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This is from Liboff Introductory quantum mechanics 1st(current is 4th) edition: enter image description here I don't understand why the distance z is measured from the bottom plate if the potential is of an electron which should gain Energy going to the top plate.

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  • $\begingroup$ Usually what you really want to know is the potential difference between the top plate and the bottom plate. It doesn't matter which way you integrate, it will just change the sign of the result: The top plate is $v$ volts above the bottom plate, or the bottom plate is $v$ volts below the top plate mean the same thing. $\endgroup$ – The Photon Sep 9 '18 at 22:48
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$E$ is the constant mechanical energy (kinetic energy plus electric potential energy) of the electron and parallel plate system.

The first term is the kinetic energy of an electron and the second term is the electric potential energy of the system.

The charge $q$ on the electron is negative so as the distance of the electron from the bottom plate $z$ increases the electric potential energy becomes more negative ie decreases which would result in the kinetic energy of an electron increasing.

Update in response to a comment from @user5389726598465

In the graphs below I have made the substitution $q=-e$ wherre $e$ is the magnitude of the charge on an electron and $V$ is the potential.

enter image description here

You will note that wherever you choose the zero of potential in going from the bottom plate to the top plate there is a decrease in the potential energy.

In both cases the potential energy is $\dfrac{q\phi_0}{d}z = - \dfrac{e\phi_0}{d}z$.
In the top example as $z$ becomes more positive the potential energy becomes more negative, ie is decreasing, whilst in the bottom example as $z$ becomes less negative the potential energy becomes less positive, ie is decreasing.

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  • $\begingroup$ Everything you said is correct! That's exactly the problem. It measures the increase of kinetic energy which is the fraction of z/d of loss in potential energy. It should measure the potential energy (1-z)/d $\endgroup$ – user5389726598465 Sep 10 '18 at 6:20
  • $\begingroup$ @user5389726598465 I have updated my answer. I think that what you meant to write in your comment is "the potential energy is (d-z)/z" otherwise it would be dimensionally incorrect? However in this example that is not necessary. $\endgroup$ – Farcher Sep 10 '18 at 7:02
  • $\begingroup$ The key point then is that the potential energy goes from zero to a negative value from the top to bottom plate. Is this a common convention? $\endgroup$ – user5389726598465 Sep 10 '18 at 18:29
  • $\begingroup$ @user5389726598465 You could change the direction of positive $z$ ie downwards. $\endgroup$ – Farcher Sep 10 '18 at 19:42
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Historical accident. From https://en.wikipedia.org/wiki/Electric_charge

One of the foremost experts on electricity in the 18th century was Benjamin Franklin, who argued in favour of a one-fluid theory of electricity. Franklin imagined electricity as being a type of invisible fluid present in all matter; for example, he believed that it was the glass in a Leyden jar that held the accumulated charge. He posited that rubbing insulating surfaces together caused this fluid to change location, and that a flow of this fluid constitutes an electric current. He also posited that when matter contained too little of the fluid it was negatively charged, and when it had an excess it was positively charged.

Eventually electrons were discovered, and it turned out that an excess of electrons was found where the charge was labelled negative. If he had chosen names the other way around, a positive number would have been used for the charge of an electron. We would have labelled the plate with excess electrons the positive plate.

Regardless of labels, electrons are repelled from the plate with excess electons, and attracted to the plate with a shortage. Electrons lose potential energy and gain kinetic energy when traveling in this direction.

Mathematically, you calculate a potential energy loss in the formula because the charge of an electron is $q$, and $q < 0$.

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  • $\begingroup$ So the electron goes to the negative plate which must be positively charged? $\endgroup$ – user5389726598465 Sep 9 '18 at 21:08
  • $\begingroup$ From the negative plate, which is negatively charged. Electrons repel each other. To the positive plate, which is positively charged. Fewer electrons means excess nucei, which are positively charged and attract electrons. $\endgroup$ – mmesser314 Sep 10 '18 at 3:47

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