# Homework Help: Capacitance of coaxial cable

1. Dec 8, 2008

### jumpboy

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

http://smg.photobucket.com/albums/v68/jumpboyb/?action=view&current=scan0001.jpg" [Broken]
Pretty much looking for a general equation for capacitance of a coaxial cable per meter length.

given:
-dielectric material
-outer conductor is grounded (V = 0V)
2. Relevant equations
C = Q/V
C=A/d$$\epsilon$$0

3. The attempt at a solution
used the second equation but won't work since its for a parallel plate capacitor.

thats it! I'm lost on this one

Last edited by a moderator: May 3, 2017
2. Dec 8, 2008

### Harenil

I think the best way to approach this problem is with Gauss's law. You need to set an arbitrary charge (it will vanish in the end.). Use a cylinder as your Gaussian surface. Hope this helps

3. Dec 8, 2008

### jumpboy

Still stuck. Here's what I did with Gauss' Law in mind. (btw, the answer is (2pi$$\epsilon$$)/ln(R2/R1) and thats not pi to the epsilon not power btw.)

So, with Gauss, you have integral of E dot dA which equals Q/$$\epsilon$$.

Capacitance - Q=VC...so subsitute that back into the gauss equation so you have....

EA = VC/epsilon $$\Rightarrow$$ C= EA[epsilon]/V

Electric potential is the negative rate of change of a electric field...ie E = -dV/ds = -V/r where r = R2 - R1

C=-rA[epsilon]=-2pi(r)3[epsilon]

and yea....nothing. I see I need to integrate something, but idk where or what logic to use for it.

4. Dec 8, 2008

### Defennder

To get V you must integrate the expression for E with respect to dr. You can't just use E = -V/r which is true only for the simplest 1D case.

5. Dec 8, 2008

### Harenil

Right, try adding the integral of E dot dl or dr (whatever you guys call it.) into your arsenal.

6. Oct 19, 2010

### Jimboslice

the expression you want to integrate is dV = -[Q/(2Pi*R)] with limits from R1 to R2 (inner to outter). This is obtained through gauss's law, saying that EA = Qin/Episilon, solving for E and substituting for the original integral of dV = -[E dot ds]. The area "A" is 2*Pi*r because after the length is taken into acount, it will account for the surface area.