Question

Consider an infinite slab of thickness 2a and uniform volume charge density ρ. This is essentially an infinite plane with a non-negligible thickness. Since the planar symmetry involves:艹-2 reflection symmetry, as well as the translation symmetry along the and y direc- tions, we place the origin at a point on the midplane of the slab. In other words, the midplane corresponds to oo = 0 (i.e., the ry plane) and the surfaces of the slab are at a (a) Use Gauss law to find the electric field produced by the slab at any point. Let z be the z-coordinate of the point. Consider the following two cases separately i, point outside the slab ( a), and ii, point inside the slab (-a-z-a) In each case, carefully choose the Gaussian surface, show it clearly, and explain how the symmetry of the proble enables you to use the Gauss' law to solve the problem. Note that this is a standard problem whose solution can be found anyuhere. It is important for you to understand how the method works (b) Check that the electric field outside and inside the slab match at the slab surfaces After that, plot E vs z graph (c) As in the case of the previous problem, we can define a charge per unit area of the slab as σ Q/A where Q is the total charge on the slab in a region having slab-surface area A. Find σ in terms of ρ and a After that, show that the electric field outside the slab can be expressed as ±o/2E0

Answer 1

Conidaing a Graunian uim bane ano. A and Roget Total flux 21E21 A ZI >

Acconding to Gouou Su 21E 21 A C o Z -C E6 Pa 2 S 0

5) Ven, it matcho. You can ahs chech it jut a 2=ta in Ez fon (Z1 20 2