# Homework Help: On Electric and Gravitational Potential Energy

1. Sep 1, 2011

### Kyoma

1. The problem statement, all variables and given/known data
I have learned that gravitational potential energy's formula is:

GPE = mgh

But I saw other formulas relating to gravitational potential energy:

$\phi$ = -G(m/r)

And then one formula like this:

F = dU(x)/dx

And then there's a formula relating to the CHANGE in gravitational potential energy:

E = m$\Delta$$\phi$

Similarly, I'm confused with the electric potential energy:

V = (coulomb's constant)(Q/r) What is this? Why is there a voltage symbol?

CHANGE in electric potential energy:

E = QV What is this?

2. Sep 1, 2011

### Pi-Bond

The context of the formulas is different. The first one you mention (mgh) is true for places with constant gravitational field strength, for example, at the surface of the earth. $\phi$ is the gravitational potential (not potential energy!) i.e. potential energy per unit mass at distance r for a body of mass m.

The general formula for potential energy (for gravity or electrostatics) is:

$U(r)=\int_r^\infty \vec F.d\vec r$

i.e. the amount of work done by the field to move a mass/charge to infinity from a point r. The differential form of this yields the formula for force you mentioned. The change in potential energy between two points can also be derived in terms of $\phi$

3. Sep 1, 2011

### kuruman

This is the gravitational potential energy near the surface of the (usually) Earth.

This is gravitational potential (not gravitational potential energy), i.e. gravitational potential energy per unit mass. It has units of Joules/kg.

With a minus sign in front, this is a force not an energy that can be derived from the potential if you know what it is.

There should be term ΔE on the left side. This gives you the change in gravitational potential energy when mass m is moved from one point in space to another such the difference in gravitational potential is Δφ.

Electric potential, i.e. electric potential energy per unit charge has units of Joules/Coulomb. One Joule/Coulomb is also known as a "Volt".

This is (most likely) the change in potential energy of charge Q when it is moved from one point in space to another such the difference in electric potential (also called voltage) is V.

I know that this is all confusing; the underlying idea is that gravitational/electric energy per unit mass/charge in a region of space is a function of x,y and z called the gravitational/electric potential. If you take the spatial derivatives of that function and you evaluate at a particular point in space, you get the components of the gravitational/electric force acting on the mass/charge at that point in space.

4. Sep 1, 2011

### vela

Staff Emeritus
As the others noted, potential energy and just plain potential are two different quantities. The gravitational potential energy of a test mass m attracted to a point mass M is
$$U_g = -\frac{GMm}{r}$$
Similarly, the electrical potential energy of a test charge q interacting with a point charge Q is
$$U_e = \frac{kQq}{r}$$
They hold for point masses and point charges. (It turns out they also work for spherically symmetric distributions of mass and charge.)

The concept of a potential comes from the idea that we can separate out the effect of the test mass or charge.
\begin{align*}
U_g &= \left(-\frac{GM}{r}\right)m = m\phi_g \\
U_e &= \left(\frac{kQ}{r}\right)q = q\phi_e \\
\end{align*}
From these, you can see that the gravitational potential at a distance r from a point mass M is equal to $\phi_g = -GM/r$. Similarly, the electric potential at a distance r from a point charge Q is equal to $\phi_e = kQ/r$. These are your second and fifth equations. They apply when you're talking about a point charge or point mass.

Note that depending on the sign of the test charge q, the electrical potential energy at a point could be positive or negative; the potential at that point, however, always has the sign of Q. This is one reason it's often more convenient to work with potentials rather than potential energy in electromagnetism. With gravity, the masses are always positive, so the concept of a gravitational potential isn't as useful. You usually just work with gravitational potential energy directly.

Now say we move a mass from a point a distance ra from M to a point a distance rb away. The change in potential energy will be equal to$$\Delta U_g = U_{g_b} - U_{g_a} = m\phi_g(r_b) - m\phi_g(r_a) = m\Delta\phi_g$$Similarly, for charges, we have $$\Delta U_e = U_{e_b} - U_{e_a} = q\phi_e(r_b) - q\phi_e(r_a) = q\Delta\phi_e$$These are your fourth and sixth equations.

The concept of a potential, however, applies more generally than to just point masses or point charges. We can calculate (in principle) the potential Φg due to some arbitrary distribution of mass. The potential energy of a mass m due to this distribution would again be given by Ug = mΦg, and the change in potential energy by ΔUg = mΔΦg. Similarly, for a charge q, we have Ue = qΦe and ΔUe = qΔΦe. In other words, these equations apply generally, not just for point masses and charges.

So where does your first equation, Ug=mgh, fit in? Near the surface of the Earth, the gravitational potential energy of a mass m at a height h is given by
$$U_g = -\frac{GMm}{R+h}$$
where M is the mass of the Earth and R is the radius of the Earth. When h is much smaller than R, we can approximate the potential energy as
$$U_g = -\frac{GMm}{R+h} = -\frac{GMm}{R(1+h/R)} \cong -\frac{GMm}{R}\left(1-\frac{h}{R}\right) = -\frac{GMm}{R} + m\left(\frac{GM}{R^2}\right)h$$
The first term is a constant and constants don't matter in potential energy, so we can toss it. The quantity GM/R2 is more familiarly known as g=9.81 m/s2, so we end up with Ug=mgh. Note that this derivation relied on the assumption that h<<R, so it's only valid when you're analyzing situations near the surface of the Earth.

Finally, your third equation is missing a minus sign, as kuruman pointed out. As Pi-Bond noted, the potential energy due to a conservative force F of an object at a point r is generally given by$$U = -\int_{\mathbf{r}_0}^\mathbf{r} \mathbf{F}\cdot d\mathbf{r}$$where r0 is where the zero of potential energy is located. If you differentiate this equation, you get$$\mathbf{F} = -\nabla U$$which in the one-dimensional case is simply F = -dU/dx. If you were to try this with the expressions above for the gravitational and electrical potential energy (using the expression for the gradient in spherical coordinates), you would recover Newton's law of gravity and Coulomb's law.