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Is the electric field inside a circuit non-conservative?

  1. May 27, 2013 #1
    As charge moves around a circuit, their energy increases by each loop.

    However, I just heard somebody say the total electric field inside a circuit is conservative.

    How is that possible? Is the field conservative or not? Why?

    Thanks for all help! :)
     
    Last edited: May 27, 2013
  2. jcsd
  3. May 27, 2013 #2
    Wires are conductors, so within them yes- the line integral of the electric field is 0. But within the emf, this is not true, so the field is not conservative there.
     
  4. May 27, 2013 #3

    Dale

    Staff: Mentor

    In circuit theory you don't deal directly with electric fields, so I don't know how this question arises. In circuit theory the length of a wire or the physical size of a lumped element is unknown. The E-field is the spatial gradient of the voltage, so to know it you would have to know the physical size of the circuit element, as well as the voltage change acorss it. So the electric field is simply something that you do not need and cannot determine within the constraints of circuit theory.

    However, energy is conserved in circuit theory. The work done on a circuit is always equal to the work done by the circuit plus the energy stored in the circuit.
     
  5. May 27, 2013 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    You can answer this very generally: Whenever and whereever there is a time-dependent magnetic field the electric field cannot be a potential field, because then and there according to Faraday's Law one has
    [tex]\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B} \neq 0.[/tex]
    For a field to have a scalar potential, i.e. to have
    [tex]\vec{E}=-\vec{\nabla} \vec \Phi,[/tex]
    its curl necessarily must vanish, which is not the case when a time-dependent magnetic field is present.
     
  6. May 27, 2013 #5
    So let's say you have a circuit without any magnetic field inside to speak of. Then the E-field inside the circuit is defined as conservative??

    However, how do you then explain the emf increasing the energy of every charge going a loop through the circuit?
     
  7. May 27, 2013 #6

    Dale

    Staff: Mentor

    What needs to be explained about it?
     
  8. May 28, 2013 #7
    isn't the emf-source part of the circuit and its electric field?
     
  9. May 28, 2013 #8

    Dale

    Staff: Mentor

    Yes. Although, as I mentioned earlier the fields are outside the scope of circuit theory.
     
  10. May 28, 2013 #9
    I'm a complete noob (1st year uni) so don't overestimate me: How can a field be conservative if it does work on a particle going one loop around?
     
  11. May 28, 2013 #10
    Note that inside the source charge carriers are pushed "uphill" against the electric field, so there must be some other force (chemical, mechanical etc) acting on them. This force does all the work.
     
  12. May 28, 2013 #11
    ohhh, so it's not an electrical field which does the work?

    Allright, then I get it. thanks for all the help, guys!
     
  13. May 28, 2013 #12

    Dale

    Staff: Mentor

    Think of another conservative force, like gravity. If water goes downhill (e.g. a river) then gravity does work on the water. The water can be used to do work on other things (e.g. a turbine). Then, if you want to continue the process you have to pump the water back up the hill.
     
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