Clarifying Gauss's Law & Electric Field Permeation

[MathewsMD]

https://www.physicsforums.com/showthread.php?t=479376

Looking at the above link, I was wondering how he found that electric field lost by removing the circular area from the sheet was proportional to $\frac {z}{(z^2 + r^2)^{1/2}}$? Why is it not just the $πr^2$? I am fairly new to Gauss's Law and having a bit of trouble understanding what the formula: $E = \frac {σ}{2ε}$ and if anyone could provide an explanation for the equation and why the problem in the link was solved with the given method, that would be great!

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html

Looking at this link, I am confused on where the 2 comes from in the denominator...is it because you are considering both sides of the surface?

To clarify, the constant ε is constant in any setting (ex. vacuum, 0K, in liquids, etc.) and it describes how an electric field permeates the space, right? Once again, I am just having a bit of trouble understanding how this formula and constant are significant, especially since I have yet to fully understand Gauss's Law, so any explanation would be amazing!

 

[Adithyan]

Actually he used the 'superposition principle' for electric field which states that the net electric field at a point is equal to the sum of individual fields from all sources.

Now, imagine that there is no circular opening. The net field experienced at point P = $\sigma$$/$2ε ( I will come to this later).

This sheet is made up of two components - The disk which is about to be removed and the portion without the disk. So, according to superposition principle,- Net field ie $\sigma$$/$2ε= Field due to disk + Field due to portion without the disk. Now try solving this.
(Be careful about finding the field due to a disk at point on its axis).Now coming to how the '2' got there, I hope you know what Gaussian surfaces are. If not, here's the link : http://en.wikipedia.org/wiki/Gaussian_surface. (Try reading through other sites as well).

Here, the most important thing to notice is, this is a sheet of charge- the charge distribution is only on one face of the sheet. But there are two circular bases of the Gaussian surface (the cylinder) and each circular area cuts the flux of that side of the sheet. Now try deriving the formula.

Hope this gives you some idea. Do notify me if you still don't get it. Happy to help you!

Regards
ADI

 

[MathewsMD]

Thank you for the response. Okay, I understand he was using the superposition principle, but don;t quite understand why he used $\frac {z}{(z^2+r^2)^{1/2}}$

It looks like this is cosθ but how is this the ratio b/w total surface area and the surface removed? I don't understand why the z is significant in finding the ratio...

 

[Adithyan]

Thank you for the response. Okay, I understand he was using the superposition principle, but don;t quite understand why he used $\frac {z}{(z^2+r^2)^{1/2}}$

It looks like this is cosθ

Yes it is the cosΘ component. I hope you know how the field has been derived by integration; if not here's the link:
http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

 

[HalfLostAndSearching]

Hello! I am happy to help clarify Gauss's Law and the concept of electric field permeation.

First, let's start with Gauss's Law. This law is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the total charge enclosed by that surface. In other words, it describes the relationship between the electric field and the charge distribution.

Now, onto the formula that was mentioned in the link: $E = \frac {σ}{2ε}$ This is known as the electric field intensity formula and it relates the electric field to the surface charge density and the electric permittivity of the medium. The constant ε, also known as the permittivity of free space, is a fundamental constant that describes how easily an electric field can permeate a given medium. It is not constant in all settings, as it depends on the properties of the medium (such as vacuum, air, or other materials).

In the link, the problem is solved using Gauss's Law and the electric field intensity formula. The 2 in the denominator comes from the fact that we are considering both sides of the surface, as you mentioned. This is because the electric field is present on both sides of the sheet, and we need to take that into account when calculating the total electric field.

Now, onto the specific question about the circular area and the electric field intensity formula. The reason why the electric field lost by removing the circular area from the sheet is proportional to $\frac {z}{(z^2 + r^2)^{1/2}}$ is because the electric field is not a constant value throughout the space. It varies with distance from the sheet, and this formula takes that into account. The circular area is essentially acting as a point charge, and the electric field intensity formula tells us that the electric field due to a point charge decreases with distance according to the inverse square law. This is why the formula includes the distance term (z) and the radius term (r).

I hope this helps to clarify things for you. Gauss's Law and the concept of electric field permeation can be difficult to understand at first, but with practice and further study, you will gain a better understanding. Keep asking questions and seeking clarification, and you will continue to improve your understanding. Best of luck in your studies!