Deriving Electric Field Inside and Outside an Infinitely Long Charged Cylinder

[wolski888]

Not sure if this is advanced. Highly doubt it but oh well

1. Homework Statement
Consider an infinitely long charged cylinder of radius R, carrying a charge whose density varies with radius as ρ(r) = ρo r. Derive expressions for the electric field (a) inside the cylinder (i.e. r<R), and (b) outside the cylinder (i.e. r>R).

2. Homework Equations
Gauss's Law
q=ρ δτ

3. The Attempt at a Solution
(a) E inside cylinder
I sketched a Gaussian surface inside of the cylinder.
I believe that E is parallel to ds ( E⃗ ||ds⃗ )
So, gauss's law becomes E∮ds = q/ϵ for the side

I believe the integral of ds is 2π r L (L being the length of the cylinder even though it is infinite.
And q = ρo r π r2 L
derived from q=ρ δτ

So we have E (2π r L) = ρo r π r2 L /ϵ
Simplifying to E = ρo r2/ 2ϵ

Is this correct for (a)?
And for (b) would it be the same idea but with a gaussian surface outside of R?

Thanks!

 

[Doc Al]

It's a bit difficult to read your notation. Consider using Latex for your equations.

(For example: q = \int \rho dv.)

Show again how you integrated to find the total charge within your gaussian surface.

 

[collinsmark]

Not sure if this is advanced. Highly doubt it but oh well

1. Homework Statement
Consider an infinitely long charged cylinder of radius R, carrying a charge whose density varies with radius as ρ(r) = ρo r. Derive expressions for the electric field (a) inside the cylinder (i.e. r<R), and (b) outside the cylinder (i.e. r>R).

2. Homework Equations
Gauss's Law
q=ρ δτ

3. The Attempt at a Solution
(a) E inside cylinder
I sketched a Gaussian surface inside of the cylinder.
I believe that E is parallel to ds ( E⃗ ||ds⃗ )
So, gauss's law becomes E∮ds = q/ϵ for the side

I believe the integral of ds is 2π r L (L being the length of the cylinder even though it is infinite.
And q = ρo r π r2 L
derived from q=ρ δτ

So we have E (2π r L) = ρo r π r2 L /ϵ
Simplifying to E = ρo r2/ 2ϵ

So close! But as Doc Al says, it's kind of hard to understand your notation. By "r2" do you mean r²? If so, try the final simplification once more. I think you forgot to cancel something out.

And for (b) would it be the same idea but with a gaussian surface outside of R?

Yes. The trick is to just be careful about determining q. When outside the cylinder, is the total charge q within the Gaussian surface a function of r or a function of R?

 

[wolski888]

Yes I meant r squared. The thing is when i reposted this problem the pasting messed up a couple of things.

 

[collinsmark]

Yes I meant r squared. The thing is when i reposted this problem the pasting messed up a couple of things.

Right. I think I figured that out when I looked at the other thread. Check that thread. I left a hint for you there.