P20S2013 – Cylindrical Cavity

The cylinder with the hollow cavity does not have enough symmetry to derive the electric field it creates from Gauss’s Law. Therefore, we will use superposition and consider the cylinder with the cavity to be the superposition of a full cylinder of radius $R$ with volume charge density $\rho $ and a cylinder of radius $R/2$ with volume charge density $-\rho $.

Indeed, superposition of the cylinder with volume charge density $–\rho $ and the cylinder with volume charge density $\rho $ will have the same effect as a cylindrical cavity of radius $R/2$ when it comes to computing the electric field.

Before using superposition, though, we recall the electric field created by an infinite cylinder with volume charge density $\rho $ at a distance $r$ from the central axis.

By Gauss’s Law, the electric field at a distance $r\le R$ from the center of such an infinite cylinder is

\overrightarrow{E}=\frac{\rho r}{2{\varepsilon }_0}\ \overrightarrow{r}\ \ \ \ \ for\ \ \ r\le R

By Gauss’s Law, the electric field at a distance $r>R$ from the center of such an infinite cylinder is

\overrightarrow{E}=\frac{\rho R^2}{2{\varepsilon }_0r}\overrightarrow{r}\ \ \ for\ \ \ r > R

Electric field at point $A$:

Point $A$ lies on the surface $\left(r=R\right)$ of the large cylinder where the electric field is equal to

{\overrightarrow{E}}_1=\frac{\rho R}{2{\varepsilon }_0}\hat{x}

Point $A$ lies on the surface $\left(r=R/2\right)$ of the small cylinder where the electric field is equal to

{\overrightarrow{E}}_2=-\frac{\rho R}{4{\varepsilon }_0}\hat{x}

By superposition, the net electric field at point $A$ is given by

\boxed{{\overrightarrow{E}}_A={\overrightarrow{E}}_1+{\overrightarrow{E}}_2=\frac{\rho R}{2{\varepsilon }_0}\hat{x}-\frac{\rho R}{4{\varepsilon }_0}\hat{x}=\frac{\rho R}{4{\varepsilon }_0}\hat{x}}

Electric field at point $B$:

Point $B$ lies a distance $r=R/2$ from the axis of the large cylinder where the electric field is equal to

{\overrightarrow{E}}_1=\frac{\rho R}{4{\varepsilon }_0}\hat{x}

Point $B$ lies at the center of the small cylinder where the electric field is equal to

{\overrightarrow{E}}_2=\overrightarrow{0}

By superposition, the net electric field at point $B$ is given by

\boxed{{\overrightarrow{E}}_B={\overrightarrow{E}}_1+{\overrightarrow{E}}_2=\frac{\rho R}{4{\varepsilon }_0}\hat{x}+\overrightarrow{0}=\frac{\rho R}{4{\varepsilon }_0}\hat{x}}

Electric field at point $C$:

Point $C$ lies on the surface $\left(r=R\right)$ of the large cylinder where the electric field is equal to

{\overrightarrow{E}}_1=-\frac{\rho R}{2{\varepsilon }_0}\hat{x}

Point $C$ lies at distance $r=3R/2$ from the axis of the small cylinder where the electric field is equal to

{\overrightarrow{E}}_2=\frac{\displaystyle{\rho {\left(\frac{R}{2}\right)}^2}}{\displaystyle{2{\varepsilon }_0\left(\frac{3R}{2}\right)}}\hat{x}=\frac{\rho R}{12{\varepsilon }_0}\hat{x}

By superposition, the net electric field at point $C$ is given by

\boxed{{\overrightarrow{E}}_C={\overrightarrow{E}}_1+{\overrightarrow{E}}_2=-\frac{\rho R}{2{\varepsilon }_0}\hat{x}+\frac{\rho R}{12{\varepsilon }_0}\hat{x}=-\frac{5\rho R}{12{\varepsilon }_0}\hat{x}}