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Conductor in a capacitor problen (E&M)

  1. Oct 9, 2015 #1
    1. The question:

    This question is from purcell's E&M book (3.71).
    (a) The plates of a capacitor have are A and separation s (assumed to be small). The plates are isolated, so the charges on them remain constant, the charge densities are +-σ. A neutral conducting slab with the same area A but thickness s/2 is initially held outside the capacitor.
    The slab is released. What is its kinetic energy at the moment it is completely inside the capacitor?

    (b) Same question, but now let the plated be connected to a battery that maintaines constant potential difference. The charge densitied are initially +-σ. (Don't forget to include the work done by the battery).

    2. Relevant equation:

    Energy stored in a capacitor:

    E = C * phi^2 / 2

    3. The attempt at a solution
    I calculated the energy within the capacitor before the conductor was released:

    U_i= ε0 E^2 A S / 2 (where A is the surface of the plates).

    Then I calculated the enerdy within the capacitor after the conductor is fully inside. Now I can look at it as a thinner capacitor with only s/2 distance between the "plates".

    U_f = ε0 E^2 A S / 4

    So my problem is - what does this energy actually mean? Is this the potential energy of the problem, that together with the kinetic energy becomes the total energy? Or is it something else?
    And in general, how do I translate this potential reduction into the work done on the slab?

    I'd really appriciate your help! Thank you.
     
  2. jcsd
  3. Oct 9, 2015 #2

    Hesch

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    Assuming that the slab is moved with no friction, the slab will pass the position where it is completely inside the capacitor and will start an oscillation obout this point. So all the difference in electrical energy has been coverterted to kinetic energy at this position.
    ½*ΔC*(ΔU)2 = ½*m*v2
     
  4. Oct 9, 2015 #3

    Hesch

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    Correction:
    ½*ΔC*(V12 - V22) = . . . . . .
     
  5. Oct 9, 2015 #4
    Thank you!
    And what about (b)?

    The problem with (b) is that using your equation results in a negative kinetic energy, which is physically impossible as far as I know...
    So I suppose we have to take the work done by the battery into consideration, but I don't know how this can be done here.
     
  6. Oct 9, 2015 #5

    Hesch

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    No, V1 > V2 and ΔC is positive → ½*ΔC*(V12 - V22) is positive → ½*m*v2 is positive.
     
  7. Oct 10, 2015 #6
    Here V1 = V2 (it is a given), but the capacity grows... And that results in a negative kinetic energy...
     
  8. Oct 10, 2015 #7

    rude man

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    Part (a): What is work? Isn't it energy? So if the field energy has decreased, as you have correctly computed, wouldn't that have to equal the work done on the slab? And if I apply mechanical work on a mass without friction, and there is no change in gravitational potential energy, where does that work have to go?

    Part (b) is a somewhat different situation since a battery is now provided to either source or sink energy. So you need to recompute the change in field energy ( is it + or - ?), the energy supplied by the battery (hint: U = volts times charge), and thus the net work done on the slab.
     
  9. Oct 10, 2015 #8

    Hesch

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    When the capacity increases, the voltage must descrease: Capacity = coulomb / volt, so if the capacity grows and coulombs are constant . . . . . ?
     
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