How to calculate the flux given the surface charge density?

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Homework Help Overview

The problem involves calculating the electric flux through a specified window surrounding a long cylinder with a uniform surface charge density. The cylinder is oriented along the z-axis, and the flux is to be determined at a distance from the cylinder, considering the geometry and symmetry of the charge distribution.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Gauss's law and the symmetry of the electric field around the cylinder. There are attempts to visualize the electric field lines and consider the implications of the uniform surface charge density on the electric field direction.

Discussion Status

The discussion includes various approaches to understanding the problem, with some participants suggesting the use of Gauss's law and symmetry arguments. There is an acknowledgment of the need for further exploration of the concepts involved, particularly regarding the electric field and flux calculations.

Contextual Notes

Some participants express uncertainty about how to begin the problem, indicating a potential gap in understanding the application of Gauss's law. There are also references to external resources for further study.

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Homework Statement


Given long cylinder of radius 0.2m lies along the z-axis and carries a uniform surface charge density if 10m C/m2. Calculate the flux passing through a window at (rho) = 2m, pie/4 <= (phi) <= 3pie/4 , 2<=z<=4.

Homework Equations


The Attempt at a Solution


not sure where to start..
 
Last edited:
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Consider Gauss's law and symmetry. (Try to picture the imaginary field lines that the charge distribution would produce.)
 
umm.. i still don't get it.. =(
 
One of the symmetries you can exploit is assuming that there isn't any E-field component which isn't parallel to the x-y plane, since it's a long cylinder. So that means the flux through any closed cylindrical surface would be due only to the E-field in the a_rho direction. And since the surface charge density is uniform, E_rho should be a function of rho only.
 
You could find the explicit function for E and integrate it over the window, but that isn't necessary thanks to our old friend Gauss. Here's how I would do it:

  1. Construct a cylindrical surface around the cylinder, with radius 2 and from z=2 to z=4
  2. Calculate the total charge Q inside this surface
  3. Invoke a symmetry argument (infinite z-dimension), to convince yourself that the E-field is everywhere perpendicular to the cylinder, so any flux through it is entirely through the curved surface and none at the plane ends.
  4. Invoke Gauss law to find the total flux out though cylindrical surface
  5. Invoke yet another symmetry argument (rotational symmetry) to convince yourself that the flux is uniform over the cylindrical surface
  6. Find the flux through the window as a fraction of the flux through the whole surface.
 
i am so sorry.. but is it ok if u show me how to solve it and get the answer?
 
You give it a try. Troels gave you a set-by-step outline of how to proceed. If you have no idea about how to use Gauss's law, then you should spend your time reviewing that before attempting this problem. I'm sure your text has several examples worked out; so does this site: http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/gaulaw.html#c4".
 
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ok. i will.
 

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