E field for an semi-infinite sheet of charge

  • Thread starter Thread starter fishingspree2
  • Start date Start date
  • Tags Tags
    Charge Field
Click For Summary

Homework Help Overview

The problem involves calculating the electric field above a semi-infinite sheet of charge with a given surface charge density. The sheet is infinite in one direction and has a finite width, leading to considerations of symmetry and integration in the analysis.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of treating the semi-infinite sheet as composed of infinite wires and the integration process involved in calculating the electric field. There are questions about the presence of a height factor in the final expression and its implications for the limit as height approaches infinity.

Discussion Status

Some participants have provided insights into the nature of the electric field and the conditions under which the height factor should disappear. There is ongoing exploration of the integration steps and the assumptions made regarding the geometry of the charge distribution.

Contextual Notes

Participants note that the problem setup involves a semi-infinite sheet rather than an infinite plane, which affects the resulting electric field expression. The discussion includes concerns about how the height of the observation point influences the outcome.

fishingspree2
Messages
138
Reaction score
0

Homework Statement


We have a sheet of charge, which is infinite in the x direction and has width d in the y direction.
Find E field at a height h above the center line of the sheet. The sheet has λs surface charge density.

My attempt:

We know that the E field for an infinite wire is λl / 2πrε in the radial direction.
We treat the semi-infinite sheet as if it was composed of infinite wires with width dy and λl = λs dy

Therefore,

dE for an infinite wire = (λs dy) / (2πrε) in r direction, where r is a unit vector from the wire to the observation point.
For each infinite wire, vector r = -y y + h z where y and z are the unit vectors.
Therefore, r = ( -y y + h z ) / sqrt(y2 + h2)
and
dE = [(λs dy) / (2πr2ε)] ( -y y + h z )

= [(λs dy) / (2πε(y2 + h2)] ( -y y + h z )

From symetry reasons I know that the E field wield be in the z direction, so I discard the y part:

dE = [(λs dy) / (2πε(y2 + h2)] (h z )

I integrate [(λs dy) / (2πε(y2 + h2)] (h z ) from y = -d/2 to d/2

= (λs h z) / (2πε) integral of (dy / y2 + h2)

completing the integration I get:

s h z) / (πε) ) atan(d/2h)

but the h factor shouldn't be there because it shouldn't depend on the height of the observation point.

Where is the mistake?

(I have done it another way here https://www.physicsforums.com/showpost.php?p=4279144&postcount=7 but I don't see where is the mistake)
 
Physics news on Phys.org
fishingspree2 said:
= (λs h z) / (2πε) integral of (dy / y2 + h2)

completing the integration I get:

s h z) / (πε) ) atan(d/2h)
Something went wrong in there. Pls post those steps in detail.
 
fishingspree2 said:
The h factor shouldn't be there because it shouldn't depend on the height of the observation point.
The h factor only goes away in the case of an infinite plane (infinite length and infinite width). What you have is a "wide" wire of infinite length, and as h ->∞, the result approaches that of a field from an "thin" infinite wire.
 
rcgldr said:
The h factor only goes away in the case of an infinite plane (infinite length and infinite width). What you have is a "wide" wire of infinite length, and as h ->∞, the result approaches that of a field from an "thin" infinite wire.
True, but the fact remains that factor h should not be there. It gives the wrong limit as h tends to infinity.
 

Similar threads

Replies
9
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 42 ·
2
Replies
42
Views
8K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
3
Views
3K
Replies
3
Views
2K
Replies
64
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K