- #1
wolski888
- 25
- 0
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!
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!