
#1
Dec509, 01:32 PM

P: 94

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.




#2
Dec509, 02:18 PM

P: 94

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



#3
Dec509, 02:19 PM

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#4
Dec509, 02:22 PM

P: 94

Dielectric Strength
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.




#5
Dec509, 02:36 PM

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#6
Dec509, 03:09 PM

P: 94

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. 



#7
Dec509, 03:44 PM

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#8
Dec509, 03:52 PM

P: 94

What would the min. value?
would i have to calculate a new er? 



#9
Dec509, 03:56 PM

P: 94

and it doesn't specify the fluid, which is kinda of the point, for us to calculate it without looking at the table.




#10
Dec509, 11:20 PM

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[tex]C=\frac{\epsilon_0 \epsilon_r A}{d}[/tex] you used that to get C, yes I get that. You know, A,d,ε_{0} and ε_{r}. You find C. What you are asking is to get ε (or ε_{0}ε_{r}) for the same C, A and d. You will just get back what you used above. Am I missing something ? Does A,d or A change? 



#11
Dec609, 07:58 AM

P: 94

Exactly thats the formula but what im 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
Dec609, 11:26 AM

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#13
Dec609, 01:27 PM

P: 94

Yep, it's positive right?




#14
Dec609, 01:28 PM

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#15
Dec609, 01:32 PM

P: 94

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
Dec609, 01:36 PM

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#17
Dec609, 01:38 PM

P: 94

2x10^3 m




#18
Dec609, 01:45 PM

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