E&M: Gauss' Law Surface Charge Density

[Wmdajt]

Homework Statement

A charge of uniform linear density 2.8nC/m is distributed along a long, thin, nonconducting rod. The rod is coaxial with a long conducting cylindrical shell (inner radius =4.8 cm, outer radius =9.40 cm). The net charge on the shell is zero. (a) What is the magnitude of the electric field 14.2 cm from the axis of the shell? What is the surface charge density on the (b) inner and (c) the outer surface of the shell?

Relevant Equations

E=λ/(2πε0r)
ε0ϕ = q

In this question I need to find the inner and outer charge density of the shell I did part A just fine, I used the formula for an electric field due to a line charge, but parts B and C is what's really confusing me. I'm not really sure how to go about it, I placed a spherical gaussian surface inside of the shell which gives me a flux of E*π*r^2. I didn't have a reason to attempt this, I was just seeing where it would take me. Any help would be appreciated, thanks!

 

[kuruman]

Homework Statement: A charge of uniform linear density 2.8nC/m is distributed along a long, thin, nonconducting rod. The rod is coaxial with a long conducting cylindrical shell (inner radius =4.8 cm, outer radius =9.40 cm). The net charge on the shell is zero. (a) What is the magnitude of the electric field 14.2 cm from the axis of the shell? What is the surface charge density on the (b) inner and (c) the outer surface of the shell?
Relevant Equations: E=λ/(2πε0r)
ε0ϕ = q

In this question I need to find the inner and outer charge density of the shell I did part A just fine, I used the formula for an electric field due to a line charge, but parts B and C is what's really confusing me. I'm not really sure how to go about it, I placed a spherical gaussian surface inside of the shell which gives me a flux of E*π*r^2. I didn't have a reason to attempt this, I was just seeing where it would take me. Any help would be appreciated, thanks!

What formula did you use? Please show your work and explain how you got E*π*r^2. A spherical Gaussian surface is not appropriate for a long wire. Do you see why?

 

[Wmdajt]

What formula did you use? Please show your work and explain how you got E*π*r^2. A spherical Gaussian surface is not appropriate for a long wire. Do you see why?

I just figured it out! I understand why a spherical Gaussian surface isn't appropriate, the way I drew my diagram was confusing me. I should've used a cylindrical surface. Negative charge is attracted to the inner wall of the shell equal to the amount of charge on the rod. From there I found the charge on the rod and divided it with the inner surface of the cylinder.

 

[kuruman]

From there I found the charge on the rod and divided it with the inner surface of the cylinder.

I don't understand this part. You are given the linear charge density on the rod. You do not explain how you found the charge density on the outer surface of the cylinder.

 

[Wmdajt]

I don't understand this part. You are given the linear charge density on the rod. You do not explain how you found the charge density on the outer surface of the cylinder.

Sorry, I wrote that in a bit of a hurry! I'll clarify.
The charge on the inner surface of the shell is the same as the opposite charge of the rod because the flux inside of a conductor must be zero.

qi = -q, Where qi is the charge of the inner shell surface and q is the charge of the rod

Charge density is defined as follows:
σ = qi/A
Charge is only on the surface of the shell due to it being a conductor, so we use the surface area of the body of a cylinder
σ = -q/(2πrh)

We clarified the outer shell charge is the same as the rod, we can find the charge using linear charge density
q = λh

Putting it all together in the equation above yields charge density of the outer surface of the shell
σ = -λ/(2πr)

The same formula is used for part C because the net charge in the shell is 0, so:
qo = qi = q, where qo is charge of the outer shell