Capacitance and Max Potential Difference

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Homework Help Overview

The discussion revolves around determining the capacitance of a Teflon-filled parallel-plate capacitor and the maximum potential difference that can be applied to it. The problem involves concepts from electrostatics, specifically capacitance and dielectric materials.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the capacitance using the formula for a parallel-plate capacitor but expresses uncertainty about the correctness of their calculation. They also seek guidance on how to approach the second part of the problem regarding the maximum potential difference.
  • Some participants clarify the formula for capacitance by including the dielectric constant for Teflon and discuss the implications of dielectric breakdown for the maximum potential difference.
  • Questions arise regarding the necessity of the capacitance value for calculating the potential difference, with some participants expressing confusion about the relationship between the two parts of the problem.

Discussion Status

The discussion is active, with participants providing insights into the formulas and concepts involved. Some guidance has been offered regarding the use of the dielectric constant and the breakdown voltage of Teflon, but there is no explicit consensus on the calculations or the necessity of the capacitance for part b.

Contextual Notes

Participants note the importance of the dielectric constant in the calculations and question the assumptions made about the capacitor's setup. There is also mention of the need to express answers in specific units, such as pF for capacitance.

sonastylol
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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 = [tex]\epsilon[/tex]0*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!
 
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sonastylol said:

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 = [tex]\epsilon[/tex]0*A all divided by d

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

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