# Gauss law for semi-infinite sheet of charge

1. Feb 20, 2013

### fishingspree2

We know that for an infinite sheet of charge, E = λ / 2ε, where λ is the surface charge density. This can be easily found using gauss law and a cylinder perpendicular to the sheet as a gaussian surface. We will end up with something like Qinc/ε = 2E ∫dS, from which we find E = λ / 2ε.

Now say that I want to find the charge for a semi-infinite sheet of charge. A sheet that is infinite in only 1 dimension. If i try to apply Gauss's Law the same way as in the previous case, I end up with exactly the same procedure and answer, which is not correct.

But I fail to see what is different in the application of Gauss Law, between the 1st and the second case.

2. Feb 20, 2013

### Staff: Mentor

With an infinite sheet of charge, you can argue from symmetry that the field must be normal to the surface, which allows you to take advantage of Gauss's law to calculate the field. Not so with a semi-infinite sheet, since there would be edge effects. No reason to expect the field to be uniform.

3. Feb 20, 2013

### fishingspree2

Alright, then what would be the way to compute the field at a point P above the centerline?

integrating dE?

also, what are edge effects?

4. Feb 20, 2013

### Staff: Mentor

Yes, you'll have to integrate.
As an example of what I mean by "edge effect", see this diagram of the field between two parallel plates. Note how the field curves at the edges of the plates (left diagram).
http://www.physics.sjsu.edu/becker/physics51/images/23_18Capacitor.JPG

Here's a schematic of the field from a charged plate (top diagram). Note how the field points outward at the edges.
https://encrypted-tbn2.gstatic.com/images?q=tbn:ANd9GcTvN7Vjo4FoMiN9xIItgqOl6i7cYSrbPEQ9h5bc5_CztBh-tHdJ [Broken]

Last edited by a moderator: May 6, 2017
5. Feb 20, 2013

### fishingspree2

Here's what I tried:

Find V at the point, then E = -∇V

dV = $\frac{dq}{4R\pi\epsilon}$, where R is the distance between the charge element and the point, R = sqrt(x2 + y2 + h2)

dq = λ dA = λ dx dy

So,
$V=\int_{-\frac{d}{2}}^{\frac{d}{2}}\int_{-\infty}^{\infty}\frac{\lambda dxdy}{4\pi\epsilon\sqrt{x^{2}+y^{2}+h^{2}}}$

But this infinite integral does not converge.

6. Feb 20, 2013

### Staff: Mentor

Break the sheet into infinitely long infinitesimal strips of charge, which can be treated as infinite lines of charge. Add up the field contribution from each strip (set up the integral), taking advantage of symmetry.

7. Feb 20, 2013

### fishingspree2

Considering that the plane is infinite in the x direction, after setting up and some manipulations, I have

dE = [λ dy cos(θ)] / 2πrε in the k direction

where θ is the angle between the z axis and the distance vector between an infinite line and the point, r is the distance.

I know that cos(θ) = h/r, where h is the z coordinate of the point (the point lies on the z axis)
and r = sqrt(h2 + y2)

substituting,

dE = (λ h dy) / 2πr2ε in the k direction
= (λ h dy) / [ 2πε (h2 + y2) ]
which I integrate from -d/2 to d/2

= λ/πε atan(d/2h) in the k direction

Last edited: Feb 20, 2013