Question 2.5: An infinite plane carries a uniform surface charge σ. Find i...

Draw a “Gaussian pillbox,” extending equal distances above and below the plane (Fig. 2.22). Apply Gauss’s lawto this surface:

In this case, Q_{enc}=\sigma A, where A is the area of the lid of the pillbox. By symmetry, E points away from the plane (upward for points above, downward for points below). So the top and bottom surfaces yield

whereas the sides contribute nothing. Thus

or

E=\frac{\sigma }{2\epsilon _{0}}\hat{n}     (2.17)

where \hat{n} is a unit vector pointing away from the surface. In Prob. 2.6, you obtained this same result by a much more laborious method.

It seems surprising, at first, that the field of an infinite plane is independent ofhow far away you are. What about the 1/r^{2} in Coulomb’s law? The point is that as you move farther and farther away from the plane, more and more charge comes into your “field of view” (a cone shape extending out from your eye), and this compensates for the diminishing influence of any particular piece. The electric field of a sphere falls off like 1/r^{2}; the electric field of an infinite line falls off like 1/r ; and the electric field of an infinite plane does not fall off at all (you cannot escape from an infinite plane).