Gauss's Law Problem: long, cylindrical charge distribution

AI Thread Summary
The discussion focuses on solving a Gauss's Law problem involving a long cylindrical charge distribution with uniform charge density ρ. For the region inside the cylinder (r < R), the electric field is correctly derived as E = ρr/2ε0. In the region outside the cylinder (r > R), participants clarify that the Gaussian surface should have a radius r, not R, to calculate the total electric flux. The confusion arises regarding the use of the charge density and volume when determining the electric field outside the cylinder. Ultimately, the correct approach involves considering the total charge within the Gaussian surface and applying Gauss's Law accordingly.
Kaleem
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Homework Statement


Consider a long, cylindrical charge distribution of radius R with uniform charge density ρ.

a) Using Gauss’s law, find the electric field at distance r from the axis, where r < R

b) Using Gauss’s law, find the electric field at distance r from the axis, where r > R

Homework Equations


∫EdA = Qinside/ε0
q = ρV
V = πr2L
A = 2πrL

The Attempt at a Solution


I've successfully solved the first part which is E = ρr/2ε0.

However for the second part I am confused as to whether or not I would need to use big R for the area, because we are now going at a distance out of the surface or to use little r.
 
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Kaleem said:
I am confused as to whether or not I would need to use big R for the area, because we are now going at a distance out of the surface or to use little r.

i think you are going out to a point r where r >R ,
so your Gaussian surface will be a cylinder of radius r (r>R) enveloping the charged cylinder and you have to calculate the total normal outward flux through this imaginary /constructed Gaussian surface-
 
drvrm said:
i think you are going out to a point r where r >R ,
so your Gaussian surface will be a cylinder of radius r (r>R) enveloping the charged cylinder and you have to calculate the total normal outward flux through this imaginary /constructed Gaussian surface-

What I got from this method so far is ρr2/2Rε0
 
Kaleem said:
What I got from this method so far is ρr2/2Rε0

R is the radius of your cylindrical charge distribtion of volume charge density rho-
so in the gaussian surface you have charges inside which will be = rho x volume of the charged cylinder ;
the flux will be passing through cylindrical surface constructed of radius r
so there will be correction in your result
 
drvrm said:
R is the radius of your cylindrical charge distribtion of volume charge density rho-
so in the gaussian surface you have charges inside which will be = rho x volume of the charged cylinder ;
the flux will be passing through cylindrical surface constructed of radius r
so there will be correction in your result

I see exactly what you mean now, thank you!
 
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