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Dielectric with a parallel-plate capacitor finding minimum area

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sonrie
#1
Apr23-08, 01:05 PM
P: 35
The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.70 and a dielectric strength of 1.9010^7 V/m . The capacitor is to have a capacitance of 1.0510^−9 F and must be able to withstand a maximum potential difference of 5600 V.

What is the minimum area the plates of the capacitor may have? use 8.85*10^-12 for permittivity of free space.

A=______ m^2.

Well i know that the capacitance equals permittivity times area times dialectric constant divided by distance between plates. The distance i got by dividing the voltage by the dialectric strength. Is that correct?

the final answer i got was 9.47*10^11 but it was wrong. Help please!
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alphysicist
#2
Apr23-08, 01:21 PM
HW Helper
P: 2,249
Hi sonrie,

That's quite large for the area of a capacitor! I think you might have just made a calculation error. What numbers did you use for both of your calculations?
atavistic
#3
Apr23-08, 01:42 PM
P: 106
The max intensity of field can be 1.9 x 10^7. Now what is the field inside a capacitor?
Relate the intensity to the potential difference.And use the capacitance formula.

sonrie
#4
Apr23-08, 03:39 PM
P: 35
Dielectric with a parallel-plate capacitor finding minimum area

For the distance i got 2.95*10^10
so the equation looks like
1.05*10^-9= 8.85*10^-12 *Area/ 2.95*10^10 = 3.50*10^-12 which is different than the one i posted before but its still wrong! Help please!
alphysicist
#5
Apr23-08, 08:27 PM
HW Helper
P: 2,249
Hi sonrie,

When you get a distance like 2.95 x 10^10 something must be wrong. That is a huge distance! What did you do to find that number?

The equation that you then used next to solve for A is missing the dielectric constant. For a parallel plate capacitor with a dielectric, we have:

[tex]
C = \kappa \epsilon_0 \frac{A}{d}
[/tex]
sonrie
#6
Apr24-08, 04:35 PM
P: 35
To find the distance if did the following 5600/1.90*10^7 = 2.94*10^7
sonrie
#7
Apr24-08, 04:37 PM
P: 35
sorry its 2.94 *10^10
sonrie
#8
Apr24-08, 04:42 PM
P: 35
With the formula that you provided by equation will look like this:

1.05*10^-9= 8.85*10^-12 *3.70* A / 2.94*10^10 so i just solve for A? then my final answer would be final answer was 9.47*10^-11 still wrong
alphysicist
#9
Apr24-08, 06:35 PM
HW Helper
P: 2,249
Quote Quote by sonrie View Post
To find the distance if did the following 5600/1.90*10^7 = 2.94*10^7
Quote Quote by sonrie View Post
sorry its 2.94 *10^10
sonrie,

That distance is huge! The radius of the earth is only about 6 x 10^6 meters or so.

It looks like you just entered it into the calculator wrong. I get:

5600 / ( 1.9 x 10^7 ) = 0.000295
sonrie
#10
Apr25-08, 12:07 PM
P: 35
your Right! so my equation will be 1.05*10^-9= 8.85*10^-12*3.70*A/2.95*10^-4 . I finally got it RIGHT!
Thank You So Much! Have a GREAT DAY!!!!!


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