at the top of the shell than it is at the bottom.

11. Jun 7, 2007

### Staff: Mentor

As long as the distribution is spherically symmetric (which is the case for concentric shells) you can apply Gauss's law to find the field at any point.

12. Jun 7, 2007

### esalihm

I don't see any reason for the charge intensity to be same at the top and bottom of a non-conducting shell. and it should not make any difference their being concentric or not shold not make any difference
or is there a point I am missing?

13. Jun 7, 2007

### Staff: Mentor

A thin "shell" has uniform density and is spherically symmetric. The fact that they are concentric shells of various densities allows you to conclude that the charge distribution is spherically symmetric--that's what allows you to find the field using Gauss's law.

Perhaps you are misinterpreting the problem:

14. Jun 7, 2007

### esalihm

a thin shell does not need to have uniform density.
as I said top part may have more negative particles and bottom could be neutral or positive. is there anything to restrict this possibility?