# Distance in Electric versus Gravitational potential energy

1. May 3, 2016

### califauna

1. The problem statement, all variables and given/known data
Consider the equations for electric potential energy:

and gravitational potential energy:

GPE=m*g*h

In the case of GPE, the potential energy increases as the distance between the two objects increases. This makes sense (to me), as the greater distance between the earth and an object for example, the greater distance there is to fall and thus more kinetic energy to build up. No problem.

With electric potential energy however, distance has the opposite effect, with potential energy decreasing as the distance between the two charged objects increases.

In the situation of two oppositely charged particles, why do we not consider the charges to be 'falling' towards each other like in the gravitational example, and thus the potential energy will be less the closer together they are when the potential is measured?

GPE and EPE are constantly being used as analogous examples, but the equations used to calculate them don't seem to be analogous at all to me, at least in the case where the charges are attracted and move towards each other.

2. Relevant equations

3. The attempt at a solution

Last edited: May 3, 2016
2. May 3, 2016

### BvU

Hi,

You are right that these two don't look similar at all. That's because you picked a rather local version of the GPE -- one that doesn't work on a scale where the earth isn't flat any more.

For force from gravity, Newton came to $$F_g =-{G m_a m_b\over r^2}$$ and the accompanying potential $$U_g = -{G m_a m_b\over r}$$ and that's more or lesss exactly the same as the expression you started with !

3. May 3, 2016

### stevendaryl

Staff Emeritus
Well, to be able to compare the two, you should consider the potential energy of oppositely-charged particles. If $q_a$ and $q_b$ are both positive, or both negative, then they repel each other, unlike gravity, which is an attractive force.

So if $q_a$ is positive, and $q_b$ is negative, then $\frac{k q_a q_b}{r}$ is a negative number. As $r$ gets bigger, it becomes the negative of a smaller number, which means that the potential is greater. Suppose $k q_a q_b$ = -1. Then with $r=1$, the potential is $-1$. With $r=2$, the potential is $-1/2$. And $-1/2 > -1$.

4. May 3, 2016

### califauna

I don't know what you mean by local version.

5. May 3, 2016

### califauna

The positive negative is the one I am considering already but its more that one I can't understand. The positive positive situation pretty much makes sense to me.

So, if the negative value is decreasing as the distance decreases, does that mean that infintely far away means infinite negative potential?

6. May 3, 2016

### stevendaryl

Staff Emeritus
The potential increases with increasing $r$, but the absolute value is decreasing, and goes to zero as $r$ goes to infinity. Once again, suppose that $k q_a q_b = -1$ for example (in whatever units). Then

1. When $r=1$, $U = -1$
2. When $r=2$, $U = -0.5$
3. When $r=3$, $U = -0.33$
4. When $r=4$, $U = -0.25$
5. etc.
Those numbers are increasing, because they are becoming less negative. The absolute values are decreasing: $1, 0.5, 0.33, 0.25, ...$. In the limit as $r \rightarrow \infty$, $U \rightarrow 0$

7. May 3, 2016

### BvU

On the surface of the earth you have, for a mass m: $$U_g(R) = -{G \, m_{\rm earth} \,m\over R}$$ with $R$ the radius of the earth. A few meters higher, say h meters higher, you are at a distance $R+h$ from the center of the earth and then the GPE is $$U_g(R+h) = -{G \,m_{\rm earth} \, m\over R+h}$$ The difference is $$U_g(R+h) - U_g(R) =\left (-{G \, m_{\rm earth} \, m\over R+h}\right ) -\left ( -{G\, m_{\rm earth} m\,\over R}\right ) = {G\, m_{\rm earth} \,m\over R} \left ( 1 - {R\over R+h} \right )$$and if we define $$g = {G\, m_{\rm earth}\over R^2}$$ (the numerical value is 9.82 m/s2 -- sound familiar ?)

and we know that with h/R << 1 we can write $1/(1+ h/R) = 1 - h/R$ to a very good approximation, then we get your GPE ! :$${\rm GPE } = U_g(R+h) - U_g(R) = m\,g\, h$$
(under the condition h << R, i.e. that you can ignore the fact that earth is not flat -- which is what I meant with 'local' : it only works if you stay close enough to the ground )​

8. May 3, 2016

### califauna

Ok, so as the distance increases the value gets closer to zero, and this means the electric potential energy is increasing. So why when we calculate the gravitational potential energy using the formula GPE=m*h*g, does the negative value get further away from zero as the distance above earth increases? Eg. if the potential energy at 1 meter above the surface is say -1 joules, then at 10 meters it will be -10 joules. Working it out that way, the greater the height the more negative the result, but working it out using the formula where you divide by distance, provided by BvU above, Ug=(-G*mamb) / r, the number doesn't get more negative but rather gets closer and closer to zero.

?

Last edited: May 3, 2016
9. May 3, 2016

### stevendaryl

Staff Emeritus
As others have pointed out, the formula $U = mhg$ is just an approximation to the formula $U = -\frac{GMm}{r}$. If you are a small distance $h$ above the surface of the Earth, and $R$ is the radius of the Earth, then $-\frac{GMm}{r}$ is appoximately $-\frac{GMm}{R} + \frac{GMm}{R^2} h$. The quantity $-\frac{GMm}{R}$ is some constant, $U_0$, and the quantity $\frac{GM}{R^2}$ is a constant, $g$. So the formula for potential energy near the Earth is approximately given by: $U = U_0 + mgh$. People just write $U = mgh$ because $U_0$ is a constant, and constants don't make any difference in potential energy, only the change in potential energy with distance.

But if you write $U = mgh$, that's a positive number, so it increases with $h$.