# E field for an semi-infinite sheet of charge

1. Apr 11, 2013

### fishingspree2

1. The problem statement, all variables and given/known data
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)

2. Apr 11, 2013

### haruspex

Something went wrong in there. Pls post those steps in detail.

3. Apr 11, 2013

### rcgldr

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.

4. Apr 11, 2013

### haruspex

True, but the fact remains that factor h should not be there. It gives the wrong limit as h tends to infinity.