An infinite plane slab, of thickness 2d,carries a uniform volumecharge density p (Fig. 2.27). Find the electric field, as a function of y,where y = 0 at the center. Plot Eversus y,calling Epositive when it points in the +ydirection and negative when it points in the -y direction.

The thickness of slab is 2d.

The uniform volumecharge density is$p.$

If there is a surface area enclosing a volume, possessing a charge inside the volume then the electric field due to the surface or volume charge is given as

$\mathbf{\oint }\mathit{E}\mathbf{.}\mathit{d}\mathit{a}\mathbf{=}\frac{\mathbf{q}}{{\mathbf{\epsilon }}_{\mathbf{0}}}$

Here $\mathit{q}$is the charge enclosed,${\mathit{\epsilon }}_{\mathbf{0}}$ is the permittivity of free surface.

The Gaussian cylinder drawn at a distance$y<\left|d\right|$from the center of the plane slab, which has volume $Ay$in +y direction, is shown as

It is known that the charge density inside the cylinder is $p$. So, the charge enclosed by the inner cylinder of volume V is obtained by integrating the charge density from 0 to $Ay$, as

$$\begin{gathered} q_{enclosed}={\int }_0^{Ay}pdV \\ =\left(p\right)\left(Ay\right) \\ =pAy \end{gathered}$$

Apply Gauss law on the Gaussian surface, by substituting $pAy\mathrm{for}{\mathrm{q}}_{\mathrm{enclosed}},$

and Afor da into $\oint E.da=\frac{q_{enclosed}}{{\epsilon }_0}$

$$\begin{gathered} \oint E.da=\frac{q_{enclosed}}{{\epsilon }_0} \\ E\left(A\right)=\frac{pAy}{{\epsilon }_0} \\ E=\frac{py}{{\epsilon }_0} \\ E=\frac{py}{{\epsilon }_0}\hat{y} \end{gathered}$$

Thus, the electric field inside the slab is $E=\frac{py}{{\epsilon }_0}\widehat{y.}$

For the Gaussian pill box drawn at a distance$\left|y\right|<d,$from the center of the plane slab , which has volume$Ad$in –y direction.

It is known that the charge density inside the cylinder is$p$. So, the charge enclosed by the pill box is obtained by integrating the charge density from 0 to

$$\begin{gathered} \mathrm{Ad},\mathrm{as} \\ {q}_{enclosed}={\int }_0^{Ad}pdV \\ =\left(p\right)\left(Ad\right) \\ =pAd \end{gathered}$$

Apply Gauss law on the Gaussian surface, by substituting$pAd\mathrm{for}{\mathrm{q}}_{\mathrm{enclosed}},$

and Afor da into $\oint E.da=\frac{q_{enclosed}}{{\epsilon }_0}$

$$\begin{gathered} \oint E.da=\frac{q_{enclosed}}{{\epsilon }_0} \\ E\left(A\right)=\frac{pAd}{{\epsilon }_0} \\ E=\frac{pd}{{\epsilon }_0} \\ =\frac{py}{{\epsilon }_0}\hat{y} \end{gathered}$$

Thus, the electric field outside the slab is $E=\frac{py}{{\epsilon }_0}\hat{y}$

Thus, the electric field, inside the slab is 0, at the center, it increases linearly with the distance, and outside the slab it remains constant. Thus electric field $\left|E\right|$is plotted against the distance yas,