Is a Gaussian surface truly arbitrary ?


by Perpendicular
Tags: arbitrary, gaussian, surface
Perpendicular
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#1
Jul22-13, 02:15 AM
P: 49
Consider a point charge q at the vertex of an arbitrary cube. If asked to consider the flux the cube experiences, q/8epsilon seems a natural answer, by constructing seven more such cubes to create an overall cube of 8 times the volume with q at its center.

But, this doesn't make sense to me. This seems to imply that every such cube is containing an octadrant of the spherical point charge. But it is a POINT charge. You can't divide it. You can have zero, or you can have q, but nothing in between, when it comes to creating your surface. So, is this an improper gaussian surface ? I am aware that , when it comes to choosing gaussian surfaces, some shapes ( like a Klein bottle ) aren't acceptable. Is this one of them ? More generally, is a discontinuity in p(r), E(r) or V(r) enough to nullify choosing a gaussian ? If I have to use a dirac delta to describe any of them , doesn't that mean my functions are mathematically improper, and that something's off ?
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mfb
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#2
Jul22-13, 01:12 PM
Mentor
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A gaussian surface "through" a point-charge is bad.

You can say that the flux through the 3 adjacent sides is zero due to symmetry, and calculate the flux through the other 3 sides with the 8-cube argument.
Alternatively, let the charge be distributed in a small sphere, and consider the limit of a size of 0.
Perpendicular
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#3
Jul23-13, 11:45 AM
P: 49
I ended up concluding that flux AT the point should be zero, because there, infinite field components in all directions cancel out to give zero field, kind of like an eye of the storm.

I dislike the sphere argument. Points aren't spheres. Points are points. So should it be, flux exists, and there is a charge enclosed, but Gauss' law doesn't hold ? That seems to be the way, right ? The alternative would be saying charge enclosed isn't well defined.

Delta Kilo
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#4
Jul23-13, 08:34 PM
P: 269

Is a Gaussian surface truly arbitrary ?


Gauss law holds. If you can't figure out how much charge you have in your volume, it's your problem, not Gauss'. And the problem is with point charge. Point charge is meant to be a charge whose size is so small compared to the distance to other charges that the exact size/shape does not matter. This is not true it your scenario: electrical field is defined by a force acting on a test charge and to measure it in your case you would have to bring test charge infinitely close to your point charge.

Now, it is up to you, you can choose to split your charge 8-ways or you can choose to put it all in one box. Mathematically it means you use a definition of a delta function (which is not a true function btw) as a limit of a sequence of functions, you just choose different base functions for that. For example you can use a sequence of uniformly charged spheres or radius ε, centered at the origin, or you can choose a sequence of cubes with diagonal (0,0,0)-(ε,ε,ε).

These 2 models give you different results for your flux figures, but it does not really matter as you cannot measure this flux in a physically realisable experiment. But if you then use those fluxes to solve a real problem, like predicting the results of a physically realisable experiment, the differences will cancel out and it won't matter which model to use.
Jano L.
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#5
Jul24-13, 06:16 AM
P: 1,027
Consider a point charge q at the vertex of an arbitrary cube. If asked to consider the flux the cube experiences, q/8epsilon seems a natural answer, by constructing seven more such cubes to create an overall cube of 8 times the volume with q at its center.
The basic situation for which the Gauss theorem applies is when the integral ##\oint \mathbf E\cdot d\mathbf S## is defined.
In your case, the field diverges on the surface and the above integral is not convergent, so one valid conclusion is to say that the flux is not defined and the Gauss theorem does not apply.


The other is to attempt to define the flux by some modification of that integral, but that does not seem to be useful.

Yet another way is to rephrase the task by replacing point charge by extended charged distribution, so there would be no infinite field and the integral will converge.

Finally, we can modify the surface so that it avoids the point charge. If ithe charge ends up inside, it contributes to the integral, otherwise it does not.

EDIT: I've edited a little bit to clarify.
Perpendicular
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#6
Jul24-13, 12:14 PM
P: 49
Why shouldn't flux AT a point be zero, though ? The outward and inward normal aren't distinct there. You can't pick one , so you must pick both, and hence whatever flux you have for one should perfectly cancel the other. Yeah, the individual quantities are undefined , but that does not necessarily mean their difference is.
WannabeNewton
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#7
Jul24-13, 12:17 PM
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The flux integral is over compact regular (or piece-wise regular) surfaces embedded in ##\mathbb{R}^{3}##. A point is neither of these (a singleton is compact but it doesn't represent a regular surface in ##\mathbb{R}^{3}##).
Perpendicular
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#8
Jul25-13, 02:26 AM
P: 49
Hmm. Could you tell me where I could look up the question of flux at a point via a point charge ? I have Griffith's and Jackson's.
WannabeNewton
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#9
Jul25-13, 02:46 AM
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Perhaps I'm misunderstanding you; what exactly do you mean by flux at a point? The closest thing I can think of is the fact that we can interpret ##\nabla \cdot E## evaluated at a point as the "infinitesimal flux of ##E## per unit volume" through an infinitesimal sphere centered at that point (so this is clearly local). I'll shift through both texts and see if I can find anything related to what you are trying to ask.
Perpendicular
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#10
Jul25-13, 11:15 AM
P: 49
Flux at a point would be referring to how the texts deal with the situation of calculating flux when a point charge is present on the surface.


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