Gauss' Law and an infinite sheet

In summary, a small circular hole with radius R = 2.03 cm has been cut in the middle of an infinite, flat, nonconducting surface with a uniform charge density σ = 4.61 pC/m^2. Using superposition, the magnitude of the electric field at point P at z = 3.05 cm can be found by adding the electric fields for the sheet and the disk with opposite charge density. The equation for the electric field along the axis for a uniformly charged disk is used, and the final result is σ/2ε.
  • #1
catie1981
45
0

Homework Statement



a small circular hole of radius R = 2.03 cm has been cut in the middle of an infinite, flat, nonconducting surface that has a uniform charge density σ = 4.61 pC/m^2. A z axis, with its origin at the hole's center, is perpendicular to the surface. What is the magnitude (in N/C) of the electric field at point P at z = 3.05 cm? (Hint: See Eq. 22-26 and use superposition.)

There is supposed to be a figure, but it really doesn't lend much to this. It is a flat sheet laid on the horizon, with the z axis going through a hole in the middle of the sheet. Point P is on the z axis, and R is just shown as the radius of the hole in the sheet. Which, is pretty much what was said in the problem...

Homework Equations



the equation mentioned above is
E = (σ / 2ε) *( 1- z/ (√z^2 + R^2))

but how I'm supposed to use that and superposition, I am not sure.


The Attempt at a Solution


I want to just plug in the given values, but I don't think that it right, because the question mentions superposition, but what am I adding together?! There was an additional hint about using a disc of equal magnitude and such, but oppostire direction. Wouldn't I then need a different equation? Help! (and Thanks)
 
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  • #2
There is a favorite trick for dealing with "holes" in charge distributions. You use the superposition of the field for an infinite sheet of uniform surface charge density σ , together with the field for a disk the size of the "hole", having uniform surface charge density -σ .
 
  • #3
oooo, so for the sheet I would use sigma/2*epsilon and the disc would have a -sigma value...then I add those numbers and viola! I have the solution. Thanks!
 
  • #4
catie1981 said:
oooo, so for the sheet I would use sigma/2*epsilon

Yes, for its electric field, since this is a non-conducting sheet.

and the disc would have a -sigma value...

Yes, in the equation you were given for the field along the axis for a uniformly charged disk at a distance z from the plane of the disk.

then I add those numbers and viola!

That should do it.
 

1. What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the total charge enclosed by that surface. It is a mathematical expression of the fact that electric charges create an electric field that extends into the space around them.

2. How does Gauss' Law apply to an infinite sheet of charge?

When applying Gauss' Law to an infinite sheet of charge, the electric field at any point above or below the sheet is constant and perpendicular to the sheet. This means that the electric flux through any closed surface that includes the sheet will be the same, regardless of the shape or size of the surface.

3. What is an electric flux?

Electric flux is a measure of the flow of an electric field through a given surface. It is defined as the dot product of the electric field and the surface area. A closed surface with a higher electric flux indicates a stronger electric field, while a closed surface with a lower electric flux indicates a weaker electric field.

4. How is an infinite sheet of charge represented in Gauss' Law?

In Gauss' Law, an infinite sheet of charge is represented by a closed surface that is parallel to the sheet, with the electric field passing through it. This allows for the electric field to be constant and perpendicular to the sheet, simplifying the calculation of the electric flux.

5. Can Gauss' Law be applied to non-infinite sheets of charge?

Yes, Gauss' Law can be applied to non-infinite sheets of charge. However, the electric field will not be constant and perpendicular to the sheet, making the calculation of the electric flux more complex. In these cases, it is often easier to use other methods, such as Coulomb's Law, to calculate the electric field.

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