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Gauss' Law -- When can it not be applied?

  1. Sep 29, 2014 #1

    MMS

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    Hello everyone. I've been asked the following question by someone and I'm not quite sure which is the correct answer. I'd appreciate some help with it.

    The question is pretty simple:

    When can't be Gauss's law applied?

    1. When the electric field of a point charge is dependent on the angles theta and phi.

    2. When the electric field of a point charge goes like 1/r^4.


    Thank you!
     
  2. jcsd
  3. Sep 29, 2014 #2

    berkeman

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    Staff: Mentor

    Can you write out Gauss's Law for us? What is the relevant equation? What do you think is the correct answer and why?

    If this is for schoolwork, I can move the thread to the Homework Help forums for you.
     
  4. Sep 29, 2014 #3

    MMS

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    I'm on my phone right now and it's pretty hard to write down equations. And no, this isn't for schoolwork. Again, someone asked me the question.

    As for what I think the correct answer is, I say it's 1.
    Gauss's law is usually used in cases of symmetry (spherical/cylindrical/planar) where we could determine that the electric field on a closed loop is constant on every point of it (dependent only on r) and fom there we could get it out of the integral. If it was dependent on an angle of some sort, then the electric field changes from point to another on it and therefore it doesn't work.

    Now the question is, am I right or wrong and why.
     
  5. Sep 30, 2014 #4
    Hi,

    You are right, and for exactly the reason you give. Well, almost. Gauss's Law applies in both cases. However, only in the case with spherical symmetry does it simplify the problem. With the electric field dependant only on r, the field will be both constant over, and perpendicular to, a spherical surface. This allows you to simplify the evaluation of Gauss's Law over the sphere and obtain a very quick expression for the magnitude of the field.
     
  6. Sep 30, 2014 #5

    MMS

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    Hey and thanks for the reply.
    You said only in the case of spherical symmetry it simplifies the problem. Not sure if you meant in general or in the particular case I've given. Anyway, I'll mention this since it doesn't hurt anyone. In case you meant in general then take for example a parallel plate capacitor. You find the electric field there by applying Gauss's law as well but the symmetry there isn't spherical.
    But it seems like we both agree on the same idea and answer so it's all good. :p

    I'd appreciate an explanation on what they mean in the 2nd answer because I don't really get it...

    Again, thank you!
     
  7. Sep 30, 2014 #6

    ZapperZ

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    In the case of an electrostatic condition, Gauss's law can always be applied. Now, it doesn't mean that you can solve it analytically every time, because that is a different issue.

    We use Gauss's law in a highly symmetric situation because we can solve for the field quicker and easier than by the brute force method of Coulomb's law. However, it doesn't mean that in other cases, it can't be applied. It still can, but you may not be able to solve it analytically, or even in a closed form.

    Zz.
     
  8. Sep 30, 2014 #7

    MMS

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    First off, thanks for the reply.

    I believe Gauss's Law can be also applied when we're not in an electrostatic condition. The charge invariance leads to the fact that the result of the surface integral on the electric field in Gauss's Law is dependent only on how much charge is confined within the closed Gaussian surface (In other words, it is not dependent on the sort of movement of the charges that are trapped within it).

    So what are you suggesting is the right answer between the two that are given? Also do you have an idea on what do they mean by answer #2?

    Thank you!
     
  9. Sep 30, 2014 #8
    Gauss' law is a consequence of the fact that the field of a point charge decreases like 1/r^2.
    Only in this case the flux through any surface enclosing the charge will be the same. The idea is that the area of the closed surface increases like r^2 and the field decreases like 1/r^2 so the two effects compensate and the flux through any closed surface is the same.

    In another universe where the field of a point charge will decrease like 1/r^4 Gauss's law will not work.
    The field will decrease much faster than the surface increases.
     
  10. Oct 1, 2014 #9

    Nabeshin

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    This is actually quite a subtle point. Which is taken as former, coulomb's law or Gauss'? (From an axiomatic point of view, either works.)

    If you take coulomb's law to now be 1/r^4, and we're still in our 3D universe, then yes gauss' law in its current form is no good.

    If you take gauss' law, now maybe we are in a 5D universe and hence the 1/r^4 dependence pops out, i.e. the 'old' coulomb's law is no good.

    As stated, I think the question #2 was meant to be simple but really is not without further elaboration...
     
  11. Oct 1, 2014 #10
    Good point about the number of dimensions of the "new" universe.:)
    I think the question means to stay in our universe (3D) and change Coulomb's law. But then, can we have a different Coulomb law in our universe?
     
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