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Energy in a shrinking spherical capacitor

  1. Feb 15, 2015 #1
    1. The problem statement, all variables and given/known data
    a conducting spherical shell has radius R and potential V. If you want, you can consider it to be part of a capacitor with the other shell at infinity. you compress the shell down to zero size while always keeping it spherical,while a battery holds the potential constant at V.

    1.What are the initial and final energy stored in the system?
    2.What is the final charge on the shell?
    3.What is the work done on/by the battery?
    4.What is the work done on/by you?

    2. The attempt at a solution

    I took the suggestion of considering as a capacitor, where the outer shell is connected to one of the terminals and the inner shell connected to the other terminal.

    1&2. I assume the question only meant electrostatic energy,which is just the energy stored in the capacitor. Hence,Einitial=(CV^2)/2 =0.5*V^2*4*pi*epsilon*R=2*pi*R*epsilonV^2

    As the sphere shrinks, C decreases as R decreases, the charges on both shells would start to recede from the surface of the shells. It leads me to think that the inner shell, which has become a point, has zero charge and zero capacitance, therefore Efinal would be zero.

    3. During the compression process, charges are forced back from the capacitors against the potential, therefore the battery receives work. However, I'm not really sure what the expression for that would be. If the charges recede then some other charge around the battery should also move as a result of electrostatic repulsion. Thus I suspect the work done on the battery is a multiple of QV, where Q is the magnitude of the charge originally on one shell.

    4. I would need to do work against the electrostatic repulsion of the charges on the sphere. By the conservation of energy, the total work I have to supply is Efinal - Einitial -work done on the battery

    There are several major things I'm confused about: How can the capacitor still maintain potential V while the inner shell shrinks to a point? How about the pattern of the electric field? The fact that there's still a potential difference means there would be an electric field between the two shells, which would probably be radial due to the geometry. However this might also imply that the charge on the 'inner shell' is non-zero.
     
  2. jcsd
  3. Feb 15, 2015 #2

    rude man

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    Right. remember, Efinal here is the energy left on the capacitor.
    Yeah, but why "a multiple"? If you have a battery of voltage V and you add Q charge to it, how much energy have you added?
    That is right. Get the work done on the battery right & you're home.
    Think of the shell shrinking to an arbitrarily small size but not zero. Zero by definition means "no shell"
    Yes, there is a remanent E field but what would it be (use Gauss) if Q is arbitrarily small?
     
  4. Feb 15, 2015 #3
    I would assume that means I have to move Q against the entire E-field within the battery. So I would say QV, which is C*V^2 or 2*Einitial.

    However, I was worried about the -Q left on the 'infinitely' large shell. Now that I think of it, once I move Q across to the other side of the battery, Q cancels the -Q, so I don't need any extra energy to move -Q.

    If I'm not wrong it would be -1.5*Einitial? Why is it negative?

    What's meant by 'arbitrarily small' ? As long as Q is non-zero I would still need to know its value in order to compute E wouldn't I?
     
  5. Feb 15, 2015 #4

    rude man

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  6. Feb 15, 2015 #5

    rude man

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    Right. Keep track of the signs. So far there has been a negative E change due to the shrunk capacitor, but a positive E added to the battery. So from that you can deduce the work done by or on you. The sum has to be zero!
    There is "no" more Q after the capacitor (shell) has shrunk to infinitesmal size.
    No, the sum of all the delta E has to be zero, see above. And it's not negative since you put twice as much energy into the battery as you lost from the capacitor.
    Q and E are infinitesimal. So E = Q = 0 after shrinking to "zero" radius. What's meant by a "differential" in Calculus? Same idea. Try not to get too worried about this.
     
  7. Feb 15, 2015 #6
    ΔECapacitor=Ef-Einitial=-0.5 *CV^2
    ΔEBattery=+CV^2

    ΔECapacitor+ΔEBattery+ΔEme=0

    ΔEme=-0.5 *CV^2

    The minus sign implies my 'energy content' has decreased, meaning I have given out energy 0.5 *CV^2..........?
     
  8. Feb 15, 2015 #7

    rude man

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    That is correct! Good work.
    Note how keeping track of the various energies keeps the problem simple, like not having to calculate mutual charge repulsions etc. or even what exactly is happening.
    Keep in mind that power = voltage x current so energy = voltage times the time integral of current (since v is constant here), which is voltage x charge.
     
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