Why does E + dA/dt = - ∇V?

[Poirot]

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.

 

[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.

 

[blue_leaf77]

∇ 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.