I have a vector potential given by:
$\mathbf{A}(x,t) = \mathbf{e}_{y}\frac{1}{2} e^{-(x-ct)^{2}/{4a^{2}}}$
Now, the question is "Determine the E and B under the condition that the scalar potential vanishes $V = 0$."
But I'm not quite sure what it means when $V=0$ ?
As far as I can see, the B-field is given by:
$\mathbf{B}=\nabla \times \mathbf{A}$
And then I have that:
$\mathbf{E} = -\nabla V$
So is it just this straightforward ? That I find the B-field from A, and since $V = 0$, the E-field is zero, or am I doing it wrong ?
Thanks in advance :)