Why is the divergence of induced electric field zero?

If someone says because

$\nabla \cdot E=\rho/\epsilon _0$ and for induced fields $\rho=0$ and hence its divergence is zero, then how do we in the first place know that this equation is valid for the induced fields?

Edit: One can prove the above gauss law from Coulomb law in statics but what about it in the dynamics?

  • $\begingroup$ You want a justification of Gauss’ Law? $\endgroup$
    – Gilbert
    Feb 3, 2021 at 14:26
  • $\begingroup$ Do you mean why physics equation? Or like why newton law? $\endgroup$
    – kyril
    Feb 3, 2021 at 14:30
  • $\begingroup$ @Dear Gilbert. Gauss law follows from coulomb in static. What about in dynamics? As Shashank points out in the comment to the answer below. $\endgroup$
    – Kashmiri
    Feb 3, 2021 at 15:30

1 Answer 1


I think you're drawing a distinction that doesn't need to be drawn. There are two equations that the electric field must satisfy, Gauss's Law & Faraday's Law: $$ \vec{\nabla} \cdot \vec{E} = \frac\rho{\epsilon_0}, \qquad \vec{\nabla} \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}. $$ The total electric field always satisfies these equations; that's what Maxwell's equations, taken as a whole, require. So in any situation (even when fields are changing), it is always the case that the divergence of $\vec{E}$ is proportional to $\rho$. And in any situation (even when there are charges), it is always the case that the curl of $\vec{E}$ is equal to the negative of the rate of change of $\vec{B}$.

If it happens to be the case that everything is static, then it follows that $\vec{E}$ is curl-free; this is the domain of electrostatics. And if it happens to be the case that there are no charges in a particular region, then $\vec{E}$ is divergence-free; this is the case of "induced fields" that you're describing above. But it's better, in general, to think about the electric field as a whole rather than as the superposition of some "static field" and some "induced field".

  • 1
    $\begingroup$ Any particular reason for Gauss’ law holding in dynamics. Like in statics one could prove (or rather verify) Gauss’ law from Coloumb’s law. Similarly in dynamics can we verify Gauss’s law for accelerating charges? I haven’t seen such a calculation though and I have no idea how it would proceed. $\endgroup$
    – Shashaank
    Feb 3, 2021 at 14:49
  • $\begingroup$ Exactly what I had at the back of my mind. $\endgroup$
    – Kashmiri
    Feb 3, 2021 at 15:27

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