Capacitance and Max Potential Difference

[sonastylol]

Homework Statement

(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

Homework Equations

I THOUGHT C = \epsilon0*A all divided by d

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 can't get part a correct. Can anyone give me the equation for part b as well?Thank you so much PhysicsForums!

 

[kamerling]

Homework Statement

(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

Homework Equations

I THOUGHT C = \epsilon0*A all divided by d

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

 

[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/

 

[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)