Gauss' Law: Cylindrical sheath

1. Apr 5, 2010

abcdmichelle

1. The problem statement, all variables and given/known data
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?

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

3. 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.

2. Apr 5, 2010

Staff: Mentor

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

3. Apr 5, 2010

abcdmichelle

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.

4. Apr 5, 2010

Staff: Mentor

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.

5. Apr 5, 2010

abcdmichelle

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