# Gauss's Law for cylindrical metal tube enclosing a wire (coaxial cable)

## Homework Statement

Immediately outside a very long cylindrical wire of radius r1 = 1mm, the electric field is 40kV/m directed towards the wire's surface.
A hollow cylindrical metal tube with inner radius r2 = 3mm is now placed around the wire, to form a coaxial cable. What will be the charge per unit length on the inner surface of this tube? Explain

## Homework Equations

The main equation that I've been using is:

E=$$\frac{\lambda}{2\pi\epsilon_{0}}$$

## The Attempt at a Solution

-In a previous part of the solution, lambda (the charge per unit length) of the wire was found to equal -2.224 nC/m
-I think that this therefore means the charge on the inner surface of the cylindrical metal tube must therefore be positive?
-My intuition says that the charge per unit length on the inner surface of this tube should be the same as that on the wire but different signs. I think that because for any given portion of the coaxial cable, the length of the wire and metal tube will both be the same. The charge enclosed will also be the same for both the wire and metal tube because only the electric field strength diminishes with radius, not the charge enclosed.
-My intuitive answer would therefore be +2.224 nC/m but I'm far from definite on it.

Is my line of thinking correct or am I a bit off?

Thanks in advance.

## Answers and Replies

I believe your intuition is wrong. The best thing to do is to draw your cylindrical guassian surfaces and do the calculations.

The field equation you have is not correct for a long cylinder. The field should depend on the distance or radius from the cylinder.

The electric field immediately outside a charged surface is the surface charge density divided by the permittivity of free space. Since you already have the electric field, you can solve for the surface charge density (and since the electric field is stated to be directed inward, you can already figure out if the charge is + or -). From Gauss's laws, the inner wire can be looked at as being thin with negligible radius. Using the surface charge density, you can condense the value per, say, every meter of length so that you may come up with a linear charge density. For the charge on the inner surface of the tube...since you know the charge of the wire, then that would imply that the charge that is on the inner surface of the cylindrical tube surrounding the wire would be those that are attracted to the wire and the charge on the outer surface of the tube would be those that repel from the wire (you seem to be right in that its positive).

Ok, I back off my original statement. Sorry for the confusion. Your intuition is correct. I was using the field just inside the outer cylinder in my guassian surface, but including both inner and outer charge densities. This was not correct. A guassian surface that includes both charge densities must be placed inside the outer metal conductor.

So this means that the charge per unit length of the inner surface is 1/3 the charge per unit length on the wire?

This is because the same equation still stands which can be proven when you consider E = surface charge density/\epsilon_{0}

E=$$\frac{\lambda}{2\pi\epsilon_{0}}$$

Since the electric field is at the inner surface is 1/3 that of the field just at the wire surface (because E diminishes linearly as radius increases) then \lambda is 1/3 * 2.224 = 0.741 nC/m?

Is this right? If not, then how do you specifically get to an answer because I've run out of ideas that sound plausibly correct.

Defennder
Homework Helper
Is this supposed to be charge/unit area instead? Because the wire has a definite radius.

Dick
Science Advisor
Homework Helper
So this means that the charge per unit length of the inner surface is 1/3 the charge per unit length on the wire?

This is because the same equation still stands which can be proven when you consider E = surface charge density/\epsilon_{0}

E=$$\frac{\lambda}{2\pi\epsilon_{0}}$$

Since the electric field is at the inner surface is 1/3 that of the field just at the wire surface (because E diminishes linearly as radius increases) then \lambda is 1/3 * 2.224 = 0.741 nC/m?

Is this right? If not, then how do you specifically get to an answer because I've run out of ideas that sound plausibly correct.

Your original intuition was correct. Though you may not have expressed it all that clearly. The field inside of the hollow metal conductor is zero. So the sum of the enclosed charges is zero. Can you correct your explanation??

Dick
Science Advisor
Homework Helper
Is this supposed to be charge/unit area instead? Because the wire has a definite radius.

Right. You'll need to change your charge per unit area to a charge per unit length before you apply the 'total enclosed charge equals zero' notion. But I was guessing you already did that since you referred to a previous problem.

Dick
Science Advisor
Homework Helper
Hey, can we go back to your 'main formula'? If lambda=coulomb/m then the units are wrong. Can we start from the beginning?

My understanding of his problem is that there is a solid inner wire (assumed to be perfect conductor) for which a radius is given. Surrounding this wire is a solid metal tube (assumed to be a perfect conductor) for which the inner radius is given. The hollow part of the description is meant to convey the idea that there is a gap between the wire and tube.

I agree with the OP's answer for linear charge density, although I don't think he got it from applying the equation he gave (something is missing in it).

I also agree (after a initial wrong answer) with his original intuition that the linear charge density on the inside of the tube is equal but opposite sign of the wire linear charge density.

I would suggest to the OP that he think about drawing a guassian surface that encloses both the inner wire with known linear charge density and the unknown charge on the inner wall. In guass problems you have to multiply (or integrate) the E field by the area of the surface. In this case, think about what the E field must be on the guassian surface that is outside both charge sources.

Dick
Science Advisor
Homework Helper
I agree with you. The answer is correct. The formula is wrong. It's missing an 'r'.