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Gauss Law, Charge, Cavity

  1. May 18, 2009 #1

    Air

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    1. The Question
    An isolated conductor of arbitrary shape has a net charge of [tex]+10\times10^{-6}[/tex]C. Inside the conductor is a cavity within which is a point charge [tex]q = +3\times10^{-6}[/tex]C. What is the charge on the outer surface of the conductor?


    2. The attempt at a solution
    On cavity wall: [tex]q=-q=-3\times10^{-6}[/tex]C
    On outer surface: Net Charge + Charge Inside Cavity [tex]=+13\times10^{-6}[/tex]C


    3. The problem I encounter
    All books I have read say that to calculate on the outer surface it's the net charge minus the charge on cavity wall. Why is it this and not how I did it which is Net Charge + Charge Inside Cavity? Surely, the charge on the cavity wall is only there when we draw the Gaussian surface.
     
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  3. May 18, 2009 #2

    Doc Al

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    Staff: Mentor

    Net charge on the conductor equals the sum of the charges on its inner and outer surfaces, Qnet = Qouter + Qinner. But Qinner + Qcavity = 0, thus Qinner = -Qcavity. Thus subtracting the inner charge is the same thing as adding the enclosed charge.
    Drawing the Gaussian surface allows you to deduce the charge on the inner surface, it doesn't create the charge. The charge doesn't know anything about what you draw or don't draw!
     
  4. May 18, 2009 #3

    Air

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    Is the charge enclosed in the Gaussian Surface zero or [tex]q = +3\times10^{-6}[/tex]C?
     
  5. May 18, 2009 #4

    Doc Al

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    Staff: Mentor

    That depends on where you draw your Gaussian surface. Assuming your Gaussian surface is within the conducting material, then the net charge enclosed will be zero.

    On the other hand, if you drew a Gaussian surface in the cavity but surrounding the point charge, the charge enclosed will equal that point charge.
     
  6. May 18, 2009 #5

    Air

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    So, Am I correct to think that between the Cavity wall and the end of the conducting material, there is no field because charge enclosed is zero hence deriving the electric field through Gauss law would give zero electric field?
     
  7. May 18, 2009 #6

    Doc Al

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    Staff: Mentor

    I would look at it the other way around. Since everywhere within the conducting material the electrostatic field is zero, any Gaussian surface contained within the conducting material must enclose zero net charge.
     
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