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Gauss Law with Charged Sphere and Infinite Parallel Plates

  1. Oct 1, 2012 #1
    Hi guys, this is more of a conceptual question, so I hope you guys can give me a detailed explanation if possible.

    1. The problem statement, all variables and given/known data

    Find the Electric Field inside a Charged Sphere (charge only on the surface) and between two Parallel Plates (oppositely charged) separated by some distance d.

    2. Relevant equations

    Gauss' Law: [itex]\oint EdA = \frac{Qencl}{\epsilon}[/itex]


    3. The attempt at a solution

    So for the charged sphere, I choose my Gaussian Surface S to be a sphere inside. I see that S encloses no charges, so Qencl = 0. Thus, E = 0.

    However, for the parallel plates, I know how to get the answer by finding E of each plate individually and summing them vectorally. The resultant E is NOT 0. I am very confused because if I choose my Gaussian surface S to be a cube or whatever in the space between the parallel plates, S contains no charges. By my logic from the charged sphere, E should equal 0, which is NOT the case.

    So basically, can anyone explain to me why I have to sum the E fields from individual plates instead of doing it with my second method?

    Thank you very much!
     
  2. jcsd
  3. Oct 1, 2012 #2

    SammyS

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    Gauss's Law only says that the net flux though the sphere is zero, not that the field is zero.
     
  4. Oct 1, 2012 #3
    Electric field is zero inside a uniformly charged sphere.
     
  5. Oct 1, 2012 #4

    SammyS

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    The charge on the sphere does not contribute to any field inside the sphere, however, ...
     
  6. Oct 1, 2012 #5
    Sorry, but i posted my reply on the basis that the OP hasn't mentioned about any other external field. :smile:
     
  7. Oct 1, 2012 #6

    SammyS

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    Good point, I edited my previous post ! :smile:
     
  8. Oct 2, 2012 #7
    Hi guys, thanks for the responses; however, I am still quite confused...

    My physics prof is not the best. He only wrote down Gauss' Law as I have written above and never told us that it's equal to the flux nor what the law meant. He merely proceeded to solve questions afterwards. Same for the sphere; we were merely told that the E field inside a uniformly charged sphere is zero, but never proven mathematically why...

    For the flux, from what I gathered from wikipedia and etc, is the amount of E lines passing through an element of area. We take the dot product between the two so we can better describe the strength of the flux due to the varying angles that E and dA can meet.

    So if the flux is zero, that means either E and dA are orthogonal, or either E or dA is zero. So for the charged sphere, since E is in the radial direction for the sphere, which is parallel to the normal of dA, it cannot be case 1. dA cannot be zero because my S has a surface area. Therefore E must be zero...

    Can you guys tell me what is wrong with my reasoning here? I also know that the Shell Theorem proves the E field inside is zero, but my prof did mention this result can also be proven with Gauss' Law. This course is really tough for me because I am never told how these equations are derived or what they mean, so I have to sometimes come up with weird explanations/interpretations for myself...
     
  9. Oct 2, 2012 #8

    SammyS

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    Yes, I'm sure many of my students have said similar things about me, even though I stated the law precisely and explained about flux, etc. What teachers say, as contrasted to what students hear them say can be two very different things. :smile:
    Gauss's Law: [itex]\displaystyle \oint \vec{E}\cdot d\vec{A} = \frac{Q_{encl}}{\varepsilon_0}[/itex]

    Yes, what you say is true for the flux passing through an element of area. That's at essentially one point. However, the integral in Gauss's Law is over an entire surface enclosing some volume. So at some places, [itex]\displaystyle \vec{E}\cdot d\vec{A}[/itex] may be positive, at other places, [itex]\displaystyle \vec{E}\cdot d\vec{A}[/itex] may be negative, and at yet other places, [itex]\displaystyle \vec{E}\cdot d\vec{A}[/itex] may be zero.

    By Integrating [itex]\displaystyle \vec{E}\cdot d\vec{A}[/itex] over an entire Gaussian surface, the total flux can be zero, without the electric field being zero at anyplace on the surface.
     
  10. Oct 2, 2012 #9
    Thank you very much Sammy!

    That line about E ⋅dA being positive or negative really helped me out!
     
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