The figure below shows a sphere of radius $a$ filled with an insulator that has a uniform, constant charge density $\rho $. Around this sphere is a conducting shell with inner radius $a$ and outer radius $b$. The electric field vanishes outside of the conducting shell $\left(r>b\right)$.

1. What is the net charge $Q_{shell}$ on the conducting shell? Express it in terms of any of the variables given.

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To determine the net charge $Q_{shell}$ on the conducting shell we use Gauss’s Law and choose a Gaussian sphere with a radius $r>b$.

Since the electric field outside of the shell is zero, we conclude that the electric flux through our Gaussian surface must be zero. By Gauss’s Law, we conclude that the net charge enclosed by the Gaussian sphere is zero and we solve for $Q_{shell}$ as follows

\begin{aligned} \oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}=0\ \ \ \ &\Rightarrow \ \ \ \ \ Q_{enc}=0 \\ \\ &\Rightarrow \ \ \ \ \ Q_{sphere}+Q_{shell}=0 \\ \\ &\Rightarrow \ \ \ \ \ \rho \frac{4}{3}\pi a^3+Q_{shell}=0 \\ \\ &\Rightarrow \ \ \ \ \ Q_{shell}=-\rho \frac{4}{3}\pi a^3 \end{aligned}

Thus, the charge on the inner shell is equal to

\boxed{Q_{shell}=-\rho \frac{4}{3}\pi a^3}

2. What is the charge per area ${\sigma }_a$ on the inner surface $\left(r=a\right)$ of the conducting shell?

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The positive charge contained in the sphere attracts the negative charges of the conducting shell (which are free to move) to the inner surface of the shell. We let $Q_{inner}$ denote that amount of charge.

To determine the charge $Q_{inner}$ we apply Gauss’s Law and choose a Gaussian sphere with a radius $r$ such that $a<r<b$.

Since the shell is a conductor, the electric field within it is zero which allows us to conclude that the electric flux through our Gaussian sphere is zero. By Gauss’s Law, we conclude that the charge enclosed by the Gaussian sphere must be zero and we solve for $Q_{inner}$ as follows

\begin{aligned} \oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}=0\ \ \ \ &\Rightarrow \ \ \ \ \ \ Q_{enc}=0 \\ \\ &\Rightarrow \ \ \ \ \ \ Q_{sphere}+Q_{inner}=0 \\ \\ &\Rightarrow \ \ \ \ \ \ Q_{inner}=-Q_{sphere} \\ \\ &\Rightarrow \ \ \ \ \ \ Q_{inner}=-\rho \frac{4}{3}\pi a^3 \end{aligned}

Finally, the charge per area ${\sigma }_a$ on the inner surface of the shell can be derived from $Q_{inner}$ and is equal to

\boxed{{\sigma }_a=\frac{Q_{inner}}{A_{inner}}=\frac{\displaystyle{-\rho \frac{4}{3}\pi a^3}}{4\pi a^2}=-\frac{\rho a}{3}}

3. What is the charge per area ${\sigma }_b$ on the outer surface $\left(r=b\right)$ of the conducting shell?

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By conservation of charge, we derive the charge $Q_{outer}$ on the outer surface of the shell as follows

Q_{shell}=Q_{inner}+Q_{outer}\ \ \ \ \Rightarrow \ \ \ \ \ Q_{outer}=Q_{shell}-Q_{inner}=0\ C

Thus, the charge per area ${\sigma }_b$ on the outer surface of the shell is equal to

\boxed{{\sigma }_b=0\ \ \ C/m^2}

which is consistent with the electric field vanishing outside of the conducting shell.

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