Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.
Sign up to join this community
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?
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".
