V to get C. Calculating Max Capacitance of Coaxial Capacitor

[seang]

Homework Statement

Consider a coaxial capacitor. If the inner radius is 5mm, the length is 3 cm, and the voltage rating of the capacitor is 2kV, what is the maximum capacitance if the dielectric between the two conductors is 2.3, and E breakdown is 15MV/m.

Homework Equations

<br /> E_r = \frac{\rho_s a}{\epsilon_r r}<br />
a is the inner radius and r is the radius between the inner and outer conductors at whch we want to find the E field. Since the E field will strongest nearest to the inner conductor, I'm using:
<br /> E_{r,breakdown} = \frac{\rho_s}{\epsilon_r}<br />
This is because I want to find the surface charge density nearest to the inner conductor that will !begin to breakdown the dielectric. I figure we never even want to start to breakdown the dielectric. I hope this makes sense.
<br /> V = \frac{\rho_s a}{\epsilon_r}[ln(a)-ln(b)]<br />
with V = 2k I used this to find b (outer radius)
Finally,
<br /> C = \frac{2\pi\epsilon_r h}{ln(b/a)}<br />

The Attempt at a Solution

I derived all the above equations, and pretty much plugged in numbers, and I'm getting about 1.6 farads which seem wrong.

Actually, thinking about it. I don't see why I can't just use gauss' law to find the enclosed charge (with E = E breakdown) and then divide by 2k

 

[seang]

I got it. If anyone is interested, the mistake I made was deriving the equations by simply replacing the permittivity of free space with the relative permittivity. In fact you are supposed to replace the permittivity of free space with the permittivity of free space times the relative permittivity.