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The connection between EMF and voltage in EM

  1. Jan 7, 2015 #1
    It's kind of embarassing but I have almost finished a introductory EM class and I'm still not sure in a formal way in which cases the EMF is equal to the voltage. EMF was defined as the work done on a charge per unit of charge by any EM originated force when the charge would take a certain path.

    From here on I kind of forgot about the real definition of the voltage which was the integral of the electric field along that path and used EMF and voltage interchangable in any situation (Faraday induction and so on). It seems to have worked out until now but now that I look at the definitions of EMF and voltage it seems that some subtleties may arise in a more formal treatment.

    I would appreciate if someone who had the same confusion in their intro EM class could elaborate a bit. Thanks.
     
  2. jcsd
  3. Jan 7, 2015 #2

    Andrew Mason

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    There is no real difference between EMF or electromotive force and voltage or "electrical potential energy per unit charge". Electromotive force is really a misnomer but old habits are hard to shake. EMF often used when referring to induction: ie. induced EMF or to the voltage produced by a battery or other source when there is no load (current). Some may try to suggest that there is a subtle difference between the terms but there really isn't - at least not one that physicists will agree on. So I wouldn't worry about it. So long as you understand that both terms refer to electric potential energy per unit charge, you can use either term interchangeably.

    AM
     
  4. Jan 8, 2015 #3
    Thanks for the answer. May I elaborate a bit on my confusion and then you tell me if I'm wrong somehwere.

    We defined EMF in class something like: ''For a charge that is being pushed by EM forces along a certain path C, the emf ##\epsilon## is defined as the work done on the charge by the EM forces divided by the charge itself.'' (Wiki uses the same definition : http://en.wikipedia.org/wiki/Electromotive_force)

    Alright so let's look at a problem where no magnetic fields are involved. A positive charged moves from a positive capactor plate towards a negative plate. ##\Delta{V}## is negative in this case since it goes from a positive potential to a negative potential. The work done on the charge is however obviously positive. This positive work is by definition equal to the emf.

    This would mean that at least in these kind of problems involving no magnetic field, there's a minus sign difference between the definition of potential difference and emf.

    The interesting part is that in electrodynamics problems that DO involve magnetic fields, this minus sign disappears and emf and voltage seem to be fully equal.
     
  5. Jan 8, 2015 #4

    Andrew Mason

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    We should start by being clear on how the convention for positive and negative works.

    Before anyone really understood electricity very well, the convention was developed that the direction given to an electric field was the direction in which a positive charge would be pushed. So the direction of the electric field between the positively charged plate and the negatively charged plate is from the positive to the negative (i.e. the direction in which a positive charge moves under the effect of the field). For a battery, the electric field direction is from + to -. If you were to connect a capacitor to a battery, the capacitor plate that is connected to the + terminal of the battery will have positive charge and the plate connected to the - terminal will have negative charge. So the field will be from the positive side to the negative side.
    I am not sure why you say the potential difference is negative. The potential difference is the difference in potential energy that a positive charge of one coulomb will have at the + plate relative to the negative plate. That potential difference is equal to the work done by the field on a unit of positive charge in passing from the + to the - plate. That is a positive number.

    AM
     
  6. Jan 8, 2015 #5
    Generally same things different causes.
     
  7. Jan 8, 2015 #6
    Ah I think we might be getting to the point of why I'm confused. When I hear potential difference I instantly think of ##\Delta{V}##. What does ##\Delta## mean? It means the final state of the ##V## function minus the initial state. So in my example above ##\Delta{V}## will be negative since the final value of the function is lower than the initial. Am I wrong using potential difference like this?
     
  8. Jan 8, 2015 #7

    Andrew Mason

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    The potential difference between two points a and b in a force field (eg. gravitational or electric) is the potential energy a unit body has at point a relative to point b. It is the amount of work that the field does on a unit body (eg. a body with a unit of mass or charge) in moving from point a to point b. It is equal to the change in kinetic energy that the body (of mass or charge) will acquire in moving from a to b. Since in a conservative field total energy is constant, ##\Delta{PE} = -\Delta{KE}##

    Think of gravity. A 1 kg object raised to a height of 1 m above the ground has 9.8 J of potential energy or is at a gravitational potential of 9.8 J/Kg relative to the ground. The potential difference between the 1 m. height and the ground is + 9.8 J/Kg. The change in kinetic energy that the 1 Kg. object will experience in falling the 1m. to the ground is +9.8 J. The change in potential energy that the 1 Kg. object will experience in falling the 1m. to the ground is -9.8 J.

    AM
     
    Last edited: Jan 8, 2015
  9. Jan 8, 2015 #8

    jtbell

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    If you like, you might mentally define a new unit: 1 "gravolt" = 1 J/kg. :oldsmile:

    I think you mean just "J" in both of these statements, instead of "J/kg".
     
  10. Jan 8, 2015 #9

    Andrew Mason

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    Yes. Thanks for pointing that out. I have corrected it.

    AM
     
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