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Capacitance and E-field of a parallel plate capacitor

  1. Feb 22, 2013 #1
    1. The problem statement, all variables and given/known data

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    3. The attempt at a solution

    I know that the E-field is definitely affected by the presence of the space charges. Since one side is positive, the positive charges in the air get pushed towards the negative plate.

    so

    Enet = Eplate - Echarges

    But the thing is I don't know how the distribution of the charges will end up as, and hence how to calculate Enet.
     
  2. jcsd
  3. Feb 22, 2013 #2

    ehild

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    The distribution of the space charge does not change, the charges do not move.

    ehild
     
  4. Feb 22, 2013 #3
    Then surely the E-field would be σ/ε0?

    But in the answer, clearly the E-field depends on x.
     
  5. Feb 22, 2013 #4
    The charges do not move. We clearly have an E field between the 2 plates due to the charge in the middle. Considering V=0 , we can see the electric field inside would be zero in the middle, positive in one side and negative in the other. This way -[itex]\int \vec{E} \vec{dx}[/itex] = ΔV = 0, so the charges do not affect the voltage of the capacitor, as well as its capacitance. Now considering V≠0, by the superposing theorem we can add the V/d to the electric field due to the charges in the middle (this electric field will only be zero in x=d/2).
     
  6. Feb 22, 2013 #5
    I see what you mean..The problem here is that I'm trying to think about how the space charges are distributed and what the E-field due to them would be.

    Let's just take all space charge to be positive Q. Thus ρ = Q/V
    At midway there are equal amount of positive charge on top and bottom, so the E-field strength would be equal and cancel.

    But as you move closer to the bottom, there are more positive charges above.

    But I still do not know how the E-field depends on the number of charges and distance...We could model them as a solid rectangular block whose E-field is only perpendicular to the surface but the question remains: How does the E-field strength change with distance?

    This question is under the topic of Capacitance...
     
  7. Feb 22, 2013 #6
    The charges are uniformly distributed. Consider each rectangle (as you sad) to be a positive plate of a capacitor. We know the E field due to a infinite plate is σ/2ε0. But the σ of each of them would be different right?
    How can we write σ in function of ρ, x and d?
     
  8. Feb 22, 2013 #7
    Ok i think i got the answer!

    Take a thin rectangular block of thickness dx and area A.

    Amount of charge in that block=
    dQ = p (dV)

    σ = (dQ)/A = (p/A)dV = p dx

    dE = σ/2ε = (p/2ε) dx


    Thus E = integral from 0 to d-x of dE - integral from 0 to x = (p/2ε)(d-x) + (p/2ε)(-x) = (p/2ε)(d-2x)

    as bottom the E-field points upwards while at the top the E-field points downwards
     
    Last edited: Feb 22, 2013
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