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?

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    $\begingroup$ 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
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    $\begingroup$ @EmilioPisanty I couldn't find it, would you point me to some book that does? $\endgroup$ – Ritik Garg Jun 27 '17 at 14:36
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    $\begingroup$ 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

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).

  • $\begingroup$ 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
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    $\begingroup$ 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

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