P20B2016 – Two Parallel Slabs

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 d\overrightarrow{A}}&=\iint_{A_1}{E{\mathrm{cos} \left(\theta \right)\ }dA_1}+\iint_{A_2}{E{\mathrm{cos} \left(\theta \right)\ }dA_2}+\iint_{A_3}{E{\mathrm{cos} \left(\theta \right)\ }dA_3} \\
\\
&=\iint_{A_1}{E{\mathrm{cos} \left(0\right)\ }dA_1}+\iint_{A_2}{E{\mathrm{cos} \left(90\right)\ }dA_2}+\iint_{A_3}{E{\mathrm{cos} \left(0\right)\ }dA_3} \\
\\
&=E{\mathrm{cos} \left(0\right)} \iint_{A_1}{dA_1}+0+E{\mathrm{cos} \left(0\right)\ }\iint_{A_3}{dA_3} \\
\\
&=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 \ \ \ \ \ E=\frac{\sigma }{2{\varepsilon }_0}

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

\boxed{
\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.
}

In a conductor, charge is free to move around. Thus, if there were a nonzero electric field inside a conductor, the free charges would experience the electric force and would accelerate, thus causing the conductor to not be in electrostatic equilibrium. The only way the free charges do not experience any electric force is if the electric field inside the conducting material is zero everywhere.

If a conductor carries a nonzero net charge, this net charge must reside somewhere. Because $E=0$ inside the conducting material, we conclude that the net charge has no choice but to be distributed on the outer surface of the conductor. If we consider a conductor such as the one shown below

we can apply Gauss’s Law to prove that the net charge must reside on the outer surface. Indeed, for ANY Gaussian surface we choose that is located inside the conducting material, the fact that $E=0$ means that the electric flux through that Gaussian surface is equal to

\oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}=\oiint_A{E{\mathrm{cos} \left(\theta \right)\ }dA}=0

By Gauss’s Law, we conclude that the charge $Q_{enc}$ enclosed in ANY such Gaussian surface must be zero since $\displaystyle{\iint_A \ \overrightarrow{E} \cdot d\overrightarrow{A} = \frac{Q_{enc}}{\varepsilon_0}}$. This means that there can be no charge inside the conducting material and that it must therefore reside on the outer surface of the conductor.

Both slabs are metallic and are therefore conductors which means that for each slab, the net charge resides on its outer surface. We let ${\sigma }_1$ denote the surface charge density of the upper surface of the top slab, ${\sigma }_2$ denote the surface charge density of the lower surface of the top slab, ${\sigma }_3$ denote the charge density of the upper surface of the bottom slab, and ${\sigma }_4$ denote the surface charge density of the lower surface of the bottom slab (see figure).

To determine these surface charge densities, we need 4 equations. To do so, we will apply Conservation of Charge to each plate, use Gauss’s Law across the gap, and then derive the electric field in the gap.

Conservation of Charge:
By Conservation of Charge of each plate, the total surface charge must equal the net charge of the corresponding plate and we write

\left\{ \begin{array}{c}
{\sigma }_1L^2+{\sigma }_2L^2=Q \\
\\
{\sigma }_3L^2+{\sigma }_4L^2=-2Q \end{array}
\right.\ \ \ \ \Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{{\sigma }_1+{\sigma }_2=\frac{Q}{L^2}} \\
\\
\displaystyle{{\sigma }_3+{\sigma }_4=-\frac{2Q}{L^2}} \end{array}
\right.

Gauss’s Law across the gap:
We choose a Gaussian cylinder with radius $R$ that stretches across the gap and has its top surface in the upper slab and its bottom surface in the lower slab as shown in the figure below.

Given that the slabs are conductors, the electric field inside of them is equal to zero and therefore there is no flux through the top or bottom surface of the cylinder. In addition, since the electric field lines between the plates are vertical, there is no flux through the lateral surface of the cylinder. Overall, the net flux through the Gaussian cylinder is zero and we have

\oiint_{cylinder}{\overrightarrow{E}\cdot d\overrightarrow{A}}=0

By Gauss’s Law, this required that the charge enclosed by the Gaussian surface must also be zero which yields

\begin{aligned}
Q_{enc}=0\ \ \ \ &\Rightarrow \ \ \ \ \ {\sigma }_2\pi R^2+{\sigma }_3\pi R^2=0 \\
\\
&\Rightarrow \ \ \ \ \ \boxed{{\sigma }_2=-{\sigma }_3}
\end{aligned}

Electric field across the gap:
The lower slab has net charge $-2Q$ and its electric field ${\overrightarrow{E}}_-$ in the gap (between the slabs) is therefore directed toward the bottom slab and is perpendicular to the slab (recall that $d\ll L$).

The upper slab has net charge $+Q$ and its electric field ${\overrightarrow{E}}_+$ in the gap (between the slabs) is therefore directed toward the bottom slab and is perpendicular to the slab (recall that $d\ll L$).

By superposition, the electric field across the gap is directed toward the bottom slab and is equal to

\overrightarrow{E}={\overrightarrow{E}}_++{\overrightarrow{E}}_-=\frac{{\sigma }_{top}}{2{\varepsilon }_0}\hat{y}+\frac{\left|{\sigma }_{bottom}\right|}{2{\varepsilon }_0}\hat{y}=\frac{Q}{2{\varepsilon }_0L^2}\hat{y}+\frac{2Q}{2{\varepsilon }_0L^2}\hat{y}=\frac{3Q}{2{\varepsilon }_0L^2}\hat{y}

This electric field must match the value of the electric field at the bottom surface of the upper slab. In other words, ${\sigma }_2$ must satisfy

\begin{aligned}
\frac{\left|{\sigma }_2\right|}{{\varepsilon }_0}=E\ \ \ \ &\Rightarrow \ \ \ \ \ \frac{\left|{\sigma }_2\right|}{{\varepsilon }_0}=\frac{3Q}{2{\varepsilon }_0L^2} \\
\\
&\Rightarrow \ \ \ \ \ \left|{\sigma }_2\right|=\frac{3Q}{2L^2}
\end{aligned}

Since ${\sigma }_2$ is positive (see above), we conclude that ${\sigma }_2$ is equal to

\boxed{{\sigma }_2=\frac{3Q}{2L^2}}

This electric field in the gap must also match the value of the electric field at the top surface of the lower slab. In other words, ${\sigma }_3$ must satisfy

\begin{aligned}
\frac{\left|{\sigma }_3\right|}{{\varepsilon }_0}=E\ \ \ \ &\Rightarrow \ \ \ \ \ \frac{\left|{\sigma }_3\right|}{{\varepsilon }_0}=\frac{3Q}{2{\varepsilon }_0L^2} \\
\\
&\Rightarrow \ \ \ \ \ \left|{\sigma }_3\right|=\frac{3Q}{2L^2}
\end{aligned}

Since ${\sigma }_3$ is negative (see above), we conclude that ${\sigma }_3$ is equal to

\boxed{{\sigma }_3=-\frac{3Q}{2L^2}}

Note: recall that, at the surface of a conducting solid, the electric field has magnitude $\sigma /{\varepsilon }_0$.

Given that ${\sigma }_2$ and ${\sigma }_3$ are now known, we substitute their expressions back into the conservation of charge equations are derive ${\sigma }_1$ and ${\sigma }_4$ as follows

\left\{ \begin{array}{c}
\displaystyle{{\sigma }_1+\frac{3Q}{2L^2}=\frac{Q}{L^2}} \\
\\
\displaystyle{{\sigma }_3-\frac{3Q}{2L^2}=-\frac{2Q}{L^2}} \end{array}
\right.\ \ \ \ \Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{{\sigma }_1=-\frac{Q}{2L^2}} \\
\\
\displaystyle{{\sigma }_4=-\frac{Q}{2L^2}} \end{array}
\right.

Finally, we conclude that the four different surface charge densities are equal to

\boxed{{\sigma }_1={\sigma }_4=-\frac{Q}{2L^2}\ \ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ {\sigma }_2=-{\sigma }_3=\frac{3Q}{2L^2}}