10.1.3: Fields in Physics
There are three fields in which we will be interested for physics 7C:
 the Gravitational Field
 the Electric Field
 the Magnetic Field
Currently, the most familiar of these is the gravitational field, so the motivation for using fields will start here.
The Gravitational Field of Earth
Let us start by making a simple statement, which is very imprecise: the Earth’s gravity is stronger than the Moon’s gravity. Justification for this statement comes from watching videos of the astronauts on the Moon and who fall more slowly and leap higher than on Earth. In science we must be more precise. If we calculate the force of the Moon on the Apollo lander, this is much greater than the force of Earth on an apple. We see that we cannot make a blanket statement that the force of gravity on Earth is always greater than the force of gravity on the Moon.
The solution to comparing the Earth's and Moon's gravities is rather simple: we need to compare apples to apples. If we ask what \(\mathbf{F}_{\text{Earth on apple}}\) is and what \(\mathbf{F}_{\text{Moon on apple}}\) is then we find \(\mathbf{F}_{\text{Earth on apple}}> \mathbf{F}_{\text{Moon on apple}}\). More generally, it is true for any object \(X\): \[ \mathbf{F}_{\text{Earth on X}} \text{(at surface of Earth)} >  \mathbf{F}_{\text{Moon on X}} \text{(at surface of Moon)}\] This is a more precise version of what we mean when we say that the Earth’s gravity is stronger than the Moon’s gravity.
We can actually do a little bit better than this. The force of gravity does not distinguish between apples, oranges or skyscrapers. If we could build a skyscraper with the mass of an apple, \(\mathbf{F}_{\text{Earth on skyscraper}}\) would be the same as \(\mathbf{F}_{\text{Earth on apple}}\); to compare the strength of the gravitational field we don’t need to use exactly the same object, just two objects with the same mass.
Now let's come back to the gravitational field. Recall from Physics 7B that the force of gravity between two masses is \[ \mathbf{F}_{\text{mass 1 on mass 2}} = \dfrac{G M_1 M_2}{r^2} = M_2 \left( \dfrac{G M_1}{r^2} \right)\] where \(r\) is the distance between the centers of mass, and the direction of the force pulls the masses together. \(G\) is known as the universal gravitational constant, and is equal to \(6.67 \times 10^{−11} \text{ Nm}^2 \text{kg}^{−2}\). \(G\) is a universal constant meaning that it takes the same value regardless of the problem we are doing. Because \(G\) is so small, we do not notice the gravitational attraction of objects around us unless the object has an enormous mass.
Now let's ask the question “what would the force of the Earth be on an object of mass 1 kg located a distance \(r\) away?" We can calculate this:\[ \mathbf{F}_{\text{Earth on 1 kg object}}  = (1 kg) \times \left( \dfrac{G M_{Earth}}{r^2} \right) \]If we agree to always compare the gravity of an object by referring to what force it would exert on a second 1 kg mass then we can do this for any second mass. We find that for any object: \[ \mathbf{F}_{\text{Earth on object}} = M_{object} \left( \dfrac{G M_{Earth}}{r^2} \right) = M_{object} \mathbf{g}_{Earth}  \]The quantity on the right, which refers to the Earth and distance, is the gravitation field of the Earth. We denote this \(\mathbf{g}_{Earth}\):\[ \mathbf{g}_{Earth}  \equiv \dfrac{G}{M_{Earth}}{r^2} \]As this definition might suggest, \(\mathbf{g}_{Earth}\) is a vector field with units of acceleration that points towards the Earth. Once we know \(\mathbf{g}_{Earth}\) we can easily calculate the force on any other mass:\[\mathbf{F}_{\text{Earth on object}}  = M_{object}  \mathbf{g}_{Earth} \]We have seen this relationship many times before (this is why we chose to call the gravitational field \(\mathbf{g}_{Earth}\) rather than some other letter). Like all fields, the value of \(\mathbf{g}_{Earth}\) depends on position; it decreases with increased \(r\), which is the distance from the center of the Earth to the position of interest.
The Direct and Field Model of Forces
In the way that we have introduced the gravitational field the field is simply a shortcut. Instead of saying “the force a 1 kg object would feel, if placed here, is 5 N” we can simply say “the gravitational field here is 5 N/kg”. The field is not necessary to determine the gravitational force between two objects, it is simply convenient. We will see later that we actually need to talk about fields if we want energy and momentum to be conserved, but for now we will simply treat them as a shortcut.
With this in mind, we have two separate ways of discussing how a gravitational force acts between two objects. The first is where we calculate the force by putting numbers into Newton’s gravitational law without considering the field: the Direct Model: \[\text{Object #1} \xrightarrow{\text{creates force on}} \text{Object #2}\] The other way we could think about this is in our new language of fields; one mass creates the field and the other feels the effects: Field Model: \[\text{Object #1} \xrightarrow{\text{creates field}} \mathbf{g}_{\text{obj #1}} \xrightarrow{\text{exerts force on}} \text{Object #2}\]
Instead of thinking of one object directly exerting a force on another we think of one object (referred to as the “source”) creating a field and then that field creates a force on the second object (sometimes referred to as the “test object”). Of course, both methods are calculations of the same thing and yield the same answer, as the next Example will show:
Example #1
A. What gravitation force does the Earth exert on a 2 kg book on its surface?
B. What gravitation force does the Earth exert on the same book 2 km above its surface?
Use both the direct and field methods (The mass and radius of the Earth can be found online)
Solution
A. Direct Method
After looking up the mass and radius of the Earth we can use Newton's law: \[\mathbf{F}_{\text{Earth on book}} = \dfrac{GM_{Earth}M_{book}}{r^2_{Earth}}\] \[=\dfrac{(6.67 \times 10^{−11} \text{ N m}^2 \text{ kg}^{−2})(5.98 \times 10^{24} \text{ kg})(2 \text{ kg})}{(6380000 \text{ m})^2}\] \[= 19.6 \text{ N}\]
A. Field Method
We know that \(\mathbf{g}_{Earth} = 9.8 \text{ N/kg}\) at the surface of the Earth. Normally we approximate this to 10 N/kg, but let us be more precise for this example. \[\mathbf{F}_{\text{Earth on book}} = M_{book} \mathbf{g}_{Earth} = (2 \text{ kg})(9.8 \text{ N/kg}) = 19.6 \text{ N}\]
Notice how this calculation was much easier, since we already knew \(\mathbf{g}_{Earth}\).
B. Direct Method
This proceeds almost exactly the same as before. The two tricky points here are that we have to recall that \(r\) is the distance from the center of the Earth, and to change 10,000 km into meters.
\[\begin{align} \mathbf{F}_{\text{Earth on book}} &= \dfrac{GM_{Earth}M_{book}}{r_{Earth}+ 10,000 \text{ km})^2} \\[5pt] &=\dfrac{(6.67 \times 10^{−11} \text{ N m}^2 \text{ kg}^{−2})(5.98 \times 10^{24} \text{ kg})(2 \text{ kg})}{(6,380,000\text{ m} + 10,000,000 \text{ m})^2} \\[5pt] &= 3.0 \text{ N} \end{align}\]
B Field method
We don't know \(\mathbf{g}_{Earth}\) at a distance of 10,000 km from the surface of the Earth off the top of our head, we calculate it first. \[\mathbf{g}_{Eart} \text{(at 10,000 km from the surface)} = \dfrac{GM_{Earth}}{(r_{Earth} + 10,000 \text{ km})^2}\] \[=\dfrac{(6.67 \times 10^{−11} \text{ N m}^2 \text{ kg}^{−2})(5.98 \times 10^{24} \text{ kg})(2 \text{ kg})}{(6,380,000\text{ m} + 10,000,000 \text{ m})^2}\] \[= 1.5 \text{ N/kg}\] We can calculate the force the Earth exerts on the book: \[\mathbf{F}_{\text{Earth on book}} = M_{book} \mathbf{g}_{Earth} = (2 \text{ kg})(1.5 \text{ N/kg}) = 3.0 N\] Because we did not know \(\mathbf{g}_{Earth}\) before starting the problem the field method was longer. But if we were asked to do the same calculation for a different mass 10,000 km above the Earth’s surface we now have \(\mathbf{g}_{Earth}\) and could do it much quicker.
Which Mass Creates the Field? (Newton's Third Law)
In the example above, when using the field method we decided that the Earth would create the field and the book would respond to it. This seems quite acceptable, as we are used to the Earth exerting a gravitational force. But what if we decided to use a book and a chair in our example? Which would be the “source” for the gravitational field, and which would be pulled by the field?
To try and work out the answer to this question, let us think about the same problem using the direct model of forces. Calculating the magnitude of the force of the book on the chair gives \[\mathbf{F}_{\text{Book on chair}} = \dfrac{GM_{book}M_{chair}}{r^2}\] where \(r\) is the distance between them. Now let us calculate the force of the chair on the book: \[\mathbf{F}_{\text{Chair on book}} = \dfrac{GM_{book}M_{chair}}{r^2} = \mathbf{F}_{\text{Book on chair}}\] These forces have the same magnitude, but pull in opposite directions. This is not a coincidence, but is a consequence of Newton’s third law that we learned in 7B: \[\mathbf{F}_{\text{A on B}} =  \mathbf{F}_{\text{B on A}}\]
In the language of the field model we see that the answer is both the chair and the book create a field. To do the complete problem in the field model we would have to look at both the book's field acting on the chair and the chair's field acting on the book \[\text{Book} \xrightarrow{\text{creates field}} \mathbf{g}_{Book} \xrightarrow{\text{exerts force on}} \text{Chair}\] \[\text{Chair} \xrightarrow{\text{creates field}} \mathbf{g}_{Chair} \xrightarrow{\text{exerts force on}} \text{Book}\]
An important consequence of this is that to be affected by a field, an object must also create a field of the same type. Note that an object does not feel its own field, only the field of all external objects. In other words, the object does not selfinteract, but to feel an external gravitational field, for example, it must also create its own gravitational field.
While an object that feels a field must also create the same field, when we are emphasizing an object's ability to create a field we refer to it as the "source" of the field. When we emphasize an object's response to a field we call it the "test" object. We can usually adopt this convention when we can ignore the response of the source object to the field of the test object. However, as we just learned, both objects create a field and are affected by the other's field. For gravity, any object with mass creates and feels a gravitational field. For the electric field, any object with (electric) charge is a source and feels the field. For the magnetic field, as discussed later, the source is moving electric charges. Likewise, charges feel magnetic fields only when they are in motion.
Exercise
In Example #1 above, would we have to worry about the force of the book on the Earth? If not, why not?
Field Lines
We learned above that the gravitational field of earth is given by \(\mathbf{g}_{Earth} = \dfrac{GM_{Earth}}{r^2}\) where \(r\) is the distance of the test object from our source, Earth. A similar argument as the one above can be made for any spherical object with mass. We can see using this method hat the gravitational field created by a spherical mass at a distance \(r\) away is \[\mathbf{g}_{spherical} = \dfrac{GM_{spherical}}{r^2}\]This equation also applies to point masses. The direction of the gravitational field is always pulling inward. To represent the field, let's create a field map by picking a set of points and drawing vectors to indicate the direction and magnitude of the gravitational field:
The vectors only refer to the value of the fields at the location that they start. The actual length of the vectors is arbitrary, because we could always define some scale to convert the vectors to the right lengths and units. The ratio of lengths between two different vectors is not arbitrary – to accurately represent the vector field, all vectors should be scaled down for the field map exactly the same way.
If we look at the previous field map of the Earth’s gravitational field, we see that far from the Earth it is almost impossible to read the direction of those arrows. We could enlarge them, but then we'd need to scale up every arrow just as much. Some of the arrows near Earth are quite big, so this can become very messy. Furthermore, the size of the arrows limits the amount of information we could put on that part of the map.
To address these shortcomings, we introduce a different representation of a vector field by using field lines. To construct field lines, we draw a continuous lines starting at a point, always going along the direction of the field. An example for the gravitational field of the Earth is shown below:
Notice that the length of the arrows no longer corresponds to the strength of the field, so the strength cannot be read directly from looking at this picture. However this picture contains all the information as the field map. While we cannot look at the length of the arrows to get the strength of the field, we can instead look at how closely packed the field lines are. The closer together we see the lines (near the Earth, in this example) the stronger the field. The further apart the field lines are (far from the Earth), the weaker the field.
If we start with the field line diagram, we can reconstruct the field at any given point the following way:
 Direction: Take a tangent to the field line at that point. This is the direction of the vector field at that point.
 Magnitude: Given by the density of the surrounding lines (denser lines mean longer vectors).
While the vector map is the most direct representation of the field, we will return to using field lines when it's convenient.

Vector Map  Field Lines 

Pros 


Cons 


Gauss's Law
One thing that has not been made explicit in the discussion of field lines so far is that they cannot just start or stop in any location. Where the field lines start or stop depends on the field.
For the gravitational field, field lines start at far distances away and can only stop when they encounter a mass. If there are no masses in a particular region then field lines cannot be created or destroyed. The number of field lines that stop at a mass is proportional to the mass of the object encountered. Thus not all the field lines in our Earth example will “stop” at the Earth (they might hit a satellite, for example). Don’t worry about field lines passing through physical objects; remember that field lines they are only a representation of the field.
For the electric field, field lines start on positive charges and end on negative charges. This makes electric examples slightly more subtle, because if the number of field lines entering is the same as the number leaving it could indicate either that the region has no charge in it or it has an equal number of positive and negative charges. For electric fields looking at the number of field lines entering and exiting a region only tells us about the net charge in the region, not how many individual charges are in that region.
In either case, in regions with no mass (for gravity) or no net charge (for electric field lines), field lines cannot be created or destroyed. Therefore if the number of field lines entering a region is different from the number leaving we must have a field source (i.e. a mass or electric charge) in that region. By knowing the difference, we can figure out exactly how much of that source is in the region. This is the essence of Gauss's Law, which we now make more precise.
Think of a region inside an imaginary closed surface. In other words, imagine a shape with an inside that you cannot get out of unless you go through the surface. A box with a lid is a closed surface, but a cup is not. We can use any imaginary region we desire for Gauss's Law, provided that it is a closed surface.
Once we pick our region, we inspect how many field lines enter it, and how many leave. The only way a field line can get in (or out) is by going through this surface. If a different number of field lines come in than go out, then field lines are being created or destroyed inside. For gravitational field lines, this would indicate a mass; for electric field lines, this would indicate a net charge. If there is no difference between the field lines entering and leaving there is no mass inside or no net charge, and overall no field lines are being created or destroyed.
The amount of field lines going through the surface is referred to as the flux. An increase in flux, for example, can either be represented with more field lines, or by field lines that are drawn closer together. Gauss’s law is simply the statement that \[\text{Net flux} \propto \text{(# field lines entering)}  \text{(# field lines leaving)}\] \[\text{Net flux} \propto \begin{cases} \text{total mass inside surface} & \quad \text{(gravity)} \\ \text{total charge inside surface} & \quad \text{(electric)} \\ \end{cases}\]
A useful analogy to a closed surface and field lines is a leaky bucket filled with water. The bucket is not closed, but you can imagine a closed surface that immediately surrounds the bucket. If water is leaking out of the bucket, it is also leaking out from the inside of this imaginary surface. Therefore the water must be passing through the surface. If we measure the flux of water through our surface, we can measure the amount of water that has leaked out of our bucket.
Of course, you did not need consider imaginary surfaces to realize that the bucket is leaking. But this familiar example displays what we do with fields; if the field lines are coming out of a surface, then something is creating them. If we are losing field lines in a region, something is destroying them.
Gauss's Law and Magnetism?
So far Gauss’s law has been discussed for gravitational fields and electric fields, but no mention of magnetic fields has been made. That is because, to the best of our knowledge magnetic monopoles (pieces of magnetic "charge") do not exist. Instead all known magnetic fields are created by moving electric fields. As a consequence magnetic fields do not “start” or “end” but instead make complete loops. Because magnetic field lines never start or end the number of magnetic field lines entering a closed surface is always equal to the number of magnetic field lines leaving that surface.