Clarifying Gauss's Law & Electric Field Permeation

AI Thread Summary
The discussion centers on understanding Gauss's Law and the electric field generated by a charged sheet, particularly the formula E = σ/(2ε). Participants explore why the electric field lost when removing a circular area from the sheet is proportional to z/√(z² + r²) instead of just πr². They clarify that the '2' in the denominator arises from considering both sides of the charged surface and the superposition principle, which states that the net electric field is the sum of individual fields. The significance of ε as a constant describing electric field permeation in various environments is also discussed. Overall, the conversation aims to clarify the mathematical relationships and principles involved in deriving the electric field from a charged sheet.
MathewsMD
Messages
430
Reaction score
7
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!
 
Physics news on Phys.org
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
 
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...
 
MathewsMD said:
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
 
  • Like
Likes 1 person
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top