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Gauss law and insulating sphere

  1. Aug 28, 2011 #1
    imagine a charge placed inside a closed conducting shell (a hollow metal box or sphere)

    Feynman says:
    - no static distribution of charges inside a closed conductor can produce any fields outside. The fields on the two sides of a closed conducting shell are completely independent.

    but, gauss law says:
    - the flux of E is proportional to the charge inside

    how is this not a contradiction ? why is the field outside the metal box is 0, doesn't the charge inside the box still has an effect, according to Gauss law ?
     
  2. jcsd
  3. Aug 28, 2011 #2
    The charge inside the sphere would induce an equal amount of opposite charge on the sphere, so if you applied Gauss' Law outside, the net charge would be zero, hence giving zero electric field.
     
  4. Aug 28, 2011 #3

    vela

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    I read the paragraph in the Feynman Lectures that I think you're referring to. It's a bit misleading. He left out some important details about how shielding is done.

    You're right that if you have an uncharged hollow metal sphere and place a charge inside the cavity, there will be a field outside of the sphere. Say the charge inside is positive. It'll induce a negative charge on the inside surface which will leave an excess of positive charges on the outside surface. We're assuming that the hollow sphere is isolated, so even though the negative charges collect on the inside surface and the positive charges collect on the outside surface, the net charge is still zero. The positive charges on the outside will result in a field outside the sphere. Gauss's Law is fine and well.

    With shielding, however, you ground the conductor so charges can flow to or from the conductor. In this case, the inner surface would again become negative, but the positive charge on the outer surface would attract negative charges which would flow in through the ground connection and cancel the positive charges. In this case, the net charge of the sphere is negative, which cancels the positive charge inside the cavity, so that the total charge inside the Gaussian surface is 0 and hence the field outside will be 0.
     
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