Variation of potential in electrical and magnetic fields question

EDIT: Not ment to say MAGNETIC, BUT GRAVITATIONAL!

Hi,

If potential in electrical and gravitational fields varies with 1/r (1 / distance), then why is it that if say you had a postive nucleus and you excite an electron to a higher energy level, it moves to a lower potential because of the increase in distance from the nucleus?

I would have thought that if you do postive work against the field, you're storing more energy.

Is it to do with the potential starting off at really negative value, right at the bottom of the potential well and so if you move to a greater distance away, you're making it more postive and so you need less energy to make it escape from that point...?

Reply 1

F1 fanatic

16

Einsteinium

EDIT: Not ment to say MAGNETIC, BUT GRAVITATIONAL!

Hi,

If potential in electrical and gravitational fields varies with 1/r (1 / distance), then why is it that if say you had a postive nucleus and you excite an electron to a higher energy level, it moves to a lower potential because of the increase in distance from the nucleus?

I would have thought that if you do postive work against the field, you're storing more energy.

Is it to do with the potential starting off at really negative value, right at the bottom of the potential well and so if you move to a greater distance away, you're making it more postive and so you need less energy to make it escape from that point...?

yes, it's to do with the sign of the potential, as the zero point is defined as at infinity. Physicists tend to be quite casual about how they use the word lower. By lower they mean it has a lower magnitude, in many cases. So, when you excite an electron, it has less negative energy (ie an increase in energy), so that its overall energy has increased. Does that make sense?

I prefer to think of these things in terms of being more or less tightly bound. I find it helps me to picture which has the lower energy. If an electron is more tightly bound, it's potential is more negative and it is deeper in the well. I then have to give it more energy to get out of the well than I would if it was less tightly bound.

Reply 2

W0nker

OP

8

Yeah, I sort of get it just my text book confuses me because it also says, for gravitational potential, that potential gets lower and lower the smaller the distance r from the mass, which is going against saying potential changes with 1/r.

Probably isn't going to make a huge difference to my understanding of the whole picture but just wanna clear it up.

Reply 3

Morbo

17

When dealing with potentials the sign is very important.

Many potentials are defined to be

V \propto -\frac{1}{r}

, with

V = 0

at

r=\infty

. So when someone says the potential energy is getting "lower", confusingly they could either mean two things:
- 1) that it's value is decreasing i.e. becoming more negative
or
- 2) that it is becoming lower in magnitude (less negative), i.e. more positve

Usually it's context-evident, but it can cause confusion.

Reply 4

W0nker

OP

8

Ahhh, my text book just says 1/r.

Well for a postive charge creating a field and a postive in the field, I'm not as confused because potential does change with 1/r because negative work is being done when it moves away, so moving away will leave it in a lower potential, since a potential hill is created for a postive charge in a field created by a postive charge, moving closer, decreasing r, increases the potential, I think.

Reply 5

F1 fanatic

16

Einsteinium

Ahhh, my text book just says 1/r.

Well for a postive charge creating a field and a postive in the field, I'm not as confused because potential does change with 1/r because negative work is being done when it moves away, so moving away will leave it in a lower potential, since a potential hill is created for a postive charge in a field created by a postive charge, moving closer, decreasing r, increases the potential, I think.

yes, because the potential is then positive since you have like charges, so that the issue of whether it is getting lower in magnitude or actual value isn't there since both amount to the same thing.

It's really just a case of thinking about it carefully and being precise in what you say, hence why I advocated the more/less tightly bound approach since that makes it more apparent. The alternative is to specifically say "The potential becomes less/more negative"