# Electron moving to the face of plate from depth d

## Homework Statement:

Attached is a pdf of the practice problem. They make it seem like the thickness L matters but I dont see it in anywhere in the calculations.

## Relevant Equations:

q = CV
I dont really understand the question. I dont understand the wording. We know the plate has a thickness L = 0.50cm. If the charge is coming from the battery wouldn't the electrons have to move the entire distance to reach the face of the plate? Because they have to move all the way from the battery terminal to the face. So the depth they have to move within the plate should be the entire thickness.

I know that a battery can only charge the plates until the capacitor has the same potential difference as the battery. So the charge q on the plate will be q = CV. And the number of electrons it can hold due to having the same potential difference is N = q/e.

I also understand how the number of electrons per unit volume is n = N/Ad.

What i dont undertand is what they mean when they say the electrons "move from a certain depth" to the face. And why they gave us the thickness L

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haruspex
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If the charge is coming from the battery wouldn't the electrons have to move the entire distance to reach the face of the plate?
No, that's a common misunderstanding. There is a vast sea of electrons in the wires and the uncharged plate, all locally neutralised by positive charges. When the switch is closed, some electrons get drawn into the battery on one side while an equal number are pushed out the other. But these individual electrons do not go all the way to the plate - they just push on the next electrons along the path. The resulting current moves much faster than the individual electrons, approaching light speed maybe.
It is almost certainly true that all the extra electrons that reach the surface of the plate were already in the plate at the start.

BvU, rtareen, Steve4Physics and 1 other person
TSny
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When the battery is connected to the capacitor, electrons do not travel all the way from the negative terminal of the battery to the lower plate of the capacitor. The lower plate acquires a net negative charge by a very tiny shift in the positions of free electrons in the system.

Consider the lower plate of the capacitor and imagine for a moment that it is not connected to the battery. It is isolated from everything else. The plate contains a "sea" of free electrons overlapping with a background of positively charged ions so that any macroscopic region of the plate is neutral.

But imagine that somehow the sea of electrons shifts as a whole upwards by a small distance d relative to the background positive charge. Then you get a layer of negative charge at the top of the plate and a corresponding layer of positive charge at the bottom of the plate.

This is sort of what happens when the capacitor is connected to the battery. However, the layer of positive charge at the bottom of the plate is neutralized by electrons that move into the plate from the wire that is connected between the plate and the battery. These neutralizing electrons from the wire only need to move a very tiny distance from the wire to the plate. Overall, the charging of the lower plate is achieved by just a very tiny shift in the free electrons of the plate and connecting wire.

BvU, rtareen and Steve4Physics
Steve4Physics
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Homework Statement:
...
They make it seem like the thickness L matters but I dont see it in anywhere in the calculations.
...
What i dont undertand is what they mean when they say the electrons "move from a certain depth" to the face. And why they gave us the thickness L
Hi. Tricky to explain but I’ll give it a go.

First, you are correct. The value of L is not used or needed.

It helps to visualise the conduction electrons (-) neatly arranged.

For an uncharged piece of copper we can visualise the electrons arranged like this (4 rows for illustration):
- - - - - - - - - - - top
- - - - - - - - - - -
- - - - - - - - - - -
- - - - - - - - - - - bottom
There are an equal number of +ve copper ions in fixed positios (not shown for clarity) so the metal is neutral.

When charged, the top surface has extra electrons so the arrangement of electrons is now like this:
------------------ top
- - - - - - - - - - -
- - - - - - - - - - -
- - - - - - - - - - - bottom

How did the extra electrons on the top row get there? Well, we can consider the change from uncharged to charged occurred in stages.

Stage 1: the battery pushed extra electrons into the bottom row:
- - - - - - - - - - - top
- - - - - - - - - - -
- - - - - - - - - - -
------------------ bottom

Stage 2: some electrons in bottom row shuffled up a row:
- - - - - - - - - - - top
- - - - - - - - - - -
------------------
- - - - - - - - - - - bottom

Stage 3: another shuffle up a row:
- - - - - - - - - - - top
------------------
- - - - - - - - - - -
- - - - - - - - - - - bottom

Stage 4: final shuffle: electrons from next-to-top row shuffled up to top row:
------------------ top
- - - - - - - - - - -
- - - - - - - - - - -
- - - - - - - - - - - bottom

It is this last stage the question as asking about. We are simply calculating the distance between our (visualised) rows, d. This is the distance from which some extra electonrs have been pushed into the top row.

Note the value of ‘L’ corresponds to the number of rows and makes no difference to d.

This is of course a gross oversimplication, because conduction electrons are moving around randomly inside the metal (like particles in a gas). But it's a simple way to think about what's happening for the purposes of answering the question.

rtareen
Thank you everybody. Things make more sense now. i appreciate each and every reply.