Capacitance of Parallel cylinders

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SUMMARY

The capacitance per unit length of two parallel cylindrical conductors is given by the formula C = πε₀(ln(d/a))⁻¹, where 'a' is the geometric mean of the two radii and 'd' is the separation distance. To achieve a capacitance of 1.2 × 10⁻¹¹ F/m with wire separations of 0.5 cm, 1.5 cm, and 5.0 cm, specific wire diameters must be calculated based on this formula. The discussion emphasizes the application of Gauss' Law to derive the electric field and potential difference necessary for capacitance calculations.

PREREQUISITES
  • Understanding of capacitance and its formulas, specifically C = Q/φ.
  • Familiarity with Gauss' Law and its application to cylindrical conductors.
  • Knowledge of logarithmic functions and geometric means.
  • Basic principles of electric fields and potential differences.
NEXT STEPS
  • Calculate the necessary wire diameters for different separations using the capacitance formula.
  • Study the derivation of electric fields for cylindrical conductors using Gauss' Law.
  • Explore the implications of varying wire separation on capacitance in transmission lines.
  • Investigate the effects of dielectric materials on capacitance in cylindrical configurations.
USEFUL FOR

Electrical engineers, physics students, and professionals involved in designing transmission lines or studying electromagnetic fields will benefit from this discussion.

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Homework Statement


Two long, cylindrical conductors of radii a_1 and a_2 are parallel and separated by a distance d, which is large compared with either radius. Show that the capacitance per unit length is given approximately by
<br /> C=\pi\epsilon_0 \left(\ln\frac{d}{a}\right)^{-1}<br />
where a is the geometric mean of the two radii.
Approximately what gauge wire (state diameter in millimeters) would be necessary to make a two-wire transmission line with a capacitance of 1.2\times10^{-11} F/m if the separation of the wires was 0.5 cm? 1.5 cm? 5.0 cm?


Homework Equations


<br /> C=\frac{Q}{\phi}<br />
<br /> \phi=-\int \vec{E}\cdot d\vec{l}<br />


The Attempt at a Solution


Honestly, I'm pretty lost on this... I'm not sure how to construct this from scratch, my only work is from backwards, and I feel shady about it...

Here it is:
We know that C=Q/\phi and we are given the capacitance per unit length C&#039;=C/l, so I did this
<br /> \phi=\frac{q}{C&#039; l}=\frac{\lambda}{\pi\epsilon_0}[\ln(d)-\ln(\sqrt{a_1 a_2})]<br />
with \lambda the charge per unit length, which can be written as
<br /> \phi=-\int_{d}^{\sqrt{a_1 a_2}} \frac{\lambda}{\pi\epsilon_0 }\frac{1}{r}dr<br />
Then the electric field would be
<br /> \vec{E}=\frac{\lambda}{\pi\epsilon_0 r}\hat{r}<br />
Now if I were to continue with this (and I'm not committed to this :) ), how could I manage to write this as a superposition of two cylindrical electric fields from two different conductors with equal and opposite charges...?


Can someone give me a hand here?

Thanks in advance,
 
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Find the electric field of the cylinders by using Gauss' Law, and use that to find the potential difference between the two cylinders. Then you can substitute your result into the formula for capacitance.
 

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