| New Reply |
Non-parallel Plate Capacitor |
Share Thread | Thread Tools |
| Feb14-08, 12:52 AM | #1 |
|
|
Non-parallel Plate Capacitor
1. The problem statement, all variables and given/known data
A capacitor has square plates, each of side a, making an angle theta with eachother. Shown that for small theta the capacitance is given by: C = [(epsilon)(a^2)/d][1-(a(theta)/2d]. 2. Relevant equations C = q/V gauss' law 3. The attempt at a solution I see how you can divide up the strip into N segments each with length a/N. But how do u get the capacitance for each strip to be C = (epsilon)a(deltax)/y ? I know how to do the rest and I know for sure that that's the right way to do it, but how do u get the capacitance for each strip?) |
| Feb14-08, 07:27 AM | #2 |
|
|
See that C = [tex]\epsilon[/tex]*S/d, in your case for each strip S = a*dx and d = y.
|
| Mar20-11, 11:38 PM | #3 |
|
|
I'm also troubled with this question, I know it has been a long time, but maybe one of you can explain me the answer?
Omer |
| New Reply |
| Thread Tools | |
Similar Threads for: Non-parallel Plate Capacitor
|
||||
| Thread | Forum | Replies | ||
| A proton is released from rest at the positive plate of a parallel-plate capacitor. | Introductory Physics Homework | 16 | ||
| Parallel Plate Capacitor | Introductory Physics Homework | 1 | ||
| Parallel-Plate Capacitor | Introductory Physics Homework | 7 | ||
| parallel plate capacitor | Introductory Physics Homework | 3 | ||
| The parallel-plate capacitor | Introductory Physics Homework | 3 | ||