# Is the E field generally conservative outside the cases when vector potential goes to zero?

In Ronald Wangsness' "Electromagnetic Fields" book, he states that $$\textbf{E}$$ generally depends on both a scalar and vector potential in the form,

$$\textbf{E} = -{\bf \nabla}\Phi -\frac{\partial \textbf{A}}{\partial t}$$

However when $$\frac{\partial \textbf{A}}{\partial t}$$ goes to zero in static cases, $$\textbf{E}$$ becomes a nonconservative electric field.

Is $$\textbf{E}$$ always conservative outside the cases when the vector potential goes to zero?

• a conservative vector field is the gradient of a scalar, so if ${\partial{A}}/{\partial t}=0$ then $E$ is conservative: $E=-\nabla \phi$. – hyportnex Dec 6 '19 at 15:28
• @hyportnex: That should be an answer. – Ben Crowell Dec 6 '19 at 15:50
• I don't understand what you're asking for here. A conservative field is by definition one of the form $E = -\nabla \Phi$. When $A=0$, your equation becomes of that form. What exactly are you confused about? – ACuriousMind Dec 7 '19 at 13:12

This can also be seen directly from Maxwell's equation, $$\nabla \times \textbf{E} = - \frac{\partial \textbf{B}}{\partial t}$$. $$\textbf{E}$$ will be a conservative field if its curl vanishes, which requires $$\frac{\partial \textbf{B}}{\partial t}= 0$$, i.e. when the magnetic field is static.
A conservative vector field $$\mathbf{v}$$ is one that $$\oint_{{L}} \mathbf{v}\cdot d\mathbf{\ell}=0$$ for all closed loops $$L$$ and this is equivalent to that there exists a scalar function $$f$$ such that $$\mathbf{v}=\nabla f$$. Therefore, if $$\frac {\partial \mathbf{A}}{\partial t }=0$$ then $$\mathbf{E}$$ is conservative because $$\mathbf{E}=-\nabla \phi$$.