Delta2
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No I don't think this is very accurate. The most accurate thing to say is that in the general case, the electric field has a scalar potential ##V## as well as a vector potential ##\vec{A}## that is it is $$\vec{E}=-\nabla V-\frac{\partial \vec{A}}{\partial t}$$vanhees71 said:The important point is that for time-dependent fields the electric field has no scalar potential, because it's curl doesn't vanish due to (SI units)
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}.$$
In the lorenz gauge, both V and A obey the non homogeneous wave equations. The source of V waves are the time varying charge densities, while the source of A waves are the time varying current densities.