Capacitance of Parallel cylinders

Click For Summary
The discussion focuses on calculating the capacitance per unit length of two parallel cylindrical conductors using the formula C = πε₀(ln(d/a))⁻¹, where 'a' is the geometric mean of the radii. A specific problem is posed to determine the necessary gauge wire diameter for a two-wire transmission line with a capacitance of 1.2 × 10⁻¹¹ F/m at various separations. Participants suggest using Gauss' Law to find the electric field between the cylinders and then derive the potential difference to calculate capacitance. The original poster expresses confusion about starting the problem and seeks guidance on how to proceed. The thread emphasizes the importance of understanding electric fields and potential differences in capacitance calculations.
jfy4
Messages
645
Reaction score
3

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,
 
Physics news on Phys.org
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.
 

Similar threads

Replies
2
Views
2K
Replies
17
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
5
Views
2K
Replies
5
Views
5K