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Moving Capacitor plates apart

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  1. Jan 26, 2016 #1
    If I have a capacitor and I maintain the potential difference between the two plates as I move them apart would I have to do any work (other than mechanical work) to move them apart? my logic tells be I shouldn't have to as I'm not doing any work again the potential field? Just wondered what you guys thought?
     
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  3. Jan 26, 2016 #2

    Hesch

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    That depends on what you include in "mechanical work".

    Besides gravitational force, the plates attract each other, due to the electric field between the plates.

    The latter force can be calculated from: Eelec = ½*V2*C , V is constant here.

    When you pull the plates apart, C will be decreased. Now say that V is maintained by some connected battery, the battery will be charged ( supplied by energy ).

    Now calculate the total change in energy in the capacitor+battery. You must add this energy to the system, when you take apart the plates.
     
    Last edited: Jan 26, 2016
  4. Jan 28, 2016 #3
    Ahh yes, I forgot that charge must be added to keep the potential constant, and I don't think the capacitance would be varied? But adding this charge requires work to be done. Silly me. Thank you :)
     
  5. Jan 28, 2016 #4

    BvU

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  6. Jan 28, 2016 #5
    Ahh okay! But would this not best be explained through the addition of charge? I accept that thee capacitance will change, but for calculation purposes?
     
  7. Jan 28, 2016 #6

    BvU

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    REad on in the link: there is even an equation for the energy stored in a capacitor !
     
  8. Jan 28, 2016 #7
    I saw- I get that this is the energy stored on the capacitor, but suppose we don't know the dielectric strength, and V is kept constant. E= 1/2 dC V^2, but how would you go about finding dC (i.e. the change in capacitance). Sorry if I'm being really dumb and missing something extremely obvious
     
  9. Jan 28, 2016 #8

    BvU

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    Correction: ##E = {1\over 2} CV^2##
    Suppose you double the distance ##d##. What happens to C ?
    You maintain V, and you have an expression for E before and after changing ##d##. The difference is the work done -- mechanical plus electrical.
    How would you go about to calculate one of the two -- and thereby determine the other as well ?
     
  10. Jan 28, 2016 #9
    If you double the distance then c would half. Okay I think I've got it. Thank you.

    From this would it be possible to calculate the force per unit area. i.e. would the use of F=-dW/dx be valid ?
     
  11. Jan 28, 2016 #10

    BvU

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    That's the mechanical one. Before embarking on that: what is the electrical work ?
     
  12. Jan 29, 2016 #11
    W=Q integral E.dr

    E= sigma/ epsilon naught?

    W= Q sigma / epsilon-naught dx ?
     
  13. Jan 29, 2016 #12

    BvU

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    I'd say ##W = \int VI\;dt = V \int I\;dt = V\Delta Q ## (since V is constant). Much easier ...
     
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