Dielectric with a parallel-plate capacitor finding minimum area

[sonrie]

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.70 and a dielectric strength of 1.90×10^7 V/m . The capacitor is to have a capacitance of 1.05×10^−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!

 

[alphysicist]

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]

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]

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]

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:

$$C = \kappa \epsilon_0 \frac{A}{d}$$

 

[sonrie]

To find the distance if did the following 5600/1.90*10^7 = 2.94*10^7

 

[sonrie]

sorry its 2.94 *10^10

 

[sonrie]

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]

To find the distance if did the following 5600/1.90*10^7 = 2.94*10^7

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]

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!