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I Gauss's law and symmetric charge distributions

  1. Aug 26, 2016 #1
    Having read several introductory notes on Gauss's law, I have found it very frustrating that when the author comes to discussing the standard examples, in which one considers symmetric charge distributions, they do not explicitly discuss the symmetries of the situation, simply stating that, "by symmetry, the electric field must be radial" ,etc.

    My question is, for each of the following cases:

    1. Spherical symmetry

    2. Cylindrical symmetry

    3. Planar symmetry

    What are the exact arguments, based on symmetry, why the electric field should point in a particular direction?


    My thoughts on the matter are as follows, please let me know if I've understood it correctly at all:

    1. In the case of a spherically symmetric charge distribution, if the electric field (at each point) did not point radially outwards (or inwards), then one could rotate the system and find that the electric field would change in each case. However, the charge distribution is symmetric, and so such a rotation will not change it's electric field, and hence, it must be that the electric field is radial at each point.

    2. In the case of cylindrical symmetry, where we assume that the cylinder is infinitely long, if the electric field had a component parallel to the cylinder, then upon reflecting the system (along the axis of the cylinder), this would change the electric field, contradicting the symmetry of the situation, hence there can be no component parallel to the cylinder. Furthermore, if the electric field pointed in a particular direction, at some angle to the surface of the cylinder, then upon rotation of the system, the electric field would change, violating the cylindrical symmetry of the charge distribution. Hence, again, the electric field is radial to the cylinder. [N.B. In this case, if the cylinder were not infinite, then we couldn't use the argument for there being no component parallel to the surface of the cylinder, since in this case, translating along the cylinder would change the amount of charge "above" and "below" and so the electric field would be stronger in one direction than the other.]

    3. Finally, in the case of planar symmetry, again assuming that the plane is infinite in extent, if the electric field pointed in a particular direction at some angle to the the plane, then one could rotate the plane and the electric field would change. Furthermore, translating along any particular direction in the plane should not change the electric field, and so there cannot be any components parallel to its surface. Accordingly, the electric field must be perpendicular to the plane. [N.B. If the plane were not infinite in extent, then translating along the plane would mean the there would be more charge in one direction than another and so the electric field would change in this case, and hence, it would not necessarily have to point perpendicularly to the surface.]
     
  2. jcsd
  3. Aug 26, 2016 #2

    jtbell

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    Yes, those are the arguments that I've always used when introducing the applications of Gauss's law to my students.
     
  4. Aug 26, 2016 #3

    Ok great, thanks for taking a look. It's a relief to know that I've understood it correctly. Would the reasoning I put for why the symmetry arguments don't hold for the cases of a finite length cylinder, and a finite sized plane be correct?

    Also, when we make these symmetry arguments are we also considering the net electric field at each point? The reason I ask is that, for example, in the case of an infinite plane of charge, couldn't one otherwise have a case in which the electric field is oriented at some angle to each point on the surface, but in such a way the the electric field is radially symmetric about each point. In this case, one could move to another point and it wouldn't look any different - the symmetry would not be broken?! Is the point that each charge produces an identical electric field and so their resultant field lines will all point in a particular direction at each point (for example, you won't have some pointing north-west and some east etc. , they will all point north-west), and hence, this direction must be perpendicular to the surface, otherwise, I could walk along the surface in one direction "against" the electric field lines, and then rotate 180 degrees and walk in the same direction as the field lines, thus violating the planar symmetry?!

    For example, I can see why it must be the case that the field must be perpendicular in the planar case

    image.jpeg

    Since, from the couple of cases on the diagram above, even if the electric field is oriented at a different angle at each point, the planar symmetry will be broken.

    But in the cases of cylindrical and spherical symmetry, I can't really argue why these configurations aren't allowed by symmetry:

    image.jpeg
     
    Last edited: Aug 26, 2016
  5. Aug 26, 2016 #4

    jtbell

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    Imagine that you have in front of you some charge distribution and its associated E field. Think of the field arrows/lines as "attached" rigidly to the charge distribution.

    You close your eyes, I rotate or translate the system, and you open your eyes again.

    If the charge distribution "looks the same" as before, i.e. you can't tell that I've done anything to the system by comparing the charge distribution to what it was before, then the field configuration must also "look the same" as before.

    For your cylindrical example, suppose I flip the cylinder end for end. Then your field arrows now point downwards instead of upwards. The field "looks different" even though the charge distribution "looks the same".

    Similarly for your sphere, rotate it 180° around any axis that lies in the plane of the paper. Now the field arrows point in a clockwise direction instead of counterclockwise.
     
  6. Aug 26, 2016 #5
    Ah, good point. I hadn't thought of those as valid symmetry operations - I've been thinking too much in terms of 2D geometry instead of 3D. So, for the sphere, is the point that whatever angle I orient electric field lines to the surface of the sphere this will always lead to inconsistencies (apart from if they are orthogonal to the surface), since we can rotate the sphere about any angle and the charge distribution doesn't change and so the direction of the electric field at each point shouldn't change. The only way this can be true for any rotation is if the field lines point radially from the centre.

    Also, is the reasoning I gave for the planar symmetry case (with the accompanying diagram) correct?

    Referring to my original post:
    Are the arguments I gave why the symmetry arguments don't work for finite length cylinders and finite size planes correct?
     
  7. Aug 26, 2016 #6

    jtbell

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    Correct. For the finite cylinder or plane, if I simply moved (translated) it some distance while your eyes were closed, you'd be able to tell when you opened your eyes that I'd moved it. So the field would also "move" along with it, and would therefore change (in general) at any particular fixed point in space.
     
  8. Aug 26, 2016 #7

    vanhees71

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    I don't understand the latter few statements. E.g., if you rotate a sphere around its center, nothing happens to it. Thus if you have a spherically symmetric charge distribution the same must hold for the electric field, i.e., it must not change under these rotations either, which immediately tells you that the vectors must point radially out from the center of the and have the same range at a given distance, i.e., ##\vec{E}=E(r) \vec{e}_r##. To determine ##E(r)## for a given charge distribution you need Gauß's Law, and the most easy way in this case is to apply the integral form to a sphere of radius ##r##.

    The cylinder of a finite length is more tricky. It's only symmetric under rotations around the cylinder axis. It's more simple to argue here in terms of the electrostatic scalar potential. The symmetry means that there's no change of the function under rotations, i.e., changing ##\phi##, which means that you must have
    $$\phi(\vec{r})=\phi(\rho,z).$$
    This implies that the field is of the form
    $$\vec{E}(\vec{r})=E_r(\rho,z) \vec{e}_{\rho} + E_z(\rho,z) \vec{e}_z.$$
    I don't see a way to figure out the field from a simple argument.

    If however, you have an infinite cylinder homogeneously charged, it's easy again, because then ##\phi=\phi(\rho)## and ##\vec{E}=E_{\rho}(\rho) \vec{E}_{\rho}##, and you can apply the trick with Gauß's Law in integral form. This shows that this method is unfortunately restricted to the very symmetric special cases.
     
  9. Aug 26, 2016 #8
    Ok cool, that's what I thought.

    Have I said anything erroneous? I really want to make sure that I understand these symmetry arguments correctly.

    This is what I was trying to allude to in my explanation for why one can't conclude that the electric field only has a radial component in the case of a finite sized cylinder, since as one moves up and down the length of the cylinder, the amount of charge "above" and "below" oneself will differ, and so the electric field must necessarily differ from point to point along the z direction.
     
    Last edited: Aug 26, 2016
  10. Aug 26, 2016 #9

    vanhees71

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    Now I see what you mean. It's of course correct. This reflects to law of how a vector field transforms by definition. If you change from one frame of reference to another by an ISO(3) transformation (i.e., a combined rotation and translation in the usual affine Euclidean 3D space), ##\vec{r}'=\hat{D} \vec{r}-\vec{a}##, it transforms like
    $$\vec{E}'(\vec{r}')=\hat{D} \vec{E}(\vec{r})=\hat{D} \vec{E}(\hat{D}^{-1} (\vec{r}'+\vec{a})).$$
    That's precisely what you describe in words.
     
  11. Aug 26, 2016 #10
    Is this in reference to my last comment:
    So have I understood the symmetry arguments correctly in the three cases that I mentioned, then?
     
  12. Aug 26, 2016 #11

    vanhees71

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  13. Aug 26, 2016 #12
    Ok, great. Thank you, and @jtbell for your help!
     
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