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Gauss's Law Problem: long, cylindrical charge distribution

  • Thread starter Kaleem
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
963
213
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-
 
  • #3
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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
 
  • #4
963
213
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
 
  • #5
21
0
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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