P20R2015 – Spherical Metallic Shell

The first thing to do in such a problem is determine how the charge contained on the metallic shell rearranges because of the inner sphere. Indeed, because the charge is free to move on the shell, a negative amount of charge $Q_{inner}$ will be pulled onto the inner surface of the shell and the remaining amount of charge (which could be positive or negative depending on the net charge of the shell initially) is repelled to the outer surface of the shell.

To derive $Q_{inner}$, we will apply Gauss’s Law with a Gaussian sphere that lies within he shell (radius $b<r<c$) and use the fact that $E=0$ inside the conducting material of the shell. To derive $Q_{outer}$, we will use conservation of charge.

Deriving $Q_{inner}$ with Gauss’s Law:

We consider a Gaussian sphere of radius $b<r<c$ as shown below and, recalling that $E=0$ inside conducting material, we conclude that the net flux through the Gaussian sphere is zero.

Indeed, everywhere on the surface of the Gaussian sphere the electric field is $E=0$ and therefore no field lines go through the Gaussian sphere and the electric flux is therefore zero. From Gauss’s Law, we then conclude that the net charge enclosed by the Gaussian sphere must be zero and write

\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}=-Q
\end{aligned}

and therefore, we conclude that the charge on the inner surface of the spherical shell is equal to

Q_{inner}=-Q_{sphere}=-Q

Finally, the surface charge density ${\sigma }_{inner}$ on the inner surface of the spherical shell is equal to

\boxed{{\sigma }_{inner}=\frac{Q_{inner}}{4\pi b^2}=-\frac{Q}{4\pi b^2}}

where $4\pi b^2$ is the area of the spherical inner surface of the shell.

Deriving $Q_{outer}$ using conservation of charge:

Now that we have derived $Q_{inner}$, we use conservation of charge to determine the charge $Q_{outer}$ on the outer surface of the spherical shell. The shell is uncharged initially and therefore $Q_{shell}=0$ (meaning it holds an equal number of positive and negative charges, not that there aren’t any charges at all on the shell). Placing the charged sphere at the center of the shell causes an amount $Q_{inner}$ to be attracted onto its inner surface and an amount $Q_{outer}$ to be repelled to its outer surface. By conservation of charge, we write

\begin{aligned}
Q_{shell}=Q_{inner}+Q_{outer}\ \ \ \ &\Rightarrow \ \ \ \ \ 0=Q_{inner}+Q_{outer} \\
&\Rightarrow \ \ \ \ \ Q_{outer}=-Q_{inner}
\end{aligned}

Thus, we conclude that charge residing on the outer surface of the shell is equal to

Q_{outer}=-Q_{inner}=Q

Finally, the surface charge density ${\sigma }_{outer}$ on the outer surface of the spherical shell is equal to

\boxed{{\sigma }_{outer}=\frac{Q_{outer}}{4\pi c^2}=\frac{Q}{4\pi c^2}}

where $4\pi c^2$ is the area of the spherical inner surface of the shell.

Given the spherical symmetry, we expect the electric field to be radial and we choose a sphere of radius $r$ centered on the insulating sphere as our Gaussian surface. With the previous work, we have determined the different amounts of charge on the inner and outer surface of the shell, making Gauss’s Law straightforward to apply in each of the four regions: inside the sphere of charge, in the gap, inside the shell, and outside of the shell.

Electric field inside the sphere $\left(r\le a\right)$

The electric flux through our Gaussian surface simplifies because everywhere on the surface of the Gaussian sphere the electric field is along $d\overrightarrow{A}$ and has constant magnitude

\begin{aligned}
\oiint_A{{\overrightarrow{E}}_I\cdot d\overrightarrow{A}}&=\iint_A{E_Icos\left(\theta \right)dA} \\
&=\int_A{E_Icos\left(0\right)dA} \\
&=E_Icos\left(0\right)\iint_A{dA} \\
&=E_I\cdot 4\pi r^2
\end{aligned}

where the area of the Gaussian sphere is equal to $A=4\pi r^2$.

The charge $Q$ is uniformly distributed inside the sphere which therefore has a volume charge density $\rho $ equal to

\rho =\frac{Q}{\displaystyle{\frac{4}{3}\pi a^3}}

The Gaussian sphere encloses a sphere of charge that carries a charge $\rho V$ where $V$ is the volume of the Gaussian sphere and we write

Q_{enc}=\rho V=\rho \cdot \frac{4}{3}\pi r^3=Q\frac{r^3}{a^3}

By Gauss’s Law, we conclude that the magnitude of the electric field created inside the sphere of charge is given by

E_I\cdot 4\pi r^2=Q\frac{r^3}{a^3}\ \ \ \ \Rightarrow \ \ \ \ \ E_I=\frac{Qr}{4\pi {\varepsilon }_0a^3}

Thus, the electric field created inside the insulating sphere is equal to

\boxed{{\overrightarrow{E}}_I=\frac{Qr}{4\pi {\varepsilon }_0a^3}\ \hat{r}\ \ \ \ \ \ \ \ \ \ \ for\ \ r < a}

Electric field in the gap $\left(a<r<b\right)$:

The electric flux through our Gaussian surface simplifies because everywhere on the surface of the Gaussian sphere the electric field is along $d\overrightarrow{A}$ and has constant magnitude

\begin{aligned}
\oiint_A{{\overrightarrow{E}}_{II}\cdot d\overrightarrow{A}}&=\iint_A{E_{II}cos\left(\theta \right)dA} \\
&=\iint_A{E_{II}cos\left(0\right)dA} \\
&=E_{II}cos\left(0\right)\iint_A{dA} \\
&=E_{II}\cdot 4\pi r^2
\end{aligned}

where the area of the Gaussian sphere is equal to $A=4\pi r^2$.

The Gaussian sphere encloses a sphere of charge $Q$ and therefore the amount of enclosed charge is equal to

Q_{enc}=Q

By Gauss’s Law, we conclude that the magnitude of the electric field created outside the sphere of charge is given by

E_{II}\cdot 4\pi r^2=\frac{Q}{{\varepsilon }_0}\ \ \ \ \Rightarrow \ \ \ \ \ E_{II}=\frac{Q}{4\pi {\varepsilon }_0r^2}

Thus, the electric field created inside the gap between the sphere and the shell is equal to

\boxed{\overrightarrow{E}_{II}=\frac{Q}{4\pi {\varepsilon }_0r^2}\ \ \hat{r} \ \ \ \ \ \ \ for \ \ a < r < b}

Electric field in the shell $\left(b<r<c\right)$:

There is no electric field inside conducting material in electrostatic material, therefore

\boxed{ E_{III}=0 \ \ \ \ \ \ \ \ for \ \ b < r < c}

Electric field outside of the shell $\left(r>c\right)$

The electric flux through our Gaussian surface simplifies because everywhere on the surface of the Gaussian sphere the electric field is along $d\overrightarrow{A}$ and has constant magnitude

\begin{aligned}
\oiint_A{{\overrightarrow{E}}_{IV}\cdot d\overrightarrow{A}}&=\iint_A{E_{IV}cos\left(\theta \right)dA} \\
&=\iint_A{E_{IV}cos\left(0\right)dA} \\
&=E_{IV}cos\left(0\right)\iint_A{dA} \\
&=E_{IV}\cdot 4\pi r^2
\end{aligned}

where the area of the Gaussian sphere is equal to $A=4\pi r^2$.

The Gaussian sphere encloses the charge $Q$ of the inner sphere and the charge $Q_{inner}+Q_{shell}=0$ and the enclosed charge is therefore equal to

Q_{enc}=Q

By Gauss’s Law, we conclude that the magnitude of the electric field created inside the sphere of charge is given by

E_{IV}\cdot 4\pi r^2=\frac{Q}{{\varepsilon }_0}\ \ \ \ \Rightarrow \ \ \ \ \ E_{IV}=\frac{Q}{4\pi {\varepsilon }_0r^2}

Thus, the electric field created inside the insulating sphere is equal to

\boxed{{\overrightarrow{E}}_{IV}=\frac{Q}{4\pi {\varepsilon }_0r^2}\ \ \hat{r} \ \ \ \ \ \ \ for \ \ r > c}