# Why is the divergence of induced electric field zero?

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?

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

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".