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Work done by a battery to move charge

  1. Sep 30, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm working my way through MIT 8.02x, the intro E&M course, on EdX. In Homework 3/Problem 3, we insert a dielectric between plates in a capacitor that is connected to a battery supplying potential V. So the charge on the capacitor increases by Q. How much work is done by the battery to add that charge to the capacitor plates?

    2. Relevant equations
    The answer given is QV.

    3. The attempt at a solution
    I get the answer, but I'm having a conceptual problem. I understand how when charge Q moves across a potential difference V the work done is QV. But, in the situation above, how is the charge moving "across" a potential difference? The charges here remain inside conductors, and there can't be a potential difference inside conductors - except that of course there must be, because otherwise the charge wouldn't move from the battery to the capacitor. Can someone help me muddle through this?

    I think my mind is still back in mechanics, where everything was force and motion. E&M is all about fields and potential energy, and I'm finding it difficult to shift my perspective.
     
  2. jcsd
  3. Oct 1, 2014 #2

    ehild

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    Homework Helper
    Gold Member

    Think of the process happening in the battery when charging the capacitor. There is some charge piled on the electrodes of the battery in stationary case.The voltage across the capacitor plates is equal to V, the voltage of the battery.
    Dynamic equilibrium is maintained at the interface between the electrode and electrolyte inside the battery: The chemical forces driving ions to the electrode or to the electrolyte are balanced by the electric forces.
    When inserting the dielectric between the capacitor plates, the voltage decreases between the plates, and electrons flow from the negative terminal of the battery to the capacitor, and from the capacitor to the positive terminal, decreasing the charge on the electrodes, so allowing the chemical forces to drive new ions to the electrodes. At the end the charge which was given to the capacitor appears on the electrodes again and equilibrium is set up.
    The net process is that some charge was driven from the negative terminal to the positive one by some chemical work. That work is equal to QV.
     
  4. Oct 1, 2014 #3
    Thanks, ehild. That was a very clear and helpful explanation.
     
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