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Homework Help: Three parallel planes - calculating charge

  1. Feb 23, 2014 #1
    1. The problem statement, all variables and given/known data
    There are three parallel identical planes of area A = 200 cm^2, and the distance between the upper and the middle one as well as the distance between the middle and the lower one is d = 3cm. The upper plane was charged to q1 = 0.5 nC. The other two were connected to a V= 100V EMF source. Calculate the charges of the middle and lower planes.

    2. Relevant equations
    [itex] V = E * d [/itex]
    [itex] E * A = q / \epsilon_0 [/itex]

    3. The attempt at a solution

    I firstly assumed that after being connected to the source, the potential of the middle plane is 0 V, and the potential of the lower plane - 100V. Then from Gaus's law, we find the electric field between the planes:
    [itex] E_{12} * A = \frac{q_1 + q_2}{\epsilon_0} [/itex]
    [itex] E_{23} * A = \frac{q_2 + q_3}{\epsilon_0} [/itex]
    Where 1st, 2nd and 3rd planes are upper, middle and lower correspondingly.
    [itex] V_{12} = E_{12}*d = \frac{(q_1 + q_2)*d}{\epsilon_0 * A} [/itex]
    [itex] V_{23} = 100 = E_{23}*d = \frac{(q_2 + q_3)*d}{\epsilon_0 * A} [/itex]
    But here we have 3 unknowns and 2 equations... Are there any other ways to calculate [itex]V_{12} [/itex], for instance?
  2. jcsd
  3. Feb 23, 2014 #2


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    Would it help if you looked at the middle plane as if it were split into two parallel planes with a conductor in between ?
  4. Feb 24, 2014 #3
    I'm sorry, but no, not really... I don't see what you are getting at :/

    Though I just came up with an idea - please correct me if I'm wrong - but could you say that [itex] V_1 = \frac{kq_2}{d} + \frac{kq_3}{2d} [/itex] ? Saying that the potential of the first plate is created by the charges of 2nd and 3rd plates?
  5. Feb 24, 2014 #4


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    The idea of my proposal is that you consider the charge difference between the top two plates separately from the voltage difference between the bottom two. A nice Gauss volume passing between 2top and 2bottom would then help you along....

    By the way, the formulation of the exercise leaves some possibilities for interpretation:
    How were they connected?
  6. Feb 24, 2014 #5
    Yes, I know the formulation is not completely clear, but there's no additional information... But I'm pretty sure we can assume that 2nd plate was connected to ground and 3rd one to 100V.

    Hmm, I thought that's similar to what I did? :?
  7. Feb 24, 2014 #6


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    This would split up the charge on the middle plate: on 2top the mirror of the 0.5 nC and on 2bottom something to deal with the 100 V.

    And I would ignore the influence of plate 1 on 3 and vice versa. With such vaguely described exercises maybe that is not directly evident.
  8. Feb 24, 2014 #7
    What do you mean by 'mirror of the 0.5nC'? The potential by 1st plate?

    So is this equation [itex] V_1 = \frac{kq_2}{d} [/itex] true then?
  9. Feb 24, 2014 #8


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    If a charge 0.5 nC is applied on the top plate, there will be a charge of -0.5 nC on the top side of the middle plate. The equation you now mention I haven't seen before. I did see a V12 that I did not believe. ANyway, they ask for charges, not voltages.

    Due to the applied voltage between plate 2 and 3 there will be a charge buildup on the bottom of plate 2 and on the top of plate 3. Same ting: equal and opposite.

    What I am trying to lure you into is to consider the top and the bottom of plate 2 separately but connected.
    I don't think I've succeeded yet, right ?

    Gotta go now for at least 5 hrs...
  10. Feb 24, 2014 #9
    Forget top plate for a while.Treat the middle and bottom plates as a parallel plate capacitor .Can you find the capacitance of the capacitor .Using that and the voltage difference applied across the plates,find the charges on the plates .

    Can you do that?
    Last edited: Feb 24, 2014
  11. Feb 24, 2014 #10
    Oh, so you mean, that the electrons of the top plate push the electrons of middle plate away? So then the total charge of middle plate is sum of middle top and middle bottom?

    You mean [itex] C = \frac{Q}{U} = \frac{\epsilon_0 A}{d}, Q = \frac{\epsilon_0 A U}{d} [/itex] ? Wow, could've checked capacitance formulas before! But why is the top plate given then? I mean, it should do some impact? Does it change the charge of the middle plate then?
  12. Feb 24, 2014 #11
    Yes..that's right .The capacitor is formed between the bottom surface of the middle plate and top surface of the bottom plate .

    The charge on the top plate will be entirely on the bottom surface .This charge will induce an equal and opposite charge on the top surface of the middle plate .

    The middle plate will have unequal charge densities on its two surfaces .The net charge is the sum of the two.
  13. Feb 24, 2014 #12
    It's clear now! Thank you a lot, both for your help and patience :)
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