Answer
The electric field is $1201021.3\hspace{2mm}N/C$ and is directed towards the centre of the cavity.
Let $q=-3.00\mu C$ be at the centre of the cavity.
$r=9.50cm$ and $r'=6.50cm$.
We draw a spherical Gaussian surface of radius $r$.
Work Step by Step
Let $\vec{E}$ be the electric field at the surface of the Gaussian surface. $\vec{dA}$ is the infinitesimal area element vector.
After applying Gauss' law, we put the value of $r$.
Finally, we have:
