- #1

- 3

- 1

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Grahamlf
- Start date

- #1

- 3

- 1

- #2

- 32,757

- 9,857

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

- #3

- 25

- 2

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.

If the first charge was at r=1, and the 2nd one was at r=2; then the Potential Energy

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 (

Last edited:

- #4

sophiecentaur

Science Advisor

Gold Member

- 27,196

- 5,906

Attractive forces involve negative PE and repulsive forces involve positive PE. As the separation increases, both approach zero.

- #5

- 944

- 675

Very correct! Attractive ones are thus just like gravity.

Attractive forces involve negative PE and repulsive forces involve positive PE. As the separation increases, both approach zero.

- #6

- 944

- 675

Welcome to PF!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?

Electrical PE = U = Kq

Last edited:

- #7

- 3

- 1

- #8

- 944

- 675

Your remark stands correct only for the possitive magnitude (or absolute value [= possitive number]) of the potential energy.

For opposite charges, e.g. (-q)q = -q

- #9

- 3

- 1

- #10

- 944

- 675

... 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.

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

Finally, U=mgh differs in the convention that

Last edited:

- #11

- 944

- 675

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

Last edited:

- #12

- 32,757

- 9,857

Note, here you explicitly state that you are assuming opposite charges.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 they are opposite then they cannot both be 1. I assume that you mean one charge is 1 and the other is -1.If both charges are just 1,

No, U=-1.then @ r=1, U=1 (ignoring k since its a constant).

No, U=-1/3If r=3 however, U=1/3.

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

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

- #13

sophiecentaur

Science Advisor

Gold Member

- 27,196

- 5,906

Share: