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Electric field in a hollow charged sphere

  1. Aug 1, 2013 #1
    I am wondering about why the electric field in a hollow sphere with charges on its surface would be zero.

    I have thought about the gaussian law argument for it. It only guarantees that the net number of electric field lines that pass the encloved surface are zero. But it says nothing about whether or not there are any field lines existing within it, only that theres a net of zero through any enclosed surface in the shell. The example I am thinking of is a shell of positive charges.

    The positive charges should all radiate an electric field in EVERY direction, so how could there be zero electric field lines at every point inside the shell.
     
  2. jcsd
  3. Aug 1, 2013 #2

    WannabeNewton

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    The key point of the Gauss's law argument is that inside the shell, ##\int _{S}E\cdot dA = 0## for any closed surface ##S##. Can this happen if ##E \neq 0## identically inside the shell?
     
  4. Aug 1, 2013 #3
    Can't it? You can have an electric field line passing through that surface and then exiting. net field is still zero but there is a field. I can see how the center would be zero due to fields cancelling.

    hm, you highlighted the word any, let me think about it a bit more.
     
  5. Aug 1, 2013 #4
    If you are talking about a conductive sphere, then you can generalize it to any geometry for the conductor or the void but I think you need to limit the argument to static electric field. First remember that if any field exists on the surface of the conductor ( here the inner surface), it is normal to the surface. Now assume a closed surface within the void nearly touching the inner surface. The integral of E.ds is zero on this surface and all the components are normal. This could be because: 1) E is zero everywhere on the void surface, 2) at some points E have opposite direction with others such that they sum is zero. But the latter cant be true because then you would find flux lines coming out of the surface at one point and going in at another point ( and zero inside the conductor) and this violate the curl-free property of static E . When field is zero on the surface, then its zero everywhere in void.
     
  6. Aug 2, 2013 #5
    If there is a field present inside a conductor it will cause a current simply because electrons will start to move when they experience a force. So in a static situation i.e. where there is no current flowing there can't be a field there.
     
  7. Aug 3, 2013 #6
    I still do not understand it. Why wouldn't there be an Electric field when the gaussian is zero? The positive charges have fields that are radial and emit everywhere. Why wouldn't those field lines radiate into the shell?
     
  8. Aug 3, 2013 #7
    well the idea is that there are still fields inside, emitted by every charge carrier residing on the conductor, it's just that the net field is zero everywhere within the hollow space. and you are right, just because there are net non-zero field lines present does not mean that the flux equals zero. the only parameter for a non-zero flux is if the gaussian surface encloses a object with the property [itex]\vec{\nabla}[/itex][itex]\bullet[/itex]E[itex]\neq[/itex]0, ie a charge carrier... so the arguments provided so far do not counter what you said, which still leaves us back to square one: why is the field everywhere inside zero?

    To everyone else: This calls for a mathematical proof, because you can keep saying "the E-fields sum up to zero", but.. this really doesn't say much about exactly how. I was/am still confused about this. So a mathematical proof would be appreciated.
     
  9. Aug 3, 2013 #8

    WannabeNewton

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    There is very simple to prove it but it does not use Gauss's law. It uses the uniqueness of solutions to Laplace's equation ##\nabla^{2} \varphi = 0## where ##\varphi## in our case is the electric potential. Assume we have a hollow conductor of any shape such that the space inside of the conductor itself is void of charge; then the potential inside satisfies ##\nabla^{2} \varphi = 0##. The boundary of this inside space is the conductor itself and the conductor is an equipotential i.e. ##\varphi = \varphi_{0} = \text{const.}## across the conductor. Hence one trivial solution for the potential everywhere inside the conductor is also ##\varphi_{0}##. But the uniqueness theorem states that this can be the only solution hence ##E = -\nabla \varphi = 0## unequivocally inside the conductor.
     
  10. Aug 3, 2013 #9
    the structure of this argument is flawed.
    and as far as the content..
    as long as the sphere is not infinitely large, the electrons will ALWAYS experience a force while residing on the sphere. It's just that the electrons keep moving around ie.. follow the path of a dynamic potential gradient, until electrostatic equilibrium is achieved. ie... until the electrons experience a net force of zero. there are still forces present.. there are still electric fields present.. just that the NET of everything is zero when in electrostatic equilibrium is reached.

    when there is no current, there are still fields present. All this says is that the net force is zero, this does not necessarily imply a zero E-field everywhere.
     
  11. Aug 3, 2013 #10

    WannabeNewton

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    The argument is by no means flawed. If the only force present is the Coloumb force then the only way the charges on the conductor can achieve an electrostatic state is if the net electric field within the material of the conductor vanishes identically for if it didn't then clearly there would be a non-vanishing Coloumb force on the charge carriers thus breaking the electrostatic state. If there are non-Coloumb forces present then yes there can be a non-vanishing net electric field within the material of the conductor such that the Coloumb force on the charge carries is balanced by the non-Coloumb forces present but clearly we are working with a system where the only force present is the Coloumb force.
     
  12. Aug 3, 2013 #11
    I think I figured out how to prove it using symmetry arguments. So we know the Net flux is zero since no charges are enclosed in the shell. So if we take any point, we know that the electric field radiates in a symmetric way, 360 degrees throughout the closed surface. This would imply there's a net flux, but gauss's law says there is no flux when there are no charges in the surface. Therefore the only way for there to be a net flux of zero would be for there to be no electric field lines.

    I wonder if this would apply if the shell wasn't circular. Would the electric field still be zero in a non-uniformly distributed shell with charge spread out evenly?
     
  13. Aug 4, 2013 #12

    WannabeNewton

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    The electric field inside of any conducting material of any geometry is zero in an electrostatic state. Anyways, if you want a general Gauss's law argument, Hassan gave one in post #4.
     
  14. Aug 4, 2013 #13
    The key point is to use both equations ∇.D=0 and ∇×E=0. The Laplace equation in WannabeNewton's reply has both equations in it, that's why he simply and rigorously proved the point.
     
  15. Aug 4, 2013 #14
    Ah, alright. I'll have to get back to the general case later when I can work out the mathematics on my own.
     
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