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Electric Fields

  1. Jan 25, 2014 #1
    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!
     
  2. jcsd
  3. Jan 26, 2014 #2
    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 = [itex]\sigma[/itex][itex]/[/itex]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 [itex]\sigma[/itex][itex]/[/itex]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
     
  4. Jan 27, 2014 #3
    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....
     
  5. Jan 27, 2014 #4
    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
     
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