6.3 Applying gauss’s law  (Page 9/13)

Magnitude at I or II: $E(z)=E_P.$

If the charge on the plane is positive, then the direction of the electric field and the area vectors are as shown in [link] . Therefore, we find for the flux of electric field through the box

where the zeros are for the flux through the other sides of the box. Note that if the charge on the plane is negative, the directions of electric field and area vectors for planes I and II are opposite to each other, and we get a negative sign for the flux. According to Gauss’s law, the flux must equal $q_{\text{enc}}\text{/}{\epsilon }_0$ . From [link] , we see that the charges inside the volume enclosed by the Gaussian box reside on an area A of the xy -plane. Hence,

Using the equations for the flux and enclosed charge in Gauss’s law, we can immediately determine the electric field at a point at height z from a uniformly charged plane in the xy -plane:

The direction of the field depends on the sign of the charge on the plane and the side of the plane where the field point P is located. Note that above the plane, $\widehat{\text{n}}=\text{+}\widehat{\text{z}}$ , while below the plane, $\widehat{\text{n}}=\text{−}\widehat{\text{z}}$ .

You may be surprised to note that the electric field does not actually depend on the distance from the plane; this is an effect of the assumption that the plane is infinite. In practical terms, the result given above is still a useful approximation for finite planes near the center.

Summary

• For a charge distribution with certain spatial symmetries (spherical, cylindrical, and planar), we can find a Gaussian surface over which $\overrightarrow{\text{E}}·\widehat{\text{n}}=E$ , where E is constant over the surface. The electric field is then determined with Gauss’s law.
• For spherical symmetry, the Gaussian surface is also a sphere, and Gauss’s law simplifies to $4\pi r^{2}E=\frac{q_{\text{enc}}}{{\epsilon }_0}$ .
• For cylindrical symmetry, we use a cylindrical Gaussian surface, and find that Gauss’s law simplifies to $2\pi rLE=\frac{q_{\text{enc}}}{{\epsilon }_0}$ .
• For planar symmetry, a convenient Gaussian surface is a box penetrating the plane, with two faces parallel to the plane and the remainder perpendicular, resulting in Gauss’s law being $2AE=\frac{q_{\text{enc}}}{{\epsilon }_0}$ .

Conceptual questions

Problems