Dielectric with a parallelplate capacitor finding minimum areaby sonrie Tags: capacitor, dielectric, minimum, parallelplate 

#1
Apr2308, 01:05 PM

P: 35

The dielectric to be used in a parallelplate 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! 



#2
Apr2308, 01:21 PM

HW Helper
P: 2,250

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? 



#3
Apr2308, 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. 



#4
Apr2308, 03:39 PM

P: 35

Dielectric with a parallelplate 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! 



#5
Apr2308, 08:27 PM

HW Helper
P: 2,250

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] 



#6
Apr2408, 04:35 PM

P: 35

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




#7
Apr2408, 04:37 PM

P: 35

sorry its 2.94 *10^10




#8
Apr2408, 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 



#9
Apr2408, 06:35 PM

HW Helper
P: 2,250

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 



#10
Apr2508, 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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