The cylinder is infinite, therefore any horizontal plane is a plane of symmetry of the charge distribution. In addition, any vertical plane containing the axis of revolution of the cylinder is a plane a of symmetry of the charge distribution as well.

We therefore conclude that the electric field created by the infinite cylinder is radial, directed outward if $\rho >0$ and directed inward if $\rho <0$.

Choice of a Gaussian surface and applying Gauss’s Law:

For our Gaussian surface, we choose a closed cylinder of length $L$ and radius $r$, centered on the axis of the cylinder as shown below. The electric field being radial, the electric flux is zero through the endcaps and there is only flux through the lateral wall of the Gaussian cylinder.

The infinite cylinder has a radius $R$ and we have no reason to assume that the electric field inside the insulating material will be the same as outside of the cylinder. Thus, we differentiate two regions: inside the insulating cylinder $\left(r\le R\right)$ and outside the insulating cylinder $\left(r>R\right)$.

Electric field inside the cylinder $\left(r\le R\right)$:

The electric flux through our Gaussian surface is equal to the sum of the electric flux through the endcaps and the lateral wall of the cylinder and we write

\begin{aligned} \oiint_A{\overrightarrow{E}\cdot \overrightarrow{dA}}&=\iint_{A_1}{Ecos\left(\theta \right)dA_1}+\iint_{A_2}{Ecos\left(\theta \right)dA_2}+\iint_{A_3}{Ecos\left(\theta \right)dA_3} \\ &=\iint_{A_1}{Ecos\left(90\right)dA_1}+\iint_{A_2}{Ecos\left(0\right)dA_2}+\iint_{A_3}{Ecos\left(90\right)dA_3} \\ &=0+Ecos\left(0\right)\iint_{A_2}{dA_2}+0 \\ &=E2\pi rL \end{aligned}

where the area of the lateral wall of the cylinder is equal to $A_2=2\pi rL$.

The Gaussian cylinder encloses a cylinder of charge with volume $V=\pi r^2L$ and therefore the amount of enclosed charge is equal to

Q_{enc}=\rho V=\rho \cdot \pi r^2L

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

E2\pi rL=\frac{\rho \cdot \pi r^2L}{{\varepsilon }_0}\ \ \ \ \Rightarrow \ \ \ \ \ E=\frac{\rho r}{2{\varepsilon }_0}

Thus, the electric field created inside the infinite cylinder is equal to

\boxed{\overrightarrow{E}=\frac{\rho r}{2{\varepsilon }_0}\ \hat{r}}

where $\rho >0$ yields radially outward electric field lines and $\rho <0$ yields radially inward electric field lines.

Outside the cylinder $\left(r>R\right)$:

The electric flux through our Gaussian surface is equal to the sum of the electric flux through the endcaps and the lateral wall of the cylinder and we write

\begin{aligned} \oiint_A{\overrightarrow{E}\cdot \overrightarrow{dA}}&=\iint_{A_1}{Ecos\left(\theta \right)dA_1}+\iint_{A_2}{Ecos\left(\theta \right)dA_2}+\iint_{A_3}{Ecos\left(\theta \right)dA_3} \\ &=\iint_{A_1}{Ecos\left(90\right)dA_1}+\iint_{A_2}{Ecos\left(0\right)dA_2}+\iint_{A_3}{Ecos\left(90\right)dA_3} \\\ &=0+Ecos\left(0\right)\iint_{A_2}{dA_2}+0 \\ &=E2\pi rL \end{aligned}

where the area of the lateral wall of the cylinder is equal to $A_2=2\pi rL$.

The Gaussian cylinder encloses a cylinder of charge with volume $V=\pi R^2L$ and therefore the amount of enclosed charge is equal to

Q_{enc}=\rho V=\rho \cdot \pi R^2L

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

E2\pi rL=\frac{\rho \cdot \pi R^2L}{{\varepsilon }_0}\ \ \ \ \Rightarrow \ \ \ \ \ E=\frac{\rho R^2}{2{\varepsilon }_0r}

Thus, the electric field created outside the infinite cylinder is equal to

\boxed{\overrightarrow{E}=\frac{\rho R^2}{2{\varepsilon }_0r}\ \hat{r}}

where $\rho >0$ yields radially outward electric field lines and $\rho <0$ yields radially inward electric field lines.

Finally, the infinite cylinder creates and electric field at any point in space that is given by

\boxed{ \left\{ \begin{array}{c} \displaystyle{\overrightarrow{E}=\frac{\rho r}{2{\varepsilon }_0}\ \hat{r}\ \ \ \ \ \ \ \ if\ \ r\le R} \\ \\ \displaystyle{\overrightarrow{E}=\frac{\rho R^2}{2{\varepsilon }_0r}\ \hat{r}\ \ \ \ \ \ \ if\ \ r>R} \end{array} \right. }

Electric field lines and graph of the magnitude $E$ vs. $r$:

The electric field lines and the graph of the electric field $\left(\rho >0\right)$ against the distance $r$ to the axis of the cylinder are sketched below

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