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Gauss' Law & Surface Charge Density

  1. Jan 26, 2010 #1
    1. The problem statement, all variables and given/known data
    A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find:
    (a) the charge density of each face of the plate and
    (b) the total charge on each face


    2. Relevant equations
    [tex]
    E = \frac{\sigma}{2\epsilon {0}}
    [/tex]

    [tex]
    \sigma = \frac{q}{a}
    [/tex]

    3. The attempt at a solution
    I missed about a week due to unforeseen circumstances, so I'm struggling with this to catch up, but so far:

    Area, converted to meters, is 0.25 m^2. Electric field, E, converted to nano units would be 80x10^-9. My understanding was that [tex]\Phi _E = 2EA[/tex] so [tex]\Phi _E = 2 \times \frac{\sigma}{2\epsilon {0}} \times \frac{q}{a}[/tex] which comes out to [tex]\Phi _E = 2 \times \frac{\sigma}{8.854\times(10^-12)}\times\frac{q}{0.50\times0.50} [/tex], however I'm a bit confused on this. How do I find the value of q, and am I correct in my setup such that [tex]\sigma[/tex] is setup within the second fraction, over 8.854*10^-12?

    Any help is greatly appreciated
     
    Last edited: Jan 26, 2010
  2. jcsd
  3. Jan 26, 2010 #2

    Gokul43201

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    What do you know to be true about a metal (conductor) when it comes to electric fields?
     
  4. Jan 26, 2010 #3
    Charge is only ever on the surface, and the interior has a 0 charge. I also know, as mentioned in the equation (and from my reading) that a conductors equation is:

    [tex]E=\frac{\sigma}{2\epsilon{0}}[/tex]

    While a nonconductor is

    [tex]E=\frac{\sigma}{\epsilon{0}}[/tex]

    I also know that finding the surface density is supposed to involve using a Gaussian cylinder.

    Correct?
     
  5. Jan 26, 2010 #4

    Gokul43201

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    Correct. Also there is no net electric field in the interior of a conductor. So when a conducting plate is placed in an external electric field, what needs to happen to the positive and negative charges in the conductor for this condition to be satisfied? (Hint: If you place freely moving positive and negative charges in an electric field, what happens to them?)
     
  6. Jan 26, 2010 #5
    I'm kind of confused by the wording, but my guess of what you are saying is referring to the charges separating out to each side of the plate, one being negative and one being positive. Yes?
     
  7. Jan 27, 2010 #6

    Gokul43201

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    Yes. You can figure out which side will get what kind of charge based on the direction of the external field. But more importantly, each of these two sheets of charge will now produce their own electric fields. The negative sheet will produce field lines towards it and the positive sheet will produce field lines away from it. The field from each of these sheets will be given by the first equation in your opening post.

    Inside the metallic plate, the total field is found by adding all three fields (including the applied external field). Can you now write an expression for the total field inside, in terms of the external field (call it Eo) and the charge density on each sheet (say, [itex]+\sigma,~-\sigma[/itex])? Draw a diagram to help yourself account for all the fields with their correct directions.
     
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