An electron in a P.P.capacitor is shot from the neg.plate to the pos. plate.
Why the book says that the negative plate is the lower potential? I thought that like charges repel thus, the electron would have a high potential when close to the neg.plate.
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An electron in a P.P.capacitor is shot from the neg.plate to the pos. plate.
Why the book says that the negative plate is the lower potential? I thought that like charges repel thus, the electron would have a high potential when close to the neg.plate.
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
No. Higher potential can be thought of as a hill and lower can be thought of as a valley FOR A POSITIVE CHARGE. Electrons have negative charge, so everything flips for them.
No. As I said, for an electron, the negative plate will have lower potential, since that means it will be a hill. The electron rolls off the hill to the valley, the positive plate.
I get you. So to escape the valley the electron would need kinetic energy exceeding the potential. An electron next to a negative plate needs less kinetic energy to escape than when its near a positive one.
Again, if a I have my negative plate at say x = 1 and my positive plate at x = 3, then voltage looks something like this:
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Since \(\vec E = - \nabla V\), and, given the charge density \(\sigma\) on the left plate (and an equivalent opposite charge on the right) we can simplify it to \(E = \frac{\sigma}{\epsilon_0} = -\frac{dV}{dx}\). So \(V=c (x-1)\) where c is some positive constant.
Put a proton anywhere on that line, and it will roll to the left. Put an electron on it, and you have to switch where it goes, since \(U=qV\), and with an electron q Is negative. Technically you have to plot \(U\), not \(V\) (which is independent of the charge in quesiton) if you want to look at kinetic energy and all that stuff about being able to roll from a valley, etc.