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Why is E field zero in a sphere of charge?

  1. Jul 26, 2006 #1
    Lets say you have a spherical or cylindrical shell (nonconducting) with a surface charge density of, say, eta. I see that virtually everywhere I look, everyone says the E field is zero at all points within the shape.

    I understand that there is no charge enclosed within the spherical shell, so gauss' law states that the net flux is zero. However, this doesn't rigorously mean that the E field inside of the shell is zero at all points. For example, if you drew an arbitrary box out in space away from the sphere, the next flux would be zero, but the E field obviously would not be. flux in = flux out. Saying the field is zero at all points because the flux is zero doesn't mean much to me. Just like drawing a gaussian surface around a dipole...there are fields, everything just sums to zero.

    Its also obvious that from the symmetry the E field would be zero in the center of the sphere, but off axis its not so obvious.

    I'm hoping that someone can give me an integral summing up the fields from each surface element for a point off center and within the sphere (or cylinder), and show me that it is zero. If not I would also appreciate a thought experiment or some not-so-nasty proof.

    Intuitively, if we had a spherical shell of mass, if I were at the center I would not move. But if I was off axis a little....apparently I wouldn't move because the g field would be zero at all points within the sphere of mass. I'm not convinced. Need math not an incorrectly applied law.

    EDIT:
    By the way I'm taking an electrical engineering EM class, which is being taught by a grad student. I asked him and he didn't know for sure. If I get the answer from him I'll post it.

    Thanks for any help.
     
    Last edited: Jul 26, 2006
  2. jcsd
  3. Jul 27, 2006 #2

    Andrew Mason

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    You don't need to do a complicated calculation. The shell must be a uniformly charged sphere with no external fields.

    Think of a concentric Gaussian sphere within that shell of radius r<R (R=radius of shell). According to Gauss' law, the net flux through that surface is 0 because there is no enclosed charge.

    That does not tell you, as you point out, that the field at any portion of that surface is 0. But symmetry does. Since all points on that Gaussian surface are identical to all others in terms of their position in relation to the charged surface, the only solution is that the field at each point is 0.

    Since you can create such a Gaussian surface for any point inside the sphere, the field everywhere inside the sphere is 0.

    AM
     
    Last edited: Jul 27, 2006
  4. Jul 27, 2006 #3

    Office_Shredder

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    jason, you can try directly calculating the field at a point through integration if you want (assuming you've done multivariable calculus). Let the surface be covered in charges of value dq, and integrate to find the net field at a specific point
     
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