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In summary, according to the formula U=(-GMm)/r, potential energy decreases as the distance between the charges increases. However, this does not happen in electrical potential energy, and in the gravitational potential energy equation U=(-GMm)/r.

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

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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 (

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

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Very correct! Attractive ones are thus just like gravity.sophiecentaur said:

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

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Welcome to PF!Grahamlf said: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

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

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

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Grahamlf said:... 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

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

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Note, here you explicitly state that you are assuming opposite charges.Grahamlf said: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.Grahamlf said:If both charges are just 1,

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

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

Likewise as r increases U also increases.Grahamlf said: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.

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