Uniform Slab-Finding Electric Field Using Gauss Law

Click For Summary

Homework Help Overview

The discussion revolves around finding the electric field due to an infinite slab of charge using Gauss's law. The problem involves analyzing the electric field both inside and outside the slab, with specific attention to the uniform charge density and variations in density as a function of position.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of cylindrical Gaussian surfaces to derive expressions for the electric field in different regions of the slab. There are attempts to clarify the orientation and positioning of the Gaussian surfaces used in the calculations.

Discussion Status

Some participants have provided guidance on the calculations, noting potential numerical errors and encouraging more detailed explanations of the reasoning behind the results. There is an ongoing exploration of the implications of varying charge density on the electric field.

Contextual Notes

Participants are considering the implications of both constant and variable charge densities, with specific reference to the mathematical forms of the electric field in each case. The discussion includes considerations of the graphical representation of the electric field as a function of position.

Arman777
Insights Author
Gold Member
Messages
2,163
Reaction score
191

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.

Homework Equations


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:
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: 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.
 
  • #10
I see thanks
 

Similar threads

Replies
6
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
2K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 12 ·
Replies
12
Views
1K
Replies
9
Views
1K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
6K
Replies
12
Views
14K
  • · Replies 20 ·
Replies
20
Views
5K