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I am currently studying the textbook A Student's Guide to Maxwell's Equations by Daniel Fleisch. In a section discussing the integral form of Gauss's law, the author presents the following electric field equations:

Conducting sphere (charge $= Q$):

$$\vec{E} = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Q}{r^2}\hat{r} \ \text{(outside, distance $r$ from center)}$$

$$\vec{E} = 0 \ \text{(inside)}$$

Uniformly charged insulating sphere (charge $= Q$, radius $= r_0)$:

$$\vec{E} = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Q}{r^2} \hat{r} \ \text{(outside, distance $r$ from center)}$$

$$\vec{E} = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Q r}{r_0^3} \hat{r} \ \text{(inside, distance $r$ from center)}$$

I have two questions regarding these equations:

  1. Why is the "outside, distance $r$ from center" equation for the "conducting sphere" and "uniformly charged insulating sphere" the same?
  2. Why is $\vec{E} = 0$ "inside" for the conducting sphere, whereas we have that $\vec{E} = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Q r}{r_0^3} \hat{r}$ "inside" for the uniformly charged insulating sphere? I am aware that, if you have a real or imaginary closed surface of any size and shape, and there is no charge inside the surface, the electric flux through the surface must be zero, but I'm unsure of exactly what about these two situations leads to this difference.

I would appreciate it if people would please take the time to clarify these points.

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The difference between the two spheres is the charge distribution. By Gauss's law any charge outside the sphere does not distinguish how the charge is distributed as long as it is spherical. Inside the sphere, of course it does matter. For the conductor all charge is on the surface of the sphere. Gauss tells us that the field inside the sphere is zero. For the uniform distribution, the charge inside r is proportional to $r^3/r_0^3$. Combining this with Coulomb's law gives the linear r-dependence.

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  • $\begingroup$ You say "For the conductor all charge is on the surface of the sphere."; why is this? Also, why is it that the charge inside $r$ is proportional to $r^3/r_0^3$? $\endgroup$ – The Pointer Jan 27 '20 at 18:34
  • $\begingroup$ Only if the charge is uniformly distributed on the surface of the sphere is the field inside the conductor zero. If the field were not zero, a current would exist to distribute the charge to make it zero. $\endgroup$ – my2cts Jan 27 '20 at 20:11
  • $\begingroup$ And what about the second part? Why is it that the charge inside $r$ is proportional to $r^3/r_0^3$? $\endgroup$ – The Pointer Jan 27 '20 at 20:34
  • $\begingroup$ The charge is proportional to the volume, hence this relation. $\endgroup$ – my2cts Jan 27 '20 at 20:42
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Gauss's law is always true but pretty much only useful when you have a symmetrical distribution of charge. With spherical symmetry it predicts that at the location of a spherical Gaussian surface, (symmetrical with the charge) the field is determined by the total charge inside the surface and is the same as if the charge were concentrated at the center of the surface. Hence, outside any symmetrical sphere of charge you can use the formula for the field of a point charge. Inside the non-conductor, you use only that part of the charge which is inside the Gaussian surface. Inside a conductor, charges will move around until the field is zero. Negative charge will collect on the inside of the conductor until it matches whatever positive charge is inside of that. Outside of the conductor, the field is still determined by the total charge inside the Gaussian surface.

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  • $\begingroup$ Thanks for the answer. Is there an error in this: "Negative charge will collect on the inside of the conductor until it matches whatever positive charge is inside of that."? It says "inside" twice, so it doesn't make sense to me. $\endgroup$ – The Pointer Jan 25 '20 at 6:23
  • $\begingroup$ Negative charge will collect on the inside surface of the conductor if there is a positive charge at a smaller radius (and visa versa). $\endgroup$ – R.W. Bird Jan 27 '20 at 14:44

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