What Is the Electric Field of a Coaxial Cable?

[Kot]

Homework Statement

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.

Homework Equations

Gauss' Law ∫EdA = Q_in/ε_o

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?

 

[tiny-tim]

Hi Kot!

Since the cable and metal sheath have the same charge per length, wouldn't they both have the same charge and cancel out?

For r > b, yes

So the electric field there is … ?

Could someone explain how I can find the Gaussian surface for the first region?

(for r < a) Why don't you just use a cylinder of radius r ?

 

[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?

 

[tiny-tim]

Hi Kot!

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

But the question says that the charges per length are equal. …

The sheath and wire are given the same charge per length +Q/L.

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?

Just call it r.

What do you get?​

 

[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 / ε_o

E2∏rL = ρ∏r²L / ε_o

E = ρ / 2 ε_o

 

[tiny-tim]

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

charge per length is what it says

the radius (a or b) has nothing to do with it

ρ∏r²L / ε_o

what is ρ ?

you seem to be calculating a volume

the charge is a surface charge, on two cylindrical shells

 

[Kot]

Oops,

EA = σA / ε_o

E(2∏rL) = σ(2∏rL) / ε_o

E = σ / ε_o

where σ is the surface charge density of the cylinders. Is this correct?

 

[tiny-tim]

σ(2∏rL)

what is this supposed to be?

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

 

[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 = Q_enclosed / ε_o . I replaced Q_enclosed with σA.

 

[tiny-tim]

But the surface charge depends only on L …

The sheath and wire are given the same charge per length +Q/L.

(and now I'm off to bed :zzz:)

 

[Kot]

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

 

[tiny-tim]

(just got up :zzz:)

If …

The sheath and wire are given the same charge per length +Q/L.

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

 

[Kot]

Would the total charge inside a cylinder be Q_enclosed = E2∏rLε_o ?

 

[tiny-tim]

why??

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

 

[Kot]

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

 

[rude man]

Homework Statement

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.

Homework Equations

Gauss' Law ∫EdA = Q_in/ε_o

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?

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

 

[Kot]

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

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

 

[rude man]

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

Right. But what is Qenclosed?

 

[Kot]

Right. But what is Qenclosed?

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

 

[rude man]

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

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

 

[rude man]

Remember, the cable's inner and outer conductors are just that - conductors.

 

[Kot]

Remember, the cable's inner and outer conductors are just that - conductors.

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

 

[rude man]

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

Yes, but my question was, where are the charges located? The problem said you have Q coulombs per meter charge sitting somewhere in/on the inside wire.

 

[Kot]

Yes, but my question was, where are the charges located? The problem said you have Q coulombs per meter charge sitting somewhere in/on the inside wire.

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

 

[rude man]

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

No. The charges all repel each other & so want to get as far away from each other as possible. So where would that be?

 

[Kot]

No. The charges all repel each other & so want to get as far away from each other as possible. So where would that be?

On the surface of the Guassian cylinder?

 

[rude man]

On the surface of the Guassian cylinder?

The Gaussian cylinder is a construct. You can place a gaussian cylinder any way you want, so that can't be the answer. Besides, the charges have to rest on some kind of matter, not some fictitious surface. Specifically, they have to be somewhere on/in the wire, since that's what you were given.

 

[Kot]

The Gaussian cylinder is a construct. You can place a gaussian cylinder any way you want, so that can't be the answer. Besides, the charges have to rest on some kind of matter, not some fictitious surface. Specifically, they have to be somewhere on/in the wire, since that's what you were given.

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?

 

[rude man]

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?

Contain it, not hold it.

Your answer is now correct: the charges all reside on the outer surface of the inner wire at r = a.

So now place your gaussian cylinder with radius r < a, how much charge does it hold?

 

[Kot]

Contain it, not hold it.

Your answer is now correct: the charges all reside on the outer surface of the inner wire at r = a.

So now place your gaussian cylinder with radius r < a, how much charge does it hold?

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.

 

[rude man]

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.

Bingo again! So what by your formula for E does that make the E field at r < a?

 

[rude man]

Back in 1 hr.

 

[Kot]

Bingo again! So what by your formula for E does that make the E field at r < a?

At r < a, E = 0.

 

[rude man]

At r < a, E = 0.

Right. And that squares with what you said before about the E field inside a conductor being zero, doesn't it.

Now, put your gaussian surface in the region a < r < b. What is E now?

 

[Kot]

Right. And that squares with what you said before about the E field inside a conductor being zero, doesn't it.

Now, put your gaussian surface in the region a < r < b. What is E now?

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).

 

[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.