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10.1.4: Potentials and Equipotentials

  • Page ID
    • Wendell Potter and David Webb et al.
    • UC Davis
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    In Physics 7A, we tied together the idea of potential energy and force. We learned that the magnitude of the force was given by the derivative (slope) of the graph of \(PE\) vs. \(r\). Just like a how forces exist between two objects, the potential energy is always an energy between two objects.

    Gravitational Potential

    What we would like to do in this unit is to talk about energy (or something like it), but only for one object. We got around this same problem with forces by introducing a new concept – fields – which told us the force on a 1 kg mass (for example) a distance \(r\) away. What to do with potential energy is now obvious; we should invent a new concept like fields that tells us what the potential energy would be for a 1 kg object a distance \(r\) from the source. The name for this new concept is potential and it is represented by \(U\). The name is an unfortunate choice because while it is closely related to the potential energy, they are not the same thing. For gravity, the relationship between the two is

    \[PE_{\text{grav between obj 1 and obj 2}} = M_1U_2\]

    This equation reminds us that it does not matter which object is considered the “source”, although in cases where one object is much bigger than the other it is conventional to treat the larger object as the source.

    For a point or spherical mass, the equation for the potential is relatively straight-forward:

    \[U_{grav} = -\dfrac{GM}{r} + U_0\]

    where\(r\) is the distance from the center of the mass creating the field. Here \(U_0\) is some arbitrary value. In Physics 7A we learned that absolute potential energy could not be measured; it was the changes in potential energy that could be observed. Likewise, Absolute potential cannot be measured either, instead only changes in potential are observable. If we add the same constant \(U_0\) to the potential at all locations, there is no experiment that can tell us what the value of \(U_0\) is. Furthermore, if we're only interested in the differences of potential energy at two different locations, then the value of \(U_0\) doesn't matter at all.

    Notice that far away from the mass (as \(r \rightarrow \infty\)) we have \(U_{grav} \approx U_0\). We will find it convenient to adopt the convention that the gravitational potential goes to zero a long way from the source. This corresponds to choosing \(U_0 = 0\), in which case the potential for our point or spherical mass is

    \[U_{grav} = - \dfrac{GM}{r}\]

    Notice that potential is negative, and because mass is always positive this tells us that the gravitational potential energy is also negative.

    Let us compare the gravitational potential energy (with zero at infinity) with the Lennard-Jones potential energy you looked at in 7A. There the \(PE\) went to zero as \(r\) became large (again, by convention), and because the potential energy was negative the total energy could be negative as well. If the total energy of a system was negative (the kinetic energy cannot be negative) this indicated the system was gravitationally bound.


    Starting with the equations presented above, can you derive the formula for the potential energy between two masses? Can you go from there to calculate the force between two objects? (See the diagrams in Relationships Between Concepts for help).


    Is the total mechanical energy of the Earth (i.e. \(KE + PE\)) positive, negative or zero? How can you tell? For this question, take \(PE = 0\)to be a very long way out of the solar system. (Hint: Remember that the Earth is orbiting the sun)


    An equipotential (i.e. “equal potential”) is the continuous curve along which every point has at the same potential. As a consequence, it takes no work to move along an equipotential; no forces pull or push in the direction of the equipotential. From this we can conclude that the force has no component along the direction of an equipotential.

    The most familiar example of equipotentials is height above the ground, as shown in the picture below. We know that the mass only gains or loses gravitational potential energy if the height changes, but moving it horizontally (i.e. along an equipotential) does not change the gravitational potential energy.


    Another example of equipotentials is the example of a topographical map from earlier. The contours show locations of constant height, and close to the Earth’s surface we have \[U = \dfrac{PE_{grav}}{m} = gh\] so lines of constant height \(h\) are also lines of constant \(U\).

    For the electric and the gravitational field, the force is always in the direction (or against the direction, for negative charges in an electric field) of the field lines. An equipotential cannot move with or against the field, as this would mean an object would gain or lose potential energy in the field. This means all equipotentials are at 90° to the field lines, and any given equipotential only intersects a given field line once. If an equipotential intersected a field line twice that would mean it was possible to move with (or against) the field and at the same time not change the potential energy of an object, which is impossible.

    We can use the fact that equipotentials and field lines are perpendicular to reconstruct one from the other. Let us take the Earth again, as it has been our example for all concepts in this chapter. We know that the equipotentials in this case (shown as dashed lines below) are all spheres because \(U\) only depends on \(r\); a sphere is a surface where each point has the same \(r\) value. Even if we did not know this, we would be able to reconstruct the equipotentials by drawing dashed lines perpendicular to the field lines as shown below:

    Screen Shot 2016-05-08 at 12.30.33 PM.png

    While every sphere we could draw this way is an equipotential, we choose to only draw selected equipotentials; we choose to draw equipotentials that are equally separated in potential, but not equally separated in space (\(r\) value). For example, the equipotentials shown in the above figure may be \(−6 \times 10^6 \text{ J/kg}\) for the one closest to the Earth, \(−5 \times 10^6 \text{ J/kg}\) for the middle equipotential and \(−4 \times 10^6 \text{ J/kg}\) for the outermost equipotential. We see that even though the equipotentials drawn change by \(1 \times 10^6 \text{ J/kg}\), they are not spaced evenly. As the equipotentials get further apart, we have to travel further with or against the field to get the same change in potential because the field gets weaker as \(r\) gets bigger. Finally, notice that if we only had the equipotentials (as on the diagram above on the right) we could completely reconstruct the field lines.

    Force and Equipotentials

    As we can see from the example of our topographical field, and our example above with Earth, if the equipotentials are close together, the field is stronger. In fact we can make this relationship precise:

    \[|\mathbf{g}| = \left| \dfrac{\text{d}U}{\text{d}r} \right| \approx \left| \dfrac{\Delta U}{\Delta r} \right|\]

    where \(\Delta U\) is the change in potential between two very close equipotentials, and \(\Delta r\) is the shortest distance between them. The direction of the field, \(\mathbf{g}\) in this case, points from high potential to low potential.

    Going back to the topographical map, contour lines with small distances \(\Delta r\) between them indicate steeper hills. Objects placed here would experience higher acceleration down the hill than objects placed elsewhere. The contour map is not an explanation of why the acceleration of an object would be greater (for that you should go back to force diagrams of a ball on a hill), but it is a convenient way of mapping the acceleration that a ball would feel. This is analogous to using gravitational equipotentials to display gravitational potential; they are a convenient description, but they do not explain why the field is the way it is. (While the hill serves as a good analogy, it is important to note that we are looking at the combined effect of gravity and the ground when discussing the acceleration of a ball. The gravitational field \(\mathbf{g}\) does not change significantly on a hill!)

    Example 2

    Two equipotentials close to the surface of the Earth have a potential difference of 1 J/kg. How far apart are they?


    We begin by taking \(\mathbf{g}_{Earth} = 10 \text{ m/s}^2\). We are interested in making steps of \(U = 1 \text{ J/kg}\) every equipotential. This tells us that the equipotentials are separated by \[\Delta r = \dfrac{|\Delta U|}{|\mathbf{g}_{Earth}|} = \dfrac{1 \text{ J/kg}}{10 \text{ m/s}^2} = 0.1 \text{ m}\]

    Screen Shot 2016-05-08 at 12.58.03 PM.png

    When we say the separation of the equipotentials is 10 cm, we mean given one equipotential, the one with the next lowest potential value is 10 cm closer to the ground. The fact that the distance traveled between equipotentials can get longer (for example, 50 cm as shown in the diagram) is completely irrelevant. The direction of the field is perpendicular to the equipotentials, going from a high equipotential to a low equipotential. In this case, the equipotentials are closest vertically and the potential decreases in height, leading to the (already known) conclusion that gravity points down.

    Example 3

    The 1 J/kg equipotentials at the surface of Pluto are separated by 1.6 meters. Is the gravitational field on the surface of Pluto stronger or weaker than the gravitational field at the surface of the Earth?


    The equipotentials are spaced further apart (larger \(\Delta r\)) for the same \(U\)). Therefore the gravitational field at the surface of Pluto is weaker than the gravitational field at the surface of the Earth.

    Notice that this does not explain why Pluto has a smaller gravitational field than Earth. To figure that out we would look at Pluto’s mass and size compared to Earth. But if someone has already calculated the field or the equipotentials for us, we can still use that information to answer useful questions.

    Potentials Do Not Always Exist

    As we learned in Physics 7A, the work done moving an object around is not a state function. This meant that the amount of work it took to move an object from one location to another could depend on more than the initial and final points; it could also depend on how you went from the initial to final point!

    If the amount of potential energy an electric charge loses while traveling through a field depends on the path it takes, then it would seem that the change in potential would depend on the path taken as well. But this makes no sense: the potential is defined at a particular point without a reference to a path. To calculate the change in potential we simply take the difference of the potential at the two end points!

    Therefore, if the change in potential energy of an object depends on the path taken, then the potential does not exist! (There are other things that can happen that can prevent a potential from existing as well). Let us show an example where it is impossible to construct a potential by showing it is impossible to construct an equipotential. Consider the vector map shown below:

    Screen Shot 2016-05-08 at 1.05.33 PM.png

    This is not completely artificial; as we learned in 7B the water velocity in a real pipe is like this. The friction on the sides reduces the velocity to zero at the edges, and the velocity is highest in the center. Our first attempt at constructing equipotentials will use the rule that the equipotentials are always perpendicular to the field lines. Because all the field lines are horizontal, our equipotentials will be vertical, as shown below.

    Screen Shot 2016-05-08 at 1.05.54 PM.png

    But there is a problem with this; as the field gets weaker (toward the edges) the equipotentials should be getting further apart. But the equipotentials cannot stay perpendicular to the field and get further apart: the first condition requires them to be vertical, the second requires them to bend. In this simple example, there are no equipotentials! This tells us that the potential does not exist either, and with a little bit of work you could show that the work required to move a charge through an electric field like this would depend on the path taken.

    The fact that we have two requirements that are not both automatically satisfied tell us that equipotentials only exist in very special circumstances. The cases that we are concerned with where potentials and equipotentials don’t exist are

    • A changing electric field (this is actually essential to induction). This is because the electric field creates closed loops, as pointed out later.
    • The magnetic field never has a potential, as the magnetic field cannot do work. We will learn that the magnetic force is always perpendicular to the direction a charge travels, so \(W = |\mathbf{F}| \Delta x\) must be zero.

    This page titled 10.1.4: Potentials and Equipotentials is shared under a not declared license and was authored, remixed, and/or curated by Wendell Potter and David Webb et al..

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