What is the potential difference in parallel plates?

[x86]

Homework Statement

So I'm confused, if you have parallel plates:

+++++++++++++++

-------------------

And there's an electron inside the plate:

+++++++++++++++
e-
-------------------

Does that mean the voltage of the negative terminal is higher than the positive terminal? i.e.+++++++++++++++ (LOW, say -20 V)
e-
------------------- (HIGH, say 50V)

And for a proton, is this correct?

+++++++++++++++ (HIGH, say 50V)
p+
------------------- (low, say -20V)

Or is it ALWAYS like this:

(call this diagram diagram sense)
+++++++++++++++ (HIGH, say 50V)
?
------------------- (low, say -20V)

Regardless of if '?' is a proton or electron?

I mean, it should always be the one in diagram sense? Right? Because V = kq/r and q will always be positive at that point?

If someone could clear this confusion for me, I would be very happy.

Homework Equations

The Attempt at a Solution

Not sure, to be honest.

To sort of rephrase my question:

Is the positive terminal ALWAYS high potential? No matter what charge is being referenced? I.e., proton, electron

 

[cepheid]

The positive terminal is always at a higher electric potential (V). So, a positive test charge between the plates would want to move from the area of high electric potential to the area of low electric potential (i.e. away from + and towards -). In contrast, a negative charge between the plates would want to move from an area of low electric potential to an area of high electric potential (i.e. away from - and towards +).

This is just a matter of definition. The second case may seem counter-intuitive at first, until you realize that in BOTH cases, the test charge is moving in a direction that reduces the potential ENERGY (U, or whatever symbol you want) of the system. That's because the potential ENERGY is the electric potential (V) multiplied by the charge (q) that you place in between the plates:

U = Vq

So, in the case of a proton as a test charge (with q = +e), the potential energy U = +Ve would be HIGHER at the + terminal (where the electric potential V is larger in magnitude) and LOWER at the - terminal, where V is smaller in magnitude. The charge moves in the direction from from high U to low U (+ to -), just as you would expect.

In the case of an electron as a test charge (with q = -e), the potential energy U = -Ve would be HIGHER at the - terminal (where the electric potential V is smaller in magnitude) and LOWER at the + terminal, where V is larger in magnitude. The charge moves in the direction from from high U to low U (- to +), just as you would expect.