Electrical potential energy question

[Grahamlf]

Why does the potential energy of a system of two charged particles decrease as the distance between them increases? It seems that the PE should increase, as in U=mgh; as h increases, the PE increases. Why does this not occur in electrical potential energy, and in the gravitational potential energy equation U=(-GMm)/r?

 

[Dale]

It seems that the PE should increase, as in U=mgh; as h increases, the PE increases. Why does this not occur in electrical potential energy, and in the gravitational potential energy equation U=(-GMm)/r?

I am not sure what you are asking. The formula U=(-GMm)/r also increases as r increases.

 

[Ibrahim Hany]

Your intuition is right; potential energy between the two opposite charges increases as the two charges get apart from each other.

If the first charge was at r=1, and the 2nd one was at r=2; then the Potential Energy between them (Potential Difference) would be Constant((1/1)-(1/2), which is (0.5) the constant.

Now if the first charge was at r=1, and the 2nd one was at r=10; then the Potential Difference would be Constant((1/1)-(1/10), which is (0.9) the constant.

 

[sophiecentaur]

Hold your horses there, folks. The sign of the PE depends on the signs of the charges involved.
Attractive forces involve negative PE and repulsive forces involve positive PE. As the separation increases, both approach zero.

 

[Stavros Kiri]

Hold your horses there, folks. The sign of the PE depends on the signs of the charges involved.
Attractive forces involve negative PE and repulsive forces involve positive PE. As the separation increases, both approach zero.

Very correct! Attractive ones are thus just like gravity.

 

[Stavros Kiri]

It seems that the PE should increase, as in U=mgh; as h increases, the PE increases. Why does this not occur in electrical potential energy, and in the gravitational potential energy equation U=(-GMm)/r?

Welcome to PF!
Electrical PE = U = Kq₁q₂/r
That sais it all! Just do the math right (with signs and everything).

 

[Grahamlf]

What I am confused on is why, according to U=(kqq)/r, the potential energy due to a charge decreases as the distance from the charge increases. This is assuming opposite charges. If both charges are just 1, then @ r=1, U=1 (ignoring k since its a constant). If r=3 however, U=1/3. This is confusing to me because in gravitational potential energy (U=mgh), as h increases, U also increases.

 

[Stavros Kiri]

What I am confused on is why, according to U=(kqq)/r, the potential energy due to a charge decreases as the distance from the charge increases. This is assuming opposite charges. If both charges are just 1, then @ r=1, U=1 (ignoring k since its a constant). If r=3 however, U=1/3. This is confusing to me because in gravitational potential energy (U=mgh), as h increases, U also increases.

Your remark stands correct only for the possitive magnitude (or absolute value [= possitive number]) of the potential energy. However, the sign is important for potential energy (just as for potentials V in batteries, + or - etc.).
For opposite charges, e.g. (-q)q = -q², and for q=1 we get a (-1) factor in PE. In other words, e.g. for equal opposite charges U= -(kqq)/r. Thus in your example you should now compare (-1) and (-1/3). Obviously (-1) < (-1/3), i.e. indeed increases, thus consistent!

 

[Grahamlf]

does that mean that potential energy is asymptotic to 0 as a function of r when the charges are opposite?

 

[Stavros Kiri]

... This is assuming opposite charges. If both charges are just 1 ...
... This is confusing to me because in gravitational potential energy (U=mgh), as h increases, U also increases.

Your mistake was simply a trivial algebraic one. I assume you are familiar with negative numbers and their algebra (i.e. the set R^- (Negative Reals)). If not, you should get familiar with the whole R set (Real Numbers [possitive, negative and zero]).

In your quote above, you should say 1 and (-1) [instead of both just 1].

Dale in his comment gave you the answer right away for the exactly similar case of Newtonian gravity.

Finally, U=mgh differs in the convention that we take as point of reference the ground (h=0) and it is always possitive (above ground). Whereas the electrical (and/or Newtonian gravitational) potential energy case has the convention of being zero at infinity (r=∞), and is always negative for opposite charges, as well as Newtonian gravity, but always positive for same sign charges. [In this very last case and only (i.e. repulsive forces), your initial remark stands and holds, but there it is expected!]

So just be careful of all signs and do the algebra right.

 

[Stavros Kiri]

does that mean that potential energy is asymptotic to 0 as a function of r when the charges are opposite?

Exactly correct!
P.S.: Nice graph! From it I conclude now that you are indeed familiar, not only with positive and negative numbers, but with all Real numbers in general, as well as Calculus.

 

[Dale]

What I am confused on is why, according to U=(kqq)/r, the potential energy due to a charge decreases as the distance from the charge increases. This is assuming opposite charges.

Note, here you explicitly state that you are assuming opposite charges.

If both charges are just 1,

If they are opposite then they cannot both be 1. I assume that you mean one charge is 1 and the other is -1.

then @ r=1, U=1 (ignoring k since its a constant).

No, U=-1.

If r=3 however, U=1/3.

No, U=-1/3

This is confusing to me because in gravitational potential energy (U=mgh), as h increases, U also increases.

Likewise as r increases U also increases.

You appear to be confused by failing to account for the sign of the charge.

 

[sophiecentaur]

Perhaps there has been a confusion between Potential and Field, from the start in this thread. Of course, as you get further away, the 'Force" gets less - because the Field gets less (+ or -). But Potential is defined as the Work done 'against' the force, over the distance form infinity to the point of interest. That is counter-intuitive to someone who has learned things like mgh but mgh only works over small ranges of height and it assumes a uniform field. Yet again, this demonstrates the problems with trying to jump into a subject half way through instead of starting at the most basic point and sticking to the definitions.