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Why does E + dA/dt = - ∇V?

  • Thread starter Poirot
  • Start date
  • #1
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Homework Statement


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."

Homework Equations


B = x A
x E = d(B)/dt
x (E + d(A)/dt)=0

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.
 

Answers and Replies

  • #2
1,344
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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.
 
  • #3
blue_leaf77
Science Advisor
Homework Helper
2,629
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x (E + d(A)/dt)=0.
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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