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Homework Help: Electricity and EMF Question

  1. Nov 9, 2012 #1

    I learned that, when a field does work on something, like when an electric field does work on an electron, it loses potential energy -- much like when a ball falls from a height (gravity is doing work on it) it loses potential energy.

    Hence, as an electron flows from the negative to the positive terminal of a battery it is losing energy, correct?

    It would then gain energy when it reaches the negative terminal. At that point the battery would do work (equal to the e.m.f) to move the electron from the negative to the positive terminal.

    Is this description correct?

    Thanks in advance!
  2. jcsd
  3. Nov 9, 2012 #2


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    This part doesn't seem quite right. Something would have to push the electron to overcome the F = qE force of the field around the terminal to move it close to the negative terminal. Nothing special would happen when it reaches the terminal (unless it is moving and crashes to a stop). The pushing work gradually gives it electric potential energy, not all of a sudden as it touches the terminal.
  4. Nov 9, 2012 #3
    Um, so:

    Is it correct to say that the e.m.f is the work done to separate the charge in the battery? This is equal to the work done by the field in moving the electron from the negative to the positive terminal? Still, according to W = -U, shouldn't that decrease the energy of the electron?
  5. Nov 9, 2012 #4


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    I think of the battery as creating an E field and the energy of the electron depends on its position in that field. Just like a capacitor's E field. The energy of an electron changes as it moves in the field, not when it contacts one of the plates. Perhaps you could think of the battery connected to a pair of parallel plates and your electron as moving close to and touching one of the plates. The potential varies with position between the plates - with no significant change happening in the last tiny bit of distance as you approach a plate.

    Now I am feeling a little confused about the "charge" on the battery. I guess it is like a capacitor with an infinitesimal plate size so the charge is essentially zero. And it has some mysterious chemical thing that maintains the potential difference.
    Last edited: Nov 9, 2012
  6. Nov 9, 2012 #5


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