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Parallel Plate Capacitor having equal charges on plates

  1. Jan 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Q .1 The plates of a parallel plate capacitor are given equal positive charges. What will be the potential difference between the plates? What will be the charges on the facing surfaces and on the outer surfaces?

    Q.2 Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a battery .Now,
    a) The facing surfaces of the capacitor have equal and opposite charges.
    b) The two plates of the capacitor have equal and opposite charges.
    c) The battery supplies equal and opposite charges to the two plates.
    d) The outer surfaces of the plates have equal charges.



    2. Relevant equations



    3. The attempt at a solution

    The set up of the two problems looks the same, say, +q, charge is given to the two plates . Let us call the two plates be A and B .The +q charge will distribute uniformly on the two surfaces. Both outer and inner surfaces of A will have +q/2 charge .Now applying Gauss Law ,the inner surface of B will have –q/2 charge and the rest +3q/2 on the outer surface .The charges q/2 on inner surface of A and –q/2 on inner surface of B will contribute to the electric field .

    Ans 1) V=Ed
    E=q/2εA
    Thus ,V=qd/2εA
    Charges on the surfaces of the plates are (q/2 ,q/2)on A and (-q/2,3q/2) on B

    Is the solution correct ?

    Ans 2) I am not clear about the role of battery in affecting the charges on the surfaces of the plates.
    Option a) ,c) looks right . Option b)is surely wrong .But not sure about option d)

    How will the battery affect the various charges on the inner and outer surfaces of the two plates ?
     
  2. jcsd
  3. Jan 9, 2013 #2

    mfb

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    Why do you expect different charge distributions for plates A and B? The problem statement does not distinguish them - you do not even have a unique way to call them "A" and "B". It does not matter in which order you add the charges.

    I agree.

    The battery should not break symmetry on the outside I think.
     
  4. Jan 9, 2013 #3
    Thanks mfb...

    Before the battery is applied , charges on outer surfaces of two plates have to be unequal as the inner surfaces have equal and opposite charges .What is the role of battery after that ?
     
  5. Jan 9, 2013 #4

    mfb

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    And -0 = 0

    It fixes the charge density at the inner sides to a value different from 0.
     
  6. Jan 9, 2013 #5

    ehild

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    Connecting the capacitor to the battery, the potential difference across the capacitor becomes the same as the voltage of the battery (U). Accordingly, the charge of the capacitor becomes Q=CU. The charge on the capacitor plates are equal in magnitude and of opposite sign.
    You can calculate the electric field outside and inside the capacitor, and you know that the surface charge density on the inner and outer surfaces of the plates are equal to ε0E.

    The charged plates can be substituted by planes of charge, and the charge of unit area is σ on one plate and -σ on the other one. (If A is the area of a plate, σ=Q/A).

    You know that the charge distribution causes the electric field strength Q/(2ε0) at both sides of a charged plane, and the electric field lines emerge from the positive charges and end in the negative ones.

    The electric field in the different regions 1, 2 and 3 are obtained by superposition from the fields of the charge distribution of both planes.
    The red arrows show the electric field lines from the positive charges on the upper plane and the blue arrows represent the electric field lines from the negative charges on the lower plane.
    You can see that the field lines cancel each other outside the capacitor and the electric field is σ/ε0 inside, pointing downward.


    ehild
     

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  7. Jan 10, 2013 #6
    I am unable to understand your explaination
     
  8. Jan 10, 2013 #7

    mfb

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    Symmetry requires that the charges on the inside are 0 on both plates in setup 1.

    The battery gives a fixed potential difference across the condensator, which can be used to determine the charge density there in setup 2 (see ehild's post for details).
     
  9. Jan 10, 2013 #8
    Ah sorry I posted in the wrong thread. Can't delete. Edit only. Sorry I'm a newbie.
     
  10. Jan 10, 2013 #9
    Thanks mfb :smile: . I have realised my mistakes as well as understood the concept.

    Thanks ehild for the wonderful explaination .
     
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