Question: A sphere of radius $a$ is made of a nonconducting material that has a uniform volume charge density $\rho$. A spherical cavity of radius $b$ is removed from sphere which is a distance $z$ from the center of the sphere. Assume that $a > z + b$. What is the electric field in the cavity?
I understand the solution but what puzzles me is that if i apply the Gauss theorem to the cavity the electric field must be 0? many thanks in advance
Best Answer
Gauss' law here is not really useful in your setup because $$ \oint \vec E\cdot d\vec S\ne \vert \vec E\vert \oint dS \tag{1} $$ i.e. the field is not radial on any Gaussian spherical surface with centre coinciding with the centre of your hole. It is only when $$ \oint \vec E\cdot d\vec S = \vert \vec E\vert \oint dS =\vert \vec E\vert S = \frac{q}{\epsilon} \tag{2} $$ that you can then invert (2) to deduce $\vert \vec E\vert$ on the Gaussian, and (2) can be obtained only if $\vert \vec E\vert$ is constant on the surface so you can "pull it out" of the integral.
It is still true that $$ \oint \vec E\cdot d\vec S=0 $$ since a Gaussian sphere inside you hole encloses no charge, but because of (1) you cannot conclude that $\vert \vec E\vert=0$.
[I take it you know the field inside the hole is constant. You can show this using the superposition principle.]