P24_033.PDF

33. (a) We use a Gaussian surface in the form of a box with rectangular sides. The cross section is shown

with dashed lines in the diagram below. It is centered at the central plane of the slab, so the left and

right faces are each a distance x from the central plane. We take the thickness of the rectangular

solid to be a , the same as its length, so the left and right faces are squares. The electric ﬁeld is

normal to the left and right faces and is uniform over them. If ρ is positive, it points outward at

both faces: toward the left at the left face and toward the right at the right face. Furthermore, the

magnitude is the same at both faces. The electric ﬂux through each of these faces is Ea 2 . The ﬁeld

is parallel to the other faces of the Gaussian surface and the ﬂux through them is zero. The total

ﬂux through the Gaussian surface is Φ = 2 Ea 2 .

←−−−−− d −−−−−→

The volume enclosed by the Gaussian surface is 2 a 2 x and the charge contained within it is q =2 a 2 xρ .

Gauss’ law yields 2 ε 0 Ea 2 =2 a 2 xρ . We solve for the magnitude of the electric ﬁeld:

E = ρx

ε 0

(b) We take a Gaussian surface of the same shape and orientation, but with x>d/ 2, so the left and

right faces are outside the slab. The total ﬂux through the surface is again Φ = 2 Ea 2 but the

charge enclosed is now q = a 2 dρ . Gauss’ law yields 2 ε 0 Ea 2 = a 2 dρ ,so

E = ρd

2 ε 0