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Induced charge - infinite plates

  1. Sep 24, 2007 #1
    No numbers here - purely conceptual.

    Two thin plates carry total surface charge densities of [tex]\sigma[/tex] and [tex]\sigma_{1}[/tex] respectively. An uncharged conducting slab is placed in between the charged plates, and then above [tex]\sigma[/tex]. What are the induced charged densities on the surfaces of the conducting slab in each of the configurations?

    The problem also states that 'polarities of those charges ([tex]\sigma[/tex]) are not specified and should be treated algebraically'.

    Every example I can find using infinite parallel plates has to do with opposite charged densities. However, this problem does not denote opposite charges of the plates, and I'm not quite sure how to treat this algebraically.

    For the first configuration (in between), since each of these plates is charged it will induce the opposite charge on the conductive surface in order to create a potential flow. So my first guess is the charge densities on the -neutral- surface will become -[tex]\sigma[/tex] and -[tex]\sigma_{1}[/tex].

    When it is placed outside the plates the neutral surface will only take on a partial charge from the nearest plate (which is actually [tex]\sigma[/tex]).

    Bit of an odd question that I could definitely use some help with.
     
  2. jcsd
  3. Sep 24, 2007 #2

    learningphysics

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    The field due to an infinite plate is charge density/2epsilon. You need the field inside the slab to be 0... you also need the net charge inside the field to be 0...

    Suppose k is the charge density on one side of the slab... then -k is the charge density on the other side (since it's neutral).

    What is the total field inside the slab (you have 4 infinite plates) in terms of the respective charge densities... they add to 0... solve for k.

    Same idea in the second part, except the directions are different.
     
  4. Sep 25, 2007 #3
    You mean the net charge of the slab needs to be 0, right?

    So what I eventually got based on what you told me (I think) is k = -[tex]\sigma[/tex] and -k = [tex]\sigma_{1}[/tex]. So basically the charge density on the surface of the slab is the inverse of the corresponding plate charge density? I hope that's right...

    When the slab is placed outside the plates, their E fields add together to make ([tex]\sigma[/tex] + [tex]\sigma_{1}[/tex])/2epsilon. The field from the slab surface with charge k must equal this for the field within it to be = 0.

    This eventually leads to k = -[tex]\sigma[/tex] -[tex]\sigma_{1}[/tex] and the opposite (sigma plus sigma one) for -k.

    Is this correct? It's somewhat similar to my original answer where for the inside k = sigma minus sigma one, and for the outside k = sigma plus sigma one.

    At least it's close!
     
  5. Sep 25, 2007 #4

    learningphysics

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    I think you're partially doing it right... we want to make sure the field inside the slab is 0.

    For the first part: The field due to [tex]\sigma[/tex] is [tex]\frac{\sigma}{2\epsilon}[/tex]

    Now the [tex]\sigma_1[/tex] is on the other side of the slab... it creates a field inside the slab of [tex]\frac{-\sigma_1}{2\epsilon}[/tex]

    The k charge creates a field of [tex]\frac{k}{2\epsilon}[/tex] the -k charge also creates a field of [tex]\frac{k}{2\epsilon}[/tex]... because it is on the opposite side...

    So the total needs to be zero. Hence [tex]\frac{\sigma}{2\epsilon}+ \frac{-\sigma_1}{2\epsilon} + 2* \frac{k}{2\epsilon} = 0[/tex]. so you can solve for k.

    In the second case we get: [tex]\frac{\sigma}{2\epsilon}+ \frac{\sigma_1}{2\epsilon} + 2* \frac{k}{2\epsilon} = 0[/tex]. so solve for k here...
     
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