Electric field of a long round bar

[Karol]

Homework Statement

I have to prove that the electric field round a long conductive round bar of radius R is equal to that of a long thin wire, meaning the charge is on the center line

Homework Equations

The Gauss theorem:
$E\times A=q$
Where A is the area of the surface and q is the charge enclosed in it

The Attempt at a Solution

To find the field E when the charge is concentrated on the axis i take a round long volume with radius r:
$E\times 2\pi rl=\lambda l\rightarrow E=\frac{\lambda}{2\pi r}$
Where $\lambda$ is the charge per unit length.
The field near a round bar:
$E\times 2\pi rl=\sigma\cdot 2\pi R l$
Where $\sigma$ is the charge per unit surface, since the charge is concentrated only on the surface.
But $\sigma\cdot 2\pi R l=\lambda l$
They both are the charge per unit length. and so is if i use the charge per unit mass.
I feel uncomfortable

 

[ehild]

Homework Statement

I have to prove that the electric field round a long conductive round bar of radius R is equal to that of a long thin wire, meaning the charge is on the center line

Homework Equations

The Gauss theorem:
$E\times A=q$
Where A is the area of the surface and q is the charge enclosed in it

It is true if you use cgs system of units where ε₀=1.

The Attempt at a Solution

To find the field E when the charge is concentrated on the axis i take a round long volume with radius r:
$E\times 2\pi rl=\lambda l\rightarrow E=\frac{\lambda}{2\pi r}$
Where $\lambda$ is the charge per unit length.

Correct.

The field near a round bar:
$E\times 2\pi rl=\sigma\cdot 2\pi R l$
Where $\sigma$ is the charge per unit surface, since the charge is concentrated only on the surface.
But $\sigma\cdot 2\pi R l=\lambda l$

Simplifying, you get that E=σ R/r and σ =λ/(2πR). Substitute the second equation into the first, what do you get? Is E the same as in case of a thin long wire at the centre?

 

[Karol]

I get $E=\frac{\lambda}{2\pi r}$ like for a long thin wire at the center, and that's good, but i feel uncomfortable with that, is that the method to prove it?

 

[ehild]

You do not need to feel uncomfortable, it was the right proof. :) Good job!

 

[Karol]

Thanks