# Electron in a uniform field

1. Aug 28, 2013

### hacker804

If an electron is placed in a uniform electric field it moves from a lower potential to a higher potential.So does it's potential energy decrease.In other words,is $PE_{f}-PE_{i}<0$?

This is confusing me.

2. Aug 28, 2013

### ZapperZ

Staff Emeritus
It's potential becomes more negative.

Zz.

3. Aug 28, 2013

### hacker804

So its potential energy decreases?

4. Aug 28, 2013

### Staff: Mentor

Correct.

Yes.

I'm guessing that you're confused because PE = qV, which seems to indicate that as V increases, PE should increase. However, the charge q of an electron is a negative number, which reverses the relationship.

If I didn't guess correctly, please describe your confusion more completely.

5. Aug 28, 2013

### ZapperZ

Staff Emeritus
What do you think? If something goes from -7 to -10, has it increased or decreased in value? Is this is money, have you increased your wealth?

Zz.

6. Aug 28, 2013

### sophiecentaur

If it ends up with increasing Kinetic Energy then it is reasonable to conclude that it loses Potential Energy. The signs take care of how that is achieved.

7. Aug 28, 2013

### hacker804

Thanks for the help.

8. Aug 28, 2013

### voko

I find this confusing. The electron has moved from a location with some external potential to a location with another external potential. Can we really say that the electron's potential has changed?

9. Aug 28, 2013

### sophiecentaur

That's what the definition of Potential implies. If work is done then its potential has changed.
What do you mean by "External Potential"? Perhaps that is something to do with your problem.

10. Aug 28, 2013

### voko

OP: "an electron is placed in a uniform electric field". This field is external with respect to the electron. The potential of this field is external to the electron.

The electron has its self-field, which is another can of worms, which has its potential. When ZZ said "It's potential becomes more negative", he apparently meant that the electron's potential changes, not the potential of the field the electron is immersed in. I find this strange, unless we are talking about the effects of motion on the electron's self-field, which I thought we were not.

11. Aug 28, 2013

### sophiecentaur

It still appears that it's a matter of definition or terminology that's giving you a problem. If you establish a field somewhere then, rather like the emf from the power supply to a circuit, that field is regarded as fixed. You then place a charge somewhere in the field and consider the force on it (Field) or the work done in bringing the charge to that place (Potential). You refer to a "uniform Electric Field" so that implies the field is not affected by the presence of your electron. The definition of a Field is the force on a Unit Charge. That would be 1 Coulomb, to be strict but where can you get a Coulomb of Charge in a bucket and actually place it somewhere? Its presence would certainly affect the local field in most circumstances. But an electron (small) is OK to consider and the uniform field would not be affected.
An electron has no 'intrinsic' Field or Potential; it has Charge. Field and potential around an electron are governed by the geometry (how far away) and maybe other nearby charges.
Your electron is a 'test charge', if you like, and you can measure the force (or work done) on it, which will tell you the Field (or Potential).

The Field is considered to be there, with or without a test charge in place. The concept of Fields is a bit abstract, actually and is worth discussion and there are approaches that just discuss the action between charges (Coulomb's Law). Have you read around this yet? I suggest you read a number of sources and it may get easier to accept.

12. Aug 28, 2013

### ZapperZ

Staff Emeritus
No, that isn't what I meant. The term "potential energy" in electrostatic is the "qV(r)", where "q" is your test charge, and "V(r)" is the external electrostatic potential at the location r of that charge. This means that for V(r) to change, the charge q has moved to another location that makes V(r) different!

Look at what the OP asked initially.

Zz.

13. Aug 28, 2013

### voko

I do not think you are understanding me. When an electron is moved from one place to another the electron's (electrostatic) potential does not "become more negative" or change in any way, except being translated in space (if that's what is meant by "becoming more negative", then I am rightly confused).

But the OP was clearly not about the electron's (electrostatic) potential. It was about the electron's (electrostatic) potential energy, which is independent of its own (electrostatic) potential, and is defined as the product of the electron's charge and the background field's electrostatic potential.

14. Aug 28, 2013

### voko

Yes, this is clear. It just not clear what that "it" was whose potential becomes more negative because V(r) is given to grow as the electron moves?

15. Aug 28, 2013

### ZapperZ

Staff Emeritus
But PE is qV(r), and if this "q" is an electron, it has a negative charge. So sure, the electron moves to a higher V(r), but this means that PE has gotten more negative! That was what I was trying to demonstrate.

Zz.

16. Aug 28, 2013

### voko

I see. You just said "potential" and not "PE" originally :)

Thanks for clearing things up.

17. Aug 28, 2013

### sophiecentaur

Is there anything in this thread that isn't printed clearly in the Electrostatics section of any A Level text book?