Recent content by Kot

I think it should be the same E as a < r < b, but in this case the r has to be greater than b. If that isn't correct then the charge enclosed would be the total of the wire and the sheath.

Now choosing the Guassian surface to be in a < r < b, the total amount of Qenclosed is equal to the charges that were described earlier (on the surface of the wire). So EA = Qenclosed / εo. A is the surface area of the Guassian cylinder 2∏*r*L and Qenclosed = λL. Using these values in Gauss's...

At r < a, E = 0.

Well since all the charges reside on the outer surface r=a, and the chosen Guassian surface has r<a. The Guassian surface does not hold any charge.

I can't see why this matters. Doesn't Guass's Law apply to a charge inside a chosen surface. If I chose the Guassian surface to be big enough, it would contain the charge. Since the charges are repelled by each other and cannot be on the Guassian surface, it would be on the surface of the wire?

On the surface of the Guassian cylinder?

The charges are located at the center of the Guassian cylinder.

Since it is a conductor, wouldn't the electric field be zero?

If I let the Guassian surface (cylinder) to be the same length as the cable, Qenclosed would be the same length as the cable L. I can rewrite Q as λL. Is that right?

The Guassian surface is a cylinder. Since it is symmetrical I don't have to do the integral, I can use EA = Qenclosed / εo. The area of the cylinder would be 2∏rL.

I used Gauss's Law. The E is the electric field and Q is the total charge enclosed inside the cylinder.

Would the total charge inside a cylinder be Qenclosed = E2∏rLεo ?

Could I get a hint on how to do this problem?

σ(2∏rL) is σA, the surface charge density of the cylinder. Since I need to find the electric field, I used Gauss' Law ∫EdA = Qenclosed / εo . I replaced Qenclosed with σA.

Oops, EA = σA / εo E(2∏rL) = σ(2∏rL) / εo E = σ / εo where σ is the surface charge density of the cylinders. Is this correct?