# Why doesn't shell theorem work here?

I think I have a misunderstanding about the shell theorem regarding electrostatics here. Since we have a conducting spherical shell, and we are looking for the $$E$$ field, shouldn't it be 0 within the conductor. However, the answer is not 0, and I fail to understand why the shell theorem doesn't apply here.

• Perhaps it means to say within the cavity – Aaron Stevens Jun 9 at 22:17

## 2 Answers

The phrase to note is "produced by the point charge". The point charge induces an equal and opposite charge on the inner wall (radius $$R_1$$) of the shell, and, at distances from the centre greater than $$R_1$$ this induced charge behaves as if it were at the centre of the cavity. Therefore inside the metal of the shell the field due to the point charge and that due to the induced charge are indeed equal and opposite and cancel to zero. But you're not being asked about the resultant field!

The charge Q resides on the outer wall (radius $$R_2$$) and produces no field at any point closer to the centre of the system. It is irrelevant to your question, as is the charge $$+q$$ on the outer wall that would maintain neutrality of the metal – if it weren't for Q!

• Or it means inside the cavity – Aaron Stevens Jun 9 at 22:18

# The Answer

In mathematics, the word "sphere" refers to the hollow $$x^2+y^z+z^2=r^2$$ and the word "ball" refers to the solid $$x^2+y^2+z^2. This question asks about a conducting spherical shell. Such an object is therefore hollow. "The electric field is zero inside of a solid conductor" applies only to a solid conductor. It doesn't apply to hollow conductors.

For a hollow conducting shell, the electric field of charges outside the conductor are negated in the cavity inside of the conducting shell. Such a conducting shell is called a Faraday cage. You can deduce for yourself that "The electric field is zero inside of a conductor" is just a generalization of how Faraday cages work.

Your hollow conducting shell is a Faraday cage. The charge $$Q$$ spreads to the outside of the Faraday cage. Thus, the points inside of the Faraday cage are shielded from $$Q$$. All that matters is $$q$$. Apply Coulomb's law to $$q$$ and you get the answer.

Another way of thinking about it is that points inside the sphere see charge $$q$$ and points outside the sphere see both charges $$q$$ and $$Q$$.

# The Shell Theorem

The shell theorem totally applies here. Charge $$Q$$ is equally distributed in a sphere. Therefore, according to the shell theorem charge $$Q$$ exerts no electric field on the inside of the sphere. All that is left is point charge $$q$$.