Timeline for Calculating the electric field in a parallel plate capacitor, being given the potential difference
Current License: CC BY-SA 3.0
6 events
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| Jun 20, 2017 at 15:51 | comment | added | Alfred Centauri | @I.Wewib, I updated my answer | |
| Jun 20, 2017 at 15:48 | history | edited | Alfred Centauri | CC BY-SA 3.0 |
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| Jun 20, 2017 at 15:43 | history | edited | Alfred Centauri | CC BY-SA 3.0 |
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| Jun 20, 2017 at 5:46 | comment | added | I. Wewib | In your answer $\phi = 0$ because $\vec{E}$ points from 0 to 1, and so does $\mathrm{d}\vec{x}$ because you integrate from 0 to 1. But in my calculations, I integrate from 1 to 0 and $\mathrm{d}\vec{x}$ points from 1 to 0 then, but the electric field still points from 0 to 1. So in my calcs, $\vec{E}$ and $\mathrm{d}\vec{x}$ oppose eachother, so $\phi = \pi$. You and I started with the same formula but I ended up with a negative $\vec{E}$ and you with a positive. This confuses me so much, where did I go wrong? Is it a reasoning mistake or a mathematical one? | |
| Jun 20, 2017 at 5:35 | comment | added | I. Wewib | This makes so much sense, thanks for your answer. But why doesn't it work the other way around? $V(0) - V(1) = - \int_1^0 E_x \mathrm{d}x = - E_x (-1 \mathrm{m})$. So $E_x = 10^5\mathrm{\frac{V}{m}}$. But $E = \frac{E_x}{cos(\phi)} = \frac{E_x}{-1}\mathrm{\frac{V}{m}}$ (because we integrated from 1 to 0, $\mathrm{d}\vec{x}$ points from 1 to 0 and $\vec{E}$ points from 0 to 1, so $\phi = \pi$). And $E =\frac{E_x}{-1}\mathrm{\frac{V}{m}}=-10^5\mathrm{\frac{V}{m}}$. So it is negative, and yours is positive. I must be doing something wrong. Can you please tell me what I did wrong there? | |
| Jun 19, 2017 at 23:05 | history | answered | Alfred Centauri | CC BY-SA 3.0 |