# Electric Field Of A Cable

1. Feb 18, 2014

### Kot

1. The problem statement, all variables and given/known data
A long coaxial cable of length L consists of a cylindrical metal sheath of radius b surrounding a conducting wire with outer radius a. The sheath and wire are given the same charge per length +Q/L. Find the electric field in terms of distance r from the central axis of the wire in the regions 0 < r < a, a < r < b, and r > b.

2. Relevant equations
Gauss' Law ∫EdA = Qino

3. The attempt at a solution
I am not really sure how to do this problem but I think I have an idea. I have to find the electric field in terms of distance r of three different regions 0 < r < a, a < r < b, and r > b. Since the cable and metal sheath have the same charge per length, wouldn't they both have the same charge and cancel out? Could someone explain how I can find the Gaussian surface for the first region?

2. Feb 18, 2014

### tiny-tim

Hi Kot!
For r > b, yes

So the electric field there is … ?
(for r < a) Why don't you just use a cylinder of radius r ?

3. Feb 18, 2014

### Kot

For r > b, the electric field wouldn't completely cancel out since I was given an arbitrary a and b.

For a < r < b, I am still not sure what radius to use. How can i express a radius that is between a and b?

4. Feb 18, 2014

### tiny-tim

Hi Kot!
But the question says that the charges per length are equal.
Just call it r.

What do you get?​

5. Feb 18, 2014

### Kot

It's charge per length is equal but doesn't the length of a and b also contribute to the electric field?

For a < r < b, if I have a Guassian surface (cylinder in this case) then I get:

EA = ρV / εo

E2∏rL = ρ∏r2L / εo

E = ρ / 2 εo

6. Feb 18, 2014

### tiny-tim

charge per length is what it says

the radius (a or b) has nothing to do with it
what is ρ ?

you seem to be calculating a volume

the charge is a surface charge, on two cylindrical shells

7. Feb 18, 2014

### Kot

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?

8. Feb 18, 2014

### tiny-tim

what is this supposed to be?

(and what does 2πr have to do with it?)

9. Feb 18, 2014

### Kot

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.

10. Feb 18, 2014

### tiny-tim

But the surface charge depends only on L
(and now i'm off to bed :zzz:)

11. Feb 18, 2014

### Kot

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

12. Feb 19, 2014

### tiny-tim

(just got up :zzz:)

If …
… then what is the total charge inside a cylinder of length L and radius r (a < r < b) ?

13. Feb 19, 2014

### Kot

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

14. Feb 19, 2014

### tiny-tim

why??

(and what is E ? and what happened to Q ?)

15. Feb 19, 2014

### Kot

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

16. Feb 19, 2014

### rude man

Start by setting up a Gaussian surface inside the inner wire ( r < a). What is the shape of the surface?

17. Feb 19, 2014

### Kot

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.

18. Feb 19, 2014

### rude man

Right. But what is Qenclosed?

19. Feb 19, 2014

### Kot

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?

20. Feb 19, 2014

### rude man

Well, where do you think the charges are located on the inner wire?