Chapter 21, Problem 59P

A charged slab extends infinitely in two dimensions and has thickness d in the third dimension, as shown in Fig. 21.36. The slab carries a uniform volume charge density ρ. Find expressions for the electric field (a) inside and (b) outside the slab, as functions of the distance x from the center plane. (Although the infinite slab is impossible, your answer is a good approximation to the field of a finite slab whose width is much greater than its thickness.)

  

  59. INTERPRET The infinitely large slab has plane symmetry, and we can apply Gauss’s law to compute the electric field.

  DEVELOP When we take the slab to be infinitely large, the electric field is everywhere normal to the slab's surface and symmetrical about Die center plane we follow the approach outlined in example 21.6 to compute the electric field. As the Gaussian surface, we choose a box that has area A on its top and bottom and that extends a distance x both up and down from the center of the slab. See figure below.

  

EVALUATE (a) For points inside the slab |x| ≤ d/2, the charge enclosed by our Gaussian box is

q_enclosed = ρV_enclosed = ρA(2x)

Thus, Gauss’s law gives

$\Phi ={\displaystyle \oint \bar{E}}\cdot \overline{dA}=E\left(2A\right)=\frac{q_{enclosed}}{{\epsilon }_0}\to E=\frac{\rho x}{{\epsilon }_0}$

The direction of $\bar{E}$is away from (toward) the central plane for positive (negative) charge density

(b) For points outside the slab |x| > d/2. the enclosed charge is

q_enclosed = ρV_enclosed = ρA(d)

Applying Gauss’s law again gives

$E=\frac{\rho d}{2{\epsilon }_0}$