Capacitance of two long parallel wire

[thedaydreamer]

Homework Statement

Consider two long parallel and oppositely charged thick wires of radius d with their central axes separate by a distance D apart. Obtain an expression for the capacitance per unit length of this pair of wires

Homework Equations

$$\oint \vec{E}\cdot d\vec{a} =\frac{Q_{enc}}{\epsilon_0}$$

$$Q = VC$$

$$V\ =\ -\int_0^d E\,dx\ =\ \frac{d\,Q}{\varepsilon\,A}$$

The Attempt at a Solution

I have the solution to this problem. It goes like this:
A problem P is considered at A distance r from the axis of the wire. Using Gauss's Law, the electric field at P is calculated and the the electric field due to the other wire is also calculated and then added. What I don't understand is which Gaussian surface is to be considered when calculating the field. Please help.

 

[Bhumble]

I would assume you are suppose to calculate the electric field in each section so that you can get the charge density.
$$E_{above} - E_{below} = \frac{\sigma}{\epsilon_0}$$

I'm assuming you use cylindrical coordinates and get a slice of the cylinder of length L to calculate the electric field then calculate the electric potential and blammo you have the capacitance.

 

[vela]

What level class is this for? The problem seems a bit complicated for an intro physics course.

 

[thedaydreamer]

Well this isn't exactly 'homework', this problem is taken from a physics competition.

^^ Actually, the solution to this problem says that " when the two wires form a capacitor charges reside only on the inner side;' Then Gauss Law has been used to calculate the field and then the electric potential and then they calculated the capacitance per unit length.

I'm assuming you use cylindrical coordinates and get a slice of the cylinder of length L to calculate the electric

The solution did not need 'a slice of the cylinder of length L' that, although I think it can done that way too. Basically I want to know which Gaussian surface to take for calculating the field at any of the two wires.

 

[vela]

To use Gauss's Law, you want to take advantage of a symmetry of the electric field. From the solution you describe, it sounds like they assumed a cylindrical symmetry for the field of each wire.

I'll note that, to me, the solution you have sounds incorrect.

 

[Bhumble]

I actually misinterpreted the problem anyway. I thought these were concentric cylinders for some reason.

$$E(r) = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \widehat{r}$$

$$\frac{q}{L} = \lambda$$

$$\widehat{r} = cos{\theta}$$

The amount of charge "seen" is where $$\theta$$ goes from 0 to $$\pi$$

This is only the Electric field due to one of the wires. And you still need to figure out what r is.

 

[Bhumble]

Hold on I think I figured out a better way to approach this problem.
$$\oint E \cdot da = \frac{Q_{enc}}{\epsilon_0}$$
$$Q_{enc} = \lambda L$$
$$\oint E \cdot da = E (2 \pi s L)$$
$$E = \frac{\lambda}{2 \pi \epsilon_0}$$

Again this is just for one of the wires.

 

[vela]

You're assuming the field is cylindrically symmetric about the axis of the wire, but the complication is that the wires have a finite size. The charge distribution on one wire will be affected by the presence of the other charged wire, which ends up breaking that symmetry.

 

[Bhumble]

They have a finite length but the problem is looking for capacitance per unit length and if the wire is sufficiently long enough we can just ignore the fringe fields at the end. Or do you mean a finite radial component so that this can't be treated as a line charge?
As for the effects of the other wire I don't see why you can't calculate the electric field of each independently and then combine them via superposition.

 

[vela]

In your application of Gauss's law, you've assumed a certain charge distribution on the wire, one that has a cylindrical symmetry. The field due to the other wire, however, will cause that charge distribution to change (because the wire has finite radial size) so you no longer have that symmetry.

 

[thedaydreamer]

Yeah, I too don't understand how the charge is distributed. Vela, I get your point, but I could not get what kind of charge distribution the solution refers to. The solution (which was very brief and without details) just mentioned the line which I mentioned above : "when the two wires form a capacitor charges reside only on the inner side"

Then went on to give mathematical relations that I'd give here only if I had learned LaTex by now.

So please help me visualize the field.

 

[vela]

What you know is that the surface of each wire is an equipotential, so what you want is a charge distribution that will give you circular equipotentials. It turns out two line charges will do that, but the locations of these line charges depends on the radius of the wires and distance between them. So your first step would be to find the field and potential due to two line charges, and then you want to adapt that to the configuration of the wires you've been given.