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## anonymous 4 years ago 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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1. anonymous

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.

2. anonymous

So the book is wrong?

3. anonymous

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.

4. anonymous

** for an electron, the negative plate will be a hill since it has lower potential, not the other way around.

5. anonymous

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.

6. anonymous

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: |dw:1349756069232:dw| 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.

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