How to Show the E Field Outside a Long, Charged Conducting Cylinder?

[Plaetean]

Homework Statement

Use the divergence theorem (and sensible reasoning) to show that the E field a distance r outside a long, charged conducting cylinder of radius r₀ which carries a charge density of σ Cm^-2 has a magnitude E=σr₀/ε₀r. What is the orientation of the field?

Homework Equations

Divergence theorem

q=∫E.dAε₀

The Attempt at a Solution

Completely lost really, not even sure where to start. The dimensions of σ mean that it can only be used with the surface integral (I think), and so every time I start playing around with the equations, I end up not using the divergence theorem at all. Not even sure if the second equation is relevant, just included it because I have a feeling it might be useful.

 

[ShayanJ]

The first thing you should do is considering a Gaussian surface, which is an imaginary surface you're going to compute the surface integral on it. Then you should calculate the amount of charge you're Gaussian surface contains.The surface integral of the field is easy to handle because of the symmetry. Calculating the charge inside the Gaussian surface isn't harder because you have the surface area of the cylinder enclosed by the Gaussian surface.Try and report the results.

 

[Plaetean]

Taking a surface integral for the sides of the cylinder I get q=2πr₀σh (where h would be an arbitrary length of the cylinder). Thanks for the reply, I'm honestly not looking for someone to just answer the question for me, but I'm completely lost.

 

[ShayanJ]

That's correct.
So you have problem with the surface integral of the field?
Just think about the direction of the field and whether is depends on any spatial variables or not.Taking symmetry into account is crucial here.

 

[HallsofIvy]

A cylinder of length h and radius r₀ has a surface area or 2\pi r_0^2h. Since you are told that the charge density on the cylinder is \sigma, the total charge is 2\pi\sigma r_0^2h. The integral of the E field over the surface of a cylinder of radius r around that must be equal to 2\pi\sigma r_0^2 h. If E is constant, that is just E times the surface area of the cylinder or radius r.

 

[Plaetean]

A cylinder of length h and radius r₀ has a surface area or 2\pi r_0^2h. Since you are told that the charge density on the cylinder is \sigma, the total charge is 2\pi\sigma r_0^2h.

The units of σ are Cm^-2 though, not Cm^-3, so you'd end up with Cm as your units for charge.