What Is the Electric Field of a Coaxial Cable?

[rude man]

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

Yes!

OK, next step: put the Gaussian surface inside the outer conductor, at r = b. What is E there?

EDIT: Never mind, I see you don't need to do this,though it would be educational for you if you did.

So OK, last step is to put your surface at r > b. What is E there?

 

[Kot]

Yes!

OK, next step: put the Gaussian surface inside the outer conductor, at r = b. What is E there?

EDIT: Never mind, I see you don't need to do this,though it would be educational for you if you did.

So OK, last step is to put your surface at r > b. What is E there?

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.

 

[rude man]

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.

It won't be the same as at a < r < b. Use your formula again for E.

Yes, the toal enclosed charge is now the sum of charges on the inside and outside conductors, i.e. 2λL.

For 'extra credit' you should assume a finite thickness for the sheath, put your surface there and again solve for E. You again know the E field is zero since the surface is inside the conductor, and if you use your formula to solve for Q with E=0 you can deduce what the charges on the inner and outer surfaces of the sheath must be.