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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?

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  • $\begingroup$ a conservative vector field is the gradient of a scalar, so if ${\partial{A}}/{\partial t}=0$ then $E$ is conservative: $E=-\nabla \phi$. $\endgroup$ – hyportnex Dec 6 '19 at 15:28
  • $\begingroup$ @hyportnex: That should be an answer. $\endgroup$ – Ben Crowell Dec 6 '19 at 15:50
  • $\begingroup$ 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? $\endgroup$ – ACuriousMind Dec 7 '19 at 13:12
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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.

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Your question is confused: "....when ∂A/∂t goes to zero in static cases, E becomes a nonconservative electric field", etc.

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 $\mathbf{E}=-\nabla \phi$.

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