# Concentric Conducting Spheres and Gauss's Law

One of the things I am struggling with in working with electrostatics is in situations where a charge is surrounded by a concentric sphere, and Gauss's Law must be applied.

Like if there is a metal sphere with charge $$Q$$ on the outside of it, with a radius $$r_1$$

I know that Gauss's Law

$$\oint\limits_SE \cdot d{\textbf A} = \frac{Q_{enc}}{e_o}$$ should apply for any closed sphere.

So, if my understanding follows correctly, the electric field anywhere inside the sphere should be $$0$$.

But what if it is surrounded by a concentric shell that has some charge on its outside. For one example, If it has $$2Q$$ charge on its outside and the shell starts at radius $$r_2$$ and ends at radius $$r_3$$.

In between the sphere and the shell, at some radius $$r$$ where $$r_1 < r < r_2$$, to find the field would it be as simple as applying Gauss's Law and solving for the field, or would you have to worry about the charge that lies on the outer shell as well, since it should also create electric field?

Also, I know that the electric field in between the conducting shell should be $$0$$, as the electric field inside a conductor has to be. Would that mean a charge $$Q$$ must be on the inner side of the shell? And if so, that means for a raidius outside of the sphere and shell, or $$r > r_2$$, the total electric field is $$0$$ because the charges cancel out?

Hence for the region $$r_1 the electric field due to charges -Q and 3Q is zero individually and electric field will be only due to charge Q. Gauss law can be applied.