A slab of insulating material has a uniform volume charge density , and a thickn

Question

A slab of insulating material has a uniform volume charge density , and a thickness d. The slab is infinite in the y and z directions, and is oriented such that it is symmetric about the y-axis (i.e. its thickness extends from -d/2 to +d/2 along the x-axis, as shown in the figure below). Derive an expression for the magnitude of the electric field for the regions listed below.  Note: Your answer should be in terms of 0, d, , and x, as needed. You may assume that d and are both of finite value. You are looking for the magnitude, so you do not need to indicate direction in your response.

(a) The region (x > d/2) , (i.e the region outside the slab for positive values of x):
E =   
(b) The region (0 < x < d/2) , (i.e the region inside the slab for positive values of x):
E =   

Now imagine that that an infinite insulating sheet of charge, with surface charge density and negligible thickness, is placed a distance a from the right edge of the infinite slab. The sheet is parallel to the slab, so both objects are infinite in the y and z directions (see the figure below). Derive an expression for the magnitude of the electric field for the regions listed below. Note: Your answer should be in terms of 0, d, a, , and x, as needed. You may assume that d, a, and are all of finite value.  You are looking for the magnitude, so you do not need to indicate direction in your response.

(c) The region (d/2 < x < a) , (i.e the region between the slab and the sheet):
E =   
(d) The region (0 < x < d/2) , (i.e the region inside the slab for positive values of x):
E =  

Explanation / Answer

We can solve this using Gauss's law. The Gaussian surface that we consider here is of a cylinder. Place the surface symmetrical to the x-axis so that the electric field lines are passing through its two bases. The height of the cylinder be h and the base area be A. Note that there will be no field lines through the curved surface.
a)
Using Gauss's law,
E x 2A = q/0
q is the charge enclosed in the cylinder = x A x d
E x 2A = Ad/0, E = d/20 [This is a constant, which is independent of x]

b)
In this case the Gaussian surface is inside the slab, its height is x
Charge enclosed by the cylinder = Ax
Using Gauss's law, E x 2A = Ax/0, E = x/20 [E is directly proportional with x]

The charged sheet will provide an electric field E = /20, this will be same everywhere because it is independent of distance.
c)
E due to the slab =  d/20 [From a)]
E due to the sheet = /20
Total E = d/20 - /20 = [d - ] / 20

d)
E due to the slab =  x/20 [From b]
E due to the sheet = /20
Total E = x/20 - /20 = [x - ] / 20

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