# Gauss' Law: Cylindrical sheath

## Homework Statement

A non-conducting, infinitely long, cylindrical sheath has inner radius r=10 m, outer radius r=15 m and a uniform charge density of 9 nC/m^3 spread throughout the sheath. Magnitude of electric field at r=5, r=12, r=17?

## Homework Equations

Q=rho(Volume) and phi=EA therefore phi=E(2pi(R)length)
phi=Q/epsilon_0

## The Attempt at a Solution

At r=5, Electric field=0
At r=12, phi=Q/epsilon_0, so E(2pi(R)length)=Q/epsilon_0
I just don't know what Q will be.
The same for r=17, I don't know how to find Q.

## Answers and Replies

Doc Al
Mentor
You are given the charge density. Figure out the volume of sheath enclosed in each case.

Thank you Doc Al!

So for r=17, if R is the outer radius of the sheath, then
Q=rho(length)piR^2, so E=(rho(length)piR^2)/(2pi(length)r(epsilon_o), thus
E=(rhoR^2)/(2r(epsilon_o))
is that correct, where i would use r=17 and R=15???????
I still don't know how I would figure out E for r=12, the area IN the sheath.

Please help! :)

Doc Al
Mentor
Thank you Doc Al!

So for r=17, if R is the outer radius of the sheath, then
Q=rho(length)piR^2, so E=(rho(length)piR^2)/(2pi(length)r(epsilon_o), thus
E=(rhoR^2)/(2r(epsilon_o))
is that correct, where i would use r=17 and R=15???????
I still don't know how I would figure out E for r=12, the area IN the sheath.

Please help! :)
The sheath only starts at r = 10m and extends to r = 15m. So to find the charge contained in a Gaussian surface with radius 12m, you need the volume of the sheath from r = 10 to 12 m. Similarly, for r = 17 you'd use the entire sheath, so from r = 10 to 15 m.

Hint: To find the volume between r1 and r2, find the volume of each full cylinder and subtract.

Thank you very very much Doc Al! This makes so much sense!