Gauss' Law for a coaxial cable

  • #1
122
4

Homework Statement



I have attached the problem

Homework Equations



E*A = Qenc/ E0

The Attempt at a Solution



At the moment I am looking at the problem more conceptually and seeing what is happening at each point and I wanted to know If I was on the right track.

r<a

As all charge would reside on the outside of the inner conductor there would be no electric field

a<r<R

The electric field would be q/(2pi*r*E1*E0)

R<r<b

The electric field would be q/(2pi*r*E2*E0)

b<r<c

This is the part that has me puzzled , I thought the inner conductor charge(q) and outer conductor charge (-q) would cancel each other , so there would be no field present? Is this part mean to represent the outer rubber shielding of a coaxial cable?

c<r

Would the electric field also be zero outside of the cable due to the two charges cancelling?


Thanks
 

Attachments

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,960
3,315
OK for the fields you've derived so far.

Homework Statement


b<r<c

This is the part that has me puzzled , I thought the inner conductor charge(q) and outer conductor charge (-q) would cancel each other , so there would be no field present? Is this part mean to represent the outer rubber shielding of a coaxial cable?
For b<r<c you are inside the outer conductor.

c<r

Would the electric field also be zero outside of the cable due to the two charges cancelling?
No, I don't think so. Can you see why the net charge of the system is not zero? [EDIT: The field might indeed be zero outside! To get the field outside of the cable use the same method you used for inside.]
 
Last edited:
  • #3
122
4
OK for the fields you've derived so far.


For b<r<c you are inside the outer conductor.



No, I don't think so. Can you see why the net charge of the system is not zero?
Do the charges not cancel as at any point beyond b you will be at two different distances from the respective charges , hence the electric fields will have a different magnitude?

For b<r<c would you add the two electric fields?
 
  • #4
TSny
Homework Helper
Gold Member
12,960
3,315
Do the charges not cancel as at any point beyond b you will be at two different distances from the respective charges , hence the electric fields will have a different magnitude?
The two conductors have equal and opposite charge. But there could be some induced charge at certain locations of the dielectric. I haven't worked it out.
Actually, for the field outside the outer conductor, I now think that the field might be zero. I'll try to find time to work on it. But I think you should think about any induced charges in the dielectrics.
For b<r<c would you add the two electric fields?
For b<r<c you are inside the outer conducting sheet. What do you know about E inside a conductor?
 
  • #5
TSny
Homework Helper
Gold Member
12,960
3,315
OK, you don't need to worry about the values of the induced charge on the surfaces of the dielectrics in order to answer the questions. You have already worked out the fields inside the dielectrics. You just need to find the field outside the outer conductor. You should be able to do this using the same method that you used for the other regions.
 
  • #6
122
4
OK, you don't need to worry about the values of the induced charge on the surfaces of the dielectrics in order to answer the questions. You have already worked out the fields inside the dielectrics. You just need to find the field outside the outer conductor. You should be able to do this using the same method that you used for the other regions.
Thanks , I have managed to solve the numerical parts as like you say , they are concerned with the fields in the dialectrics.I have checked the answers and the electric fields are shown to be 0 at the point r>b but I don't understand why this would be the case , is it as simple as drawing a Gaussian surface at r=c and saying the enclosed charge would be +q and -q therefore 0?
 
  • #7
TSny
Homework Helper
Gold Member
12,960
3,315
Thanks , I have managed to solve the numerical parts as like you say , they are concerned with the fields in the dialectrics.I have checked the answers and the electric fields are shown to be 0 at the point r>b but I don't understand why this would be the case , is it as simple as drawing a Gaussian surface at r=c and saying the enclosed charge would be +q and -q therefore 0?
There are two separate regions where r > b that you should consider. First, the region inside the outer conducting sheet: b < r < c. Points in this region are inside the conducting material of the sheet. So, deducing E and D in this region is easy.

The other region is r > c which consists of all points outside the cable. Here you can use your Gaussian surface and use the fact that the total charge enclosed is zero. Note that there are induced charges on the surfaces of the dielectrics as well as the "free" charge q and -q on the conductors. But, the total induced charge must be zero because the net charge of the dielectrics is zero.
 
  • #8
122
4
There are two separate regions where r > b that you should consider. First, the region inside the outer conducting sheet: b < r < c. Points in this region are inside the conducting material of the sheet. So, deducing E and D in this region is easy.

The other region is r > c which consists of all points outside the cable. Here you can use your Gaussian surface and use the fact that the total charge enclosed is zero. Note that there are induced charges on the surfaces of the dielectrics as well as the "free" charge q and -q on the conductors. But, the total induced charge must be zero because the net charge of the dielectrics is zero.
Could you expand on the first part ,where b < r < c , what is the significance of points in this region being inside the conducting sheet?

Thanks
 
  • #11
TSny
Homework Helper
Gold Member
12,960
3,315
OK, you are very welcome.
 

Related Threads on Gauss' Law for a coaxial cable

  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
9K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
6
Views
13K
Replies
5
Views
19K
  • Last Post
Replies
0
Views
9K
  • Last Post
Replies
0
Views
3K
Top