# Electric PE - am I not getting something?

1. Oct 17, 2012

### miaou5

So I know the law of energy conservation applies when dealing with electric force, but I don't quite understand what electric PE is. From my understanding of energy conservation,

Electric PE (initial) + KE (initial) = Electric PE (final) + KE (final) (ignoring gravity)

and thus, KE (final) - KE (initial) = -(Electric PE (final) - Electric PE (initial))

For instance, if a negative test charge was to be released from rest and started traveling toward a positive source charge, and I were to calculate the velocity of the test charge as it reaches the source charge, Electric PE (final) would be close to infinity, thus velocity(final) would be infinity as well? However, from the homework problems I'm doing, they're only using Electric PE (initial) and setting that equal to KE (final), which I don't understand -- how come they're not accounting for Electric PE (final)? Thank you so much guys! This forum rocks!

2. Oct 17, 2012

### JustinRyan

your first equation is a statement of the conservation of energy and holds. But that energy can change forms from potential to kinetic so the final ratio may not equal the initial ratio..
a negative charge's electric potential is maximum when further from the positive source and falls as it gets closer, converting potential energy to kinetic energy. That may be why potential final is neglected as it may be zero once it reaches the positive source.

3. Oct 17, 2012

### sophiecentaur

If you take a point negative charge and a point positive charge then, if you calculate, classically, the potential energy between them when they hit each other it is, indeed, -∞ and they would be travelling infinitely fast towards each other.
PE is proportional to Q1 Q2 / d, where d is the spacing and Q are the charges.

A massive positive charge would represent an infinitely deep potential well for a tiny negative charge to fall into. However, any real object has a finite radius which puts a 'flat bottom' on this well so it no longer has infinite depth because d is never zero. A charged object always has a finite size and its potential at the surface is what counts in your calculations. Once inside it (imagine this were possible by drilling a hole)) the potential changes with a different law and reaches a minimum at the centre (if it's a sphere) and there is no longer any attractive force on the negative test charge. The same thing applies in the 'hole in the Earth' problem which is discussed frequently here.

There is a further reason why this doesn't happen for protons and electrons and that is due to the quantum mechanics of the situation, which imposes a minimum energy for the lowest state.

4. Oct 18, 2012

### JustinRyan

hmm my bad. true I am thinking in gravitational potential in which mass of falling object is mostly ignored. I guess that can't be ignored when charges are of comparable values. PE will increase. but must be stopped by something before it reaches infinite.

5. Oct 18, 2012

### sophiecentaur

Hang on. GPE is all about mass.

6. Oct 18, 2012

### nasu

This may happen when the final PE is zero.
Can you show an actual example?

7. Oct 18, 2012

### sophiecentaur

It doesn't usually matter what the potential is, so much as the Change in potential.

8. Oct 18, 2012

### JustinRyan

sorry i meant the mass of the falling object is ignored when calculating the increase in attractive force due to thier increasing proximity to the surface. yep GPE is mass dependant.

9. Oct 18, 2012

### sophiecentaur

So is the force. How can it be otherwise? GPE is the integral of force over the distance.

10. Oct 21, 2012

### JustinRyan

I have not said it was otherwise? but I see I did not say my bad ....enough. and the invariance is entirely due to the distances in relation to the source, and representing the source of charge or mass as a mathematical point when it isn't. so in conclusion, don't post when you're toast.

11. Oct 21, 2012

### sophiecentaur

12. Oct 21, 2012

### AJ Bentley

Whoa! Start again.

Electric potential energy for opposite charges works exactly the same as gravity.
If you hold a negative charge a long way from a positive, it has a large PE. Like an object suspended high in the sky.

If it's released the negative charge falls down the potential. When it gets close to the the positive charge, all of the PE has been converted into KE - just as a falling stone picks up speed.

So final KE = initial PE. Simple as that.

(You can always ignore gravity in electrostatic problems because it's a gadzillion times weaker - there's no need to even mention it)

13. Oct 22, 2012

### JustinRyan

When it gets close to the the positive charge, all of the PE has been converted into KE

How close?

So final KE = initial PE. Simple as that.

Not really. As PE is inversly proportional to the distance between the two sources it will increase sources approach each other. As d -> 0 , PE -> infinite.

A mathematical anomaly for sources that do not have any size. GPE is usually only considered for d > radius of earth but if earth were squeezed into a point, GPE of the stone would increase toward infinity as it approached the point.