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

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])

  • $\begingroup$ Read where? Which page? $\endgroup$ – Qmechanic Jun 23 '19 at 18:43
  • $\begingroup$ @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} $\endgroup$ – user232243 Jun 23 '19 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.

1Subject to some typical restrictions on smoothness and the like.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy