Why does E + dA/dt = - ∇V?

Poirot · Apr 24, 2016

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.

spaghetti3451 · Apr 24, 2016

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 · Apr 24, 2016

Poirot said: ↑

∇ 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.

Log in or Sign up

Why does E + dA/dt = - ∇V?

Poirot

spaghetti3451

blue_leaf77

Solving for dA in Gauss's Law (Replies: 1)

How to calculate dA of a hemisphere. (Replies: 1)