Capaciatance of two parallel wires

In summary: I don't know what expression you would use in that case, but if the wires are relatively close then the voltage would be smaller.
  • #1
Brian-san
45
0

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
[tex]\vec{E}=-\nabla\Phi[/tex]

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:
[tex]\vec{E_1}=\frac{2\lambda}{r_1}\vex{\hat{r_1}}, \vec{E_2}=\frac{-2\lambda}{r_2}\vex{\hat{r_2}}[/tex]

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

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

Going from a1 to d-a2, gives the potential difference (I assume this is safe, since d is much greater than both wire radii)
[tex]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})[/tex]

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

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

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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  • #2
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
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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  • #4
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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  • #5
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
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?
 

What is capacitance?

Capacitance is the ability of a system to store electrical charge. It is measured in farads (F) and is directly proportional to the amount of charge that can be stored for a given potential difference.

How is capacitance calculated for two parallel wires?

The capacitance of two parallel wires can be calculated using the formula C = (ε0 * εr * A)/d, where ε0 is the permittivity of free space, εr is the relative permittivity of the material between the wires, A is the area of overlap between the wires, and d is the distance between the wires.

What factors affect the capacitance of two parallel wires?

The capacitance of two parallel wires is affected by the distance between the wires, the area of overlap between the wires, and the relative permittivity of the material between the wires. It is also affected by the length and thickness of the wires.

How does capacitance change if the distance between the wires is increased?

If the distance between the wires is increased, the capacitance decreases. This is because the electric field between the wires is weaker, resulting in less charge being stored for a given potential difference.

Why is the capacitance of two parallel wires important?

The capacitance of two parallel wires is important because it is a fundamental property of electrical circuits and is used in many applications such as capacitors, transmission lines, and high-speed data communication. It also plays a crucial role in the design and performance of electronic devices.

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