Find the electric flux through the parabolodial surface

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Discussion Overview

The discussion revolves around calculating the electric flux through a parabolodial surface in the context of electric fields and Gauss's law. Participants explore different interpretations of the problem, particularly whether the parabolodial surface can be treated as a closed surface for the purposes of applying Gauss's law.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • One participant notes that in their textbook, the electric flux through a parabolodial surface due to a uniform electric field is calculated using the formula ##\pi r^2 * E = \Phi##.
  • Another participant argues that a paraboloid is not a Gaussian surface, suggesting that Gauss's law does not apply in this case.
  • A further clarification states that a paraboloid is an open surface and that Gaussian surfaces must be closed to use Gauss's law.
  • One participant proposes that they might be discussing a closed paraboloid and suggests that the flux through the parabolodial surface could be related to the flux through the circular end cap.
  • Another participant speculates that the question may have been misinterpreted, indicating that they are considering only the flux through the parabolodial portion.
  • There is a suggestion that the net flux is zero, but the flux through the parabolodial region might equal the flux through the circular region.

Areas of Agreement / Disagreement

Participants express differing views on whether the parabolodial surface can be treated as a closed surface for the purposes of applying Gauss's law. While some agree on the relationship between the flux through the parabolodial region and the circular region, there is no consensus on the application of Gauss's law to this scenario.

Contextual Notes

There are limitations regarding the assumptions about the nature of the parabolodial surface and its applicability as a Gaussian surface. The discussion also reflects uncertainty about the interpretation of the problem and the conditions under which the flux calculations are made.

oneplusone
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In my textbook there are two types of electric flux problems:

1. Find the electric flux through the parabolodial surface due to a uniform electric field of magnitude E_0 in the direction as shown.

Here they use ##\pi r^2 * E = \Phi ##


2. An infinitely long line charge having a uniform charge per unit length ##\laambda ## lies a distance d from point O. construct a sphere with center O, and radius less than d. Find the flux thru the surface of the sphere.

Here the answer is 0 since q enclosed is 0.


But for #1, wouldn't it be zero also since the paraboloid has no charge contained inside of it?
 
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A paraboloid is not a Gaussian surface, so Gauss' law doesn't apply.
 
More explicitly, a paraboloid is an open surface, not a closed one. A Gaussian surface for use in Gauss's Law must be a closed surface.
 
I think he/she is talking about a closed paraboloid such as this one.

pse6.24.p13.png


They may be talking about the single surface of the paraboloid, not the net total of all the surfaces. Since the flux coming into the cap is negative, and the sum of all the flux passing through the surfaces must be zero:

0 = \PhiEndCircle + \PhiParaboloid = -E0∏r2 + \PhiParaboloid

Add the E0∏r2 to both sides:

\PhiParaboloid = E0∏r2

I don't know if you can do this; someone else should verify as I have had no formal education on gauss's first law.

Edited:
oneplusone: We are both in tenth grade (I think). cool.
 
Last edited:
As in, perhaps the question was misinterpreted and we are asking for just the flux passing through the paraboloid portion, basically equivalent to the flux passing through the end circle.
 
@Above, yes that is the same question! (where'd you find the diagram??)

So am i correct in saying that the NET flux is zero, but the flux thru the parabolodial region equals the flux thru the circular region?
 
oneplusone said:
the NET flux is zero, but the flux thru the parabolodial region equals the flux thru the circular region?

As far as I can tell, yes.

I got the image by going onto google images and searching "paraboloid flux".
 

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