# Field inside hollow conducting sphere

If a conducting hollow sphere is under the influence of an electric field, by electrostatic induction positive and negative charges should align as shown in the picture. But then, what happens inside the sphere?

By Gauss's law, the field inside should be zero? However, the positive charges at the inner wall should create a field pointing to the center? That contradiction confuses me, can anybody please explain what is my fault here?

• Have you tried here @ What is difference between conducting hollow sphere and charged hollow sphere? and here @ Gaussian surface Sep 25, 2022 at 11:14
• Sep 25, 2022 at 11:16
• Yes, there more or less refer to Gauss's law. I see, it must be zero inside. But we have charges on the interior wall, should they not create a field (feels like contradicting Gauss's law)? Sep 25, 2022 at 11:24
• Have you got the book "classical electromagnetism" by Robert H. Good ? It deals with hollow conducting spheres very well, see pages p.183 to 227. Sep 25, 2022 at 11:48
• Unfortunately, I do not have access to that book... Sep 25, 2022 at 12:07

The field outside the sphere has negative “divergence,” because more of the arrows point towards the sphere than away from it. The mathematical version of the previous sentence is

$$\oint \mathrm d\vec A\cdot\vec E = Q_\text{inside}$$

where

• $$\mathrm d\vec A$$ means “a little bit of some surface”
• $$\mathrm d\vec A \cdot \vec E$$ means “does the field point toward the inside or the outside”
• $$\oint$$ means “add up all the little bits until you have a closed shape”

and $$Q_\text{inside}$$ is the total electric charge inside the surface you’ve made. (Dear fellow pedants: I’m using units where there’s no annoying constant here.) The surface doesn’t have to be a material surface (but it can be); you can make any imaginary surface with your brain and this relationship should hold.

So, suppose you have arrows pointing away from the inner surface of the conductor into the void. What are they pointing towards? Either some of the arrows meet in the void somewhere (in which case there must be a negative charge at that location), or you have the same number of arrows entering the void as exiting.

Note also that for the field outside to have negative divergence, you should have drawn more $$-$$ than $$+$$ symbols. I haven’t counted yours. You are correct that, if you have a surface with net positive charge, the electric field should be different on the one side than on the other side. But that suggests that you just shouldn’t have drawn your inner ring of $$+$$ charges, or that you should have included your “extra” negative charge in the void someplace.