Applicability of Gauss law

In summary, Gauss's law can be used to find the electric field inside a spherical charge distribution, as long as the distribution is spherically symmetric.
  • #1
neelakash
511
1

Homework Statement



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?

Homework Equations


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.
 
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  • #2
neelakash said:

Homework Statement



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?

Homework Equations


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.

Gauss law is
[tex] \int \vec{E}\cdot d\vec{a} = \frac{1}{\epsilon_{0}} Q_{enc} [/tex]

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
neelakash said:
(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.
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
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
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
Actually by 2 pi^2 i think.
 
  • #7
Niishi said:
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?
Why multiply by anything? The integral of the charge density is the charge.
 
  • #8
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
Doc Al said:
Why multiply by anything? The integral of the charge density is the charge.

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

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.

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
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
esalihm said:
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.
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
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
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:

neelakash said:
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?
 
  • #14
Doc Al said:
A thin "shell" has uniform density

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?
 

1. What is Gauss law and why is it important?

Gauss law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It is important because it allows us to easily calculate the electric field in situations with high symmetry, and it is a key tool in understanding the behavior of electric charges and fields.

2. What types of surfaces can Gauss law be applied to?

Gauss law can be applied to any closed surface, whether it is a simple geometric shape like a sphere or a more complex shape like a cube. The only requirement is that the surface is closed, meaning there are no holes or gaps in it.

3. Can Gauss law be used for both point charges and continuous charge distributions?

Yes, Gauss law can be used for both point charges and continuous charge distributions. For point charges, the closed surface is chosen to enclose the charge, while for continuous charge distributions, the surface is chosen to enclose the entire distribution.

4. How does Gauss law relate to Coulomb's law?

Coulomb's law is a mathematical expression of the force between two point charges, while Gauss law is a statement about the relationship between the electric field and the charge distribution. Gauss law can be derived from Coulomb's law, and it provides a more general and powerful way of understanding electric fields.

5. What are some real-life applications of Gauss law?

Gauss law has numerous real-life applications, including calculating the electric field inside a capacitor, analyzing the behavior of charged particles in a particle accelerator, and understanding the electric field around a charged conductor. It also has practical applications in the design of electronic devices and in the study of atmospheric electricity.

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