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Say for the sake of this question we have a microscopic 3D sphere of radius $r$. This disk has a charge $q$ which is uniformly distributed.

What is the electric field within the sphere? Does it vary as a function of radius?

Edit: I've spent of bit of time on this over the last few days, trying to extend the approach that Griffiths used for point charges to this system. If there's an error in this approach please let me know.

The flux of $E$ through an enclosed space is,

$$\oint \textbf{E} \cdot d\textbf{a} = \frac{q}{\epsilon_0}$$

Making the charge just the flux times $\epsilon_0$. Because of the superposition principle the electric field at any point can be determined by adding together the contribution of all individual fields.

$$\textbf{E} = \sum_i^n \textbf{E}_i$$

Where, $\textbf{E} = \frac{q}{\epsilon_{0} r^{2}} = \frac{\Phi}{r^{2}}$.

If we subdivide the sphere into infinitesimally small sections we can in principle sum the various electric fields. However, since the charge is uniformly distributed, for any point within the charge there will be a $\textbf{E}$ where $r = 0$.

The electric field within a charge is therefore infinite.

WaveInPlace
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  • Do you mean a circle or a disk? A disk includes the interior of the circle. Also, it is much easier to work with a sphere or a ball, which have very simple fields. – Ghoster Apr 24 '23 at 23:21
  • Depends what the sphere is made off. The field inside a conducting sphere is zero. For a charged insulator it would depend on the charge density distribution in the interior. – FlatterMann Apr 24 '23 at 23:25
  • $\vec{E} \propto R$ – jensen paull Apr 24 '23 at 23:43
  • Sorry, my mistake on the terminology. By your definition this would be a disk. – WaveInPlace Apr 24 '23 at 23:53
  • Related: https://physics.stackexchange.com/q/536983/ – Ghoster Apr 25 '23 at 00:10
  • @jensenpaull That’s true for a ball, but false for a disk. The field everywhere in the plane of the disk is given in an answer to the question I linked to in the comment above. – Ghoster Apr 25 '23 at 00:14
  • @FlatterMann, since we know the charge is distributed uniformly over the region, we can infer that the region is not filled with a conductive material. – The Photon Apr 25 '23 at 00:52
  • Hi WaveInPlace. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. – Qmechanic Apr 25 '23 at 08:39
  • @ThePhoton It is simply not clear to me what geometry the OP is talking about since the terminology seems to change from 2d to 3d. – FlatterMann Apr 25 '23 at 13:02
  • @FlatterMann, that may be from my attempts to make this as simple as possible. Ultimately I'd like to know what the electric field is within a non-point-like charge. The disk is my attempt at defining the structure of the charge, to make the math more concrete. – WaveInPlace Apr 25 '23 at 13:20
  • We can calculate the field of an arbitrary charge distribution by integrating over Coulomb's law. This is usually done for the electrostatic potential, since the field follows from it by taking the gradient: https://physics.stackexchange.com/questions/73586/potential-of-arbitrary-charge-distribution – FlatterMann Apr 25 '23 at 13:26
  • Isn't that the exact opposite of what I'm asking? That's an empty sphere surrounded by charge, whereas I've got a charged space surrounded by nothing. – WaveInPlace Apr 25 '23 at 13:32
  • Thanks, @Qmechanic! If the question still isn't fit for purpose I'll take some time later to overhaul it/start fresh with an extensive description. – WaveInPlace Apr 25 '23 at 13:40

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