P20-070 – Application: infinite sheet of charge

Applying Gauss’s Law: infinite sheet of charge

Consider an infinite sheet with uniform surface charge density $\sigma =Q/A$ (charge per unit area).

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

The sheet is infinite, therefore any vertical plane is a plane of symmetry of the charge distribution and the electric field is vertical. If $\sigma >0$ it points away from the sheet above and below, and if $\sigma <0$ it points toward the sheet above and below.

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

For our Gaussian surface, we choose a vertical closed cylinder of length $L$ and radius $r$ with its axis perpendicular to the sheet and centered on the sheet. The electric field being vertical, the electric flux is zero through the lateral cylinder wall and there is only flux through the top and bottom endcaps.

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(0\right)dA_1}+\iint_{A_2}{Ecos\left(90\right)dA_2}+\iint_{A_3}{Ecos\left(0\right)dA_3} \\
&=Ecos\left(0\right)\iint_{A_1}{dA_1}+0+Ecos\left(0\right)\iint_{A_1}{dA_1} \\\
&=E\pi r^2+E\pi r^2 \\
&=2E\pi r^2
\end{aligned}

where the area of the top and bottom endcaps of the cylinder is equal to $A_1=A_3=\pi r^2$.

The Gaussian cylinder encloses a surface area $A=\pi r^2$ with surface charge density $\sigma $ and therefore the amount of enclosed charge is equal to

Q_{enc}=\sigma A=\sigma \cdot \pi r^2

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

2E\pi r^2=\frac{\sigma \cdot \pi r^2}{{\varepsilon }_0}\ \ \ \ \Rightarrow \ \ \ \ \ \boxed{E=\frac{\sigma }{2{\varepsilon }_0}}

Thus, the electric field created above and below the sheet is given by

\begin{aligned}
\left\{ \begin{array}{c}
\displaystyle{\overrightarrow{E}=+\frac{\sigma }{2{\varepsilon }_0}\ \hat{z}\ \ \ \ \ \ \ \ \ \ \ \ if\ \ z>0} \\
\\
\displaystyle{\overrightarrow{E}=-\frac{\sigma }{2{\varepsilon }_0}\ \hat{z}\ \ \ \ \ \ \ \ \ \ \ \ if\ \ z<0} \end{array}
\right.
\end{aligned}

It is worth pointing out that the electric field created by the infinite sheet is uniform which means that it is the same everywhere and its magnitude does not decrease with the distance from the sheet. This, of course, is not realistic but neither is the assumption that the sheet of charge is infinite.

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

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

The electric field lines and the graph of the electric field $\left(\sigma >0\right)$ against the distance $z$ to the sheet are sketched below