Gauss' Law on a Coaxial Cable

  • #1

Homework Statement


A coaxial cable consists of two concentric cylindrical conductors as shown here: http://puu.sh/luFZx/7f3e1ccb07.png . The inner cylinder has a linear charge density of +λ1, meaning that each meter of the cable will have λ C of charge on the inner cylinder. The outer cylinder has a linear charge density of +λ2. Various regions of this structure are labelled as follows: (1) inside the inner cylinder, (2) within the inner conductor, (3) between the cylinders, (4) within the outer conductor, and (5) outside the cable.

a. Find an expression for the electric field in each of the five regions as a function of the distance from the center of the cable, r.
b. If the inner cylinder has a charge density of +10 nC/m and the outer cylinder has a charge density of +15 nC/m, what is the magnitude of the electric field at 30 mm from the center? (Assume that r=30 mm is located in section #3).

***This is in electrostatic equilibrium btw.

Homework Equations



E = λ/(2πrϵ0)

3. The Attempt at a Solution

For part A, I tried drawing a Gaussian surface and using equations.. but then I realized it was a conductor in electrostatic equilibrium. We were taught a week or two ago that the electric field of a conductor in electrostatic equilibrium would always equal 0, as would any holes in it. So why don't the first four regions have E = 0?

I was waiting to do part B til I finish part A.
 

Answers and Replies

  • #2
264
26
The electric field inside of a conductor (in equilibrium) is zero because if
there were a field inside of the conductor an electric current would flow.
So E is immediately zero inside the conductors; regions 2 and 4.
Now you need to apply Gauss' Law for the other regions by examining the net charge
enclosed within the appropriate Gaussian (imaginary) surface (here cylinders).
 
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  • #3
Thanks! I guess what we were taught applied to something else.
For part B, I just plug the two values into the equations I get and add them?
 
  • #4
264
26
What is the enclosed charge if you draw a Gaussian cylinder thru section 3?
Gauss' Law refers to enclosed charges.
 
  • #5
What is the enclosed charge if you draw a Gaussian cylinder thru section 3?
Gauss' Law refers to enclosed charges.

I got E = +λ1/(2πrϵ0) for region 1, E = +λ2/(2πrϵ0) for region 2, and E = N (weird N i can't draw sorry) / 2ϵ0 for region 3. I have a feeling this is wrong but I never really learned this and I am having a hard time understanding what part of it is screwy.
 
  • #6
264
26
Gauss Law says that Flux = Q / e0 where e0 is epsilon 0 and Q is the "Total" enclosed charge.
Then E * A = Q / e0 = y L / e0 where y here represents the linear charge density
The Area of a Gaussian cylinder is just A = 2 * pi * r * L
That gives the equation you need for this problem which you are using for E.
Now Q = y * L which is the total charge on a cylinder of length L.
Region 1 contains no charge so the electric flux everywhere within that region (within any imaginary Gaussian cylinder) must be zero.
Now look at flux within any of the other regions to find the field within each of these regions.
Some things to remember are that there cannot be in an electric field within an electrostatic conductor (else a current would flow)
and that electric field lines start on positive charges and end on negative charges so the charge in region 3 would have to
induce an equal magnitude of charge on the inner surface of region 5.
 
  • #7
Region 1 contains no charge so the electric flux everywhere within that region (within any imaginary Gaussian cylinder) must be zero.
I remember reading that any hole within a conductor in electrostatic equilibrium has an electric field that must equal zero unless it has a charge in it. This one does have a charge. But would it still equal zero?

Thanks for all the help so far by the way!
 
  • #8
264
26
As I read the problem any charge was on the cylinders themselves and no charge in regions 1, 3, or 5.
So regions 1, 3, and 5 are just empty space.
 

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