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Parallel plate capacitance

  1. Jun 7, 2008 #1
    1. The problem statement, all variables and given/known data

    two parallel plates have equal and opposite charges of 0.01 coulombs. the electric field between the plates is 10^5 newtons/coulomb and the separation between the plates is 2mm. what is the capacitance of the plates?

    2. Relevant equations

    capacitance C = Q/V where Q is charge in coulombs, V is electric potential in volts
    electric field E = V/d where d is distance in meters
    Q = [epsilon_0*A/d]V where A is area and epsilon_0 is a constant = 8.85*10^-12

    3. The attempt at a solution

    i solved the electric field equation for V, so that V = dE and substituted that dE for V in the capacitance equation to get:

    C = Q/dE = Q/(0.002*10^5) = Q/200

    my questions are how do i factor in two, opposite charges or two equal charges for that matter, into the equation or am i interpreting the problem incorrectly?

    is the problem asking me to find the capacitance in two different cases?
    case1 --> two plates, with the same charge
    case2---> two plates, opposite charges

    how can i use the equation C=Q/200 and incorporate the charges of TWO plates? there is only one Q in the equation.
    if that is the case, will what i attempted earlier work to find the capacitance, it seems it will?

    thanks
     
    Last edited: Jun 7, 2008
  2. jcsd
  3. Jun 7, 2008 #2

    alphysicist

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    Hi scholio,

    I believe a capacitor always has charges on it's two plates that are equal in magnitude and opposite in polarity. The Q in the equation refers to the magnitude of the charge that is on either plate.
     
  4. Jun 7, 2008 #3
    so do i just use 0.01 coulombs for Q, and was i correct on how i interpreted the problem, finding the capacitance with the two cases i mentioned? how would Q change for case2 since they are opposite charges?

    thanks
     
  5. Jun 7, 2008 #4

    alphysicist

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    The original problem stated that the plates had equal and opposite charges, so that would be case 2 and I think you approached it correctly.

    Case 1 would not be a capacitor (at least by the definition I learned), and so that problem would not come up.
     
  6. Jun 7, 2008 #5

    alphysicist

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    I reread your post and I think I see your question. When they say "equal and opposite charges" they mean charges that are equal in magnitude and opposite in polarity.
     
    Last edited: Jun 7, 2008
  7. Jun 7, 2008 #6
    still doesn't that mean there are two charges, and how does that work with the one Q in the equation C=Q/200, how do i handle the two charges now? would i do two calculations using the same equation just with different signs for Q such that.

    calculation 1, where Q is positive

    C = Q/ 200 = +/200 = +1/200 farad

    calculation 2, where Q is negative

    C = Q/200 = -/200 = -1/200 farad

    is that how i'm supposed to answer the question?
     
    Last edited: Jun 7, 2008
  8. Jun 7, 2008 #7

    alphysicist

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    A capacitor always has two charges; a positive charge one plate and a negative charge on the other. The capacitor is the group of two plates together, and the Q refers to the magnitude of the charge on either plate (because the magnitude will be the same for both).


    So if you have two plates that are both postive, or both negative, or if only one is charged, or if the magnitude on them is not the same, then they would not be a capacitor.

    It may seem like this is unnecessarily restrictive, but remember, the normal way to charge up two plates is to connect them to a battery; then all you are doing is taking some negative charge from one plate and moving it to the other. Then you automatically have equal and opposite charges on the two plates, and they function as a capacitor.

    -----

    I just saw your edits on your post. You would say that plate 1 has charge Q and plate 2 has charge -Q. The capacitance of the two plate system is then:

    C=Q/V,

    where Q is the magnitude (positive).

    So we treat the group of two plates as one capacitor, and find the capacitance of that.
     
  9. Jun 7, 2008 #8
    i'm still somewhat confused, i understand that a capacitor is composed of two plates that are oppositely charged, but since we take two plates as one capacitor and if the magnitude of the Q's on both plates are the same, wouldn't the equation become C =2Q/V = 2/200 = 1/100 farad?

    thanks for your patience
     
  10. Jun 7, 2008 #9

    alphysicist

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    I understand what you're saying, but I think the way this particular problem is worded it is having you think about the fundamental formula C=Q/V from the wrong point of view.

    Here's the way to think about it: If you have two plates with capacitance C, and you hook them to a battery V, then Q is the amount of charge that is transferred from one plate to another.

    (So from the way the equation is set up, you always have Q on one plate and -Q on the other, but you still only use Q in the equation since that is them magnitude transferred from one plate to the other.)
     
  11. Jun 7, 2008 #10
    so in otherword, in the case of this problem since a numerical value for Q is not specified, i should assume it to be a constant thus my answer for the capacitance would be as such

    C = Q constant/ dE = Q constant/(0.002*10^5) = Q constant/ 200 = 1/ 200 farads

    right?
     
  12. Jun 7, 2008 #11

    alphysicist

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    I'm not sure what you're referring to; do you mean for the original problem? In the original problem the charge was specified; it was 0.01 coulombs.
     
  13. Jun 8, 2008 #12
    oh you are correct, so instead it would be C = Q / dE = 0.01/(0.002*10^5) = 0.01/ 200 = 5*10^-5 farads

    correct?
     
  14. Jun 8, 2008 #13

    alphysicist

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    That looks right to me.
     
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