What are the Electric Field Lines and Flux in Coaxial Cables?

[ussrasu]

I don't really understand Gauss' law - any help with this question would be appreciated?

Coaxial cables are made of a copper wire in the center and a concentric cylindrical shell of copper outside, with insulating material in between and outside the shell. The charge per unit length of the cable are given as λ1 (positive) in the inner wire and −λ2 (negative) in the outer shell.

Sketch qualitatively the electric field lines on a plane perpendicular to the cable assuming it is very long.
Consider three cases:
(a) λ1 > λ2,
(b) λ1 = λ2 and
(c) λ1 < λ2.

Explain your diagrams using the concepts of flux and Gauss' law.

 

[FluxCapacitator]

The key to understanding Gauss's law is that a Gaussian surface is a completely made up surface that you use as a tool. Since we have a wire in this problem, a good surface that perfectly encloses the wire is a cylinder(minus the bases). The cylinder's surface area would be 2\pi r^2 where r is the radial distance from the wire.

Now Gauss's Law states that \int E \bullet dA=q/\epsilon_0 , where q is the charge enclosed by the imaginary surface you just created.

So if you want the electric field, but you have charge that's enclosed by any area, even one you just made up, you can use that equation to find the electric field.

It's really useful, and it makes these types of problems a breeze once you understand it.

 

[ussrasu]

What happens in each of the situations though? i.e in parts a) b) and c) - what is different in each of them? And how do you explain these observations in terms of Gauss' Law and Flux?

Thanks

 

[mukundpa]

(Charges are considered per unit length)

The E field within the conductor is zero therefore the flux originated from charge + \lambda_1 = \frac{\lambda_1} {\epsilon_0} is to be terminated on the inner surface of the outer cylinder and thus charge on the inner surface of outer cylinder must be - \lambda_1

But the total charge on the outer cylinder is - \lambda_2 hence the rest of the charge remain on the outer surface and will be

- \lambda_2 + \lambda_1

now I think you may find the correct answer for all three cases.