P20S2014 – Two Concentric Insulators

Physics 8B – Fall 2024 (Lowhorn) Gauss’s Law – Problems P20S2014 – Two Concentric Insulators

Two Concentric Insulators

Consider two concentric solid spheres made from insulators. The small sphere has a radius $R_1$ and a positive uniform charge density $\rho $, while the shell surrounding it has a radius $R_2$ and a negative uniform charge density $–\rho $.

1. You are told that the electric field outside the spheres is zero for $r > R_2$. What is $R^3_2/R^3_1$?

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To derive the ratio $R^3_2/R^3_1$ we use the fact that the electric field for $r>R_2$ is zero and apply Gauss’s Law with a Gaussian sphere of radius $r>R_2$.

Indeed, on the surface of such a sphere, the electric field is zero which, in turn, causes the electric flux through the sphere to be zero. By Gauss’s Law, we then conclude that the charge enclosed by the Gaussian sphere must be zero and we derive the ratio $R^3_2/R^3_1$ as follows

\begin{aligned} \oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}=0\ \ \ \ &\Rightarrow \ \ \ \ \ Q_{enc}=0 \\ \\ &\Rightarrow \ \ \ \ \ Q_1+Q_2=0 \\ \\ &\Rightarrow \ \ \ \ \ \rho \cdot \frac{4}{3}\pi R^3_1-\rho \cdot \left[\frac{4}{3}\pi R^3_2-\frac{4}{3}\pi R^3_1\right]=0 \\ \\ &\Rightarrow \ \ \ \ \ R^3_1-\left[R^3_2-R^3_1\right]=0 \\ \\ &\Rightarrow \ \ \ \ \ 2R^3_1-R^3_2=0 \\ \\ &\Rightarrow \ \ \ \ \ \boxed{\frac{R^3_2}{R^3_1}=2} \end{aligned}

Note: the charge enclosed by the inner sphere is equal to $\displaystyle{Q_1=\rho \cdot \frac{4}{3}\pi R^3_1}$ while the charge enclosed by the outer sphere (which is not a full sphere) is equal to $\displaystyle{Q_2=-\rho \cdot \left[\frac{4}{3}\pi R^3_2-\frac{4}{3}\pi R^3_1\right]}$.

2. What is the electric field ${\overrightarrow{E}}_1\left(r\right)$ in the region where $0 < r < R_1$?

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Given the spherical symmetry of the charge distribution, we expect the electric field to be radially outward and we choose a sphere of radius $r$ as our Gaussian surface for each region.

Inside the inner sphere $(0<r<R_1$):

The electric flux through the Gaussian surface is equal to

\oiint_{sphere}{{\overrightarrow{E}}_1\cdot d\overrightarrow{A}}=\oiint_{sphere}{E_1{\mathrm{cos} \left(0\right)\ }dA}=E_1\cdot 4\pi r^2

The charge enclosed by the Gaussian surface is the product of $\rho $ by the volume enclosed by the Gaussian surface and we have

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

By Gauss’s Law, we derive the magnitude of the electric field ${\overrightarrow{E}}_1$ as follows

\begin{aligned} \oiint_{sphere}{{\overrightarrow{E}}_1\cdot d\overrightarrow{A}}=\frac{Q_{enc}}{{\varepsilon }_0}\ \ \ \ &\Rightarrow \ \ \ \ \ E_1\cdot 4\pi r^2=\frac{4\rho }{3{\varepsilon }_0}\pi r^3 \\ \\ &\Rightarrow \ \ \ \ \ E_1=\frac{\rho r}{3{\varepsilon }_0} \end{aligned}

The electric field inside the inner sphere is therefore given by

\boxed{{\overrightarrow{E}}_1=\frac{\rho r}{3{\varepsilon }_0}\ \hat{r}}

3. What is the electric field ${\overrightarrow{E}}_2\left(r\right)$ in the region where $R_1\le r\le R_2$?

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Outside the inner sphere $(0\le r\le R_1$):

The electric flux through the Gaussian surface is equal to

\oiint_{sphere}{{\overrightarrow{E}}_2\cdot d\overrightarrow{A}}=\oiint_{sphere}{E_2{\mathrm{cos} \left(0\right)\ }dA}=E_2\cdot 4\pi r^2

The charge enclosed by the Gaussian surface is equal to

Q_{enc}=\rho \cdot \frac{4}{3}\pi R^3_1-\rho \cdot \left[\frac{4}{3}\pi r^3-\frac{4}{3}\pi R^3_1\right]

By Gauss’s Law, we derive the magnitude of the electric field ${\overrightarrow{E}}_2$ as follows

\begin{aligned} \oiint_{sphere}{{\overrightarrow{E}}_2\cdot d\overrightarrow{A}}=\frac{Q_{enc}}{{\varepsilon }_0}\ \ \ \ &\Rightarrow \ \ \ \ \ E_2\cdot 4\pi r^2=\frac{\rho }{{\varepsilon }_0}\cdot \frac{4}{3}\pi R^3_1-\frac{\rho }{{\varepsilon }_0}\cdot \left[\frac{4}{3}\pi r^3-\frac{4}{3}\pi R^3_1\right] \\ \\ &\Rightarrow \ \ \ \ \ E_2\cdot 4\pi r^2=\frac{\rho }{{\varepsilon }_0}\left[\frac{8}{3}\pi R^3_1-\frac{4}{3}\pi r^3\right] \\ \\ &\Rightarrow \ \ \ \ \ E_2=\frac{\rho }{3{\varepsilon }_0}\left[\frac{2R^3_1}{r^2}-r\right] \end{aligned}

The electric field inside the outer spherical shell is therefore given by

\boxed{\overrightarrow{E_2}=\frac{\rho }{3{\varepsilon }_0}\left[\frac{2R^3_1}{r^2}-r\right]\ \hat{r}}