# Physics of the Coaxial Cable

## Homework Statement

"A coaxial cable consists of an inner wire and a concentric cylindrical outer conductor. If the conductors carry equal but opposite charges, show that there is no net charge on the outside of the outer conductor.

Gauss's Law

## The Attempt at a Solution

I tried to approach this problem as logically as possible before laying any numbers to paper or manipulating equations, but was not able to even get that far. Our teacher brushed over very quickly the concept of uniform volume charge distribution and conductors in electric fields. I know there can be no electric field within a conductor. Therefore, if we assume the wire carries a negative charge, then that means there is an excess of electrons that will move themselves to the surface of the conductor, correct? ("If a conductor in electrostatic equilibrium carries a net charge, it must reside on the conductor surface"). So we have a negative field resulting in the cavity thus far. However, if we assume the outside cylindrical shell to be positive in charge, then there will be a positive net charge on the surface of the shell, both on the inside by the wire and the outside as well, correct? Now because of uniform charge distribution, the charge on the inside of the shell will in effect cancel its own field out, still leaving a negative field inside the cavity, but how is the net charge on the outside of the shell in any way negated by the wire? It has to have a positive net charge according to Gauss's law? Thanks for any tips, suggestions or hints!

-Confused student

## Answers and Replies

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learningphysics
Homework Helper
Hint... looking at the outer conductor... what is the field inside that conductor... what is the field outside of that conductor... uses gauss' law and symmetry for this part...

Then finally... when you know the field inside and outside the conductor... do gauss law using a tiny volume that cross the outer boundary of this conductor...

learningphysics
Homework Helper
So we have a negative field resulting in the cavity thus far. However, if we assume the outside cylindrical shell to be positive in charge, then there will be a positive net charge on the surface of the shell, both on the inside by the wire and the outside as well, correct?
No... you're right about the inside, but not the outside.

Hint... looking at the outer conductor... what is the field inside that conductor... what is the field outside of that conductor... uses gauss' law and symmetry for this part...

Then finally... when you know the field inside and outside the conductor... do gauss law using a tiny volume that cross the outer boundary of this conductor...
Thanks for the speedy reply!

Actually INSIDE the material, there is no field. In the cavity of the outer conductor, there is a negative field resulting from the wire, E = k(-q)/r^2, r being the distance from the wire and -q being the charge on the wire. If the outer conductor has a charge of +q, then the total enclosed charge by a gaussian surface surrounding the entire cable would be 0, meaning there can be no field outside of the cable, and if theres no field outside the cable, there can be no net charge on the outer surface?

If that is the logic, then what is going on in terms of electron arrangement and such to make this occur? I think I get it, but it's still a little hazy and I want to make sure I fully understand the concept.

learningphysics
Homework Helper
Actually here's another way to think about it... actually simpler this way... since the inner and outer conductor have equal charges... and since the field inside the outer conductor is 0... that means the net charge inside a gaussian surface within the outer conductor is 0... That means that all the positive charge must be on the inner surface... hence the outer surface must have a net charge of 0, and by symmetry a charge density of 0...

I understand! Thanks for the explanation!

learningphysics
Homework Helper
I understand! Thanks for the explanation!
You're welcome.