# Capacitance of coaxial cable

## Homework Statement

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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)

## Homework Equations

C = Q/V
C=A/d$$\epsilon$$0

## 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

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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

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.

Defennder
Homework Helper
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.

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

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.