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Calculating force between capactor plates using virtual work

  1. May 24, 2017 #1
    1. The problem statement, all variables and given/known data
    Use “virtual work” to calculate the attractive force between conductors in the parallel plate capacitor (area A, separation z). That is, use conservation of energy to determine how much work must be done to move one plate by an infinitesimal amount, and then use the value of the work to determine the force. Do your virtual work computations in two ways:
    (a)
    keeping fixed the charges on the plates, and,
    (b)
    keeping a fixed the voltage between the plates.

    2. Relevant equations
    Wby me=P.E.=(1/2)(Aε0/z)V2=(Q2/2Aε0)z
    dW= F.dz

    3. The attempt at a solution
    Let's consider the case when the negative plate is up and the positive plate is down along the z- axis.
    The electrostatic force acting on the negative plate is along the -ve z-axis.
    To move negative plate along +ve z-axis by an infinitesimal amount,
    (a)
    keeping fixed the charges on the plates,
    Wby me = (Q2/2Aε0)z
    dWby me = (Q2/2Aε0)dz, (1)

    dW= F.dz, (2)
    From (1) and (2),
    Fby me=(Q2/2Aε0) along +ve z-axis
    Fattractive= - Fby me=(Q2/2Aε0) along -ve z-axis

    where dWby me is work done by me in moving negative plate along +ve z-axis by an infinitesimal amount dz,
    Fby me is forced applied by me on the system
    Fattractive is the attractive force between conductors in the parallel plate capacitor



    (b)
    keeping a fixed the voltage between the plates
    Wby me=
    (1/2)(Aε0/z)V2
    dWby me = (-1/2)(Aε0)(V/z)2dz, (3)
    dW= F.dz, (4)
    From (1) and (2),
    Fby me=
    (1/2)(Aε0)(V/z)2 along negative z-axis
    Fattractive= - Fby me=
    (1/2)(Aε0)(V/z)2
    along +ve z-axis

    While I think that should be Fby me in positive z-axis ( as I am pulling the negative plate upward , so both Fby me and the displacement dz is in the upward direction and hence dWby me in (3) should be +ve).
    I can't understand physically how dWby me in (3) can be negative?
    Where am I wrong?
     
  2. jcsd
  3. May 24, 2017 #2

    rude man

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    The case of fixed V is complicated by the fact that some of the energy is supplied by or given to the source of emf (the battery for example).
    So the energy balance is work done by you + energy supplied by battery = change in E field energy
     
  4. May 24, 2017 #3
    Thank you for the reply.
    By definition of work,
    dWby me= Fby me.dz
    Here, both Fby me and dz are in the +ve z direction, so dWby me is supposed to be +ve.
    What is wrong with this argument?
     
  5. May 24, 2017 #4

    rude man

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    Nothing! If you do it right and take the battery into account, you will find that you did do positive work. The field energy went negative but the battery got charged by twice that amount in magnitude, so bottom line you charged up the battery by the work you did, with the (loss in) field energy contributing an equal amount.
     
  6. May 26, 2017 #5
    After moving negative plate along +ve z-axis by an infinitesimal amount dz,keeping V constant,
    the charge on the capacitor has got decreased to Q' = (Aε0/(z+dz))V i.e. Q'<Q,
    So, moving negative plate along +ve z-axis by an infinitesimal amount dz, keeping V constant, discharges the capacitor.
    Hence, the potential energy of the capacitor gets decreased.
    Now,according to you this loss in P.E. and work done by me goes to the battery
    and as a consequence, the battery gets charged up.
    What is meant by charging up the battery , here, the battery is still supplying the same potential difference V?
    Experimentally, how will I verify that the battery has got charged or not?
    Does it mean that lifetime of the battery has got increased?
    How do you know that the battery got charged by twice (how twice?) that amount (which amount?)in magnitude?


    If I am right, then, Wby me is not equal to loss in P.E..
    Wby me = (total energy given to the battery - loss in P.E)
    How to calculate total energy given to the battery?
    How to solve the second part of the question?
     
    Last edited: May 26, 2017
  7. May 26, 2017 #6

    rude man

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    Why "experimentally"? You are supposed to be analyzing on paper.
    the battery lifetime is not an issue. And it doesn't have to be a battery; it could be a rectified ac generator output or many other kinds of dc emf sources.
    If I force charge Q into a source of emf, by how much have I increased its energy? Use this answer and the energy balance equation I gave you before. That is all you need.
    this is correct.
    see above
     
    Last edited: May 26, 2017
  8. May 26, 2017 #7
    Thank you, Thanks a lot.

    For calculating energy given to the battery :
    The charge (Q-Q') goes to the positive terminal of the battery at constant potential V and -(Q-Q') goes to the negative terminal of the battery at constant potential 0V. Let's assume that inside the battery the negative charge remains at the negative terminal and the positive charge moves to the negative terminal to make it neutral and thus to maintain the constant potential difference between the two terminals. In doing so, the work done by me in this case is (-V)(Q-Q') = the energy given to the battery = Ebatt. I am getting Ebatt negative here. Where am I wrong?

    But battery's mechanism ,in general , is such that it collects positive charge at negative terminal and send it to the positive terminal hence work done on transporting a positive charge Q by the battery is positive i.e.QV.
    I think ,here in this case, it is work done by the battery which is negative, i.e. (-V)(Q-Q') is work done by the battery and work done by me = - (work done by the battery ) = V(Q-Q') = Ebatt
    So, assuming that total energy given to the battery is positive,
    total energy given to the battery = V(Q-Q') =dWby me + loss in P.E. of the capacitor,
    where loss in P.E. of the capacitor = -(P.E.final - P.E.initial) = (V/2)(Q-Q')
    dWby me = total energy given to the battery - (loss in P.E. of the capacitor)
    = V(Q-Q') - (V/2)(Q-Q')
    = (V/2) [ (Aε0/(z))V -(Aε0/(z+dz))V]
    =(V2/2)Aε0[(1/z)-(1/z+dz)]
    =(V2/2)Aε0[(1/z2)dz]
    So, Fby me = (V2/2)Aε0[(1/z2) along positive z-axis


    Is this o.k.?

    Now, work done by attractive electrostatic force is = dWelec = Felecdz,
    where Felec <0, dz>0
    (P.E.final - P.E.initial) = - dWelec
    (V/2)(Q-Q') = - dWelec = - Felecdz,
    Felec = (V2/2)Aε0[(1/z2) along negative z-axis on negative plate
    Hence, Felec = - Fby me
     
    Last edited: May 26, 2017
  9. May 26, 2017 #8

    rude man

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    You are getting closer but not there yet.
    I move a charge Q from a potential of zero (the battery negative terminal) to +V (the battery positive terminal). What is the work done by me?
    You're worrying too much time about battery details instead of just sticking to fundamentals. What's important is potential difference, not battery "innards". The voltage source could just as easily have been a large capacitor, large in comparison to the capacitance of your plates, such that effectively for that capacitor, dV = 0 for adding or subtracting charge dQ.
    Keep it up, you're making good progress.
     
  10. May 27, 2017 #9
    The work done by you is QV.
    But in this problem,(Q-Q') comes to the positive terminal from the positive plate and reaches to the negative plate via negative terminal and thus the work done by me is negative i.e.-(Q-Q') V.
    Is this correct?
     
  11. May 27, 2017 #10

    rude man

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    [QUOTE="Pushoam, post: 5771099, member: 619344[/quote]
    OK, how about this: you know that battery energy W is reduced if I shunt the battery with a resistor R. Define positive current as current entering the battery at the + end and leaving at the - end, so we have negative battery current, and Q in the battery decreases. The battery energy change is ΔW = -Vt/R where t = time of current flow. Or, ΔW = -QV where Q = ∫i dt = -Vt/R.

    Getting back to your problem, you know you lose charge dQ when you move the plates apart dz, right? Where does that charge go? Is there positive or negative battery current as consequence? Compare with the resistor case.

    Or again, if you think of a large capacitor instead of a battery, and you know charge was lost on your plates, then did you gain or lose charge on your large capacitor? Did the large capacitor gain the charge lost by the plates? And if so, did the large capacitor not increase its stored energy?
     
  12. May 28, 2017 #11
    Assume that the battery is nothing but a big capacitor with charge Q and potential difference V.
    Now as I shunt the battery with a resistor R for time Δt, a +ve charge ΔQ goes from the positive plate to the negative plate of this capacitor and hence the charge of the capacitor gets reduced to Q - ΔQ. Consequently, the potential energy of the capacitor gets reduced to V(Q - ΔQ)/2 from V(Q /2).
    According to conservation of energy ,
    loss in the potential energy i.e. V(ΔQ /2) = work done on ΔQ by the capacitor = kinetic energy of the ΔQ + the thermal energy developed in the resistor
    Hence, the work done on the charge ΔQ is ΔQ V= ΔQ V i.e.positive.
    Looking it in another way,
    +ve charge ΔQ goes from the positive plate at potential V to the negative plate at zero potential .
    Hence, the work done on the charge ΔQ is ΔQ(0 - V)= -ΔQ V i.e. negative.

    What is wrong here?
     
  13. May 28, 2017 #12

    vela

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    The work W = V(Q-Q') is the work needed to get the charge to the positive terminal of the battery from the positive plate of the capacitor.


    In the scenario you describe, the electrical energy lost when the charge subsequently is moved down to the negative terminal is converted to chemical potential energy inside the battery. As rude man said, you don't really need to worry about that in this problem. All you care about is the energy required to get the charge to the positive terminal of the battery.
     
  14. May 28, 2017 #13

    rude man

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    OK.
    This is numerically incorrect. The large capacitor will also lose a small amount of voltage ΔV. The result is that the large capacitor will lose energy = V ΔQ, not V ΔQ/2. I show that calculation below.

    Anyway, I'm not sure of the relevancy of what you're discussing to the problem at hand. We are trying to convince ourselves that (1) when you move the plates away from each other a distance dz, charge dQ is moved from the plates and deposited on the large capacitor, and (2) we need to determine the amount dQ. Hooking up a resistor to the large capacitor and trying to figure out what the work was done on the charge ΔQ seems irrelevant, and I don't know how to answer your question. I only invoked the resistor-large capacitor picture to convince you that if charge flows into the large capacitor's + terminal, that large capacitor has gained energy, and vice-versa. Which you probably knew anyway.

    Let's stick to the problem at hand:
    When you displace the plates by dz, charge dQ flows out of the plates:
    Q = CV (of the plates)
    dQ = C dV + V dC = V dC since V is effectively constant for the plates;
    with dC = -ε0A/z2 dz
    And energy loss in the plate field is d(CV2/2) = V dQ/2.
    so charge dQ = V dC has left the plates and, by charge conservation, moved to the large capacitor, which gained energy, as follows:
    The large capacitor has gained energy = (Q + dQ)(V + dV)/2 - QV/2 = (V dQ + Q dV)/2
    but dV = dQ/C and C = Q/V so Q dV = V dQ
    so energy change of large capacitor = V dQ for a plate displacement of dz.
    So of the large-capacitor energy gain V dQ, half came from the plate field and so the other half had to come from F dz.
     
    Last edited: May 28, 2017
  15. May 28, 2017 #14
    The work needed to get the charge (Q-Q') to the positive terminal of the battery from the positive plate of the capacitor is zero as the potential difference between positive terminal of the battery positive plate of the capacitor is almost zero(as the charge (Q-Q') is infinitesimally small). .
     
  16. May 28, 2017 #15
    But,we have already assumed that the battery i.e.large capacitor maintains constant potential difference i.e.V and so, the large capacitor has gained energy = (Q + dQ)(V )/2 - QV/2 = (V dQ )/2.
    If I assume that potential difference of large capacitor has got increased to V + dV, then the potential difference of small capacitor has also got increased to V + dV, as the two capacitors are connected by wire.
    Then,
    it seems to me that you are taking dV=0 for small capacitor and dV= non-zero for large capacitor.
    This I didn't understand. Will you please explain it?
     
  17. May 29, 2017 #16

    rude man

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    That is a very good question and I will answer it later today. If you want to work on it in the meantime, expand d(1/2 CV2) which is the differential change in energy of the plate capacitor. Hint: dQ/Q' << dC/2C and dV = V dQ/Q'. Here Q' = large capacitor charge and C and V refer to the plates.
     
  18. May 29, 2017 #17

    rude man

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    You can easily show that that's wrong.
     
  19. May 29, 2017 #18

    rude man

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    You move a charge dQ from the plates to the large capacitor. You have lost plate energy of V dQ/2 but have gained capacitor energy of V dQ. And you have done this as you say with zero work, You have just created a perpetual-motion machine!
     
  20. May 29, 2017 #19

    rude man

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    As promised in post 16:
    plate capacitor energy U = 1/2 CV2
    We change dz differentially which incurs V + dV, C + dC and U + dU:
    So differential change in U is dU = 1/2 (2CV dV + V2 dC) = CV dV + V2 dC/2
    but dV = V dQ/Q' where Q' is charge on large capacitor
    so dU = CV2 dQ/Q' + V2 dC/2
    divide both terms by C: V2 dQ/Q' + V2 dC/2C
    Now, I can make Q' as large as I want, so the first term can be << second term.
    But the 1st term is in dV so dV has negligible effect on dU.
    So dU = V2 dC/2.
    I will leave it to you to show that V2 dC/2 = V dQ/2.
    Remember, V is constant since that's the voltage before the displacements dz, dQ, dV etc.
    Note also that dV > 0 but dU, dC < 0. dQ < 0 for the plates but > 0 for the large capacitor C'.
    QED
     
    Last edited: May 29, 2017
  21. May 29, 2017 #20
    I was wrong here.
    Change in the potential energy of the bigger capacitor = 1/2[ (Q'+dQ') (V'+dV') - (Q'V')], (1)
    where Q' is the charge of the bigger capacitor,
    V' is the potential of the bigger capacitor,
    dQ' is the increase in charge of the bigger capacitor,
    dV' is the increase in potential of the bigger capacitor,

    dV' = (V'/Q')dQ' (2)


    Now, here V' = V(potential of the smaller capacitor) (3)
    dQ' = dQ (charge lost by the smaller capacitor) (4)
    So, dV' = (V/Q')dQ, is very small as Q' could be as large as we want (5)
    So, change in the potential difference across the bigger capacitor is effectively negligible.
    Consequently , change in the potential difference across the smaller capacitor is also negligible.
    But, this does not imply,
    1/2[ (Q'+dQ') (V'+dV') - (Q'V')] = 1/2[ (Q'+dQ') (V'+0) - (Q'V')] =1/2[ (Q'+dQ') V' - (Q'V')] = 1/2 (dQ') V'
    We can not substitute dV' = 0 here as substituting dV' =0 makes Q'dV' =0, which is wrong.
    Hence , even though dV' is negligible, its contribution in the increase of potential energy of the bigger capacitor is not negligible because of Q' being large.

    Hence,
    1/2[ (Q'+dQ') (V'+dV') - (Q'V')] = 1/2[(dQ') V' + (dV') Q' ]
    Using eqns. 3,4,5
    = 1/2[(dQ) V + (dQ) V ]
    =V dQ
    ( Learning : Before substituting dV' = 0 , I must check whether it is proper to do so.
    I should not substitute blindly.)



    Thank you ,thanks a lot.
     
    Last edited: May 29, 2017
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