# 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.

Related Introductory Physics Homework Help News on Phys.org
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.

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.