| Thread Closed |
Dielectric with a parallel-plate capacitor finding minimum area |
Share Thread |
| Apr23-08, 01:05 PM | #1 |
|
|
Dielectric with a parallel-plate capacitor finding minimum area
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! |
| Apr23-08, 01:21 PM | #2 |
|
Recognitions:
|
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? |
| Apr23-08, 01:42 PM | #3 |
|
|
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. |
| Apr23-08, 03:39 PM | #4 |
|
|
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! |
| Apr23-08, 08:27 PM | #5 |
|
Recognitions:
|
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] |
| Apr24-08, 04:35 PM | #6 |
|
|
To find the distance if did the following 5600/1.90*10^7 = 2.94*10^7
|
| Apr24-08, 04:37 PM | #7 |
|
|
sorry its 2.94 *10^10
|
| Apr24-08, 04:42 PM | #8 |
|
|
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 |
| Apr24-08, 06:35 PM | #9 |
|
Recognitions:
|
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 |
| Apr25-08, 12:07 PM | #10 |
|
|
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!!!!! |
| Thread Closed |
Similar Threads for: Dielectric with a parallel-plate capacitor finding minimum area
|
||||
| Thread | Forum | Replies | ||
| Simple Harmonic Motion of a Dielectric Slab in a Parallel Plate Capacitor | Classical Physics | 6 | ||
| parallel plate capacitor and dielectric | Introductory Physics Homework | 10 | ||
| [SOLVED] Parallel-plate capacitor: Two dielectric materials | Introductory Physics Homework | 8 | ||
| More than one dielectric ibetween a parallel plate capacitor | Introductory Physics Homework | 13 | ||
| Parallel-plate capacitor with two dielectric between them | Advanced Physics Homework | 1 | ||