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# 10.4: Energy In Fields

### 10.4.1 Electric field energy

Fields possess energy, as argued on page 559, but how much energy? The answer can be found using the following elegant approach. We assume that the electric energy contained in an infinitesimal volume of space $$dv$$ is given by $$dU_e=f(\mathbf{E})dv$$, where $$f$$ is some function, which we wish to determine, of the field E. It might seem that we would have no easy way to determine the function $$f$$, but many of the functions we could cook up would violate the symmetry of space. For instance, we could imagine $$f(\mathbf{E})=aE_y$$, where $$a$$ is some constant with the appropriate units. However, this would violate the symmetry of space, because it would give the $$y$$ axis a different status from $$x$$ and $$z$$. As discussed on page 212, if we wish to calculate a scalar based on some vectors, the dot product is the only way to do it that has the correct symmetry properties. If all we have is one vector, E, then the only scalar we can form is $$\mathbf{E}\cdot\mathbf{E}$$, which is the square of the magnitude of the electric field vector.

a / Two oppositely charged capacitor plates are pulled apart.

In principle, the energy function we are seeking could be proportional to $$\mathbf{E}\cdot\mathbf{E}$$, or to any function computed from it, such as $$\sqrt{\mathbf{E}\cdot\mathbf{E}}$$ or $$(\mathbf{E}\cdot\mathbf{E})^7$$. On physical grounds, however, the only possibility that works is $$\mathbf{E}\cdot\mathbf{E}$$. Suppose, for instance, that we pull apart two oppositely charged capacitor plates, as shown in figure a. We are doing work by pulling them apart against the force of their electrical attraction, and this quantity of mechanical work equals the increase in electrical energy, $$U_e$$. Using our previous approach to energy, we would have thought of $$U_e$$ as a quantity which depended on the distance of the positive and negative charges from each other, but now we're going to imagine $$U_e$$ as being stored within the electric field that exists in the space between and around the charges. When the plates are touching, their fields cancel everywhere, and there is zero electrical energy. When they are separated, there is still approximately zero field on the outside, but the field between the plates is nonzero, and holds some energy. Now suppose we carry out the whole process, but with the plates carrying double their previous charges. Since Coulomb's law involves the product $$q_1q_2$$ of two charges, we have quadrupled the force between any given pair of charged particles, and the total attractive force is therefore also four times greater than before. This means that the work done in separating the plates is four times greater, and so is the energy $$U_e$$ stored in the field. The field, however, has merely been doubled at any given location: the electric field $$\mathbf{E}_+$$ due to the positively charged plate is doubled, and similarly for the contribution $$\mathbf{E}_-$$ from the negative one, so the total electric field $$\mathbf{E}_++\mathbf{E}_-$$ is also doubled. Thus doubling the field results in an electrical energy which is four times greater, i.e., the energy density must be proportional to the square of the field, $$dU_e\propto(\mathbf{E}\cdot\mathbf{E})dv$$. For ease of notation, we write this as $$dU_e\propto E^2dv$$, or $$dU_e=aE^2dv$$, where $$a$$ is a constant of proportionality. Note that we never really made use of any of the details of the geometry of figure a, so the reasoning is of general validity. In other words, not only is $$dU_e=aE^2dv$$ the function that works in this particular case, but there is every reason to believe that it would work in other cases as well.

It now remains only to find $$a$$. Since the constant must be the same in all situations, we only need to find one example in which we can compute the field and the energy, and then we can determine $$a$$. The situation shown in figure a is just about the easiest example to analyze. We let the square capacitor plates be uniformly covered with charge densities $$+\sigma$$ and $$-\sigma$$, and we write $$b$$ for the lengths of their sides. Let $$h$$ be the gap between the plates after they have been separated. We choose $$h\ll b$$, so that the field experienced by the negative plate due to the positive plate is $$E_+=2\pi k\sigma$$. The charge of the negative plate is $$-\sigma b^2$$, so the magnitude of the force attracting it back toward the positive plate is $$(\text{force})=(\text{charge})(\text{field})=2\pi k\sigma^2 b^2$$. The amount of work done in separating the plates is $$(\text{work})=(\text{force})(\text{distance})=2\pi k\sigma^2 b^2h$$. This is the amount of energy that has been stored in the field between the two plates, $$U_e=2\pi k\sigma^2 b^2h=2\pi k\sigma^2 v$$, where $$v$$ is the volume of the region between the plates.

We want to equate this to $$U_e=aE^2v$$. (We can write $$U_e$$ and $$v$$ rather than $$dU_e$$ and $$dv$$, since the field is constant in the region between the plates.) The field between the plates has contributions from both plates, $$E=E_++E_-=4\pi k\sigma$$. (We only used half this value in the computation of the work done on the moving plate, since the moving plate can't make a force on itself. Mathematically, each plate is in a region where its own field is reversing directions, so we can think of its own contribution to the field as being zero within itself.) We then have $$aE^2v=a\cdot 16\pi^2k^2\sigma^2 \cdot v$$, and setting this equal to $$U_e=2\pi k\sigma^2 v$$ from the result of the work computation, we find $$a=1/8\pi k$$. Our final result is as follows:

The electric energy possessed by an electric field E occupying an infinitesimal volume of space $$dv$$ is given by

$\begin{equation*} dU_e = \frac{1}{8\pi k}E^2 dv , \end{equation*}$

where $$E^2=\mathbf{E}\cdot\mathbf{E}$$ is the square of the magnitude of the electric field.

This is reminiscent of how waves behave: the energy content of a wave is typically proportional to the square of its amplitude.

self-check:

We can think of the quantity $$dU_{e}/dv$$ as the energy density due to the electric field, i.e., the number of joules per cubic meter needed in order to create that field. (a) How does this quantity depend on the components of the field vector, $$E_x$$, $$E_y$$, and $$E_z$$? (b) Suppose we have a field with $$E_x\neq0$$, $$E_y$$=0, and $$E_z$$=0. What would happen to the energy density if we reversed the sign of $$E_x$$?

Example 14: A numerical example

$$\triangleright$$ A capacitor has plates whose areas are $$10^{-4}\ \text{m}^2$$, separated by a gap of $$10^{-5}$$ m. A 1.5-volt battery is connected across it. How much energy is sucked out of the battery and stored in the electric field between the plates? (A real capacitor typically has an insulating material between the plates whose molecules interact electrically with the charge in the plates. For this example, we'll assume that there is just a vacuum in between the plates. The plates are also typically rolled up rather than flat.)

$$\triangleright$$ To connect this with our previous calculations, we need to find the charge density on the plates in terms of the voltage we were given. Our previous examples were based on the assumption that the gap between the plates was small compared to the size of the plates. Is this valid here? Well, if the plates were square, then the area of $$10^{-4}\ \text{m}^2$$ would imply that their sides were $$10^{-2}$$ m in length. This is indeed very large compared to the gap of $$10^{-5}$$ m, so this assumption appears to be valid (unless, perhaps, the plates have some very strange, long and skinny shape).

Based on this assumption, the field is relatively uniform in the whole volume between the plates, so we can use a single symbol, $$E$$, to represent its magnitude, and the relation $$E=d V/d x$$ is equivalent to $$E=\Delta V/\Delta x=(\text{1.5 V})/(\text{gap})= 1.5\times10^5\ \text{V}/\text{m}$$.

Since the field is uniform, we can dispense with the calculus, and replace $$d U_{e} = (1/8\pi k) E^2 d v$$ with $$U_{e} = (1/8\pi k) E^2 v$$. The volume equals the area multiplied by the gap, so we have

\begin{align*} U_{e} &= (1/8\pi k) E^2(\text{area})(\text{gap})\\ &= \frac{1}{8\pi\times9\ \times10^9\ \text{N}\!\cdot\!\text{m}^2/\text{C}^2}( 1.5\times10^5\ \text{V}/\text{m})^2(10^{-4}\ \text{m}^2)(10^{-5}\ \text{m})\\ &= 1\times10^{-10}\ \text{J} \end{align*}

self-check:

Show that the units in the preceding example really do work out to be joules.

Example 15: Why $$k$$ is on the bottom

It may also seem strange that the constant $$k$$ is in the denominator of the equation $$d U_{e} = (1/8\pi k) E^2 d v$$. The Coulomb constant $$k$$ tells us how strong electric forces are, so shouldn't it be on top? No. Consider, for instance, an alternative universe in which electric forces are twice as strong as in ours. The numerical value of $$k$$ is doubled. Because $$k$$ is doubled, all the electric field strengths are doubled as well, which quadruples the quantity $$E^2$$. In the expression $$E^2/8\pi k$$, we've quadrupled something on top and doubled something on the bottom, which makes the energy twice as big. That makes perfect sense.

Example 16: Potential energy of a pair of opposite charges

Imagine taking two opposite charges, b, that were initially far apart and allowing them to come together under the influence of their electrical attraction.

b / Example 16.

According to our old approach, electrical 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 electrical 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 17: A spherical capacitor

$$\triangleright$$ A spherical capacitor, c, consists of two concentric spheres of radii $$a$$ and $$b$$. Find the energy required to charge up the capacitor so that the plates hold charges $$+ q$$ and $$- q$$.

c / Example B. Part of the outside sphere has been drawn as if it is transparent, in order to show the inside sphere.

$$\triangleright$$ On page 102, I proved that for gravitational forces, the interaction of a spherical shell of mass with other masses outside it is the same as if the shell's mass was concentrated at its center. On the interior of such a shell, the forces cancel out exactly. Since gravity and the electric force both vary as $$1/ r^2$$, the same proof carries over immediately to electrical forces. The magnitude of the outward electric field contributed by the charge $$+ q$$ of the central sphere is therefore

$\begin{equation*} |\mathbf{E}_+| = \left\{ \begin{array}{lr} 0, & r\lt a \\ kq/ r^2, & r> a \end{array} \right. , \end{equation*}$

where $$r$$ is the distance from the center. Similarly, the magnitude of the inward field contributed by the outside sphere is

$\begin{equation*} |\mathbf{E}_-| = \left\{ \begin{array}{lr} 0, & r\lt b \\ kq/ r^2, & r> b \end{array} \right. . \end{equation*}$

In the region outside the whole capacitor, the two fields are equal in magnitude, but opposite in direction, so they cancel. We then have for the total field

$\begin{equation*} |\mathbf{E}| = \left\{ \begin{array}{lr} 0, & r\lt a \\ kq/ r^2, & a\lt r\lt b \\ 0, & r> b \end{array} \right. , \end{equation*}$

so to calculate the energy, we only need to worry about the region $$a\lt r\lt b$$. The energy density in this region is

\begin{align*} \frac{d U_{e}}{d v} &= \frac{1}{8\pi k} E^2 \\ &= \frac{ kq^2}{8\pi} r^{-4} . \end{align*}

This expression only depends on $$r$$, so the energy density is constant across any sphere of radius $$r$$. We can slice the region $$a\lt r\lt b$$ into concentric spherical layers, like an onion, and the energy within one such layer, extending from $$r$$ to $$r+dr$$ is

\begin{align*} d U_{e} &= \frac{d U_{e}}{d v} dv \\ &= \frac{d U_{e}}{d v} (\text{area of shell}) (\text{thickness of shell}) \\ &= (\frac{ kq^2}{8\pi} r^{-4}) (4\pi r^2) (dr) \\ &= \frac{ kq^2}{2} r^{-2}dr . \end{align*}

Integrating over all the layers to find the total energy, we have

\begin{align*} U_{e} &= \int d U_{e} \\ &= \int_{a}^{b} \frac{ kq^2}{2} r^{-2}dr \\ &= \left.-\frac{ kq^2}{2} r^{-1}\right|_{a}^{b} \\ &= \frac{ kq^2}{2}\left(\frac{1}{a}-\frac{1}{b}\right) \\ \end{align*}
##### Discussion Questions

◊ The figure shows a positive charge in the gap between two capacitor plates. 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?

d / Discussion question A.

◊

​ ​

e / Discussion question B.

The figure shows a spherical capacitor. In the text, the energy stored in its electric field is shown to be

$\begin{equation*} U_{e} = \frac{ kq^2}{2}\left(\frac{1}{a}-\frac{1}{b}\right) . \\ \end{equation*}$

What happens if the difference between $$b$$ and $$a$$ is very small? Does this make sense in terms of the mechanical work needed in order to separate the charges? Does it make sense in terms of the energy stored in the electric field? Should these two energies be added together?

Similarly, discuss the cases of $$b\rightarrow\infty$$ and $$a\rightarrow0$$.

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

◊ In example 16 on page 585, I argued that for the charges shown in the figure, the fields contain less energy when the charges are closer together, because the region of cancellation expanded, while the region of reinforcing fields shrank. 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.

### 10.4.2 Gravitational field energy

Example B depended on the close analogy between electric and gravitational forces. In fact, every argument, proof, and example discussed so far in this section is equally valid as a gravitational example, provided we take into account one fact: only positive mass exists, and the gravitational force between two masses is attractive. This is the opposite of what happens with electrical forces, which are repulsive in the case of two positive charges. As a consequence of this, we need to assign a negative energy density to the gravitational field! For a gravitational field, we have

$\begin{equation*} dU_g = -\frac{1}{8\pi G}g^2 dv , \end{equation*}$

where $$g^2=\mathbf{g}\cdot\mathbf{g}$$ is the square of the magnitude of the gravitational field.

### 10.4.3 Magnetic field energy

So far we've only touched in passing on the topic of magnetic fields, which will deal with in detail in chapter 11. Magnetism is an interaction between moving charge and moving charge, i.e., between currents and currents. Since a current has a direction in space,2 while charge doesn't, we can anticipate that the mathematical rule connecting a magnetic field to its source-currents will have to be completely different from the one relating the electric field to its source-charges. However, if you look carefully at the argument leading to the relation $$dU_e/dv = E^2/8\pi k$$, you'll see that these mathematical details were only necessary to the part of the argument in which we fixed the constant of proportionality. To establish $$dU_e/dv \propto E^2$$, we only had to use three simple facts:

• The field is proportional to the source.
• Forces are proportional to fields.
• Field contributed by multiple sources add like vectors.

All three of these statements are true for the magnetic field as well, so without knowing anything more specific about magnetic fields --- not even what units are used to measure them! --- we can state with certainty that the energy density in the magnetic field is proportional to the square of the magnitude of the magnetic field. The constant of proportionality is given on p. 665.

### Contributors

Benjamin Crowell (Fullerton College). Conceptual Physics is copyrighted with a CC-BY-SA license.