How to calculate the flux given the surface charge density?

In summary, the flux through a window at (rho) = 2m, pie/4 <= (phi) <= 3pie/4 , 2<=z<=4 is Q/(4*pi*rho^2)
  • #1
kloong
36
0

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..
 
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  • #2
Consider Gauss's law and symmetry. (Try to picture the imaginary field lines that the charge distribution would produce.)
 
  • #3
umm.. i still don't get it.. =(
 
  • #4
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.
 
  • #5
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.
 
  • #6
i am so sorry.. but is it ok if u show me how to solve it and get the answer?
 
  • #7
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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  • #8
ok. i will.
 

1. What is surface charge density?

Surface charge density is a measure of the amount of electric charge present per unit area on the surface of a material. It is usually denoted by the symbol σ and is expressed in units of coulombs per square meter (C/m²).

2. How is surface charge density related to flux?

Surface charge density is directly related to the electric flux passing through a surface. The electric flux passing through a surface is equal to the product of the surface charge density and the area of the surface. In mathematical terms, this can be expressed as Φ = σA.

3. What is the equation for calculating flux given surface charge density?

The equation for calculating flux (Φ) given surface charge density (σ) is Φ = σA, where A is the area of the surface. This equation is based on Gauss's Law, which states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space.

4. How do I determine the direction of the flux?

The direction of the flux is determined by the direction of the electric field passing through the surface. The electric field lines always point in the direction of the flux, so if the electric field is entering the surface, the flux will be positive, and if the electric field is leaving the surface, the flux will be negative.

5. Can flux be negative?

Yes, flux can be negative. This occurs when the electric field lines are leaving the surface, resulting in a negative value for the flux. Negative flux is often associated with a net outward flow of electric charge from the surface, while positive flux represents a net inward flow of charge.

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