P20-050 – Application: infinite line of charge

Applying Gauss’s Law: infinite wire

Consider an infinite wire with uniform linear charge density $\lambda $.

Charge distribution symmetry and electric field $\overrightarrow{E}$:

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

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

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 wire 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 cylinder.

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 length of wire $L$ and therefore an amount of enclosed charge equal to

Q_{enc}=\lambda L

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

E2\pi rL=\frac{\lambda L}{{\varepsilon }_0}\ \ \ \ \Rightarrow \ \ \ \ \ E=\frac{\lambda }{2\pi {\varepsilon }_0r}

Thus, the electric field created by the infinite wire is equal to

\boxed{\overrightarrow{E}=\frac{\lambda }{2\pi {\varepsilon }_0r}\hat{r}}

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

Note: the arbitrary length $L$ cancels out of the expression for $E$ eventually which makes sense since the wire is infinite and its electric field should therefore not depend on the arbitrary length $L$.

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

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