What is the Capacitance per Unit Length of Two Parallel Wires?

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The discussion focuses on calculating the capacitance per unit length of two parallel wires using Gauss' Law and potential difference equations. The user derives the potential difference and capacitance formula, ultimately finding a diameter for the wires that achieves a specified capacitance. There is clarification on the nature of linear charge densities on the wires, noting that one can be positive and the other negative if currents flow in opposite directions. The user expresses uncertainty about their assumptions but receives confirmation that their mathematical approach is correct. The conversation also touches on a related problem involving closer wire separation, prompting a request for further assistance.
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Homework Statement


Two long cylindrical conductors (wires) of radii a1 and a2 respectively are separated by a distance d >> a1, a2. Find the capacitance per unit length for the system. If d=0.5cm and a1=a2=a, what must be the diameter of the wires to give a capacitance per unit length of 9x10^-3? (CGS units)

Homework Equations


Gauss' Law
Q=CV
\vec{E}=-\nabla\Phi

The Attempt at a Solution


Assuming wire one has linear charge density λ and wire 2 has linear charge density -λ, then by Gauss' Law:
\vec{E_1}=\frac{2\lambda}{r_1}\vex{\hat{r_1}}, \vec{E_2}=\frac{-2\lambda}{r_2}\vex{\hat{r_2}}

Integrating with respect to the radial coordinates gives:
\Phi_1=2\lambda ln(r_1), \Phi_2=-2\lambda ln(r_2)

So at any point, the potential is a superposition of these two,
\Phi=2\lambda ln(\frac{r_1}{r_2})

Going from a1 to d-a2, gives the potential difference (I assume this is safe, since d is much greater than both wire radii)
V=2\lambda ln(\frac{d-a_2}{a_2})-2\lambda ln(\frac{a_1}{d-a_1})=2\lambda ln(\frac{(d-a_1)(d-a_2)}{a_1a_2})

Finally,
C=\frac{1}{2ln(\frac{(d-a_1)(d-a_2)}{a_1a_2})}

For the above case in the problem, I get
a=\frac{d}{1+e^{\frac{1}{4C}}}

and a solution of a=4.318x10^-13 cm

Does everything look about right? I've never dealt with capacitance in this manner before, usually it was stuff like parallel plates, concentric cylindrical/spherical shells, or circuits. Thanks.
 
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It looks about right. I redid the capacitance by superposing the electric fields first, then integrating and my expression for C came out the same.
 
I see no problem in the mathematics process, but there is one thing I did not fully understand- the linear charge density on one wire is positive while the linear charge on the other wire is negative. Also both wires have the same linear charge density.

As far as I know the linear charge density can be +ve and -ve when the current flowing through the conductors are flowing in the opposite direction. And the linear charge density on both the wires will have equal magnitude if equal current is flowing through them.

If these all assumption are correct, which I don't know for sure, the answer should be correct.

http://www.esd-inc.com"
 
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georgeryder9 said:
As far as I know the linear charge density can be +ve and -ve when the current flowing through the conductors are flowing in the opposite direction. And the linear charge density on both the wires will have equal magnitude if equal current is flowing through them.
http://www.esd-inc.com"

Current carrying wires are uncharged.
 
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Thanks, I'm glad it came out right on the first go. It's taken a while to get back into working mode since I had a year off before grad school, so I'm always second guessing my answers.
 
I am currently having a very similar problem, except that I can't assume the wires are relatively distant. What expression would i have for the voltage in that case?
 
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