# $\nabla. \vec E =\frac{\rho}{\epsilon}$ for a non conservative field

$$\nabla\cdot \vec E =\frac{\rho}{\epsilon}$$

In this formula, suppose I have a non conservative $$\vec E$$ field, and I put this formula into this expression I will get a corresponding value of $$\rho$$.

But I have read that no amount of constant charge can produce a non conservative $$\vec E$$ field.

So what does the value of $$\rho$$ I get correspond to?

(I don't have much knowledge about this as it is a bit out of syllabus for my highschool course, but my teacher told us about it, as it makes calculating easier and faster[So if this might appear a bit naive to you, well it is :P])

• Read where? Which page? – Qmechanic Jun 23 at 18:43
• @Qmechanic Griffith , Introduction to Electrodynamics {But I haven't read the complete book. I read it only for theory as the vector formula/format is mostly out at syllabus for highschool physics} – user232243 Jun 23 at 18:46

Every field1 $$\vec{F}$$ can be decomposed into two parts: one that is conservative (i.e. $$\nabla\times\vec A=0$$) and one that is divergenceless (i.e. $$\nabla\cdot\vec B=0$$), where $$\vec F=\vec A+\vec B$$. If you take the divergence of $$\vec F$$, you get: $$\nabla\cdot\vec F=\nabla\cdot\vec A+\nabla\cdot\vec B=\nabla\cdot\vec A$$
Note that changing the non-conservative part $$\vec B$$ does not change the divergence and thus does not affect $$\rho$$. In other words, if you solve for $$\rho$$ you will get the density corresponding to just the conservative part of the field.