# Uniform Slab-Finding Electric Field Using Gauss Law

• Arman777
In summary, we considered an infinite slab of charge with thickness 2a and a uniform charge density. We used Gauss' law to find the electric field at a point inside and outside the slab, using a cylindrical Gaussian surface. We found that the electric field was proportional to the height of the point from the origin, and that it increased until z=a and then remained constant. We also considered the case where the charge density was not constant and found that the electric field was proportional to the cube of the height inside and outside the slab, and that the graph followed a similar pattern.
Arman777
Gold Member

## Homework Statement

Uniform Slab: Consider an inﬁnite slab of charge with thickness 2a. We choose the origin inside the slab at an equal distance from both faces (so that the faces of the slab are at z = +a and z = −a). The charge density ρ inside the slab is uniform (i.e., ρ =const). Consider a point with coordinates (x,y,z). Using Gauss’ law, ﬁnd the electric ﬁeld
(a) when the point is inside the slab (−a < z < +a),
(b) and when the point is outside the slab (z > a or z < −a).
(c) Sketch the Ez vs z graph.
(d) If the density was not constant at its a function of z like ##ρ=Bz^2## then calculate the upper steps again.

Gauss Law

## The Attempt at a Solution

a) I took a cylinder Gaussian surface inside the slab forand from that I found ##E=\frac {ρz} {2ε_0}## .z is the height of the point that we choose from the origin.
b)I took a cylinder again and from that I found ##E=\frac {ρa} {2ε_0}##
c)The field will be constant cause ρ and a is constant also ##ε_0## so As z increases it inrease until a.And from that its constant.
d)Then Electric field will be ##E=\frac {Bz^3} {6ε_0}## for inside , ##E=\frac {Ba^3} {6ε_0}## for outside ?
Is these true ?

Last edited:
Arman777 said:

## The Attempt at a Solution

a) I took a cylinder Gaussian surface inside the slab forand from that I found ##E=\frac {ρz} {2ε_0}## .z is the height of the point that we choose from the origin.
You have the right approach. But your answer is off by a numerical factor. Can you show in more detail how you got your result? In particular, how was your cylindrical Gaussian surface oriented and positioned?

I took a cylinder like height is z and radius is x.So ##Eπx^2=Q/ε_0## ##Q=ρπx^2z## But there's two sides so we should multiply this by 2 so ##2Eπx^2=\frac {Q} {ε_0}## and then ##E=\frac {pz} {2ε_0}##

Oh wait

Height is ##2z##

##E=\frac {ρz} {ε_0}## ?

Arman777 said:
##E=\frac {ρz} {ε_0}## ?
Yes.

Arman777
Then (b) is ##E=\frac {pa} {ε_0}##
(c) will be the same (it increase until ##z=a## then its constant)
for (d) inside ##E=\frac {Bz^3} {3ε_0}## and outside ##E=\frac {Ba^3} {3ε_0}## and the graph again rises at z rises then its constant ??

Looks good. Make sure your graphs include negative values of z as well as positive values of z. So you will need to think about the sign of Ez for negative z.

I see thanks

## 1. What is Gauss Law and how is it used to find the electric field of a uniform slab?

Gauss Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the enclosed charge. To find the electric field of a uniform slab, we can use Gauss Law by imagining a Gaussian surface that encloses the slab and applying the formula E = Q/ε0A, where E is the electric field, Q is the charge enclosed by the Gaussian surface, and ε0 is the permittivity of free space.

## 2. What is a uniform slab and why is it important in electric field calculations?

A uniform slab is a material with a constant charge density and thickness. It is important in electric field calculations because it simplifies the problem and allows us to use Gauss Law to find the electric field. In real-life situations, objects may not have a uniform charge distribution, but understanding the electric field of a uniform slab can help us approximate the electric field of more complex objects.

## 3. How do the dimensions of the slab affect the electric field calculated using Gauss Law?

The dimensions of the slab, specifically the thickness and surface area, directly affect the electric field calculated using Gauss Law. As the thickness of the slab increases, the electric field decreases, and as the surface area of the slab increases, the electric field increases. This is because the electric field is inversely proportional to the distance from the charge and directly proportional to the amount of charge enclosed.

## 4. Can Gauss Law be used to find the electric field of non-uniform objects?

Yes, Gauss Law can be used to find the electric field of non-uniform objects, but it may require more complex mathematical techniques. In these situations, the Gaussian surface may need to be divided into smaller parts to account for varying charge densities or non-uniform shapes. The principle of Gauss Law still applies, but the calculations may be more challenging.

## 5. Are there any limitations to using Gauss Law to find the electric field of a uniform slab?

Yes, there are some limitations to using Gauss Law in this context. One limitation is that it assumes the slab is infinitely large, which may not be true in real-life situations. Additionally, Gauss Law only applies to stationary charges, so it cannot be used for objects with moving charges. Finally, the electric field calculated using Gauss Law may not be accurate for objects with highly irregular shapes or non-uniform charge distributions.

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