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# Chapter 22. The nonmechanical universe

“Okay. Your duties are as follows: Get Breen. I don't care how you get him, but get him soon. That faker! He posed for twenty years as a scientist without ever being apprehended. Well, I'm going to do some apprehending that'll make all previous apprehending look like no apprehension at all. You with me?”

“Yes,” said Battle, very much confused. “What's that thing you have?”

“Piggy-back heat-ray. You transpose the air in its path into an unstable isotope which tends to carry all energy as heat. Then you shoot your juice light, or whatever along the isotopic path and you burn whatever's on the receiving end. You want a few?”

“No,” said Battle. “I have my gats. What else have you got for offense and defense?” Underbottam opened a cabinet and proudly waved an arm. “Everything,” he said.

“Disintegraters, heat-rays, bombs of every type. And impenetrable shields of energy, massive and portable. What more do I need?”1 }

Cutting-edge science readily infiltrates popular culture, though sometimes in garbled form. The Newtonian imagination populated the universe mostly with that nice solid stuff called matter, which was made of little hard balls called atoms. In the early twentieth century, consumers of pulp fiction and popularized science began to hear of a new image of the universe, full of x-rays, N-rays, and Hertzian waves. What they were beginning to soak up through their skins was a drastic revision of Newton's concept of a universe made of chunks of matter which happened to interact via forces. In the newly emerging picture, the universe was made of force, or, to be more technically accurate, of ripples in universal fields of force. Unlike the average reader of Cosmic Stories in 1941, you now have enough technical background to understand what a “force field” really is.

## 22.1 The stage and the actors

a / This Global Positioning System (GPS) system, running on a smartphone attached to a bike's handlebar, depends on Einstein's theory of relativity. Time flows at a different rates aboard a GPS satellite than it does on the bike, and the GPS software has to take this into account.

b / The clock took up two seats, and two tickets were bought for it under the name of “Mr. Clock.”

c / Newton's laws do not distinguish past from future. The football could travel in either direction while obeying Newton's laws.

d / Fields carry energy.

e / Discussion question B.

### Newton's instantaneous action at a distance

The Newtonian picture has particles interacting with each other by exerting forces from a distance, and these forces are imagined to occur without any time delay. For example, suppose that super-powerful aliens, angered when they hear disco music in our AM radio transmissions, come to our solar system on a mission to cleanse the universe of our aesthetic contamination. They apply a force to our sun, causing it to go flying out of the solar system at a gazillion miles an hour. According to Newton's laws, the gravitational force of the sun on the earth will immediately start dropping off. This will be detectable on earth, and since sunlight takes eight minutes to get from the sun to the earth, the change in gravitational force will, according to Newton, be the first way in which earthlings learn the bad news --- the sun will not visibly start receding until a little later. Although this scenario is fanciful enough to be at home in the pages of Cosmic Stories, it shows a real feature of Newton's laws: that information can be transmitted from one place in the universe to another with zero time delay, so that transmission and reception occur at exactly the same instant.

Newton was sharp enough to realize that this required a nontrivial assumption, which was that there was some completely objective and well-defined way of saying whether two things happened at exactly the same instant. He stated this assumption explicitly: “Absolute, true, and mathematical time, of itself, and from its own nature flows at a constant rate without regard to anything external...”

### No absolute time

Ever since Einstein, we've known that this assumption was false. When Einstein first began to develop the theory of relativity, around 1905, the only real-world observations he could draw on were ambiguous and indirect. Today, the evidence is part of everyday life. For example, every time you use a GPS receiver, figure a, you're using Einstein's theory of relativity. Somewhere between 1905 and today, technology became good enough to allow conceptually simple experiments that students in the early 20th century could only discuss in terms like “Imagine that we could...” A good jumping-on point is 1971. In that year, J.C. Hafele and R.E. Keating brought atomic clocks aboard commercial airliners, b, and went around the world, once from east to west and once from west to east. Hafele and Keating observed that there was a discrepancy between the times measured by the traveling clocks and the times measured by similar clocks that stayed home at the U.S. Naval Observatory in Washington. The east-going clock lost time, ending up off by $$-59\pm10$$ nanoseconds, while the west-going one gained $$273\pm7$$ ns. Although this example is particularly dramatic, a large number of other experiments have also confirmed that time is not absolute, as Newton had imagined.

Nevertheless, the effects that Hafele and Keating observed were small. This makes sense: Newton's laws have already been thoroughly tested by experiments under a wide variety of conditions, so a new theory like relativity must agree with Newton's to a good approximation, within the Newtonian theory's realm of applicability. This requirement of backward-compatibility is known as the correspondence principle.

### Causality

It's also reassuring that the effects on time were small compared to the three-day lengths of the plane trips. There was therefore no opportunity for paradoxical scenarios such as one in which the east-going experimenter arrived back in Washington before he left and then convinced himself not to take the trip. A theory that maintains this kind of orderly relationship between cause and effect is said to satisfy causality.

Causality is like a water-hungry front-yard lawn in Los Angeles: we know we want it, but it's not easy to explain why. Even in plain old Newtonian physics, there is no clear distinction between past and future. In figure c, number 18 throws the football to number 25, and the ball obeys Newton's laws of motion. If we took a video of the pass and played it backward, we would see the ball flying from 25 to 18, and Newton's laws would still be satisfied. Nevertheless, we have a strong psychological impression that there is a forward arrow of time. I can remember what the stock market did last year, but I can't remember what it will do next year. Joan of Arc's military victories against England caused the English to burn her at the stake; it's hard to accept that Newton's laws provide an equally good description of a process in which her execution in 1431 caused her to win a battle in 1429. There is no consensus at this point among physicists on the origin and significance of time's arrow, and for our present purposes we don't need to solve this mystery. Instead, we merely note the empirical fact that, regardless of what causality really means and where it really comes from, its behavior is consistent. Specifically, experiments show that if an observer in a certain frame of reference observes that event A causes event B, then observers in other frames agree that A causes B, not the other way around. This is merely a generalization about a large body of experimental results, not a logically necessary assumption. If Keating had gone around the world and arrived back in Washington before he left, it would have disproved this statement about causality.

### Time delays in forces exerted at a distance

Relativity is closely related to electricity and magnetism, and we will go into relativity in more detail in chapters 24-27. What we care about for now is that relativity forbids Newton's instantaneous action at a distance. For suppose that instantaneous action at a distance existed. It would then be possible to send signals from one place in the universe to another without any time lag. This would allow perfect synchronization of all clocks. But the Hafele-Keating experiment demonstrates that clocks A and B that have been initially synchronized will drift out of sync if one is in motion relative to the other.2 With instantaneous transmission of signals, we could determine, without having to wait for A and B to be reunited, which was ahead and which was behind. Since they don't need to be reunited, neither one needs to undergo any acceleration; each clock can fix an inertial frame of reference, with a velocity vector that changes neither its direction nor its magnitude. But this violates the principle that constant-velocity motion is relative, because each clock can be considered to be at rest, in its own frame of reference. Since no experiment has ever detected any violation of the relativity of motion, we conclude that instantaneous action at a distance is impossible.

Since forces can't be transmitted instantaneously, it becomes natural to imagine force-effects spreading outward from their source like ripples on a pond, and we then have no choice but to impute some physical reality to these ripples. We call them fields, and they have their own independent existence. Chapters 22-24 are mainly about the electric and magnetic fields, although we'll also talk about the gravitational field. Ripples of the electric and magnetic fields turn out to be light waves. Fields don't have to wiggle; they can hold still as well. The earth's magnetic field, for example, is nearly constant, which is why we can use it for direction-finding.

Even empty space, then, is not perfectly featureless. It has measurable properties. For example, we can drop a rock in order to measure the direction of the gravitational field, or use a magnetic compass to find the direction of the magnetic field. This concept made a deep impression on Einstein as a child. He recalled that as a five-year-old, the gift of a magnetic compass convinced him that there was “something behind things, something deeply hidden.”

### More evidence that fields of force are real: they carry energy.

The smoking-gun argument for this strange notion of traveling force ripples comes from the fact that they carry energy. In figure d/1, Alice and Betty hold positive charges A and B at some distance from one another. If Alice chooses to move her charge closer to Betty's, d/2, Alice will have to do some mechanical work against the electrical repulsion, burning off some of the calories from that chocolate cheesecake she had at lunch. This reduction in her body's chemical energy is offset by a corresponding increase in the electrical potential energy $$q\Delta V$$. Not only that, but Alice feels the resistance stiffen as the charges get closer together and the repulsion strengthens. She has to do a little extra work, but this is all properly accounted for in the electrical potential energy.

But now suppose, d/3, that Betty decides to play a trick on Alice by tossing charge B far away just as Alice is getting ready to move charge A. We have already established that Alice can't feel charge B's motion instantaneously, so the electric forces must actually be propagated by an electric field. Of course this experiment is utterly impractical, but suppose for the sake of argument that the time it takes the change in the electric field to propagate across the diagram is long enough so that Alice can complete her motion before she feels the effect of B's disappearance. She is still getting stale information about B's position. As she moves A to the right, she feels a repulsion, because the field in her region of space is still the field caused by B in its old position. She has burned some chocolate cheesecake calories, and it appears that conservation of energy has been violated, because these calories can't be properly accounted for by any interaction with B, which is long gone.

If we hope to preserve the law of conservation of energy, then the only possible conclusion is that the electric field itself carries away the cheesecake energy. In fact, this example represents an impractical method of transmitting radio waves. Alice does work on charge A, and that energy goes into the radio waves. Even if B had never existed, the radio waves would still have carried energy, and Alice would still have had to do work in order to create them.

##### Discussion Questions

Amy and Bill are flying on spaceships in opposite directions at such high velocities that the relativistic effect on time's rate of flow is easily noticeable. Motion is relative, so Amy considers herself to be at rest and Bill to be in motion. She says that time is flowing normally for her, but Bill is slow. But Bill can say exactly the same thing. How can they both think the other is slow? Can they settle the disagreement by getting on the radio and seeing whose voice is normal and whose sounds slowed down and Darth-Vadery?

The rod in the figure is perfectly rigid. At event A, the hammer strikes one end of the rod. At event B, the other end moves. Since the rod is perfectly rigid, it can't compress, so A and B are simultaneous. In frame 2, B happens before A. Did the motion at the right end cause the person on the left to decide to pick up the hammer and use it?

## 22.2 The gravitational field

f / The wind patterns in a certain area of the ocean could be charted in a “sea of arrows” representation like this. Each arrow represents both the wind's strength and its direction at a certain location.

g / The gravitational field surrounding a clump of mass such as the earth.

h / The gravitational fields of the earth and moon superpose. Note how the fields cancel at one point, and how there is no boundary between the interpenetrating fields surrounding the two bodies.

Given that fields of force are real, how do we define, measure, and calculate them? A fruitful metaphor will be the wind patterns experienced by a sailing ship. Wherever the ship goes, it will feel a certain amount of force from the wind, and that force will be in a certain direction. The weather is ever-changing, of course, but for now let's just imagine steady wind patterns. Definitions in physics are operational, i.e., they describe how to measure the thing being defined. The ship's captain can measure the wind's “field of force” by going to the location of interest and determining both the direction of the wind and the strength with which it is blowing. Charting all these measurements on a map leads to a depiction of the field of wind force like the one shown in the figure. This is known as the “sea of arrows” method of visualizing a field.

Now let's see how these concepts are applied to the fundamental force fields of the universe. We'll start with the gravitational field, which is the easiest to understand. As with the wind patterns, we'll start by imagining gravity as a static field, even though the existence of the tides proves that there are continual changes in the gravity field in our region of space. Defining the direction of the gravitational field is easy enough: we simply go to the location of interest and measure the direction of the gravitational force on an object, such as a weight tied to the end of a string.

But how should we define the strength of the gravitational field? Gravitational forces are weaker on the moon than on the earth, but we cannot specify the strength of gravity simply by giving a certain number of newtons. The number of newtons of gravitational force depends not just on the strength of the local gravitational field but also on the mass of the object on which we're testing gravity, our “test mass.” A boulder on the moon feels a stronger gravitational force than a pebble on the earth. We can get around this problem by defining the strength of the gravitational field as the force acting on an object, divided by the object's mass.

##### definition of the gravitational field

The gravitational field vector, $$\mathbf{g}$$, at any location in space is found by placing a test mass $$m_t$$ at that point. The field vector is then given by $$\mathbf{g}=\mathbf{F}/m_t$$, where $$\mathbf{F}$$ is the gravitational force on the test mass.

The magnitude of the gravitational field near the surface of the earth is about 9.8 N/kg, and it's no coincidence that this number looks familiar, or that the symbol $$\mathbf{g}$$ is the same as the one for gravitational acceleration. The force of gravity on a test mass will equal $$m_t\mathbf{g}$$, where $$\mathbf{g}$$ is the gravitational acceleration. Dividing by $$m_t$$ simply gives the gravitational acceleration. Why define a new name and new units for the same old quantity? The main reason is that it prepares us with the right approach for defining other fields.

The most subtle point about all this is that the gravitational field tells us about what forces would be exerted on a test mass by the earth, sun, moon, and the rest of the universe, if we inserted a test mass at the point in question. The field still exists at all the places where we didn't measure it.

##### Example 1: Gravitational field of the earth
$$\triangleright$$ What is the magnitude of the earth's gravitational field, in terms of its mass, $$M$$, and the distance $$r$$ from its center?

$$\triangleright$$ Substituting $$|\mathbf{F}|= GMm_{t}/ r^2$$ into the definition of the gravitational field, we find $$|\mathbf{g}|= GM/ r^2$$. This expression could be used for the field of any spherically symmetric mass distribution, since the equation we assumed for the gravitational force would apply in any such case.

### Sources and sinks

If we make a sea-of-arrows picture of the gravitational fields surrounding the earth, g, the result is evocative of water going down a drain. For this reason, anything that creates an inward-pointing field around itself is called a sink. The earth is a gravitational sink. The term “source” can refer specifically to things that make outward fields, or it can be used as a more general term for both “outies” and “innies.” However confusing the terminology, we know that gravitational fields are only attractive, so we will never find a region of space with an outward-pointing field pattern.

Knowledge of the field is interchangeable with knowledge of its sources (at least in the case of a static, unchanging field). If aliens saw the earth's gravitational field pattern they could immediately infer the existence of the planet, and conversely if they knew the mass of the earth they could predict its influence on the surrounding gravitational field.

### Superposition of fields

A very important fact about all fields of force is that when there is more than one source (or sink), the fields add according to the rules of vector addition. The gravitational field certainly will have this property, since it is defined in terms of the force on a test mass, and forces add like vectors. Superposition is an important characteristics of waves, so the superposition property of fields is consistent with the idea that disturbances can propagate outward as waves in a field.

##### Example 2: Reduction in gravity on Io due to Jupiter's gravity

$$\triangleright$$ The average gravitational field on Jupiter's moon Io is 1.81 N/kg. By how much is this reduced when Jupiter is directly overhead? Io's orbit has a radius of $$4.22\times10^8$$ m, and Jupiter's mass is $$1.899\times10^{27}$$ kg.

$$\triangleright$$ By the shell theorem, we can treat the Jupiter as if its mass was all concentrated at its center, and likewise for Io. If we visit Io and land at the point where Jupiter is overhead, we are on the same line as these two centers, so the whole problem can be treated one-dimensionally, and vector addition is just like scalar addition. Let's use positive numbers for downward fields (toward the center of Io) and negative for upward ones. Plugging the appropriate data into the expression derived in example 1, we find that the Jupiter's contribution to the field is $$-0.71$$ N/kg. Superposition says that we can find the actual gravitational field by adding up the fields created by Io and Jupiter: $$1.81-0.71$$ N/kg = 1.1 N/kg. You might think that this reduction would create some spectacular effects, and make Io an exciting tourist destination. Actually you would not detect any difference if you flew from one side of Io to the other. This is because your body and Io both experience Jupiter's gravity, so you follow the same orbital curve through the space around Jupiter.

### Gravitational waves

Looking back at the argument given on p. 618 for the existence of energy-bearing ripples in the electric field, we see that nowhere was it necessary to appeal to any specific properties of the electrical interaction. We therefore expect energy-carrying gravitational waves to exist, and Einstein's general theory of relativity does describe such waves and their properties. J.H. Taylor and R.A. Hulse were awarded the Nobel Prize in 1993 for giving indirect evidence to confirm Einstein's prediction. They discovered a pair of exotic, ultra-dense stars called neutron stars orbiting one another very closely, and showed that they were losing orbital energy at the rate predicted by general relativity.

i / The part of the LIGO gravity wave detector at Hanford Nuclear Reservation, near Richland, Washington. The other half of the detector is in Louisiana.

A Caltech-MIT collaboration has built a pair of gravitational wave detectors called LIGO to search for more direct evidence of gravitational waves. Since they are essentially the most sensitive vibration detectors ever made, they are located in quiet rural areas, and signals will be compared between them to make sure that they were not due to passing trucks. The project began operating at full sensitivity in 2005, and is now able to detect a vibration that causes a change of $$10^{-18}$$ m in the distance between the mirrors at the ends of the 4-km vacuum tunnels. This is a thousand times less than the size of an atomic nucleus! There is only enough funding to keep the detectors operating for a few more years, so the physicists can only hope that during that time, somewhere in the universe, a sufficiently violent cataclysm will occur to make a detectable gravitational wave. (More accurately, they want the wave to arrive in our solar system during that time, although it will have been produced millions of years before.)

## 22.3 The electric field

j / Example 3.

k / A Geiger-Müller tube.

l / Proof that the electric field of a line of charge is proportional to $$1/r$$, example 4.

n / A dipole field. Electric fields diverge from a positive charge and converge on a negative charge.

o / A water molecule is a dipole.

p / Example 7.

q / Discussion question H.

### Definition

The definition of the electric field is directly analogous to, and has the same motivation as, the definition of the gravitational field:

##### definition of the electric field

The electric field vector, $$\mathbf{E}$$, at any location in space is found by placing a test charge $$q_t$$ at that point. The electric field vector is then given by $$\mathbf{E}=\mathbf{F}/q_t$$, where $$\mathbf{F}$$ is the electric force on the test charge.

Charges are what create electric fields. Unlike gravity, which is always attractive, electricity displays both attraction and repulsion. A positive charge is a source of electric fields, and a negative one is a sink.

The most difficult point about the definition of the electric field is that the force on a negative charge is in the opposite direction compared to the field. This follows from the definition, since dividing a vector by a negative number reverses its direction. It's as though we had some objects that fell upward instead of down.

self-check:

Find an equation for the magnitude of the field of a single point charge $$Q$$.

##### Example 3: Superposition of electric fields
$$\triangleright$$ Charges $$q$$ and $$- q$$ are at a distance $$b$$ from each other, as shown in the figure. What is the electric field at the point P, which lies at a third corner of the square?

$$\triangleright$$ The field at P is the vector sum of the fields that would have been created by the two charges independently. Let positive $$x$$ be to the right and let positive $$y$$ be up.

Negative charges have fields that point at them, so the charge $$-q$$ makes a field that points to the right, i.e., has a positive $$x$$ component. Using the answer to the self-check, we have

\begin{align*} E_{-q,x} &= \frac{ kq}{ b^2} \\ E_{-q,y} &= 0 . \end{align*}

Note that if we had blindly ignored the absolute value signs and plugged in $$- q$$ to the equation, we would have incorrectly concluded that the field went to the left.

By the Pythagorean theorem, the positive charge is at a distance $$\sqrt{2} b$$ from P, so the magnitude of its contribution to the field is $$E= kq/2 b^2$$. Positive charges have fields that point away from them, so the field vector is at an angle of 135° counterclockwise from the $$x$$ axis.

\begin{align*} E_{q,x} &= \frac{ kq}{2 b^2} \text{cos}\ 135° \\ &= -\frac{ kq}{2^\text{3/2} b^2} \\ E_{q,y} &= \frac{ kq}{2 b^2} \text{sin}\ 135° \\ &= \frac{ kq}{2^\text{3/2} b^2} \end{align*}

The total field is

\begin{align*} E_\text{x} &= \left(1-2^{-\text{3/2}}\right)\frac{ kq}{ b^2} \\ E_{y} &= \frac{ kq}{2^\text{3/2} b^2} . \end{align*}

##### Example 4: A line of charge
In a complete circuit, there is typically no net charge on any of the wires. However, there are some devices in which a circuit is intentionally left open in order to produce a nonzero net charge on a wire. One example is a type of radiation detector called a Geiger-Müller tube, figure k. A high high voltage is applied between the outside of the cylinder and the wire that runs along the central axis. A net positive charge builds up on the wire and a negative one on the cylinder's wall. Electric fields originate from the wire, spread outward from the axis, and terminate on the wall. The cylinder is filled with a low-pressure inert gas. An incoming particle of radioactivity strikes an atom of the gas, ionizing it, i.e., splitting it into positively and negatively charged parts, known as ions. These ions then accelerate in opposite directions, since the force exerted by an electric field on a charged particle flips directions when the charge is reversed. The ions accelerate up to speeds at which they are capable of ionizing other atoms when they collide with them. The result is an electrical avalanche that causes a disturbance on the voltmeter.

Motivated by this example, we would like to find how the field of a long, uniformly charged wire varies with distance. In figure l/1, the point P experiences a field that is the vector sum of contributions such as the one coming from the segment $$q$$. The field $$\mathbf{E}_q$$ arising from this segment has to be added to similar contributions from all other segments of the wire. By symmetry, the total field will end up pointing at a right angle to the wire. We now consider point $$\text{P}'$$, figure l/2, at twice the distance from the wire. If we reproduce all the angles from l/1, then the new triangle is simply a copy of the old one that has been scaled up by a factor of two. The left side's length has doubled, so $$q'=2q$$, and this would tend to make $$\mathbf{E}_{q'}$$ twice as big. But all the distances have also been doubled, and the $$1/r^2$$ in Coulomb's law therefore contributes an additional factor of 1/4. Combining these two factors, we find $$\mathbf{E}_{q'}=\mathbf{E}_q/2$$. The total field is the sum of contributions such as $$\mathbf{E}_{q'}$$, so if all of these have been weakened by a factor of two, the same must apply to the total as well. There was nothing special about the number 2, so we conclude that in general the electric field of a line of charge is proportional to $$1/r$$.

Applying this to the Geiger-Müller tube, we can see the reason why the device is built with a wire. When $$r$$ is small, $$1/r$$ is big, and the field is very strong. Therefore the device can be sensitive enough to trigger an avalanche in the gas when only a single atom has been ionized.

We have only shown that the field is proportional to $$1/r$$, but we haven't filled in the other factors in the equation. This is done in example 14 on p. 637.

### Dipoles

The simplest set of sources that can occur with electricity but not with gravity is the dipole, consisting of a positive charge and a negative charge with equal magnitudes. More generally, an electric dipole can be any object with an imbalance of positive charge on one side and negative on the other. A water molecule, o, is a dipole because the electrons tend to shift away from the hydrogen atoms and onto the oxygen atom.

m / 1. A uniform electric field created by some charges “off-stage.” 2. A dipole is placed in the field. 3. The dipole aligns with the field.

Your microwave oven acts on water molecules with electric fields. Let us imagine what happens if we start with a uniform electric field, m/1, made by some external charges, and then insert a dipole, m/2, consisting of two charges connected by a rigid rod. The dipole disturbs the field pattern, but more important for our present purposes is that it experiences a torque. In this example, the positive charge feels an upward force, but the negative charge is pulled down. The result is that the dipole wants to align itself with the field, m/3. The microwave oven heats food with electrical (and magnetic) waves. The alternation of the torque causes the molecules to wiggle and increase the amount of random motion. The slightly vague definition of a dipole given above can be improved by saying that a dipole is any object that experiences a torque in an electric field.

What determines the torque on a dipole placed in an externally created field? Torque depends on the force, the distance from the axis at which the force is applied, and the angle between the force and the line from the axis to the point of application. Let a dipole consisting of charges $$+q$$ and $$-q$$ separated by a distance $$\ell$$ be placed in an external field of magnitude $$|\mathbf{E}|$$, at an angle $$\theta$$ with respect to the field. The total torque on the dipole is

\begin{align*} \tau &= \frac{\ell}{2}q|\mathbf{E}|\sin \theta+\frac{\ell}{2}q|\mathbf{E}|\sin \theta \\ &= \ell q|\mathbf{E}|\sin \theta . \end{align*}

(Note that even though the two forces are in opposite directions, the torques do not cancel, because they are both trying to twist the dipole in the same direction.) The quantity $$\ell q$$ is called the dipole moment, notated $$D$$. (More complex dipoles can also be assigned a dipole moment --- they are defined as having the same dipole moment as the two-charge dipole that would experience the same torque.)

##### Example 5: Dipole moment of a molecule of NaCl gas

$$\triangleright$$ In a molecule of NaCl gas, the center-to-center distance between the two atoms is about 0.6 nm. Assuming that the chlorine completely steals one of the sodium's electrons, compute the magnitude of this molecule's dipole moment.

$$\triangleright$$ The total charge is zero, so it doesn't matter where we choose the origin of our coordinate system. For convenience, let's choose it to be at one of the atoms, so that the charge on that atom doesn't contribute to the dipole moment. The magnitude of the dipole moment is then

\begin{align*} D &= (6\times10^{-10}\ \text{m})( e) \\ &= (6\times10^{-10}\ \text{m})( 1.6\times10^{-19}\ \text{C}) \\ &= 1\times10^{-28}\ \text{C}\cdot\text{m} \end{align*}

### Alternative definition of the electric field

The behavior of a dipole in an externally created field leads us to an alternative definition of the electric field:

##### alternative definition of the electric field

The electric field vector, $$E$$, at any location in space is defined by observing the torque exerted on a test dipole $$D_t$$ placed there. The direction of the field is the direction in which the field tends to align a dipole (from $$-$$ to +), and the field's magnitude is $$|\mathbf{E}|=\tau/D_t\sin\theta$$.

The main reason for introducing a second definition for the same concept is that the magnetic field is most easily defined using a similar approach.

### Voltage related to electric field

Voltage is potential energy per unit charge, and electric field is force per unit charge. We can therefore relate voltage and field if we start from the relationship between potential energy and force,

$\begin{multline*} \Delta PE = -Fd , \shoveright{\text{[assuming constant force and}}\\ \text{motion parallel to the force]} \end{multline*}$

and divide by charge,

$\begin{multline*} \frac{\Delta PE}{q} = -\frac{F}{q}d , \end{multline*}$

giving

$\begin{multline*} \Delta V = -Ed , \shoveright{\text{[assuming constant force and}}\\ \text{motion parallel to the force]} \end{multline*}$

In other words, the difference in voltage between two points equals the electric field strength multiplied by the distance between them. The interpretation is that a strong electric field is a region of space where the voltage is rapidly changing. By analogy, a steep hillside is a place on the map where the altitude is rapidly changing.

##### Example 6: Field generated by an electric eel

$$\triangleright$$ Suppose an electric eel is 1 m long, and generates a voltage difference of 1000 volts between its head and tail. What is the electric field in the water around it?

$$\triangleright$$ We are only calculating the amount of field, not its direction, so we ignore positive and negative signs. Subject to the possibly inaccurate assumption of a constant field parallel to the eel's body, we have

$\begin{equation*} |\mathbf{E}| = \frac{\Delta V}{\Delta x} = 1000\ \text{V/m} . \end{equation*}$

##### Example 7: The hammerhead shark
One of the reasons hammerhead sharks have their heads shaped the way they do is that, like quite a few other fish, they can sense electric fields as a way of finding prey, which may for example be hidden in the sand. From the equation $$E=\Delta V/\Delta x$$, we can see that if the shark is sensing the voltage difference between two points, it will be able to detect smaller electric fields if those two points are farther apart. The shark has a network of sensory organs, called the ampullae of Lorenzini, on the skin of its head. Since the network is spread over a wider head, the $$\Delta x$$ is larger. Some sharks can detect electric fields as weak as 50 picovolts per meter!
##### Example 8: Relating the units of electric field and voltage

From our original definition of the electric field, we expect it to have units of newtons per coulomb, N/C. The example above, however, came out in volts per meter, V/m. Are these inconsistent? Let's reassure ourselves that this all works. In this kind of situation, the best strategy is usually to simplify the more complex units so that they involve only mks units and coulombs. Since voltage is defined as electrical energy per unit charge, it has units of J/C:

$\begin{equation*} \frac{\text{V}}{\text{m}} = \frac{\text{J/C}}{\text{m}} = \frac{\text{J}}{\text{C}\cdot\text{m}} . \end{equation*}$

To connect joules to newtons, we recall that work equals force times distance, so $$\text{J}=\text{N}\cdot\text{m}$$, so

$\begin{equation*} \frac{\text{V}}{\text{m}} = \frac{\text{N}\cdot\text{m}}{\text{C}\cdot\text{m}} = \frac{\text{N}}{\text{C}} \end{equation*}$

As with other such difficulties with electrical units, one quickly begins to recognize frequently occurring combinations.

##### Discussion Questions

In the definition of the electric field, does the test charge need to be 1 coulomb? Does it need to be positive?

Does a charged particle such as an electron or proton feel a force from its own electric field?

Is there an electric field surrounding a wall socket that has nothing plugged into it, or a battery that is just sitting on a table?

In a flashlight powered by a battery, which way do the electric fields point? What would the fields be like inside the wires? Inside the filament of the bulb?

Criticize the following statement: “An electric field can be represented by a sea of arrows showing how current is flowing.”

The field of a point charge, $$|\mathbf{E}|=kQ/r^2$$, was derived in the self-check above. How would the field pattern of a uniformly charged sphere compare with the field of a point charge?

The interior of a perfect electrical conductor in equilibrium must have zero electric field, since otherwise the free charges within it would be drifting in response to the field, and it would not be in equilibrium. What about the field right at the surface of a perfect conductor? Consider the possibility of a field perpendicular to the surface or parallel to it.

Compare the dipole moments of the molecules and molecular ions shown in the figure.

Small pieces of paper that have not been electrically prepared in any way can be picked up with a charged object such as a charged piece of tape. In our new terminology, we could describe the tape's charge as inducing a dipole moment in the paper. Can a similar technique be used to induce not just a dipole moment but a charge?

The earth and moon are fairly uneven in size and far apart, like a baseball and a ping-pong ball held in your outstretched arms. Imagine instead a planetary system with the character of a double planet: two planets of equal size, close together. Sketch a sea of arrows diagram of their gravitational field.

## 22.4 Calculating energy in fields

r / Example 9.

s / Example 10.

t / Discussion question A.

We found on p. 618 that fields have energy, and we now know as well that fields act like vectors. Presumably there is a relationship between the strength of a field and its energy density. Flipping the direction of the field can't change the density of energy, which is a scalar and therefore has no direction in space. We therefore expect that the energy density to be proportional to the square of the field, so that changing $$\mathbf{E}$$ to $$-\mathbf{E}$$ has no effect on the result. This is exactly what we've already learned to expect for waves: the energy depends on the square of the amplitude. The relevant equations for the gravitational and electric fields are as follows:

\begin{align*} (\text{energy stored in the gravitational field per \text{m}^3}) &= -\frac{1}{8\pi G}|\mathbf{g}|^2\\ (\text{energy stored in the electric field per \text{m}^3}) &= \frac{1}{8\pi k}|\mathbf{E}|^2\\ \end{align*}

A similar expression is given on p. 679 for the magnetic field.

Although funny factors of $$8\pi$$ and the plus and minus signs may have initially caught your eye, they are not the main point. The important idea is that the energy density is proportional to the square of the field strength in all cases. We first give a simple numerical example and work a little on the concepts, and then turn our attention to the factors out in front.

In chapter 22 when we discussed the original reason for introducing the concept of a field of force, a prime motivation was that otherwise there was no way to account for the energy transfers involved when forces were delayed by an intervening distance. We used to think of the universe's energy as consisting of

\begin{align*} &\text{kinetic energy}\\ + &\text{gravitational potential energy based on the distances between}\\ &\text{objects that interact gravitationally}\\ + &\text{electric potential energy based on the distances between}\\ &\text{objects that interact electrically}\\ + &\text{magnetic potential energy based on the distances between}\\ &\text{objects that interact magnetically} ,\\ \text{but in nonstatic situations we must use a different method:} &\text{kinetic energy}\\ + &\text{gravitational potential energy stored in gravitational fields}\\ + &\text{electric potential energy stored in electric fields}\\ + &\text{magnetic potential stored in magnetic fields} \end{align*}

Surprisingly, the new method still gives the same answers for the static cases.

##### Example 9: Energy stored in a capacitor
A pair of parallel metal plates, seen from the side in figure r, can be used to store electrical energy by putting positive charge on one side and negative charge on the other. Such a device is called a capacitor. (We have encountered such an arrangement previously, but there its purpose was to deflect a beam of electrons, not to store energy.)

In the old method of describing potential energy, 1, we think in terms of the mechanical work that had to be done to separate the positive and negative charges onto the two plates, working against their electrical attraction. The new description, 2, attributes the storage of energy to the newly created electric field occupying the volume between the plates. Since this is a static case, both methods give the same, correct answer.

##### Example 10: Potential energy of a pair of opposite charges
Imagine taking two opposite charges, s, that were initially far apart and allowing them to come together under the influence of their electrical attraction.

According to the old method, potential energy is lost because the electric force did positive work as it brought the charges together. (This makes sense because as they come together and accelerate it is their potential energy that is being lost and converted to kinetic energy.)

By the new method, we must ask how the energy stored in the electric field has changed. In the region indicated approximately by the shading in the figure, the superposing fields of the two charges undergo partial cancellation because they are in opposing directions. The energy in the shaded region is reduced by this effect. In the unshaded region, the fields reinforce, and the energy is increased.

It would be quite a project to do an actual numerical calculation of the energy gained and lost in the two regions (this is a case where the old method of finding energy gives greater ease of computation), but it is fairly easy to convince oneself that the energy is less when the charges are closer. This is because bringing the charges together shrinks the high-energy unshaded region and enlarges the low-energy shaded region.

##### Example 11: Energy transmitted by ripples in the electric and magnetic fields

We'll see in chapter 24 that phenomena like light, radio waves, and x-rays are all ripples in the electric and magnetic fields. The old method would give zero energy for a region of space containing a light wave but no charges. That would be wrong! We can only use the old method in static cases.

Now let's give at least some justification for the other features of the expressions for energy density, $$-\frac{1}{8\pi G}|\mathbf{g}|^2$$ and $$\frac{1}{8\pi k}|\mathbf{E}|^2$$, besides the proportionality to the square of the field strength.

First, why the different plus and minus signs? The basic idea is that the signs have to be opposite in the gravitational and electric cases because there is an attraction between two positive masses (which are the only kind that exist), but two positive charges would repel. Since we've already seen examples where the positive sign in the electric energy makes sense, the gravitational energy equation must be the one with the minus sign.

It may also seem strange that the constants $$G$$ and $$k$$ are in the denominator. They tell us how strong the three different forces are, so shouldn't they be on top? No. Consider, for instance, an alternative universe in which gravity is twice as strong as in ours. The numerical value of $$G$$ is doubled. Because $$G$$ is doubled, all the gravitational field strengths are doubled as well, which quadruples the quantity $$|\mathbf{g}|^2$$. In the expression $$-\frac{1}{8\pi G}|\mathbf{g}|^2$$, we have quadrupled something on top and doubled something on the bottom, which makes the energy twice as big. That makes perfect sense.

##### Discussion Questions

The figure shows a positive charge in the gap between two capacitor plates. First make a large drawing of the field pattern that would be formed by the capacitor itself, without the extra charge in the middle. Next, show how the field pattern changes when you add the particle at these two positions. Compare the energy of the electric fields in the two cases. Does this agree with what you would have expected based on your knowledge of electrical forces?

Criticize the following statement: “A solenoid makes a charge in the space surrounding it, which dissipates when you release the energy.”

In example 10, I argued that the fields surrounding a positive and negative charge contain less energy when the charges are closer together. Perhaps a simpler approach is to consider the two extreme possibilities: the case where the charges are infinitely far apart, and the one in which they are at zero distance from each other, i.e., right on top of each other. Carry out this reasoning for the case of (1) a positive charge and a negative charge of equal magnitude, (2) two positive charges of equal magnitude, (3) the gravitational energy of two equal masses.

## 22.5 Voltage for nonuniform fields (optional calculus-based section)

The calculus-savvy reader will have no difficulty generalizing the field-voltage relationship to the case of a varying field. The potential energy associated with a varying force is

$\begin{equation*} \Delta PE = -\int F \: dx , \text{[one dimension]} \end{equation*}$

so for electric fields we divide by $$q$$ to find

$\begin{equation*} \Delta V = -\int E \: dx , \text{[one dimension]} \end{equation*}$

Applying the fundamental theorem of calculus yields

$\begin{equation*} E = -\frac{dV}{dx} . \text{[one dimension]} \end{equation*}$

##### Example 12: Voltage associated with a point charge

$$\triangleright$$ What is the voltage associated with a point charge?

$$\triangleright$$ As derived previously in self-check A on page 623, the field is

$\begin{equation*} |\mathbf{E}| = \frac{ kQ}{ r^2} \end{equation*}$

The difference in voltage between two points on the same radius line is

\begin{align*} \Delta V &= \int d V \\ &= -\int E_{x} d x \end{align*}

In the general discussion above, $$x$$ was just a generic name for distance traveled along the line from one point to the other, so in this case $$x$$ really means $$r$$.

\begin{align*} \Delta V &= -\int_{ r_1}^{ r_2} E_{r} d r \\ &= -\int_{ r_1}^{ r_2} \frac{ kQ}{ r^2} d r \\ &= \left.\frac{ kQ}{ r}\right]_{ r_1}^{ r_2} \ &= \frac{ kQ}{ r_2}-\frac{ kQ}{ r_1} . \end{align*}

The standard convention is to use $$r_1=\infty$$ as a reference point, so that the voltage at any distance $$r$$ from the charge is

$\begin{equation*} V = \frac{ kQ}{ r} . \end{equation*}$

The interpretation is that if you bring a positive test charge closer to a positive charge, its electrical energy is increased; if it was released, it would spring away, releasing this as kinetic energy.

self-check:

Show that you can recover the expression for the field of a point charge by evaluating the derivative $$E_{x}=-d V/d x$$.

u / Left: A topographical map of Stowe, Vermont. From one constant-height line to the next is a height difference of 200 feet. Lines far apart, as in the lower village, indicate relatively flat terrain, while lines close together, like the ones to the west of the main town, represent a steep slope. Streams flow downhill, perpendicular to the constant-height lines. Right: The same map has been redrawn in perspective, with shading to suggest relief.

## 22.6 Two or three dimensions

v / The constant-voltage curves surrounding a point charge. Near the charge, the curves are so closely spaced that they blend together on this drawing due to the finite width with which they were drawn. Some electric fields are shown as arrows.

w / Self-check C.

The topographical map shown in figure u suggests a good way to visualize the relationship between field and voltage in two dimensions. Each contour on the map is a line of constant height; some of these are labeled with their elevations in units of feet. Height is related to gravitational potential energy, so in a gravitational analogy, we can think of height as representing voltage. Where the contour lines are far apart, as in the town, the slope is gentle. Lines close together indicate a steep slope.

If we walk along a straight line, say straight east from the town, then height (voltage) is a function of the east-west coordinate $$x$$. Using the usual mathematical definition of the slope, and writing $$V$$ for the height in order to remind us of the electrical analogy, the slope along such a line is $$\Delta V/\Delta$$x. If the slope isn't constant, we either need to use the slope of the $$V-x$$ graph, or use calculus and talk about the derivative $$dV/dx$$.

What if everything isn't confined to a straight line? Water flows downhill. Notice how the streams on the map cut perpendicularly through the lines of constant height.

It is possible to map voltages in the same way, as shown in figure v. The electric field is strongest where the constant-voltage curves are closest together, and the electric field vectors always point perpendicular to the constant-voltage curves.

Figure x shows some examples of ways to visualize field and voltage patterns.

Mathematically, the calculus of section 22.5 generalizes to three dimensions as follows:

\begin{align*} E_x &= -dV/dx \\ E_y &= -dV/dy \\ E_z &= -dV/dz \end{align*}

self-check:

Imagine that the topographical map in figure w represents voltage rather than height. (a) Consider the stream that starts near the center of the map. Determine the positive and negative signs of $$dV/dx$$ and $$dV/dy$$, and relate these to the direction of the force that is pushing the current forward against the resistance of friction. (b) If you wanted to find a lot of electric charge on this map, where would you look?

x / Two-dimensional field and voltage patterns. Top: A uniformly charged rod. Bottom: A dipole. In each case, the diagram on the left shows the field vectors and constant-voltage curves, while the one on the right shows the voltage (up-down coordinate) as a function of x and y. Interpreting the field diagrams: Each arrow represents the field at the point where its tail has been positioned. For clarity, some of the arrows in regions of very strong field strength are not shown --- they would be too long to show. Interpreting the constant-voltage curves: In regions of very strong fields, the curves are not shown because they would merge together to make solid black regions. Interpreting the perspective plots: Keep in mind that even though we're visualizing things in three dimensions, these are really two-dimensional voltage patterns being represented. The third (up-down) dimension represents voltage, not position.

## 22.7 Field lines and Gauss's law (optional)

z / Example 13.

aa / The number of field lines coming in and out of each region depends on the total charge it encloses.

ab / Example 14.

When we look at the “sea of arrows” representation of a field, y/1, there is a natural visual tendency to imagine connecting the arrows as in y/2. The curves formed in this way are called field lines, and they have a direction, shown by the arrowheads.

y / Two different representations of an electric field.

##### Example 13: An avalanche transceiver
The dipole field pattern is universal: it always has the same mathematical shape, and this holds regardless of whether the field is electric or magnetic, static or oscillating. One of its universal properties is that the field lines always pass through the source, and this is taken advantage of in a device called an avalanche transceiver used by backcountry skiers and hikers in winter. The device is worn on the body and is left in transmitting mode while the user is traveling. It creates a dipole pattern of electric and magnetic fields, oscillating at 457 kHz. If a member of the party is buried in an avalanche, his companions must find and locate him within about 30-90 minutes, or he will suffocate. The rescuers switch their own transceivers to receive mode, allowing them to determine the direction of the field line; when the antenna is parallel to the field line, the oscillating electric field drives electrons up and down it, creating the maximum possible current.

In the figure, the rescuer moves along the field line from A to B. At B, she verifies that the strength of the signal has increased. (If it hadn't, she would have had to turn around and go in the opposite direction.) She redetermines the direction of the field and continues in the new direction. Continuing the process, she proceeds along an approximation to the field line. At D she finds that the field strength has fallen off again, so she knows that she has just passed over the victim's position.

Electric field lines originate from positive charges and terminate on negative ones. We can choose a constant of proportionality that fixes how coarse or fine the “grain of the wood” is, but once this choice is made the strength of each charge is shown by the number of lines that begin or end on it. For example, figure y/2 shows eight lines at each charge, so we know that $$q_1/q_2=(-8)/8=-1$$. Because lines never begin or end except on a charge, we can always find the total charge inside any given region by subtracting the number of lines that go in from the number that come out and multiplying by the appropriate constant of proportionality. Ignoring the constant, we can apply this technique to figure aa to find $$q_A=-8$$, $$q_B=2-2=0$$, and $$q_C=5-5=0$$.

Figures y and aa are drawn in two dimensions, but we should really imagine them in three dimensions, so that the regions A, B, and C in figure aa are volumes bounded by surfaces. It is only because our universe happens to have three dimensions that the field line concept is as useful as it is. This is because the number of field lines per unit area perpendicularly piercing a surface can be interpreted as the strength of the electric field. To see this, consider an imaginary spherical surface of radius $$r$$, with a positive charge $$q$$ at its center. The field at the surface equals $$kq/r^2$$. The number of field lines piercing the surface is independent of $$r$$, but the surface's area is proportional to $$r^2$$, so the number of piercings per unit area is proportional to $$1/r^2$$, just like the electric field. With this interpretation, we arrive at Gauss's law, which states that the field strength on a surface multiplied by the surface area equals the total charge enclosed by the surface. (This particular formulation only works in the case where the field pierces the surface perpendicularly and is constant in magnitude over the whole surface. It can be reformulated so as to eliminate these restrictions, but we won't require that reformulation for our present purposes.)

##### Example 14: The field of a line of charge
In example 4 on p. 625, we found by simple scaling arguments that the electric field of a line of charge is proportional to $$1/r$$, where $$r$$ is the distance from the line. Since electric fields are always proportional to the Coulomb constant $$k$$ and to the amount of charge, clearly we must have something of the form
[4] $$E=(...)k(q/L)/r$$, where $$q/L$$ is the charge per unit length and $$(...)$$ represents an unknown numerical constant (which you can easily verify is unitless). Using Gauss's law we can fill in the final piece of the puzzle, which is the value of this constant.

Figure ab shows two surfaces, a cylindrical one A and a spherical one B. Every field line that passes through A also passes through B. By making A sufficiently small and B sufficiently large, we can make the field on each surface nearly constant and perpendicular to the surface, as required by our restricted form of Gauss's law. If radius $$r_B$$ is made large, the field at B made by the line of charge becomes indistinguishable from that of a point charge, $$kq/r_B^2$$. By Gauss's law, the electric field at each surface is proportional to the number of field lines divided by its area. But the number of field lines is the same in both cases, so each field is inversely proportional to the corresponding surface area. We therefore have

\begin{align*} E_A &= E_B\left(\frac{A_B}{A_A}\right) \\ &= \left(\frac{kq}{r_B^2}\right)\left(\frac{4\pi r_B^2}{2\pi r_A L}\right) \\ &= \frac{2kq}{Lr_A} . \end{align*}

The unknown numerical constant equals 2.

##### Example 15: No charge on the interior of a conductor
I asserted on p. 577 that for a perfect conductor in equilibrium, excess charge is found only at the surface, never in the interior. This can be proved using Gauss's law. Suppose that a charge $$q$$ existed at some point in the interior, and it was in stable equilibrium. For concreteness, let's say $$q$$ is positive. If its equilibrium is to be stable, then we need an electric field everywhere around it that points inward like a pincushion, so that if the charge were to be perturbed slightly, the field would bring it back to its equilibrium position. Since Newton's third law forbids objects from making forces on themselves, this field would have to be the field contributed by all the other charges, not by $$q$$ itself. But by Gauss's law, an external set of charges cannot form this kind of inward-pointing pincushion pattern; for such a pattern, Gauss's law would require there to be some negative charge inside the pincusion.

## 22.8 Electric field of a continuous charge distribution (optional calculus-based section)

ac / Example 16.

Charge really comes in discrete chunks, but often it is mathematically convenient to treat a set of charges as if they were like a continuous fluid spread throughout a region of space. For example, a charged metal ball will have charge spread nearly uniformly all over its surface, and in for most purposes it will make sense to ignore the fact that this uniformity is broken at the atomic level. The electric field made by such a continuous charge distribution is the sum of the fields created by every part of it. If we let the “parts” become infinitesimally small, we have a sum of an infinite number of infinitesimal numbers, which is an integral. If it was a discrete sum, we would have a total electric field in the $$x$$ direction that was the sum of all the $$x$$ components of the individual fields, and similarly we'd have sums for the $$y$$ and $$z$$ components. In the continuous case, we have three integrals.

##### Example 16: Field of a uniformly charged rod
$$\triangleright$$ A rod of length $$L$$ has charge $$Q$$ spread uniformly along it. Find the electric field at a point a distance $$d$$ from the center of the rod, along the rod's axis. (This is different from examples 4 on p. 625 and 14 on p. 637, both because the point is on the axis of the rod and because the rod is of finite length.)

$$\triangleright$$ This is a one-dimensional situation, so we really only need to do a single integral representing the total field along the axis. We imagine breaking the rod down into short pieces of length $$d z$$, each with charge $$d q$$. Since charge is uniformly spread along the rod, we have $$d q=\lambdad z$$, where $$\lambda= Q/ L$$ (Greek lambda) is the charge per unit length, in units of coulombs per meter. Since the pieces are infinitesimally short, we can treat them as point charges and use the expression $$kd q/ r^2$$ for their contributions to the field, where $$r= d- z$$ is the distance from the charge at $$z$$ to the point in which we are interested.

\begin{align*} E_{z} &= \int \frac{ kd q }{ r^2} \\ &= \int_{- L/2}^{+ L/2} \frac{ k\lambdad z }{ r^2} \\ &= k \lambda \int_{- L/2}^{+ L/2} \frac{d z}{( d- z)^2} \end{align*}

The integral can be looked up in a table, or reduced to an elementary form by substituting a new variable for $$d- z$$. The result is

\begin{align*} E_{z} &= k\lambda\left(\frac{1}{ d- z}\right)_{- L/2}^{+ L/2} \\ &= \frac{ kQ}{ L} \left(\frac{1}{ d- L/2}-\frac{1}{ d+ L/2}\right) . \end{align*}

For large values of $$d$$, this expression gets smaller for two reasons: (1) the denominators of the fractions become large, and (2) the two fractions become nearly the same, and tend to cancel out. This makes sense, since the field should get weaker as we get farther away from the charge. In fact, the field at large distances must approach $$kQ/ d^2$$, since from a great distance, the rod looks like a point.

It's also interesting to note that the field becomes infinite at the ends of the rod, but is not infinite on the interior of the rod. Can you explain physically why this happens?

## Vocabulary

field — a property of a point in space describing the forces that would be exerted on a particle if it was there

sink — a point at which field vectors converge

source — a point from which field vectors diverge; often used more inclusively to refer to points of either convergence or divergence

electric field — the force per unit charge exerted on a test charge at a given point in space

gravitational field — the force per unit mass exerted on a test mass at a given point in space

electric dipole — an object that has an imbalance between positive charge on one side and negative charge on the other; an object that will experience a torque in an electric field

## Notation

\notationitem{$$\mathbf{g}$$}{the gravitational field}

\notationitem{$$\mathbf{E}$$}{the electric field}

$$D$$ — an electric dipole moment

## Other Notation

$$d$$, $$p$$, $$m$$ — other notations for the electric dipole moment

## Summary

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Experiments show that time is not absolute: it flows at different rates depending on an observer's state of motion. This is an example of the strange effects predicted by Einstein's theory of relativity. All of these effects, however, are very small when the relative velocities are small compared to $$c$$. This makes sense, because Newton's laws have already been thoroughly tested by experiments at such speeds, so a new theory like relativity must agree with the old one in their realm of common applicability. This requirement of backwards-compatibility is known as the correspondence principle.

Since time is not absolute, simultaneity is not a well-defined concept, and therefore the universe cannot operate as Newton imagined, through instantaneous action at a distance. There is a delay in time before a change in the configuration of mass and charge in one corner of the universe will make itself felt as a change in the forces experienced far away. We imagine the outward spread of such a change as a ripple in an invisible universe-filling field of force.

We define the gravitational field at a given point as the force per unit mass exerted on objects inserted at that point, and likewise the electric field is defined as the force per unit charge. These fields are vectors, and the fields generated by multiple sources add according to the rules of vector addition.

When the electric field is constant, the voltage difference between two points lying on a line parallel to the field is related to the field by the equation $$\Delta V=-Ed$$, where $$d$$ is the distance between the two points.

Fields of force contain energy. The density of energy is proportional to the square of the magnitude of the field. In the case of static fields, we can calculate potential energy either using the previous definition in terms of mechanical work or by calculating the energy stored in the fields. If the fields are not static, the old method gives incorrect results and the new one must be used.

## Homework Problems

ae / Problem 4.

af / Problem 13.

ag / Problem 15.

1. In our by-now-familiar neuron, the voltage difference between the inner and outer surfaces of the cell membrane is about $$V_{out}-V_{in}=-70\ \text{mV}$$ in the resting state, and the thickness of the membrane is about 6.0 nm (i.e., only about a hundred atoms thick). What is the electric field inside the membrane?(answer check available at lightandmatter.com)

2. The gap between the electrodes in an automobile engine's spark plug is 0.060 cm. To produce an electric spark in a gasoline-air mixture, an electric field of $$3.0\times10^6$$ V/m must be achieved. On starting a car, what minimum voltage must be supplied by the ignition circuit? Assume the field is uniform.(answer check available at lightandmatter.com)
(b) The small size of the gap between the electrodes is inconvenient because it can get blocked easily, and special tools are needed to measure it. Why don't they design spark plugs with a wider gap?

3. (a) At time $$t=0$$, a positively charged particle is placed, at rest, in a vacuum, in which there is a uniform electric field of magnitude $$E$$. Write an equation giving the particle's speed, $$v$$, in terms of $$t$$, $$E$$, and its mass and charge $$m$$ and $$q$$.(answer check available at lightandmatter.com)
(b) If this is done with two different objects and they are observed to have the same motion, what can you conclude about their masses and charges? (For instance, when radioactivity was discovered, it was found that one form of it had the same motion as an electron in this type of experiment.)

4. (solution in the pdf version of the book) Three charges are arranged on a square as shown. All three charges are positive. What value of $$q_2/q_1$$ will produce zero electric field at the center of the square?

5. Show that the magnitude of the electric field produced by a simple two-charge dipole, at a distant point along the dipole's axis, is to a good approximation proportional to $$D/r^3$$, where $$r$$ is the distance from the dipole. [Hint: Use the approximation $$(1+\epsilon)^p\approx 1+p\epsilon$$, which is valid for small $$\epsilon$$.]

6. Consider the electric field created by a uniform ring of total charge $$q$$ and radius $$b$$. (a) Show that the field at a point on the ring's axis at a distance $$a$$ from the plane of the ring is $$kqa(a^2+b^2)^{-3/2}$$. (b) Show that this expression has the right behavior for $$a=0$$ and for $$a$$ much greater than $$b$$.

7. Example 4 on p. 625 showed that the electric field of a long, uniform line of charge falls off with distance as $$1/r$$. By a similar technique, show that the electric field of a uniformly charged plane has no dependence on the distance from the plane.

8. Given that the field of a dipole is proportional to $$D/r^3$$ (problem 5), show that its voltage varies as $$D/r^2$$. (Ignore positive and negative signs and numerical constants of proportionality.) ∫

9. A carbon dioxide molecule is structured like O-C-O, with all three atoms along a line. The oxygen atoms grab a little bit of extra negative charge, leaving the carbon positive. The molecule's symmetry, however, means that it has no overall dipole moment, unlike a V-shaped water molecule, for instance. Whereas the voltage of a dipole of magnitude $$D$$ is proportional to $$D/r^2$$ (problem 8), it turns out that the voltage of a carbon dioxide molecule along its axis equals $$k/r^3$$, where $$r$$ is the distance from the molecule and $$k$$ is a constant. What would be the electric field of a carbon dioxide molecule at a distance $$r?$$(answer check available at lightandmatter.com) ∫

10. A proton is in a region in which the electric field is given by $$E=a+bx^3$$. If the proton starts at rest at $$x_1=0$$, find its speed, $$v$$, when it reaches position $$x_2$$. Give your answer in terms of $$a$$, $$b$$, $$x_2$$, and $$e$$ and $$m$$, the charge and mass of the proton.(answer check available at lightandmatter.com) ∫

11. Consider the electric field created by a uniformly charged cylinder that extends to infinity in one direction. (a) Starting from the result of problem 8, show that the field at the center of the cylinder's mouth is $$2\pi k\sigma$$, where $$\sigma$$ is the density of charge on the cylinder, in units of coulombs per square meter. [Hint: You can use a method similar to the one in problem 9.] (b) This expression is independent of the radius of the cylinder. Explain why this should be so. For example, what would happen if you doubled the cylinder's radius? ∫

12. In an electrical storm, the cloud and the ground act like a parallel-plate capacitor, which typically charges up due to frictional electricity in collisions of ice particles in the cold upper atmosphere. Lightning occurs when the magnitude of the electric field builds up to a critical value, $$E_c$$, at which air is ionized.
(a) Treat the cloud as a flat square with sides of length $$L$$. If it is at a height $$h$$ above the ground, find the amount of energy released in the lightning strike.(answer check available at lightandmatter.com)
(b) Based on your answer from part a, which is more dangerous, a lightning strike from a high-altitude cloud or a low-altitude one?
(c) Make an order-of-magnitude estimate of the energy released by a typical lightning bolt, assuming reasonable values for its size and altitude. $$E_c$$ is about $$10^6$$ V/m.

See problem 16 for a note on how recent research affects this estimate.

13. The neuron in the figure has been drawn fairly short, but some neurons in your spinal cord have tails (axons) up to a meter long. The inner and outer surfaces of the membrane act as the “plates” of a capacitor. (The fact that it has been rolled up into a cylinder has very little effect.) In order to function, the neuron must create a voltage difference $$V$$ between the inner and outer surfaces of the membrane. Let the membrane's thickness, radius, and length be $$t$$, $$r$$, and $$L$$. (a) Calculate the energy that must be stored in the electric field for the neuron to do its job. (In real life, the membrane is made out of a substance called a dielectric, whose electrical properties increase the amount of energy that must be stored. For the sake of this analysis, ignore this fact.) [Hint: The volume of the membrane is essentially the same as if it was unrolled and flattened out.] (answer check available at lightandmatter.com)
(b) An organism's evolutionary fitness should be better if it needs less energy to operate its nervous system. Based on your answer to part a, what would you expect evolution to do to the dimensions $$t$$ and $$r?$$ What other constraints would keep these evolutionary trends from going too far?

14. To do this problem, you need to understand how to do volume integrals in cylindrical and spherical coordinates. (a) Show that if you try to integrate the energy stored in the field of a long, straight wire, the resulting energy per unit length diverges both at $$r\rightarrow 0$$ and $$r\rightarrow \infty$$. Taken at face value, this would imply that a certain real-life process, the initiation of a current in a wire, would be impossible, because it would require changing from a state of zero magnetic energy to a state of infinite magnetic energy. (b) Explain why the infinities at $$r\rightarrow 0$$ and $$r\rightarrow \infty$$ don't really happen in a realistic situation. (c) Show that the electric energy of a point charge diverges at $$r\rightarrow 0$$, but not at $$r\rightarrow \infty$$.

A remark regarding part (c): Nature does seem to supply us with particles that are charged and pointlike, e.g., the electron, but one could argue that the infinite energy is not really a problem, because an electron traveling around and doing things neither gains nor loses infinite energy; only an infinite change in potential energy would be physically troublesome. However, there are real-life processes that create and destroy pointlike charged particles, e.g., the annihilation of an electron and antielectron with the emission of two gamma rays. Physicists have, in fact, been struggling with infinities like this since about 1950, and the issue is far from resolved. Some theorists propose that apparently pointlike particles are actually not pointlike: close up, an electron might be like a little circular loop of string. ∫

15. The figure shows cross-sectional views of two cubical capacitors, and a cross-sectional view of the same two capacitors put together so that their interiors coincide. A capacitor with the plates close together has a nearly uniform electric field between the plates, and almost zero field outside; these capacitors don't have their plates very close together compared to the dimensions of the plates, but for the purposes of this problem, assume that they still have approximately the kind of idealized field pattern shown in the figure. Each capacitor has an interior volume of 1.00 $$\text{m}^3$$, and is charged up to the point where its internal field is 1.00 V/m. (a) Calculate the energy stored in the electric field of each capacitor when they are separate. (b) Calculate the magnitude of the interior field when the two capacitors are put together in the manner shown. Ignore effects arising from the redistribution of each capacitor's charge under the influence of the other capacitor. (c) Calculate the energy of the put-together configuration. Does assembling them like this release energy, consume energy, or neither?(answer check available at lightandmatter.com)

16. In problem 12 on p. 643, you estimated the energy released in a bolt of lightning, based on the energy stored in the electric field immediately before the lightning occurs. The assumption was that the field would build up to a certain value, which is what is necessary to ionize air. However, real-life measurements always seemed to show electric fields strengths roughtly 10 times smaller than those required in that model. For a long time, it wasn't clear whether the field measurements were wrong, or the model was wrong. Research carried out in 2003 seems to show that the model was wrong. It is now believed that the final triggering of the bolt of lightning comes from cosmic rays that enter the atmosphere and ionize some of the air. If the field is 10 times smaller than the value assumed in problem 12, what effect does this have on the final result of problem 12?(answer check available at lightandmatter.com)

\begin{handson}{}{Field vectors}{\onecolumn} Apparatus:

3 solenoids

DC power supply

compass

ruler

cut-off plastic cup

At this point you've studied the gravitational field, $$\mathbf{g}$$, and the electric field, $$\mathbf{E}$$, but not the magnetic field, $$\mathbf{B}$$. However, they all have some of the same mathematical behavior: they act like vectors. Furthermore, magnetic fields are the easiest to manipulate in the lab. Manipulating gravitational fields directly would require futuristic technology capable of moving planet-sized masses around! Playing with electric fields is not as ridiculously difficult, but static electric charges tend to leak off through your body to ground, and static electricity effects are hard to measure numerically. Magnetic fields, on the other hand, are easy to make and control. Any moving charge, i.e. any current, makes a magnetic field.

A practical device for making a strong magnetic field is simply a coil of wire, formally known as a solenoid. The field pattern surrounding the solenoid gets stronger or weaker in proportion to the amount of current passing through the wire.

1. With a single solenoid connected to the power supply and laid with its axis horizontal, use a magnetic compass to explore the field pattern inside and outside it. The compass shows you the field vector's direction, but not its magnitude, at any point you choose. Note that the field the compass experiences is a combination (vector sum) of the solenoid's field and the earth's field.

2. What happens when you bring the compass extremely far away from the solenoid?

What does this tell you about the way the solenoid's field varies with distance?

Thus although the compass doesn't tell you the field vector's magnitude numerically, you can get at least some general feel for how it depends on distance.

3. The figure below is a cross-section of the solenoid in the plane containing its axis. Make a sea-of-arrows sketch of the magnetic field in this plane. The length of each arrow should at least approximately reflect the strength of the magnetic field at that point.

Does the field seem to have sources or sinks?

4. What do you think would happen to your sketch if you reversed the wires?

Try it.

5. Now hook up the two solenoids in parallel. You are going to measure what happens when their two fields combine in the at a certain point in space. As you've seen already, the solenoids' nearby fields are much stronger than the earth's field; so although we now theoretically have three fields involved (the earth's plus the two solenoids'), it will be safe to ignore the earth's field. The basic idea here is to place the solenoids with their axes at some angle to each other, and put the compass at the intersection of their axes, so that it is the same distance from each solenoid. Since the geometry doesn't favor either solenoid, the only factor that would make one solenoid influence the compass more than the other is current. You can use the cut-off plastic cup as a little platform to bring the compass up to the same level as the solenoids' axes.

a)What do you think will happen with the solenoids' axes at 90 degrees to each other, and equal currents? Try it. Now represent the vector addition of the two magnetic fields with a diagram. Check your diagram with your instructor to make sure you're on the right track.

b) Now try to make a similar diagram of what would happen if you switched the wires on one of the solenoids.

After predicting what the compass will do, try it and see if you were right.

c)Now suppose you were to go back to the arrangement you had in part a, but you changed one of the currents to half its former value. Make a vector addition diagram, and use trig to predict the angle.

Try it. To cut the current to one of the solenoids in half, an easy and accurate method is simply to put the third solenoid in series with it, and put that third solenoid so far away that its magnetic field doesn't have any significant effect on the compass.

\end{handson}

(c) 1998-2013 Benjamin Crowell, licensed under the Creative Commons Attribution-ShareAlike license. Photo credits are given at the end of the Adobe Acrobat version.

##### Footnotes
[1] From “The Reversible Revolutions,” by Cecil Corwin, Cosmic Stories, March 1941. Art by Morey, Bok, Kyle, Hunt, Forte. Copyright expired.
[2] As discussed in ch. 24, there are actually two different effects here, one due to motion and one due to gravity.