Calculate Dielectric Strength Breakdown Voltage

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Homework Help Overview

The discussion revolves around calculating the dielectric strength breakdown voltage for a system involving two parallel plates with an applied voltage. Participants are exploring the relationship between capacitance, dielectric strength, and the properties of the materials involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for capacitance and its dependence on area, dielectric strength, and distance. Questions arise regarding the calculation of dielectric strength without specific information about the fluid between the plates. There is also an exploration of the units involved in measuring dielectric strength.

Discussion Status

Participants are actively engaging with the problem, raising questions about the assumptions made regarding the dielectric material and the calculations involved. Some guidance has been offered regarding the relationship between voltage, distance, and electric field strength, but no consensus has been reached on the specific values or methods to use.

Contextual Notes

There is a noted lack of specification regarding the fluid between the plates, which is central to determining the dielectric strength. Participants are also navigating the challenge of converting units for dielectric strength as presented in their resources.

theman408
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I know it has to do something with the Breakdown voltage but I've looked everywhere on my book and i have no idea how to calculate it.
 
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i would reallly like to know, without this i can't do my problem.

this is all i need.
 
theman408 said:
i would reallly like to know, without this i can't do my problem.

this is all i need.

What is the problem exactly?
 
It's two parallell plates which 12 volts are applied, i already calculated the capacitance with the Area and distance between the plates. Now i must calculate it's dielectric strength.
 
theman408 said:
already calculated the capacitance with the Area and distance between the plates. Now i must calculate it's dielectric strength.

The capacitance depends on the area, dielectric strength and distance. How did you find C with only two?
 
Sorry, forgot to specify. i used the formula of

C= eoerS/D

where Eo is the permitivity constant, Er is the relative permitivity which was given in the exercise, S is the area of the surface and D is the distance between the plates.

The Problem is which is the min. dielectric strength that it has.
 
theman408 said:
Sorry, forgot to specify. i used the formula of

C= eoerS/D

where Eo is the permitivity constant, Er is the relative permitivity which was given in the exercise, S is the area of the surface and D is the distance between the plates.

The Problem is which is the min. dielectric strength that it has.

but the dielectric strength depends on the fluid between the plates. The dielectric strength which is given by e0er
 
What would the min. value?

would i have to calculate a new er?
 
and it doesn't specify the fluid, which is kinda of the point, for us to calculate it without looking at the table.
 
  • #10
theman408 said:
and it doesn't specify the fluid, which is kinda of the point, for us to calculate it without looking at the table.

What you are doing sort of looks counter-intuitive to me.

C=\frac{\epsilon_0 \epsilon_r A}{d}

you used that to get C, yes I get that. You know, A,d,ε0 and εr. You find C.


What you are asking is to get ε (or ε0εr) for the same C, A and d. You will just get back what you used above. Am I missing something :confused:? Does A,d or A change?
 
  • #11
Exactly that's the formula but what I am asked to calculate is the min. dielectric strength V/M.

They give us Er which i looked up on a table and it's the Dielectric constant of Barium titanate.
 
  • #12
theman408 said:
Exactly that's the formula but what I am asked to calculate is the min. dielectric strength V/M.

V/M ? as in volt per metre as units? If that is the case then those units mean you need to find the electric field strength. Which is simply E=V/d
 
  • #13
Yep, it's positive right?
 
  • #14
theman408 said:
Yep, it's positive right?

I would think so.
 
  • #15
The table in the back of my book that gives the different dielectric strengths it's unit is expressed x10^6 V/M and with that formula it only gives me kv/m.
 
  • #16
theman408 said:
The table in the back of my book that gives the different dielectric strengths it's unit is expressed x10^6 V/M and with that formula it only gives me kv/m.

What is the distance between the plates?
 
  • #17
2x10^-3 m
 
  • #18
theman408 said:
2x10^-3 m

Well the dielectric strength gives the maximum electric field that can be applied before breakdown occurs. I doubt 12V is the maximum voltage, but that would be how to find it.
 
  • #19
Im in the crossroads in using this formula E= Q/EoArea that yields 7.2 x10^6 V/M or E= V/D that yields 6 kv/M

A= 1m^2
Q= 63.72 uC
D= 2mm
V= 12 V

all the dielectric strength values in the table appear in x10^6 V/M
 

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