Electric Potential Homework: Gauss's Law, Integrals, Bounds

[Aero6]

Homework Statement

A long coaxial cable carries a volume charge density rho=alpha*s on the inner cylinder (radius a) and a uniform surface charge density on the outer cylindrical shell (radius b). This surface charge is negative and of just the right magnitude so that the cable as a whole is electrically neutral.
a)Find the electric field in each of the three regions:
s<a, a<s<b, s>b

b)find the potential difference in each of these regions with a refernce point at infinity

Homework Equations

Gauss's law integral of E*da = Qencl/epsilon
V=-integral E*dl

The Attempt at a Solution

b) a<s<b
I'm confused about integrating to find Qencl. Qencl=integral rho*dtao where dtao=s*ds*dtheta*dz, but when I set up the bounds on the integral for s, I don't understand which bounds I am supposed to include. Since there is a less than sign and not a less than or equal to sign when the problem says that s is greater than a and less than b, how is it okay to integrate so from a to an arbitrary distance that is less than b? Isn't this still including the distance a, which we shouldn't do because of the strict greater than sign? Also, how would this problem change if I was asked to find the electric field in the region: s is greater than OR equal to a and less than or equal to b?

Thank you

 

[Delphi51]

What shape is your Gaussian enclosure? I'm thinking of an infinitely long cylinder with the cable in the center. For the s > b, the total charge inside this is zero so an easy answer.
For a < s < b, there is charge and there may be some difficulty with charge and area being infinite but if you think "very long" instead of infinite, all that should cancel out. I don't even see the need for an integral - just the formula for the surface area of a cylinder.

 

[Aero6]

The shape I'm using is a gaussian cylinder. Right, so the electric field for s> b is 0 because the coaxial cable is neutral, so we do not see any charge outside. The electric field for s<a can be solved using: E-field=Qencl/2*pi*r*L and Qencl can be found by using: Qencl=integral of rho*dtao, where dtao is rdr*dtheta*dz and plugging in Qencl into Gauss's law. I don't think I phrased my question correctly. in some coaxial problems we are told to find the electric field in the region between the cables where a<s<b where s is the radius of our Gaussian surface. In another problem (also dealing with a coaxial cable), the question has asked to find the electric field in the region where s is greater than OR EQUAL TO a and s is LESS THAN OR EQUAL TO b. Does the less than or equal to make a difference from a problem that does not ask less than or equal to? I know Gauss's law says that no points outside of our gaussian surface will act on the electric field inside our Gaussian surface, but for questions that ask less than or equal to, is it necessary to use 2 gaussian surfaces? one that integrates from a to s and one that integrates from s to b?