Physics Help Forum Electric field inside non-uniform shells

 May 9th 2011, 04:33 PM #1 Junior Member   Join Date: Oct 2010 Posts: 21 Electric field inside non-uniform shells Consider a spherical shell where the charge is distributed only on the outer surface of the shell but the charge is not uniformly distributed. Is the electric field zero everywhere inside the shell? explain why. Sky
May 9th 2011, 05:17 PM   #2

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 Originally Posted by Skyrim Consider a spherical shell where the charge is distributed only on the outer surface of the shell but the charge is not uniformly distributed. Is the electric field zero everywhere inside the shell? explain why. Sky
You might say "please explain why."

Two words: Gauss' Law.

I'm sure you have seen the derivation of the 0 E field inside an empty cavity based on the uniform charge density on the surface and the fact that the surface area depends on r^2 and the force law is an inverse square law, etc, etc. Obviously that doesn't apply here so think in terms of the electric flux, which measures a sort of "density" of the electric field. Since there are no charges inside the cavity, the flux through the outer area of the cavity is still 0. In fact this holds for any Gaussian surface inside the cavity. So there is no electric field inside the cavity.

Or just use Gauss' Law. It's much faster.

-Dan
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May 9th 2011, 05:37 PM   #3
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 Originally Posted by topsquark You might say "please explain why." Two words: Gauss' Law. I'm sure you have seen the derivation of the 0 E field inside an empty cavity based on the uniform charge density on the surface and the fact that the surface area depends on r^2 and the force law is an inverse square law, etc, etc. Obviously that doesn't apply here so think in terms of the electric flux, which measures a sort of "density" of the electric field. Since there are no charges inside the cavity, the flux through the outer area of the cavity is still 0. In fact this holds for any Gaussian surface inside the cavity. So there is no electric field inside the cavity. Or just use Gauss' Law. It's much faster. -Dan
Are you sure about that? As far as I know, Gauss' shell theorems apply only to uniform spherical charge distributions. With non-uniform charge distributions, there is no reason for the field on the Gaussian surface to be equal everywhere? Or is there something fishy?

May 9th 2011, 06:23 PM   #4

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 Originally Posted by Skyrim Are you sure about that? As far as I know, Gauss' shell theorems apply only to uniform spherical charge distributions. With non-uniform charge distributions, there is no reason for the field on the Gaussian surface to be equal everywhere? Or is there something fishy?
Gauss' theorem calculates the flux through a surface, not the distribution itself, so we don't care what the charge distribution is doing. Notice that the flux through a surface can only be non-zero when there is a charge density inside the Gaussian surface. For charge distributions outside the surface it is fairly easy to show that any flux coming in is matched by flux leaving the surface. (Else flux would accumulate inside the surface over time, which is absurd.)

-Dan
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May 9th 2011, 06:29 PM   #5
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 Originally Posted by topsquark Gauss' theorem calculates the flux through a surface, not the distribution itself, so we don't care what the charge distribution is doing. Notice that the flux through a surface can only be non-zero when there is a charge density inside the Gaussian surface. For charge distributions outside the surface it is fairly easy to show that any flux coming in is matched by flux leaving the surface. (Else flux would accumulate inside the surface over time, which is absurd.) -Dan
Yes, it is true that the total flux of any Gaussian surface inscribed within the shell is zero. That is not what I asked. I asked whether the electric field would be zero everywhere within the sphere if the charge distribution was non-uniform. Remember that most shell theorems are proved assuming uniform charge distributions. That is why you shouldn't disregard the non-uniformity in my question.

You are correct of course in saying that the total flux is zero, but my question is whether you are fully correct in concluding that the total flux being zero implies that the field everywhere is zero. That is dubious considering that this charge distribution is non-symmetric.

I still do not understand your explanation of the electric field being zero everywhere. Given your reasoning it does not follow smoothly from the fact that the flux is zero, given that the charge distribution is non-uniform.

May 9th 2011, 07:45 PM   #6

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 Originally Posted by Skyrim Yes, it is true that the total flux of any Gaussian surface inscribed within the shell is zero. That is not what I asked. I asked whether the electric field would be zero everywhere within the sphere if the charge distribution was non-uniform. Remember that most shell theorems are proved assuming uniform charge distributions. That is why you shouldn't disregard the non-uniformity in my question. You are correct of course in saying that the total flux is zero, but my question is whether you are fully correct in concluding that the total flux being zero implies that the field everywhere is zero. That is dubious considering that this charge distribution is non-symmetric. I still do not understand your explanation of the electric field being zero everywhere. Given your reasoning it does not follow smoothly from the fact that the flux is zero, given that the charge distribution is non-uniform.
Okay I think I can do this in one post. Hopefully the attachment is clear.

This proof of Gauss' theorem for electrostatics given here is based on the proof in Jackson's "Classical Electrodynamics," 2nd ed., section 1.3 pg. 30 - 31. The text is graduate level, but the Math is undergraduate level. The diagrams come from the same source.

Note that no charge distributions on or outside of the Gaussian surface are used in the proof.

Consider a point charge q inside a closed surface S. (See the top figure of the attachment.) Let r be the distance from the charge to a point on the surface, let n be the outward unit normal to S at that point, and let da be an area element on the surface S. If E at this point due to q makes an angle (theta) with n then the normal component of E times da is
$\vec{E} \cdot \hat{n} ~da = q \frac{cos(\theta)}{r^2}da$

Since E is directed along the line from the surface element to the charge q, cos(theta) da = r^2 d(Omega), where d(Omega) is the element of solid angle subtended by da at the position of the charge. (See top diagram in the attachment.) Thus
$\vec{E} \cdot \hat{n} ~da = q~d \Omega$

If we now integrate the normal component of E over the whole surface it is easy to see that
$\oint _S \vec{E} \cdot \hat{n}~da = \begin{cases} 4 \pi q & \text{ if q lies inside S} \\ 0 & \text{ if q lies outside S} \end{cases}$

This is for a single point charge. If there is a charge density outside the surface S there is no contribution to the E field inside. If there is a charge density inside the surface S Gauss' Law becomes
$\oint _S \vec{E} \cdot \hat{n}~da = 4 \pi \int _V \rho (\vec{x}) d^3x$

Notice that any charges outside the Gaussian surface S are ignored due to the flux argument I gave earlier. (The surface integral is equal to the flux and the net flux due to a charge outside the surface S is identically 0.)

-Dan
Attached Thumbnails

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May 9th 2011, 09:25 PM   #7
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 Originally Posted by topsquark Okay I think I can do this in one post. Hopefully the attachment is clear. This proof of Gauss' theorem for electrostatics given here is based on the proof in Jackson's "Classical Electrodynamics," 2nd ed., section 1.3 pg. 30 - 31. The text is graduate level, but the Math is undergraduate level. The diagrams come from the same source. Note that no charge distributions on or outside of the Gaussian surface are used in the proof. Consider a point charge q inside a closed surface S. (See the top figure of the attachment.) Let r be the distance from the charge to a point on the surface, let n be the outward unit normal to S at that point, and let da be an area element on the surface S. If E at this point due to q makes an angle (theta) with n then the normal component of E times da is $\vec{E} \cdot \hat{n} ~da = q \frac{cos(\theta)}{r^2}da$ Since E is directed along the line from the surface element to the charge q, cos(theta) da = r^2 d(Omega), where d(Omega) is the element of solid angle subtended by da at the position of the charge. (See top diagram in the attachment.) Thus $\vec{E} \cdot \hat{n} ~da = q~d \Omega$ If we now integrate the normal component of E over the whole surface it is easy to see that $\oint _S \vec{E} \cdot \hat{n}~da = \begin{cases} 4 \pi q & \text{ if q lies inside S} \\ 0 & \text{ if q lies outside S} \end{cases}$ This is for a single point charge. If there is a charge density outside the surface S there is no contribution to the E field inside. If there is a charge density inside the surface S Gauss' Law becomes $\oint _S \vec{E} \cdot \hat{n}~da = 4 \pi \int _V \rho (\vec{x}) d^3x$ Notice that any charges outside the Gaussian surface S are ignored due to the flux argument I gave earlier. (The surface integral is equal to the flux and the net flux due to a charge outside the surface S is identically 0.) -Dan
Being a high school junior, I don't understand vector calculus and I don't doubt those calculations one bit. Although I'm keen to argue that my question has nothing to do with flux. I understand it's vital to Gauss' law, but again, my question really is not about Gauss' Law's proof but about its applicability to non-uniform charge distributions in solving for electric fields.

IF what you say is completely correct, then the electric field on any Gaussian sphere that encloses no charge will always be zero. This would imply that non-zero electric fields do not exist.

Clearly that is not the case.

The lack of symmetry entails by nature that there be some non-zero field at some point in the sphere closer to the region that has higher average charge density??

May 10th 2011, 05:24 AM   #8

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 Originally Posted by Skyrim Being a high school junior, I don't understand vector calculus and I don't doubt those calculations one bit. Although I'm keen to argue that my question has nothing to do with flux. I understand it's vital to Gauss' law, but again, my question really is not about Gauss' Law's proof but about its applicability to non-uniform charge distributions in solving for electric fields. IF what you say is completely correct, then the electric field on any Gaussian sphere that encloses no charge will always be zero. This would imply that non-zero electric fields do not exist. Clearly that is not the case. The lack of symmetry entails by nature that there be some non-zero field at some point in the sphere closer to the region that has higher average charge density??
You posted this in the College/University Physics forum so I had assumed you were at least taking Introductory E and M.

Again, the proof does not specify anything about the nature of the charge distributions. Consider: the proof I gave mentions that the electric field due to a point charge outside the Gaussian surface S is 0. When we integrate over the charge distribution outside S we are essentially just adding up a bunch of zeros. So there is still no contribution from sources outside S.

As to there never being a 0 E field inside S you have a point. All matter is made up of charges. However it is the net charge inside S that matters and for all practical purposes, unless we induce a charge on a material, all matter is neutral. So even if we have matter inside S we can still have no (measurable) E field in there.

I had a thought last night about non-uniform charge distributions inside the Gaussian surface in case you were wondering about that as well. Again the proof uses first point charges, then integrates over the charge distribution so it still holds. On the other hand there is nothing that says that we have to be able to actually solve for E...the integral might be too intractable to solve analytically. We might need to resort to numerical methods.

I'm not certain if this is able to answer your questions. If not please let me know what you know...with you at the High School level (AP Calc or not) I don't know what your Math level is.

-Dan
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May 10th 2011, 05:54 AM   #9
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Hmm. Thanks for all the feedback man. This is very strange though you must admit. The problem was on the AP exam this year which is why it's bothering me so much. None of the other problems gave me as much confusion. I understand only single variable calculus (up to BC). But it is strange if the field should be zero everywhere inside a non-uniform shell as well as a uinform shell.

Consider the case:
Suppose you have a point charge P somewhere in space. Suppose now that you draw a Gaussian surface anywhere in space that does NOT enclose the point charge P. There will be no net flux through that surface since that surface does not enclose the charge. However, if I am correct in understanding what you said, that would also mean that there is no field within that Gaussian surface. That is not the case because then it means we can construct a Gaussian surface anywhere in space and claim that because it encloses no charge, it forces a zero field inside and on its surface.

May 10th 2011, 08:37 AM   #10

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 Originally Posted by Skyrim Hmm. Thanks for all the feedback man. This is very strange though you must admit. The problem was on the AP exam this year which is why it's bothering me so much. None of the other problems gave me as much confusion. I understand only single variable calculus (up to BC). But it is strange if the field should be zero everywhere inside a non-uniform shell as well as a uinform shell. Consider the case: Suppose you have a point charge P somewhere in space. Suppose now that you draw a Gaussian surface anywhere in space that does NOT enclose the point charge P. There will be no net flux through that surface since that surface does not enclose the charge. However, if I am correct in understanding what you said, that would also mean that there is no field within that Gaussian surface. That is not the case because then it means we can construct a Gaussian surface anywhere in space and claim that because it encloses no charge, it forces a zero field inside and on its surface.
You appear to be right about at least some of what you are saying. The thing that's bugging me is that I can't find where I'm wrong. Give me a couple of days on this and I'll get back to you.

-Dan
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