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?


1 Answer 1


According to the situation described by you,
-Q charge will be induced on the inner surface of the shell and 3Q will be present on the outer surface of the shell.
All the charges will distribute themselves uniformly over the surface.
Hence for the region $r_1<r<r_2$ 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.
The charges present on the outer surface of shell or outside the shell always create zero electric field inside the shell even if the outer charges had ununiform distribution.


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