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Applicability of Gauss law

  • Thread starter neelakash
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
1,444
2

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
Doc Al
Mentor
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1,206
(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
11
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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
11
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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
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Actually by 2 pi^2 i think.
 
  • #7
Doc Al
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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
511
1
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
11
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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
90
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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
Doc Al
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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
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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
Doc Al
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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:

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
90
0
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?
 

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