Suppose we have a sphere of radius $R$ with a uniform charge density $\rho$ that has a cavity of radius $R/2$, the surface of which touches the outer surface of the sphere. The question was to calculate the field inside the cavity.
Naively, I used Gauss' law to determine that $\mathbf{E}=0$ inside the cavity. However, the solution I have stated that the field is actually the superposition of the field of the sphere without the cavity, and the field of the cavity, wherein the charge density is the negative of that of the original sphere. So,
\begin{align}
\mathbf{E}_{\text{inside sphere}}&=\frac{\rho}{3}(x,y,z)\\
\mathbf{E}_{\text{inside cavity}}&=-\frac{\rho}{3}(x+R/2,y,z)\\
\\
\Longrightarrow \ \mathbf{E}_{\text{net inside cavity}}&=\mathbf{E}_{\text{inside sphere}}+\mathbf{E}_{\text{inside cavity}}\\
&=-\frac{\rho R}{6}(1,0,0).
\end{align}
I am confused by the rationale of this approach, and also why Gauss' law gives the incorrect answer.
Best Answer
The field inside the cavity is not 0. The flux through the cavity is 0, but there is still an electric field.
Gauss's Law works great in situations where you have symmetry. for a sphere with no cavity, you have perfect spherical symmetry. When you include the cavity, you change the charge distribution on the sphere to be asymmetrical so Gauss's Law doesn't work the easy way we're used to. Instead, we can use superposition of electric fields to calculate the field inside the cavity.