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$V$ is a scalar field, then

$$\nabla\times \nabla V = 0$$

Faraday's law:

$$\nabla \times \mathbf{E} = -\frac{d\mathbf{B}}{dt},\\ \mathbf{E} = -\nabla V\, .$$


marked as duplicate by Qmechanic Feb 15 '17 at 11:17

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  • $\begingroup$ There is no contradiction in Maxwell's equations, but you're correct: If there is a time varying magnetic field, you can no longer write $\vec{E}=-\nabla V$. $\endgroup$ – user12029 Feb 15 '17 at 6:54
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    $\begingroup$ Possible duplicate of How is the curl of the electric field possible? $\endgroup$ – dasdingonesin Feb 15 '17 at 9:21

The theorem is about fields, not about physics, of course. The fact that dB/dt induces a curl in E does not mean that there is an underlying scalar field V which corresponds to that E-field. Only conservative electric fields have a representation as gradient of the scalar potential.

In the presence of a changing B field, E is not conservative, and V is undefined (well, at least poorly defined, and not easy to measure).


The correct result is $\mathbf{E}=-\boldsymbol{\nabla}V-\frac{\partial \mathbf{A}}{\partial t}$ with $\mathbf{B}=\boldsymbol{\nabla}\times \mathbf{A}$.


$\vec{E} = -\vec\nabla V$ is true only in electrostatics, in general you cannot write the electric field as a gradient of some scalar function.

Faraday's law is more fundamental than $\vec{E} = -\vec\nabla V$. Faraday's law is always correct, $\vec{E} = -\vec\nabla V$ is not.


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