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theman408

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- Thread starter theman408
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In summary: What is the distance between the plates?2x10^-3 m2x10^-3 mSorry, I don't understand.You are looking for the dielectric strength for a given voltage and current. 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.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

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theman408

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- #2

theman408

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

this is all i need.

this is all i need.

- #3

rock.freak667

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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?

- #4

theman408

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- #5

rock.freak667

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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?

- #6

theman408

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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

rock.freak667

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theman408 said:

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 e

- #8

theman408

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What would the min. value?

would i have to calculate a new er?

would i have to calculate a new er?

- #9

theman408

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- #10

rock.freak667

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theman408 said:

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

[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

What you are asking is to get ε (or ε

- #11

theman408

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They give us Er which i looked up on a table and it's the Dielectric constant of Barium titanate.

- #12

rock.freak667

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theman408 said:Exactly that's the formula but what I am asked to calculate is themin. 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

theman408

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Yep, it's positive right?

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rock.freak667

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theman408 said:Yep, it's positive right?

I would think so.

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theman408

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- #16

rock.freak667

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theman408 said:

What is the distance between the plates?

- #17

theman408

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2x10^-3 m

- #18

rock.freak667

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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

theman408

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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

Dielectric strength breakdown voltage is the maximum voltage that an insulating material can withstand before it breaks down and allows electricity to pass through it.

Dielectric strength breakdown voltage is calculated by dividing the breakdown voltage by the thickness of the insulating material. The result is measured in volts per unit of thickness (V/m).

The factors that affect dielectric strength breakdown voltage include the type and quality of the insulating material, the thickness and uniformity of the material, and the presence of any impurities or defects.

Dielectric strength breakdown voltage is important because it determines the maximum voltage that an insulating material can safely withstand, which is crucial for preventing electrical failures and ensuring the safe operation of equipment.

Dielectric strength breakdown voltage is used to select appropriate insulating materials for various applications, such as in electrical transformers, cables, and electronic devices. It is also used to test and certify the safety and reliability of these materials.

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