# On proving the $E$-field is null inside the layer of a conductor having a cavity

Let us consider a conductor having a cavity (both can have an arbitrary shape) wherein a positive charge $$+q\,$$ is located:

At the electrostatic equilibrium, the inner wall of the conductor will have a negative charge $$-q_{1}$$, and the outer wall a positive charge $$q_{1}$$.

The $$E$$-field in the region between the inner and outer wall is determined only by $$+q$$ and $$-q_{1}$$. How can one prove the total $$E$$-field in the layer is null?

The electrostatic equilibrium argument only tell us the potential is constant on the outer and inner surface of the conductor, it does not give any additional information about the $$E$$-field or the potential inside the thickness since there is no charges in it. Also, we cannot claim $$q_{1}= q$$, because, since we don't know the $$E$$-field within the layer, we cannot apply Gauss's law. How can we get out of this impasse?

• Are you not satisfied with the argument showing that the field inside any conductor is $0$? Sep 7, 2019 at 1:39
• @Aaron Stevens Which argument? Sep 7, 2019 at 1:45
• That if there was a field the charges would then move until the field is $0$. So it just has to be $0$. Sep 7, 2019 at 1:59
• @Aaron Stevens But there are no charges inside the layer, $Q=0$, charges, in this case, are located only on the inner and outer wall of the conductor. Sep 7, 2019 at 2:04
• I never said absence of charge means no field. You seemed to be saying to me that the field cannot be $0$ because there is only charge on the surface. That's why I was trying to show you how you can still have a zero field when this is the case. I agree, you need to start with $0$ field for the entire argument. I have told you why it must be zero, and if you don't believe that then if you look up any resource they will say the same thing. Sep 7, 2019 at 10:07