Capacitance and Max Potential Difference

1. Feb 26, 2008

sonastylol

1. The problem statement, all variables and given/known data
(a) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.90 cm2 and insulation thickness of 0.0600 mm.

(b) Determine the maximum potential difference that can be applied to the capacitor.
kV

2. Relevant equations
I THOUGHT C = $$\epsilon$$0*A all divided by d

3. The attempt at a solution

I did 8.85x10^-12 * 1.90 / .06x10^-3
it came up wrong.

I have to answer in pF btw.

I also have no idea how to answer part b since I cant get part a correct. Can anyone give me the equation for part b as well?

Thank you so much PhysicsForums!

2. Feb 26, 2008

kamerling

This is only valid if there is vacuum between the plates. (or air with a very small error).

3. Mar 7, 2009

Edict

C = epsilon_0 * A * K / d where K is the dielectric constant of Teflon (2.1)
(so just multiply the answer you got by 2.1)

the dielectric breakdown of Teflon according to the website below is 60 kV/mm so I think your answer to b) is:
V = Ed --> (60 kV/mm) * (0.06 mm) = 3.6 kV. Could you please check this answer for me? I don't want to use up my last guess.

got my info from:
http://hypertextbook.com/physics/electricity/dielectrics/

4. Mar 9, 2009

Edict

I was right. V = Ed ; where E is the electric field needed to conduct teflon and d is the distance between the parallel plates. the electric field needed to break down teflon is 60 kV/mm so (60 kV/mm) * (0.06 mm) = 3.6 kV. You must raise the potential difference of the parallel plates to 3600 Volts in order to cause the teflon to conduct and the capacitor to short circuit. This is strange because part a) of the question asks you to find the capacitance and that value is not needed for part b)