# Homework Help: Capaciatance of two parallel wires

1. Sep 29, 2009

### Brian-san

1. The problem statement, all variables and given/known data
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)

2. Relevant equations
Gauss' Law
Q=CV
$$\vec{E}=-\nabla\Phi$$

3. 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.

2. Sep 30, 2009

### kuruman

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.

3. Sep 30, 2009

### georgeryder9

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" [Broken]

Last edited by a moderator: May 4, 2017
4. Sep 30, 2009

### kuruman

Current carrying wires are uncharged.

Last edited by a moderator: May 4, 2017
5. Sep 30, 2009

### Brian-san

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.

6. Oct 6, 2009

### noodle_snacks

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?