2
$\begingroup$

The cross product between nabla and electric field equals to zero ($\nabla \times \vec{E}=0$) so that means that the field lines are not curved. The relation that gives as the $\vec{E}$ if we use spherical coordinates includes only the $r$ component. Why does it not include the $\phi$ and $\theta$ coordinates too?

$\endgroup$
  • $\begingroup$ What you call "cross product between nabla and electric field" is called the curl of the electric field. That's a good term to learn. $\endgroup$ – Bill N Mar 19 '19 at 15:28
  • $\begingroup$ Thank you Bill!! $\endgroup$ – Katerina Mar 19 '19 at 16:37
  • $\begingroup$ I think you should revise your understanding of the consequence of the curl of a vector field being zero. As far as I remember Purcell's textbook on electromagnetism should contain enough material (and pictures) to develop your own qualitative inderstanding about curl and divergence of vector fields. $\endgroup$ – GiorgioP Mar 19 '19 at 19:45
2
$\begingroup$

The electric field $\textbf{E}_{\rm point}$ due to a point charge is spherically symmetric i.e. it does only depend upon how far you are from that charge. Such force fields are called central force fields. But the electric field due to a dipole $\textbf{E}_{\rm dip}$ (for example) will, in general, depend on $\theta$ and $\phi$. But central or not, electrostatic field is always irrotational or conservative i.e., $\nabla\times\textbf{E}=0$.

A non-rotating vector field may also have a nonzero curl. For instance, see this. Also as G. Smith pointed out in his comment below, a curved vector field does not ensure a nonzero curl.

$\endgroup$
  • 1
    $\begingroup$ And the field lines of the dipole are curved despite the fact that the field has no curl. $\endgroup$ – G. Smith Mar 19 '19 at 16:54
  • $\begingroup$ Good point! @G.Smith $\endgroup$ – SRS Mar 19 '19 at 16:55
  • $\begingroup$ The curl doesn't measure curvature - it measures rotation. The link to example where $\nabla \times F \neq 0$ is just not obvious. $\endgroup$ – Cinaed Simson Mar 19 '19 at 18:49

Not the answer you're looking for? Browse other questions tagged or ask your own question.