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In practice, one need not make these adjustments because presumably ##\rho_0\approx. 0##.

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Hey andrewkirk,

In practice, one need not make these adjustments because presumably ##\rho_0\approx. 0##.

I am hesitant to think that Gauss's law is an approximation. I agree that what you have done seems to force out the correct solution but is the inclusion of ##\rho_{0}## justified? Basically, it seems that ##\rho_{0}## is highly artificial because it is zero until the charge distribution becomes infinite, then ##\rho_{0}=\rho## to fix the problem.

I was pointing out a problem with the differential form of Gauus's law so I did not specify a Gaussian surface. From what I can tell however, the integral form has the same problem. If you consider the Gaussian surface to be a spherical shell and take the radius out to infinity, it predicts a non-vanishing electric field.

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anorlunda

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BruceW

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ooh, this is an interesting question. I believe the system you are talking about is called a "non-neutral Coulomb gas" in statistical physics. It is maybe quite a niche subject. I found a couple of papers talking about this problem in 2D http://www.sciencedirect.com/science/article/pii/092145349390271Q and https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.59.1001, which has application for superconducting films. From what I can work out, they include both ultraviolet and infrared cutoffs. In particular, the infrared cutoff means the electric potential obeys $$( \nabla^2 - {\lambda_c}^{-2}) V_{\lambda_c}(r) = -2 \pi \delta (r)$$ and this has something to do with the screening of the electric field. I guess the 2 cutoffs allows for calculations to be made, and then at the end of calculations, you can eliminate the cutoffs in a self-consistent way. Something along those lines, I don't know the detail really.

Edit : For clarity, $$\nabla^2 V(r) = -2 \pi \delta (r)$$ is the potential of point charge without using the cutoff.

Edit : For clarity, $$\nabla^2 V(r) = -2 \pi \delta (r)$$ is the potential of point charge without using the cutoff.

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In the case of a spherical uniform charge distribution, you can use symmetry to argue that the electric field is radially symmetric at the Gaussian surface. Gauss's law then states that the Electric field at the surface is only a function of the charge enclosed by the surface. Now take the radius of the sphere to infinity. Apparently something about Gauss's law fails in this limit. My question is which part of the argument is failing here?But that does not forbid other fields from charges outside the surface.

Not really. I'm just assuming all the charges are locked in place as if all space was filled with a good insulator with uniform charge density.I believe the system you are talking about is called a "non-neutral Coulomb gas"

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Why do you think it would be zero? It will be negligible because charge only exists where there are particles and the density of particles in the universe in negligible - because of all the empty inter-galactic space. But there's no reason why it need be zero. Since it's so small, we can adopt an assumption that it is zero for the purpose of calculation, and that's what Gauss's law does. We only need to drop that assumption in the case of a thought experiment that (I expect) is unrealisable in practice.it seems that ##\rho_{0}## is highly artificial because it is zero until the charge distribution becomes infinite, then ##\rho_{0}=\rho## to fix the problem.

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This is a huge and unverified assumption. You are saying that there is some unidentified quantity which is so small it has never been detected but fixes the problems encountered when dealing with an infinite charge distribution.But there's no reason why it need be zero. Since it's so small, we can adopt an assumption that it is zero for the purpose of calculation, and that's what Gauss's law does.

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What exactly is the assumption that you are concerned about, and why do you believe it to be untenable?This is a huge and unverified assumption.

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It sounds like you are saying that Gauss's law is only an approximation and that to get the exact answer, you would need to invoke a new term ##\rho_{0}##. The way you would calculate this term and when it becomes important is somewhat mysterious to me. As far as I know, Gauss's law is already exact without this corrective term.What exactly is the assumption that you are concerned about, and why do you believe it to be untenable?

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The ##\rho_0## term never becomes important in this universe because, if we accept the cosmological principle that the universe is homogeneous and isotropic at the large scale (on which most cosmology is based), it follows that the term is negligible because the universe is almost entirely empty space - and hence chargeless.

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This is a classical problem though and in classical ED, Gauss's law is exact.Yes that's right, the law is not exact. Quite apart from this issue, it is an approximation that only works when relativistic and quantum effects are small enough to be ignored.

What does this mean? I still don't know how you are defining ##\rho_{0}## so It's not clear to me if it is a physically or mathematically meaningful variable.The ρ0ρ0\rho_0 term never becomes important in this universe

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See post #2.I still don't know how you are defining ##\rho_{0}##

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jasonRF

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I tend to take a pragmatic engineering approach, so I would just solve the equation. One solution to ##\nabla \cdot \mathbf{E} = \rho / \epsilon_0## for constant ##\rho## is ##\mathbf{E} = \frac{\rho }{3 \epsilon_0}\mathbf{r}##. Since there is no preferred direction the fact that the field is radial seems reasonable by symmetry.

EDIT: in case you were wondering, I used no process to find this solution. When I first learned that ##\nabla \cdot \mathbf{r} = 3 ## I found it amusing for some reason and never forgot it. So when looking for a vector field that had a constant divergence it immediately came to mind. I also solved the PDE before attempting the integral form; the solution of the PDE was so simple and had such great symmetry that it was clear the integral form had to work.

Using the integral form of Gauss's law to solve for the field requires symmetry and a proper exploitation of that symmetry. I think there are very few problems that can be solved that way and I have goofed in this regards before, but this problem may be one . The charge density is uniform so is certainly spherically symmetric. If we assume the field is radial and only a function of ##r## and use a Gaussian surface that is a sphere of radius r, then the surface integral of the radial field yields ##4 \pi r^2 E_r## and the charge enclosed is ##\frac{4}{3} \pi r^3 \rho##. Combining these results, I get that Gauss's integral law gives ##E_r = \frac{\rho r}{3 \epsilon_0}##, in agreement with the solution to the PDE.

Does that make sense?

Jason

EDIT: in case you were wondering, I used no process to find this solution. When I first learned that ##\nabla \cdot \mathbf{r} = 3 ## I found it amusing for some reason and never forgot it. So when looking for a vector field that had a constant divergence it immediately came to mind. I also solved the PDE before attempting the integral form; the solution of the PDE was so simple and had such great symmetry that it was clear the integral form had to work.

Using the integral form of Gauss's law to solve for the field requires symmetry and a proper exploitation of that symmetry. I think there are very few problems that can be solved that way and I have goofed in this regards before, but this problem may be one . The charge density is uniform so is certainly spherically symmetric. If we assume the field is radial and only a function of ##r## and use a Gaussian surface that is a sphere of radius r, then the surface integral of the radial field yields ##4 \pi r^2 E_r## and the charge enclosed is ##\frac{4}{3} \pi r^3 \rho##. Combining these results, I get that Gauss's integral law gives ##E_r = \frac{\rho r}{3 \epsilon_0}##, in agreement with the solution to the PDE.

Does that make sense?

Jason

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jasonRF

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Jason

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Right! This is the issue I'm having.

Jason

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BruceW

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Oh, I see. I believe you should still be able to use the cut-offs to get an answer. I think that for electrostatics of charges that continue to infinity, you need to use the more complicated equations, rather than the usual Gauss' law. Sorry I don't know much about it, so I can't give a better answer. I guess it should be not so surprising that the usual method doesn't work, since trying to use the usual method, you get an infinite electric potential, which doesn't decay at infinity.Not really. I'm just assuming all the charges are locked in place as if all space was filled with a good insulator with uniform charge density.

Edit: well, we should expect that it doesn't decay at infinity. The point I was meaning is that the usual derivation for uniqueness of the electric potential makes use of the fact that it decays sufficiently fast at infinity https://en.wikipedia.org/wiki/Uniqueness_theorem_for_Poisson's_equation

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tech99

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E is defined as the force experienced by a unit charge. Force on an object can only be measured if we have a reference frame so we can measure acceleration, or alternatively, use the reaction of the force on another object. As your space is homogenous, we cannot measure force and we cannot measure an E. We cannot have a force if there is nothing for it to react against.

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This would seem another example of the age-old problem of infinities. I'd say the answer is there is no such thing as an infinite volume of charge (in all directions). For any finite volume the theorem gives the correct value of ∫∫

So similarly the potential of an infinite line of charge of finite charge density λ is also infinite. And so on.

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I like Serena

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It seems to me that your solution is correct for a uniformly charged ball.I tend to take a pragmatic engineering approach, so I would just solve the equation. One solution to ##\nabla \cdot \mathbf{E} = \rho / \epsilon_0## for constant ##\rho## is ##\mathbf{E} = \frac{\rho }{3 \epsilon_0}\mathbf{r}##. Since there is no preferred direction the fact that the field is radial seems reasonable by symmetry.

EDIT: in case you were wondering, I used no process to find this solution. When I first learned that ##\nabla \cdot \mathbf{r} = 3 ## I found it amusing for some reason and never forgot it. So when looking for a vector field that had a constant divergence it immediately came to mind. I also solved the PDE before attempting the integral form; the solution of the PDE was so simple and had such great symmetry that it was clear the integral form had to work.

Using the integral form of Gauss's law to solve for the field requires symmetry and a proper exploitation of that symmetry. I think there are very few problems that can be solved that way and I have goofed in this regards before, but this problem may be one . The charge density is uniform so is certainly spherically symmetric. If we assume the field is radial and only a function of ##r## and use a Gaussian surface that is a sphere of radius r, then the surface integral of the radial field yields ##4 \pi r^2 E_r## and the charge enclosed is ##\frac{4}{3} \pi r^3 \rho##. Combining these results, I get that Gauss's integral law gives ##E_r = \frac{\rho r}{3 \epsilon_0}##, in agreement with the solution to the PDE.

Does that make sense?

Jason

At the center of the ball the electric field is zero.

And the electric fields builds up radially to bigger and bigger values until we reach the surface of the ball.

This matches with the principle of super position, where we consider the electric field as a super position of the contributions from point charges.

If we take it to the limit, the electric field would become bigger and bigger ad infinitum, which is an impossibility.

This actually makes sense -- we have proven that an infinite universe with constant charge density is not possible.

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I'd say the answer is there is no such thing as an infinite volume of charge (in all directions).

I think you are both missing the point. It's not about whether or not an infinite charge density exists, its about why the solution provided by Gauss's law fails. Remember that Gauss's law works for other infinite charge distributions such as an infinite rod or infinite sheet of charge even though these things don't actually exist.we have proven that an infinite universe with constant charge density is not possible.

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Here's a discussion about this question:

http://www.sbfisica.org.br/rbef/pdf/332701.pdf

http://www.sbfisica.org.br/rbef/pdf/332701.pdf

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In the differential equation, ## \nabla \cdot E =\rho/\epsilon_o ##, there is always a homogeneous solution to ## \nabla \cdot E=0 ## that may need to get added to it. ## \\ ## A similar thing occurs when you consider ## B=\mu_o H+M ## and take ## \nabla \cdot B=0 ##. (Take divergence of both sides of the equation). This gives ## \nabla \cdot H=-\nabla \cdot M/\mu_o ## which gives ## H(x)=-\int \frac{(\nabla' \cdot M(x')) (x-x')}{4 \pi \mu_o |x-x'|^3} \, d^3 x' ##. The question is, where is the contribution to ## B ## from any currents in conductors? And the answer is that it shows up in the homogeneous solution ## \nabla \cdot H=0 ##.

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Vanadium 50

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The implicit assumption is that the electric field from a configuration of charges is unique. How do we know that? The proof involves something called a Helmholtz Decomposition. One of the conditions of this proof is that fields far away ("at infinity") vanish. This is true for a line of charge (symmetry reduces this to a 2-d problem) and sheet of charge (symmetry reduces this to a 1-d problem) but not a universe of charge.

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