A spherical copper shell with net charge $Q_{copper}$, inner radius $2R$, and outer radius $4R$ is surrounded by a concentric spherical aluminum shell with net charge $Q_{aluminum}$, inner radius $6R$, and outer radius $8R$.

A. Multiple Choice

The charge on the inner surface of the copper shell is

a. positive
b. negative
c. zero
d. it depends on the sign of $Q_{copper}$
e. it depends on the sign of $Q_{aluminum}$

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Answer: c

The copper shell is a conductor and therefore the charge $Q_{copper}$ on the copper shell lives on its outer surface which means that the charge on the inner surface of the copper shell is zero.

In addition, for later convenience let’s figure out the charge residing on the inner surface and the outer surface of the copper shell:

The charge $Q_{copper}$ on the outer surface of the copper shell pulls onto the inner surface of the aluminum shell a charge $Q_{inner}=-Q_{copper}$.

The net charge of the aluminum shell being equal to $Q_{aluminum}$, a total amount of charge $Q_{outer}=Q_{copper}+Q_{aluminum}$ then lives on the outer surface of the aluminum shell.

B. You are told $Q_{aluminum}=+3Q$. Also it is found that a proton released from rest at a distance of $r=10\ R$ (from center) remains at rest.

Find the force on the proton (magnitude and direction) when instead it is released at a distance of $r=5R$ (from center). Make a qualitatively accurate sketch of the electric field lines in all regions for this arrangement.

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If the electric force acting on a proton at $r=10R$ is zero then the electric field at that location must be zero. Therefore, we choose a Gaussian sphere of radius $r=10R$ centered on the shells (as shown below) and apply Gauss’s Law.

Since $E=0$ on the Gaussian surface, the electric flux through the surface is equal to zero i.e.

\oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}=0

By Gauss’s Law, it then follows that the net charge enclosed by the Gaussian sphere must equal zero i.e. that we have

\begin{aligned} Q_{enc}=0\ \ \ \ &\Rightarrow \ \ \ \ \ Q_{copper}+Q_{aluminum}=0 \\ \\ &\Rightarrow \ \ \ \ \ Q_{copper}=-Q_{aluminum}=-3Q \end{aligned}

With this information, we can then compute the electric field at $r=5R$ by choosing a Gaussian sphere of radius $5R$ as drawn below.

The electric flux through the Gaussian sphere is equal to

\oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}=\oiint_A{E{\mathrm{cos} \left(180\right)\ }dA}=-E\cdot 4\pi r^2=-100E\pi R^2

Since only the charge from the copper shell is enclosed, we have

\frac{Q_{enc}}{{\varepsilon }_0}=\frac{Q_{copper}}{{\varepsilon }_0}=-\frac{3Q}{{\varepsilon }_0}

By Gauss’s Law, the magnitude $E$ of the electric field at location $r=5R$ is equal to

\begin{aligned} \oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}=\frac{Q_{enc}}{{\varepsilon }_0}\ \ \ \ \ &\Rightarrow \ \ \ \ -\ 100E \cdot \pi R^2=-\frac{3Q}{{\varepsilon }_0} \\ \ &\Rightarrow \ \ \ \ \ E=\frac{3Q}{100\pi {\varepsilon }_0R^2} \end{aligned}

The electric force acting on the proton at $r=5R$ is therefore directed inward (since $Q_{enc} < 0$) and has magnitude

\boxed{F_E=eE=\frac{3Qe}{100\pi {\varepsilon }_0R^2}}

The electric field lines of this arrangement are drawn below

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