P20B2020 – Concentric Spheres

Because of the spherical symmetry of the charge distribution and the fact that the charge density $\rho \left(r\right)$ is only a function of $r$ (and therefore for a given $r$, you have the same $\rho \left(r\right)$ at every point), we conclude that the electric field created by the charge distribution is radial, directed outward (because ${\rho }_0 > 0$).

With spherical symmetry, we can use Gauss’s Law to derive the electric field created by the distribution and we separately treat the region inside the sphere (region I) from the region outside of the sphere (region II).

Note: we need to proceed with caution we computing the enclosed charge, because the charge density is not constant and we cannot simply multiply it by the volume to get $Q_{enc}$. Rather, we will have to integrate $\rho \left(r\right)$ over the volume enclosing charge.

Region I: inside the sphere

We consider a Gaussian sphere of radius $r_0$ centered on the solid sphere with $r\le R_1$.

The electric flux through the Gaussian sphere is equal to

\oiint_{sphere}{{\overrightarrow{E}}_I\cdot d\overrightarrow{A}}=\oiint_{sphere}{E_I{\mathrm{cos} \left(0\right)\ }dA}=E_I\cdot 4\pi r^2_0

To find the charge enclosed by the Gaussian sphere, we consider an infinitesimal spherical shell of radius $r < r_0$ and thickness $dr$. The infinitesimal volume of that shell is then equal to $dV=4\pi r^2dr$ and the amount of charge that it encloses is equal to

dQ_{enc}=\rho \left(r\right)dV=4\pi r^2\cdot \frac{{\rho }_0R^2_1}{r^2}dr=4\pi {\rho }_0R^2_1dr

To derive the total amount of charge $Q_{enc}$ enclosed by the Gaussian sphere of radius $r_0$, we integrate $dQ_{enc}$ over the Gaussian sphere from $r=0$ to $r=r_0$ as follows

Q_{enc}=\int^{r=r_0}_{r=0}{dQ_{enc}}=\int^{r=r_0}_{r=0}{4\pi {\rho }_0R^2_1dr}=4\pi {\rho }_0R^2_1r_0

By Gauss’s Law, we derive the magnitude $E_I$ of the electric field as follows

\begin{aligned}
\oiint_{sphere}{{\overrightarrow{E}}_I\cdot d\overrightarrow{A}}=\frac{Q_{enc}}{{\varepsilon }_0}\ \ \ \ &\Rightarrow \ \ \ \ \ E_I\cdot 4\pi r^2_0=\frac{\pi {\rho }_0R^2_1r_0}{{\varepsilon }_0} \\
\\
&\Rightarrow \ \ \ \ \ E_I=\frac{{\rho }_0R^2_1}{{\varepsilon }_0r_0}
\end{aligned}

Thus, the electric field inside the solid sphere is given by

\boxed{{\overrightarrow{E}}_I=\frac{{\rho }_0R^2_1}{{\varepsilon }_0r_0}\ \ \hat{r}}

Region II: outside the sphere

We consider a Gaussian sphere of radius $r_0$ centered on the solid sphere with $r>R_1$.

The electric flux through the Gaussian sphere is equal to

\oiint_{sphere}{{\overrightarrow{E}}_{II}\cdot d\overrightarrow{A}}=\oiint_{sphere}{E_{II}{\mathrm{cos} \left(0\right)\ }dA}=E_I\cdot 4\pi r^2_0

To find the charge enclosed by the Gaussian sphere, we consider an infinitesimal spherical shell of radius $r < r_0$ and thickness $dr$. The infinitesimal volume of that shell is then equal to $dV=4\pi r^2dr$ and the amount of charge that it encloses is equal to

dQ_{enc}=\rho \left(r\right)dV=4\pi r^2\cdot \frac{{\rho }_0R^2_1}{r^2}dr=4\pi {\rho }_0R^2_1dr

To derive the total amount of charge $Q_{enc}$ enclosed by the Gaussian sphere of radius $r_0$, we integrate $dQ_{enc}$ over the Gaussian sphere from $r=0$ to $r=R_1$ as follows

Q_{enc}=\int^{r=R_1}_{r=0}{dQ_{enc}}=\int^{r=R_1}_{r=0}{4\pi {\rho }_0R^2_1dr}=4\pi {\rho }_0R^3_1

By Gauss’s Law, we derive the magnitude $E_I$ of the electric field as follows

\begin{aligned}
\oiint_{sphere}{{\overrightarrow{E}}_{II}\cdot d\overrightarrow{A}}=\frac{Q_{enc}}{{\varepsilon }_0}\ \ \ \ &\Rightarrow \ \ \ \ \ E_{II}\cdot 4\pi r^2_0=\frac{\pi {\rho }_0R^3_1}{{\varepsilon }_0} \\
\\
&\Rightarrow \ \ \ \ \ E_{II}=\frac{{\rho }_0R^3_1}{{\varepsilon }_0r^2_0}
\end{aligned}

Thus, the electric field inside the solid sphere is given by

\boxed{{\overrightarrow{E}}_{II}=\frac{{\rho }_0R^3_1}{{\varepsilon }_0r^2_0}\ \ \hat{r}}

To derive the charge $Q_c$ carried by the conducting sphere, we consider a Gaussian sphere of radius $r_0 > R_2$ centered on the two spheres.

Recalling that $E_{III}=0$ in the region $r>R_2$, we conclude that the electric flux through the Gaussian sphere must be zero and write

\oiint_{sphere}{{\overrightarrow{E}}_{III}\cdot d\overrightarrow{A}}=0

By Gauss’s Law, we conclude that the charge enclosed by the Gaussian sphere must be zero. Indeed,

\begin{aligned}
\oiint_{sphere}{{\overrightarrow{E}}_{III}\cdot d\overrightarrow{A}}=\frac{Q_{enc}}{{\varepsilon }_0}\ \ \ \ \ &\Rightarrow \ \ \ \ \ 0=\frac{Q_{enc}}{{\varepsilon }_0} \\ 
\\
&\Rightarrow \ \ \ \ \ Q_{enc}=0
\end{aligned}

Thus, the charge $Q_c$ carried by the conducting sphere can be derived as follows

\begin{aligned}
Q_{enc}=0\ \ \ \ &\Rightarrow \ \ \ \ \ Q_c+Q_{solid\ sphere}=0 \\
\\
&\Rightarrow \ \ \ \ \ Q_c+4\pi {\rho }_0R^3_1=0 \\
\\
&\Rightarrow \ \ \ \ \ Q_C=-4\pi {\rho }_0R^3_1
\end{aligned}

The charge $Q_c$ carried by the conducting sphere is equal to

\boxed{Q_c=-4\pi {\rho }_0R^3_1}

The shell is a conductor and therefore the excess charge it carries lives on its inner and outer surfaces.

We let $Q_{inner}$ denote the charge on the inner surface $\left(r=R_1\right)$ of the shell and $Q_{outer}$ denote the charge on the outer surface $\left(r=R_2\right)$ of the shell. We then consider a Gaussian sphere of radius $r$ such that $R_1 < r < R_2$ centered on the solid sphere. \textit{}

Since the electric field inside the conducting shell is zero, we conclude that the electric flux through the Gaussian sphere must be zero as well. We then proceed the same way we did in the previous question and conclude that the charge enclosed by the Gaussian sphere must be zero by Gauss’s Law. This allows us to solve for $Q_{inner}$ as follows

\begin{aligned}
\oiint_{sphere}{{\overrightarrow{E}}_{II}\cdot d\overrightarrow{A}}=\frac{Q_{enc}}{{\varepsilon }_0}\ \ \ \ &\Rightarrow \ \ \ \ \ 0=Q_{enc} \\
\\
&\Rightarrow \ \ \ \ \ Q_{solid\ sphere}+Q_{inner}=0 \\
\\
&\Rightarrow \ \ \ \ \ Q_{inner}=-Q_{solid \ sphere} \\
\\
&\Rightarrow \ \ \ \ \ Q_{inner}=-4\pi {\rho }_0R^3_1
\end{aligned}

Thus, we conclude that the surface charge density ${\sigma }_1$ on the inner surface of the shell is equal to

\boxed{{\sigma }_1=\frac{Q_{inner}}{4\pi R^2_1}=-\frac{4\pi {\rho }_0R^3_1}{4\pi R^2_1}=-{\rho }_0R_1}

Now that we know $Q_{inner}$, we derive $Q_{outer}$ by Conservation of Charge. Indeed, since the conducting shell carries a net charge $Q_{net}=Q_c$, we solve for $Q_{outer}$ as follows

\begin{aligned}
Q_{net}=Q_c\ \ \ \ &\Rightarrow \ \ \ \ \ Q_{inner}+Q_{outer}=Q_c \\
\\
&\Rightarrow \ \ \ \ \ Q_{outer}=Q_c-Q_{inner} \\
\\
&\Rightarrow \ \ \ \ \ Q_{outer}=-4\pi {\rho }_0R^3_1-\left(-4\pi {\rho }_0R^3_1\right) \\ 
\\
&\Rightarrow \ \ \ \ \ Q_{outer}=0
\end{aligned}

Thus, we conclude that the surface charge density ${\sigma }_2$ on the outer surface of the shell is equal to

\boxed{{\sigma }_2=\frac{Q_{outer}}{4\pi R^2_2}=0}

Finally, since the charge carried by the shell is located on its inner surface exclusively, we conclude that the volume charge density ${\rho }_c$ of the shell is equal to

\boxed{{\rho }_c=0}

since no charge is distributed throughout its volume.