# Conductor with cavity: net charge?

Suppose that we have a uncharged spherical conductor that carries a cavity with a charge $$+q$$ in it. The problem I am dealing with consists in answering this question: What is the field outside the sphere? Well, in the solution, it is said that, because the conductor carries no net charge and there is an induced charge $$-q$$ that is on the outer surface of the cavity, a charge $$+q$$ will distribute uniformly over the surface of the sphere, and will produce a electric field around the sphere. What I don't understand is, how can a sphere with no net charge can produce an electric field? Maybe I should understand no net charge as the charge INSIDE the conductor, which does not include the surface charge?

• Do you what a dipole is? Because if so, here you have another system with no net charge that does generate an electric field: i.stack.imgur.com/z17OF.gif – Ofek Gillon Dec 25 '19 at 19:07
• Meaning: no net charge doesn't mean no electric field. – Ofek Gillon Dec 25 '19 at 19:09

But in your question the field outside the uncharged shell is due to the system containing the charge($$+q$$) inside the cavity and the uncharged shell rather than the uncharged shell only. So saying that the field outside is due to an uncharged shell is completely wrong.
If you remove the shell, then the electric field will change(it will not change in the case where the charge $$+q$$ was initially at the centre of the shell). And similarly if you remove the charge $$+q$$, the field everywhere will become zero and thus will have changed from its initial value. So clearly, the field is not due to a single component, rather it is "the field of the system".