Gauss's Law Problem: long, cylindrical charge distribution

[Kaleem]

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.

 

[drvrm]

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-

 

[Kaleem]

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

 

[drvrm]

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

 

[Kaleem]

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!