A long, non conducting, solid cylinder of radius 4.2 cm has a nonuniform volume charge density ρ = Ar^2, a function of the radial distance r from the cylinder axis. A = 2.5 µC/m5.
(a) What is the magnitude of the electric field at a radial distance of 3.2 cm from the axis of the cylinder?
(b) What is the magnitude of the electric field at a radial distance of 5.2 cm from the axis of the cylinder?
e0 Ø = charge enclosed ; e0 is the permittivity constant 8.85e-12 and Ø can be the flux
e0 E(2π r L) = λ L ; λ is the linear charge density
The Attempt at a Solution
I thought that I could just integrate Ar^2 twice, which would give me λ. Then I could multiply by the height of the cylinder, L. After I did all that I arrived at this:
E=(Ar^3) / (24e0) with e0 being the permittivity constant.
The correct form would have me do this though:
charge enclosed = ∫ 2π r L (Ar^2) dr ; from 0 to r
= 2π L A ∫ r^3 dr ; from 0 to r
If you integrate that and solve for E you get this:
E=(Ar^3) / (4e0) <----- This gives the correct answer.
Can someone explain to me why the method I'm using wrong? Is there some constant I am missing in my method? Or is it just wrong?
And this is also my first post here on Physics Forums? Please tell me if I am doing something wrong so I can correct it in any future post.