Why does E + dA/dt = - ∇V?

1. Apr 24, 2016

Poirot

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. Apr 24, 2016

spaghetti3451

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. Apr 24, 2016

blue_leaf77

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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