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Why does E + dA/dt = - ∇V?

  1. Apr 24, 2016 #1
    1. The problem statement, all variables and given/known data
    This isn't necessarily a question but something I found in my notes that I didn't understand.
    It says that from B = x A one can plug this into Faraday-Lenz's law x E = d(B)/dt and it leads to the equation:
    x (E + d(A)/dt)=0.
    From this is says "Using Helmholtz theorem, one can represent the curl free field as the gradient of a scalar potential."

    2. Relevant equations
    B = x A
    x E = d(B)/dt
    x (E + d(A)/dt)=0

    3. The attempt at a solution
    I know that in electrostatics the curl of the electric field is zero and hence there is a scalar potential associated such that E = - V but I don't see the jump for the case involving the vector potential.

    Any guidance/explanation would be greatly appreciated.
     
  2. jcsd
  3. Apr 24, 2016 #2
    In electrostatics, there are no electric currents, i.e., ##\vec{B}=\vec{0}## so that ##\nabla \times \vec{E} = \vec{0}##.

    However, in general, ##\vec{B}=\vec{B}(\vec{x},t)## so that ##\nabla \times \vec{E} \neq \vec{0}##

    See if you can work out the consequence of having a non-zero magnetic field on the form of the electric field ##\vec{E}## in terms of the scalar potential ##V## and the vector potential ##\vec{A}##.

    P.S.: This does not really strike me as being a homework problem, so it's probably better suited for the 'Classical Physics' subforum.
     
  4. Apr 24, 2016 #3

    blue_leaf77

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    The curl of a gradient is always zero ##\nabla \times \nabla \Phi = 0## (this is an identity) where ##\Phi## a scalar function.
     
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