The slab extends infinitely in the $xy$-plane and therefore we expect the electric field to be perpendicular to the slab, directed outward (assuming $\rho >0$, otherwise inward). In addition, the electric field at the center of the slab is zero by symmetry.

To derive the electric field created by the slab, we must distinguish the inside of the slab and the outside of the slab. For each region, we apply Gauss’s Law to derive the magnitude of the electric field and use a similar method to the one used in the case of the infinite sheet of charge.

Electric field inside the slab: $\displaystyle{z < \frac{d}{2}}$

We use a Gaussian cylinder of radius $R$ and height $z$ centered on the middle of the slab as shown below. The volume of slab material that the cylinder encloses is therefore $V=\pi R^2\left(2z\right)$ and the top and bottom surface have area $A=\pi R^2$.

The flux through the cylinder is limited to the flux through its top and bottom surface because the electric field lines are parallel to its length. The net flux through the cylinder is therefore equal to

\begin{aligned} \oiint_{cylinder}{\overrightarrow{E}\cdot d\overrightarrow{A}}&=\iint_{A_1}{\overrightarrow{E}\cdot d\overrightarrow{A_1}}+\iint_{A_2}{\overrightarrow{E}\cdot d\overrightarrow{A_2}}+\iint_{A_3}{\overrightarrow{E}\cdot d\overrightarrow{A_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\cdot \pi R^2+0+E\cdot \pi R^2 \\ \\ &=2E\cdot \pi R^2 \end{aligned}

The charge enclosed by the cylinder is equal to the product of its volume charge density $\rho $ and the volume $V$ of slab enclosed by the cylinder and we have

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

By Gauss’s Law, we derive that the magnitude $E$ of the electric field inside the slab is given by

2E\cdot \pi R^2=\frac{2\rho \pi R^2z}{{\varepsilon }_0}\ \ \ \ \ \Rightarrow \ \ \ \ \ \ E=\frac{\rho z}{{\varepsilon }_0}

Therefore, the electric field inside the slab is given by

\left\{ \begin{array}{c} \displaystyle{\overrightarrow{E}=\frac{\rho z}{{\varepsilon }_0}\ \hat{z}\ \ \ \ \ \ \ \ \ \ \ \ for\ \ \ \ \ 0 < z < \frac{d}{2}} \ \ \ \\ \\ \displaystyle{\overrightarrow{E}=-\frac{\rho z}{{\varepsilon }_0}\ \hat{z}\ \ \ \ \ \ \ \ for\ \ \ \ -\frac{d}{2} < z < 0} \end{array} \right.

2. What is the value of the electric field outside the slab (above and below)?

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Electric field outside the slab: $\displaystyle{z > \frac{d}{2}}$

We use a Gaussian cylinder of radius $R$ and height $z$ centered on the middle of the slab as shown below. The volume of slab material that the cylinder encloses is therefore $V=\pi R^2d$ and the top and bottom surface have area $A=\pi R^2$.

\begin{aligned} \oiint_{cylinder}{\overrightarrow{E}\cdot d\overrightarrow{A}}&=\iint_{A_1}{\overrightarrow{E}\cdot d\overrightarrow{A_1}}+\iint_{A_2}{\overrightarrow{E}\cdot d\overrightarrow{A_2}}+\iint_{A_3}{\overrightarrow{E}\cdot d\overrightarrow{A_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\cdot \pi R^2+0+E\cdot \pi R^2 \\ \\ &=2E\cdot \pi R^2 \end{aligned}

The charge enclosed by the cylinder is equal to the product of its volume charge density $\rho $ and the volume $V$ of slab enclosed by the cylinder and we have

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

By Gauss’s Law, we derive that the magnitude $E$ of the electric field inside the slab is given by

2E\cdot \pi R^2=\frac{2\rho \pi R^2z}{{\varepsilon }_0}\ \ \ \ \ \Rightarrow \ \ \ \ \ \ E=\frac{\rho d}{2{\varepsilon }_0}

Therefore, the electric field outside of the slab is given by

\left\{ \begin{array}{c} \displaystyle{\overrightarrow{E}=\frac{\rho d}{2{\varepsilon }_0}\ \hat{z}\ \ \ \ \ \ \ \ for\ \ \ \ \ \ \ z > \frac{d}{2}} \\ \\ \displaystyle{\overrightarrow{E}=-\frac{\rho d}{{2\varepsilon }_0}\ \hat{z}\ \ \ \ \ \ for\ \ \ \ \ \ z < -\frac{d}{2}} \end{array} \right.

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