1
$\begingroup$

What would electric field lines in the interior of a uniformly charged dielectric solid sphere (charge spread throughout the volume with a uniform charge density) look like? How do we even go about visualising field lines?

$\endgroup$
  • 4
    $\begingroup$ This is a standard exercise and it is covered in depth in most (all?) introductory textbooks. $\endgroup$ – Emilio Pisanty Jun 27 '17 at 14:32
  • 1
    $\begingroup$ @EmilioPisanty I couldn't find it, would you point me to some book that does? $\endgroup$ – Ritik Garg Jun 27 '17 at 14:36
  • 2
    $\begingroup$ An exercise in Chapter 4 of Griffiths's Introduction to Electrodynamics involves calculating the electric field of a uniformly charged dielectric sphere. I don't have the fourth edition in front of me, but it's Exercise 4.20 in the third edition. $\endgroup$ – Michael Seifert Jun 27 '17 at 21:30
4
$\begingroup$

If you have a sphere with uniformly distributed charge, the solution must be spherically symmetrical. Specifically, we know that the field intensity at a radius $r$ is proportional to the charge inside the sphere with radius $r$, and scaled by the dielectric constant:

$$\nabla\cdot \mathbf E=\frac{\rho}{\epsilon}$$.

It follows that the field will increase linearly with $r$ (because it will scale as $$\rm\frac{volume}{area}=\frac{\frac43 \pi r^3}{4\pi r^2}\propto r$$

Once you get to the edge of the sphere, the field will drop off in the usual $1/r^2$ manner.

This is a tricky thing to visualize with lines - it's a bit easier with colors:

enter image description here

(note - in this picture I assumed the dielectric constant of the sphere was 3; this suppresses the electric field inside compared to outside. Tip of the hat to Michael Seifert for pointing out that I had shown the case for $\epsilon_r=1$ without mentioning this explicitly).

$\endgroup$
  • $\begingroup$ Thanks, but colors don't do the trick. With colors one can't make out the direction of the field as with lines. $\endgroup$ – Ritik Garg Jun 27 '17 at 17:16
  • 2
    $\begingroup$ The field lines are purely radial from the center out. But since charge is distributed, field lines would have to "appear" at different distances - this is hard to do cleanly when the charge distribution is continuous (rather than discrete). $\endgroup$ – Floris Jun 27 '17 at 17:32
  • $\begingroup$ The electric field magnitude should be discontinuous at the surface of the sphere, as there is a bound surface charge there. $\endgroup$ – Michael Seifert Jun 27 '17 at 21:26
  • $\begingroup$ (For a dielectric medium, that is. Your answer is correct for $\epsilon_r = 1$.) $\endgroup$ – Michael Seifert Jun 27 '17 at 21:32
  • $\begingroup$ @MichaelSeifert that is an important point I completely omitted to mention - fixed now. Thank you! $\endgroup$ – Floris Jun 27 '17 at 21:48

Your Answer

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

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