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2 equations for electric field

  1. Oct 14, 2015 #1
    Hi guys!
    I have noticed that the electric field has 2 equations to it- the negative gradient of the potential, and the negative partial derivative of A with respect to time. Is this like e=mc^2 and e=hν, where only one equation can be used at a time? Can an electric field be described by -∇v OR -∂A/∂t (in which case you are getting an electric field from changing a magnetic field).
     
  2. jcsd
  3. Oct 14, 2015 #2

    BvU

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    Good observation !

    At school we use Gauss's law to find E (or the potential V) for given charge distributions (that's usually the electrostatics chapters) and use ##\vec E = -\nabla V## to link the two.

    And in the magnetism chapters we encounter Faraday's law of induction to find an e.m.f. (again, a V and an E!) as a consequence of a changing magnetic field.
    So yes, the corny answer is: yes the field can be described by both.

    Further on in the curriculum we have electromagnetism and things are heavily intertwined and we have the Maxwell equations

    But you are right: the electric field has to satisfy all the Maxwell equations at all times.
     
  4. Oct 14, 2015 #3
    If it is an electrostatic field, only then it can be equal to the negative gradient of a potential. If it is changing with time, then you need a term -∂A/∂t to calculate the electric field. A is called the vector potential, which is also related to the changing magnetic field that you talked about.
     
  5. Oct 14, 2015 #4
    Is there another way to define the magnetic field other than the curl of A. I know from Ampere's law, the curl of B relies on two things: the change in E with time, and the current density J. Is it like the electric field, where only one of those can describe any given magnetic field. Would that mean there is another part to B=∇XA? I may be completely off, but it seems as though there are 2 mutually exclusive ways to define each of the fields in electromagnetism. I'm sorry if this is a stupid observation.
     
  6. Oct 14, 2015 #5
    There are two "equations" here. The dependence of the curl of B on dE/dt and the current density J, is one of Maxwell's equations and is always satisfied. This has nothing to do with writing B as the curl of A. The relation B = curl A follows from another Maxwell equation, which says that the divergence of B is always zero. This is also always satisfied, though there is a lot of discussion about it, because that Maxwell equation says that there are no magnetic monopoles. If we accept that, then the divergence of B is zero, and B can always be written as the curl of a vector field, which is called the vector potential A. This part has nothing to do with time dependence. Whether B is time dependent or not, you can always write it as the curl of A.
     
  7. Oct 14, 2015 #6
    But is ∇X(∇XA) equal to J OR ∂E/∂t or J PLUS ∂E/∂t?
     
  8. Oct 14, 2015 #7

    BvU

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    The plus sign. See the Maxwell equations link.
     
  9. Oct 14, 2015 #8
    Correct. Maxwell equations are always correct, so curl (curl A) = J + ∂E/∂t. Particular case of that equation: If the fields are time independent, then the second term is zero and curl (curl A) = curl B = J, which is Ampere's law.
     
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