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Unformatted text preview: Electrostatic Energy 8—1 The electrostatic energy of charges. A uniform sphere In the study of mechanics, one of the most interesting and useful discoveries
was the law of the conservation of energy. The expressions for the kinetic and
potential energies of a mechanical system helped us to discover connections between
the states of a system at two different times without having to look into the details
of what was occurring in between. We wish now to consider the energy of electro
static systems. In electricity also the principle of the conservation of energy will
be useful for discovering a number of interesting things. The law of the energy of interaction in electrostatics is very simple; we have,
in fact, already discussed it. Suppose we have two charges ql and £12 separated by
the distance r12. There is some energy in the system, because a certain amount of
work was required to bring the charges together. We have already calculated the
work done in bringing two charges together from a large distance. It is (1142 .
41reor12 (8.1) We also know, from the principle of superposition, that if we have many charges
present, the total force on any charge is the sum of the forces from the others. It
follows, therefore, that the total energy of a system of a number of charges is the
sum of terms due to the mutual interaction of each pair of charges. If q, and q,
are any two of the charges and r,,— is the distance between them (Fig. 8—1), the
energy of that particular pair is 414:“ _
41reorij (8'2) The total electrostatic energy U is the sum of the energies of all possible pairs of charges:
_ 4in _
U — E 4760)}; all pairs (8.3) If we have a distribution of charge speciﬁed by a charge density p, the sum of Eq.
(8.3) is, of course, to be replaced by an integral. We shall concern ourselves with two aspects of this energy. One is the applica
tion of the concept of energy to electrostatic problems; the other is the evaluation
of the energy in different ways. Sometimes it is easier to compute the work done
for some special case than to evaluate the sum in Eq. (8.3), or the corresponding
integral. As an example, let us calculate the energy required to assemble a sphere
of charge with a uniform charge density. The energy is just the work done in
gathering the charges together from inﬁnity. Imagine that we assemble the sphere by building up a succession of thin
spherical layers of inﬁnitesimal thickness. At each stage of the process, we gather
a small amount of charge and put it in a thin layer from r to r + dr. We continue
the process until we arrive at the ﬁnal radius a (Fig. 8—2). If Q, is the charge of the
sphere when it has been built up to the radius r, the work done in bringing a charge
dQ to it is dU = Q’dQ
41reor (8.4)
8—1 81 The electrostatic energy of
charges. A uniform sphere 8—2 The energy of a condenser.
Forces on charged conductors 8—3 The electrostatic energy of an
ionic crystal 84 Electrostatic energy in nuclei
8—5 Energy in the electrostatic ﬁeld
8—6 The energy of a point charge Review: Chapter 4, Vol. I, Conservation
of Energy
Chapters 13 and 14, Vol. I,
Work and Potential Energy Fig. 81. The electrostatic energy of
a system of particles is the sum of the
electrostatic energy of each pair. Fig. 82. The energy of a uniform
sphere of charge can be computed by
imagining that it is assembled from
successive spherical shells. If the density of charge in the sphere is p, the charge Q, is Qr = 10; W3,
and the charge dQ is
dQ = p ° 47rr2 dr. Equation (8.4) becomes = 41rp2r4 dr . dU 360 (8.5) The total energy required to assemble the sphere is the integral of dU from r =
0 to r = a, or
_ 41rpza5 U 1560 (8.6) Or if we wish to express the result in terms of the total charge Q of the sphere, U =% 92 . (8.7) The energy is proportional to the square of the total charge and inversely pro
portional to the radius. We can also interpret Eq. (8.7) as saying that the average
of (1 /r,,) for all pairs of points in the sphere is 3/5a. 8—2 The energy of a condenser. Forces on charged conductors We consider now the energy required to charge a condenser. If the charge Q
has been taken from one of the conductors of a condenser and placed on the other,
the potential difference between them is V = , (8.8) ('3th where C is the capacity of the condenser. How much work is done in charging
the condenser? Proceeding as for the sphere, we imagine that the condenser has
been charged by transferring charge from one plate to the other in small increments
dQ. The work required to transfer the charge dQ is dU = VdQ.
Taking V from Eq. (8.8), we write _QdQ
““1“?" Or integrating from zero charge to the ﬁnal charge Q, we have U = 9C:  (8.9) l
2
This energy can also be written as
U = ;CV2. (8.10)
Recalling that the capacity of a conducting sphere (relative to inﬁnity) is Cspherc = 47"5 00, we can immediately get from Eq. (8.9) the energy of a charged sphere, 92 (3.11) 1
U _ i 41reoa' 8—2 This, of course, is also the energy of a thin spherical shell of total charge Q and is
just 5/6 of the energy of a uniformly charged sphere, Eq. (8.7). We now consider applications of the idea of electrostatic energy. Consider
the following questions: What is the force between the plates of a condenser? Or
what is the torque about some axis of a charged conductor in the presence of an—
other with opposite charge? Such questions are easily answered by using our
result Eq. (8.9) for electrostatic energy of a condenser, together with the principle
of virtual work (Chapters 4, 13, and 14 of Vol. I). Let’s use this method for determining the force between the plates of a
parallelplate condenser. If we imagine that the spacing of the plates is increased
by the small amount Az, then the mechanical work done from the outside in
moving the plates would be AW = FAz, (8.12) where F is the force between the plates. Thiswork must be equal to the change
in the electrostatic energy of the condenser.
By Eq. (8.9), the energy of the condenser was originally _192
”—5? The change in energy (if we do not let the charge change) is _ 1 2 1
AU — 5 Q A(E')  (8.13)
Equating (8.12) and (8.13), we have
_ g2 (l) FAz — 2 A C  (814) This can also be written as
2
FAz = —§a AC. (8.15) The force, of course, results from the attraction of the charges on the plates, but
we see that we do not have to worry in detail about how they are distributed;
everything we need is taken care of in the capacity C. It is easy to see how the idea is extended to conductors of any shape, and for
other components of the force. In Eq. (8.14), we replace F by the component we
are looking for, and we replace Az by a small displacement in the corresponding
direction. Or if we have an electrode with a pivot and we want to know the torque
r, we write the virtual work as AW = 7A0, where M is a small angular displacement. Of course, A(1 / C) must be the change in
1 / C which corresponds to A0. We could, in this way, ﬁnd the torque on the mov
able plates in a variable condenser of the type shown in Fig. 8—3.
Returning to the special case of a parallelplate condenser, we can use the
formula we derived in Chapter 6 for the capacity:
1 d —C = as (8.16) where A is the area of each plate. If we increase the separation by Az, 1 A2
46) r 232'
From Eq. (8.14) we get that the force between the plates is 2
F = ng (8.17) 8—3 Fig. 8—3. What is the torque on a
variable capacitor? \'
cououcrms \\ PLATE LAYER OF SURFACE
CHARGE a' 50 lEI E0 Fig. 8—4. The ﬁeld at the surface of
a conductor varies from zero to E0 =
a/eo, as one passes through the layer of
surface charge. Let’s look at Eq. (8.17) a little more closely and see if we can tell how the force
arises. If for the charge on one plate we write Q = 0A.
Eq. (8.17) can be rewritten as
1 0'
F — 5 Q 6—0 
Or, since the electric ﬁeld between the plates is
0'
E0 — 6—0 ’
then
F = §QE0. (8.18) One would immediately guess that the force acting on one plate is the charge
Q on the plate times the ﬁeld acting on the charge. But we have a surprising factor
of onehalf. The reason is that E 0 is not the ﬁeld at the charges. If we imagine that
the charge at the surface of the plate occupies a thin layer, as indicated in Fig. 8—4,
the ﬁeld will vary from zero at the inner boundary of the layer to E0 in the space
outside of the plate. The average ﬁeld acting on the surface charges is Eo/2. That
is why the factor onehalf is in Eq. (8.18). You should notice that in computing the virtual work we have assumed that
the charge on the condenser was constant—that it was not electrically connected
to other objects, and so the total charge could not change. Suppose we had imagined that the condenser was held at a constant potential
diﬁ‘erence as we made the virtual displacement. Then we should have taken U = aerV2
and in place of Eq. (8.15) we would have had
FAz = 13V2 AC, which gives a force equal in magnitude to the one in Eq. (8.15) (because V = Q/ C),
but with the opposite sign! Surely the force between the condenser plates doesn’t
reverse in sign as we disconnect it from its charging source. Also, we know that
two plates with opposite electrical charges must attract. The principle of virtual
work has been incorrectly applied in the second case—we have not taken into
account the virtual work done on the charging source. That is, to keep the po
tential constant at V as the capacity changes, a charge VAC must be supplied by
a source of charge. But this charge is supplied at a potential V, so the work done
by the electrical system which keeps the potential constant is V2 AC. The mechan
ical work F Az plus this electrical work V2 AC together make up the change in the
total energy in AC of the condenser. Therefore F A2 is —1}V2 AC, as before. 8—3 The electrostatic energy of an ionic crystal We now consider an application of the concept of electrostatic energy in atomic
physics. We cannot easily measure the forces between atoms, but we are often
interested in the energy differences between one atomic arrangement and another,
as, for example, the energy of a chemical change. Since atomic forces are basically
electrical, chemical energies are in large part just electrostatic energies. Let’s consider, for example, the electrostatic energy of an ionic lattice. An
ionic crystal like NaCl consists of positive and negative ions which can be thought
of as rigid spheres. They attract electrically until they begin to touch; then there is
a repulsive force which goes up very rapidly if we try to push them closer together. For our ﬁrst approximation, therefore, we imagine a set of rigid spheres
that represent the atoms in a salt crystal. The structure of the lattice has been
determined by xray diﬂ‘raction. It is a cubic lattice—~like a threedimensional 84 checkerboard. Figure 8—5 shows a crosssectional view. The spacing of the ions is
2.81 A (=2.81 X 10‘8 cm). If our picture of this system is correct, we should be able to check it by asking
the following question: How much energy will it take to pull all these ions apart—
that is, to separate the crystal completely into ions? This energy should be equal
to the heat of vaporization of NaCl plus the energy required to dissociate the
molecules into ions. This total energy to separate NaCl to ions is determined experi
mentally to be 7.92 electron volts per molecule. Using the conversion lev = 1.602 X 10‘19joule,
and Avogadro’s number for the number of molecules in a mole,
N0 = 6.02 X 1023,
the energy of vaporization can also be given as
W = 7.64 X 105 joules/mole. Physical chemists prefer for an energy unit the kilocalorie, which is 4190 joules;
so that l ev per molecule is 23 kilocalories per mole. A chemist would then say
that the dissociation energy of NaCl is W = 183 kcal/mole. Can we obtain this chemical energy theoretically by computing how much
work it would take to pull apart the crystal? According to our theory, this work is
the sum of the potential energies of all the pairs of ions. The easiest way to ﬁgure
out this sum is to pick out a particular ion and compute its potential energy with
each of the other ions. That will give us twice the energy per ion, because the energy
belongs to the pairs of charges. If we want the energy to be associated with one
particular ion, we should take half the sum. But we really want the energy per
molecule, which contains two ions, so that the sum we compute will give directly
the energy per molecule. The energy of an ion with one of its nearest neighbors is e2/a, where e2 =
qE/41reo and a is the centertocenter spacing between ions. (We are considering
monovalent ions.) This energy is 5.12 ev, which we already see is going to give us
a result of the correct order of magnitude. But it is still a long way from the inﬁnite
sum of terms we need. Let’s begin by summing all the terms from the ions along a straight line.
Considering that the ion marked Na in Fig. 8—5 is our special ion, we shall consider
ﬁrst those ions on a horizontal line with it. There are two nearest Cl ions with
negative charges, each at the distance 0. Then there are two positive ions at the
distance 2a, etc. Calling the energy of this sum U1, we write II a e2 2 2 2 2
”1 “(‘t+2“§+z+'“) 2e2 1 1 1
‘7(1"2+§‘Z+'”)' (8.19) The series converges slowly, so it is difﬁcult to evaluate numerically, but it is known
to be equal to In 2. So 2 2
U1 = —39—1n2 = —1.386€— (8.20)
a a Now consider the next adjacent line of ions above. The nearest is negative
and at the distance 0. Then there are two positives at the distance V2 a. The next
pair are at the distance V3 a, the next at V 10 a, and so on. So for the whole line we get the series
_e_2 __ l + . . . (3 21)
a 1 ' 85 + SIN
§l~
a Fig. 8—5. Cross section of a so“
crystal on an atomic scale. The checker
board arrangement of Na and CI ions is
the same in the two cross sections per pendicular to the one shown. (See Vol. I,
Fig. 1—7.) There are four such lines: above, below, in frant, and in back. Then there are the
four lines which are the nearest lines on diagonals, and on and on. If you work patiently through for all the lines, and then take the sum, you
ﬁnd that the grand total is which is just somewhat more than what we obtained in (8.20) for the ﬁrst line.
Using e2/a = 5.12 ev, we get U = 8.94 ev. Our answer is about 10% above the experimentally observed energy. It shows that
our idea that the whole lattice is held together by electrical Coulomb forces is
fundamentally correct. This is the ﬁrst time that we have obtained a speciﬁc
property of a macroscopic substance from a knowledge of atomic physics. We
Will do much more later. The subject that tries to understand the behavior of
bulk matter in terms of the laws of atomic behavior is called solidstate physics. Now what about the error in our calculation? Why is it not exactly right?
It is because of the repulsion between the ions at close distances. They are not
perfectly rigid spheres, so when they are close together they are partly squashed.
They are not very soft, so they squash only a little bit. Some energy, however, is
used in deforming them, and when the ions are pulled apart this energy is released.
The actual energy needed to pull the ions apart is a little less than the energy that
we calculated; the repulsion helps in overcoming the electrostatic attraction. Is there any way we can make an allowance for this contribution? We could
if we knew the law of the repulsive force. We are not ready to analyze the details
of this repulsive mechanism, but we can get some idea of its characteristics from
some largescale measurements. From a measurement of the compressibility of
the whole crystal, it is possible to obtain a quantitative idea of the law of repulsion
between the ions and therefore of its contribution to the energy. In this Way it
has been found that this contribution must be 1/9.4 of the contribution from the
electrostatic attraction and, of course, of opposite sign. If we subtract this contribu
tion from the pure electrostatic energy, we obtain 7.99 ev for the dissociation energy
per molecule. It is much closer to the observed result of 7.92 ev, but still not in
perfect agreement. There is one more thing we haven’t taken into account: we
have made no allowance for the kinetic energy of the crystal vibrations. If a cor
rection is made for this effect, very good agreement with the experimental number
is obtained. The ideas are then correct; the major contribution to the energy of a
crystal like NaCl is electrostatic. 8—4 Electrostatic energy in nuclei We will now take up another example of electrostatic energy in atomic
physics, the electrical energy of atomic nuclei. Before we do this we will have to
discuss some ,properties of the main forces (called nuclear forces) that hold the
protons and neutrons together in a nucleus. In the early days of the discovery of
nuclei—and of the neutrons and protons that make them up—it was hoped that
the law of the strong, nonelectrical part of the force between, say, a proton and
another proton would have some simple law, like the inverse square law of elec
tricity. For once one had determined this law of force, and the corresponding ones
between a proton and a neutron, and a neutron and a neutron, it would be possible
to describe theoretically the complete behavior of these particles in nuclei. There
fore a big program was started for the study of the scattering of protons, in the
hope of ﬁnding the law of force between them; but after thirty years of effort,
nothing simple has emerged. A considerable knowledge of the force between proton
and proton has been accumulated, but we ﬁnd that the force is as complicated as
it can possibly be. What we mean by “as complicated as it can be” is that the force depends on
as many things as it possibly can. 8—6 First, the force is not a simple function of the distance between the two protons.
At large distances there is an attraction, but at closer distances there is a repulsion.
The distance dependence is a complicated function, still imperfectly known. Second, the force depends on the orientation of the protons’ spin. The protons
have a spin, and any two interacting protons may be spinning with their angular
momenta in the same direction or in opposite directions. And the force is different
when the spins are parallel from what it is when they are antiparallel, as in (a)
and (b) of Fig. 8—6. The difference is quite large; it is not a small effect. Third, the force is considerably different when the separation of the two
protons is in the direction parallel to their spins, as in (c) and (d) of Fig. 8—6, than
it is when the separation is in a direction perpendicular to the spins, as in (a) and (b). Fourth, the force depends, as it does in magnetism, on the velocity of the
protons, only much more strongly than in magnetism. And this velocity—dependent
force is not a relativistic effect; it is strong even at speeds much less than the speed
of light. Furthermore, this part of the force depends on other things besides the
magnitude of the velocity. For instance, when a proton is moving near another
proton, the force is different when the orbital motion has the same direction of
rotation as the spin, as in (e) of Fig. 8—6, than when it has the opposite direction
of rotation, as in (f). This is called the “spin orbit” part of the force. The force between a proton and a neutron and between a neutron and a
neutron are also equally complicated. To this day we do not know the machinery
behind these forces—that is to say, any simple way of understanding them. There is, however, one important way in which the nucleon forces are simpler
than they could be. That is that the nuclear force between two neutrons is the same
as the force between a proton and a neutron, which is the same as the force between
two protons! If, in any nuclear situation, we replace a proton by a neutron (or vice
versa), the nuclear interactions are not changed. The “fundamental reason” for
this equality is not known, but it is an example of an important principle that can
be extended also to the interaction laws of other strongly interacting particles~
such as the 7rmesons and the “strange” particles. This fact is nicely illustrated by the locations of the energy levels in similar
nuclei. Consider a nucleus like B11 (boroneleven), which is composed of ﬁve
protons and six neutrons. In the nucleus the eleven particles interact with one
another in a most complicated dance. Now, there is one conﬁguration of all the
possible interactions which has the lowest possible energy; this is the normal state
of the nucleus, and is called the ground state. If the nucleus is disturbed (for exam
ple, by being struck by a highenergy proton or other particle) it can be put into
any number of other conﬁgurations, called excited states, each of which will have
a characteristic energy that is higher than that of the ground state. In nuclear
physics research, such as is carried on with Van de Graaff generator (for example,
in Caltech’s Kellogg and Sloan Laboratories), the energies and other properties
of these excited states are determined by experiment. The energies of the ﬁfteen
lowest known excited states of B11 are shown in a onedimensional graph on the
left half of Fig. 8—7. The lowest horizontal line represents the ground state.
The ﬁrst excited state has an energy 2.14 Mev higher than the ground state,
the next an energy 4.46 Mev higher than the ground state, and so on. The study
of nuclear physics attempts to ﬁnd an explanation for this rather complicated
pattern of energies; there is as yet, however, no complete general theory of
such nuclear energy levels. If we replace one of the neutrons in B1 1 with a proton, we have the nucleus
of an isotope of carbon, C1 1. The energies of the lowest sixteen excited states of
C11 have also been measured; they are shown in the right half of Fig. 8—7.
(The broken lines indicate levels for which the experimental information is
questionable.) Looking at Fig. 8—7, we see a striking similarity between the pattern of the
energy levels in the two nuclei. The ﬁrst excited states are about 2 Mev above the
ground states. There is a large gap of about 2.3 Mev to the second excited state,
then a small jump of only 0.5 Mev to the third level. Again, between the fourth
and ﬁfth levels, a big jump; but between the ﬁfth and sixth a tiny separation of the 8—7 Fig. 8—6. The force between two
protons depends on every possible
parameter. 2l4 BII :\l982 Fig. 8—7. The energy levels of BH
and CH (energies in Mev). The ground
state of C” is 1.982 Mev higher than
that of B". order of 0.1 Mev. And so on. After about the tenth level, the correspondence
seems to become lost, but can still be seen if the levels are labeled with their other
deﬁning characteristics—for instance, their angular momentum and what they do
to lose their extra energy. The striking similarity of the pattern of the energy levels of B11 and C11 is
surely not just a coincidence. It must reveal some physical law. It shows, in fact,
that even in the complicated situation in a nucleus, replacing a neutron by a proton
makes very little change. This can mean only that the neutronneutron and proton
proton forces must be nearly identical. Only then would we expect the nuclear
conﬁgurations with ﬁve protons and six neutrons to be the same as with six protons
and ﬁve neutrons. Notice that the properties of these two nuclei tell us nothing about the neutron
proton force; there are the same number of neutron—proton combinations in both
nuclei. But if we compare two other nuclei, such as C“, which has six protons and
eight neutrons, with N 1‘, which has seven of each, we ﬁnd a similar correspondence
of energy levels. So we can conclude that the pp, nn, and pn forces are identical
in all their complexities. There is an unexpected principle in the laws of nuclear
forces. Even though the force between each pair of nuclear particles is very compli
cated, the force between the three possible different pairs is the same. But there are some small differences. The levels do not correspond exactly;
also, the ground state of C1 1 has an absolute energy (its mass) which is higher than
the ground state of B11 by 1.982 Mev. All the other levels are also higher in
absolute energy by this same amount. So the forces are not exactly equal. But
we know very well that the complete forces are not exactly equal; there is an elec
trical force between two protons because each has a positive charge, while between
two neutrons there is no such electrical force. Can we perhaps explain the differ
ences between B1 1 and C1 1 by the fact that the electrical interaction of the protons
is diﬂ'erent in the two cases? Perhaps even the remaining minor differences in the
levels are caused by electrical effects? Since the nuclear forces are so much stronger
than the electrical force, electrical effects would have only a small perturbing effect
on the energies of the levels. In order to check this idea, or rather to ﬁnd out what the consequences of this
idea are, we ﬁrst consider the difference in the groundstate energies of the two
nuclei. To take a very simple model, we suppose that the nuclei are spheres of
radius r (to be determined), containing Z protons. If we consider that a nucleus
is like a sphere with uniform charge density, we would expect the electrostatic
energy (from Eq. 8.7) to be U = $2492. (8.22) where (16 is the elementary charge of the proton. Since Z is ﬁve for B1 1 and six for
C1 1, their electrostatic energies would be different. With such a small number of protons, however, Eq. (8.22) is not quite correct.
If we compute the electrical energy between all pairs of protons, considered as points
which we assume to be nearly uniformly distributed throughout the sphere, we
ﬁnd that in Eq. (8.22) the quantity Z 2 should be replaced by Z(Z — 1), so the
energy is U_ 3Z(Z—1)q§ 32(2—1)e2
_§_.__~__.__. 41reor — 5 r (8'23) If we knew the nuclear radius r, we could use (8.23) to ﬁnd the electrostatic energy
difference between B11 and C“. But let’s do the opposite; let’s instead use the
observed energy difference to compute the radius, assuming that the energy differ
ence is all electrostatic in origin. That is, however, not quite right. The energy diﬁ‘erence of 1.982 Mev between
the ground states of B11 and C11 includes the rest energies—that is, the energy
mcz—of all the particles. In going from B1 1 to C1 1, we replace a neutron by a
proton, which has less mass. So part of the energy difference is the difference in
the rest energies of a neutron and a proton, which is 0.784 Mev. The difference, 8—8 to be accounted for by electrostatic energy, is thus more than 1.982 Mev; it is 1.982 + 0.784 = 2.786 Mev. Using this energy in Eq. (8.23), for the radius of either B11 or Cl 1 we ﬁnd
r = 3.12 X 1013cm. (8.24) Does this number have any meaning? To see whether it does, we should
compare it with some other determination of the radius of these nuclei. For
example, we can make another measurement of the radius of a nucleus by seeing
how it scatters fast particles. From such measurements it has been found, in fact,
that the density of matter in all nuclei is nearly the same, i.e., their volumes are
proportional to the number of particles they contain. If we let A be the number of
protons and neutrons in a nucleus (a number very nearly proportional to its mass),
it is found that its radius is given by r = A1/3ro, (8.25)
where r0 = 1.2 X 1013cm. (8.26) From these measurements we ﬁnd that the radius of a B1 1 (or a Cl 1) nucleus
is expected to be r = (1.2 x 10*”)(11)Ila = 2.7 x lo—lscm. Comparing this result with (8.24), we see that our assumptions that the
energy difference between B11 and C11 is electrostatic is fairly good; the dis
crepancy is only about 15% (not bad for our ﬁrst nuclear computation!). The reason for the discrepancy is probably the following. According to the
current understanding of nuclei, an even number of nuclear particles—in the case
of B1 1, ﬁve neutrons together with ﬁve protons—makes a kind of core; when one
more particle is added to this core, it reVolves around on the outside to make a new
spherical nucleus, rather than being absorbed. If this is so, we should have taken
a diﬂ'erent electrostatic energy for the additional proton. We should have taken
the excess energy of C1 1 over B1 1 to be just 2'qu
41600 ’ which is the energy needed to add one more proton to the outside of the core.
This number is just 5/6 of what Eq. (8.23) predicts, so the new prediction for the
radius is 5/6 of (8.24), which is in much closer agreement with what is directly
measured. We can draw two conclusions from this agreement. One is that the electrical
laws appear to be working at dimensions as small as 10"13 cm. The other is that
we have veriﬁed the remarkable coincidence that the nonelectrical part of the forces
between proton and proton, neutron and neutron, and proton and neutron are
all equal. 85 Energy in the electrostatic ﬁeld We now consider other methods of calculating electrostatic energy. They can
all be derived from the basic relation Eq. (8.3), the sum, over all pairs of charges,
of the mutual energies of each charge—pair. First we wish to write an expression
for the energy of a charge distribution. As usual, we consider that each volume
element dV contains the element of charge p dV. Then Eq. (8.3) should be written _ 1 p(1)p(2)
U — 2 !; 41reor12 dV1 dV2. (8.27) space 89 Notice the factor 4;, which is introduced because in the double integral over dVl
and d V2 we have counted all pairs of charge elements twice. (There is no convenient
way of writing an integral that keeps track of the pairs so that each pair is counted
only once.) Next we notice that the integral over d V2 in (8.27) is just the potential at (1). That is,
p(2) =
[4W60712 de ¢(1), so that (8.27) can be written as U = % /p(l)¢(l)dV1. Or, since the point (2) no longer appears, we can simply write = :—/p¢ (W. (8.28) This equation can be interpreted as follows. The potential energy of the charge
p dV is the product of this charge and the potential at the same point. The total
energy is therefore the integral over ¢p dV. But there is again the factor 4;. It is
still required because we are counting energies twice. The mutual energies of two
charges is the charge of one times the potential at it due to the other. Or, it can be
taken as the second charge times the potential at it from the ﬁrst. Thus for two
point charges we would write U = ql¢(1)= q. 412:,” or U = mm = q2 4W2,” Notice that we could also write U = ﬂqr¢(1) + (12¢(2)] (829) The integral in (8.28) corresponds to the sum of both terms in the brackets of
(8.29). That is why we need the factor i. An interesting. question is: Where is the electrostatic energy located? One
might also ask: Who cares? What is the meaning of such a question? If there is
a pair of interacting charges, the combination has a certain energy. Do we need
to say that the energy is located at one of the charges or the other, or at both, or in
between? These questions may not make sense because we really know only that
the total energy is conserved. The idea that the energy is located somewhere is
not necessary. Yet suppose that it did make sense to say, in general, that energy is located
at a certain place, as it does for heat energy. We might then extend our principle
of the conservation of energy with the idea that if the energy in a given volume
changes, we should be able to account for the change by the ﬂow of energy into
or out of that volume. You realize that our early statement of the principle of the
conservation of energy is still perfectly all right if some energy disappears at one
place and appears somewhere else far away without anything passing (that is, with
out any special phenomena occurring) in the space between. We are, therefore,
now discussing an extension of the idea of the conservation of energy. We might
call it a principle of the local conservation of energy. Such a principle would say
that the energy in any given volume changes only by the amount that ﬂows into or
out of the volume. It is indeed possible that energy is conserved locally in such a
way. If it is, we would have a much more detailed law than the simple statement
of the conservation of total energy. It does turn out that in nature energy is
conserved locally. We can ﬁnd formulas for where the energy is located and how it
travels from place to place. There is also a physical reason why it is imperative that we be able to say
where energy is located. According to the theory of gravitation, all mass is a source 8—10 of gravitational attraction. We also know, by E = mc 2, that mass and energy are
equivalent. All energy is, therefore, a source of gravitational force. If we could not
locate the energy, we could not locate all the mass. We would not be able to say
where the sources of the gravitational ﬁeld are located. The theory of gravitation
would be incomplete. If we restrict ourselves to electrostatics there is really no way to tell where the
energy is located. The complete Maxwell equations of electrodynamics give us
much more information (although even then the answer is, strictly speaking, not
unique.) We will therefore discuss this question in detail again in a later chapter.
We will give you now only the result for the particular case of electrostatics.
The energy is located in space, where the electric ﬁeld is. This seems reasonable
because we know that when charges are accelerated they radiate electric ﬁelds.
We would like to say that when light or radiowaves travel from one point to another,
they carry their energy with them. But there are no charges in the waves. So we
would like to locate the energy where the electromagnetic ﬁeld is and not at the
charges from which it came. We thus describe the energy, not in terms of the
charges, but in terms of the ﬁelds they produce. We can, in fact, show that Eq.
(8.28) is numerically equal to = 3’ E EdV. (8.30) We can then interpret this formula as saying that when an electric ﬁeld is present,
there is located in space an energy whose density (energy per unit volume) is “=2E.E=9£2. 2 2 (8.31) This idea is illustrated in Fig. 8—8. To show that Eq. (8.30) is consistent with our laws of electrostatics, we begin
by introducing into Eq. (8.28) the relation between p and «1: that we obtained in
Chapter 6: We get
U = —3 a W: dV. (8.32) Writing out the components of the integrand, we see that 2 _ 62¢ 932 614:)
”Mtg—£5 ay2 622 a M a¢2 a 04> 64:2 a(a¢>_(a_¢)2
«The—Vb“) +3<¢a—y)‘(‘a;) W W az = V ' (¢ V45)  (V45) ' (V40 (5333)
Gur energy integral is then ll U = %/(v¢)(v¢)dV— % v(¢ v¢)dV. We can use Gauss’ theorem to change the second integral into a surface integral: / V ' (d) V42) dV = f (d) V¢)  n da. (8.34) vol . surface We evaluate the surface integral in the case that the surface goes to inﬁnity
‘ (so the volume integrals become integrals over all space), supposing that all the
charges are located within some ﬁnite distance. The simple way to proceed is to
take a spherical surface of enormous radius R whose center is at the origin of
coordinates. We know that when we are very far away from all charges, ¢ varies
as l/R and V¢ as l/Rz. (Both will decrease even faster with R if there the net 8—11 r
h Fig. 8—8. Each volume element dV =
dx dy dz in an electric ﬁeld contains the energy (co/2)E2 dV. charge in the distribution is zero.) Since the surface area of the large sphere in
creases as Rz, we see that the surface integral falls oﬁ‘ as (l/R)(l/R2)R2 = (UK)
as the radius of the sphere increases. So if we include all space in our integration
(R + co), the surface integral goes to zero and we have that U = g / (v¢)  (V¢)dV = % EEdV. (8.35)
33.1%.. .52}... We see that it is possible for us to represent the energy of any charge distribution
as being the integral over an energy density located in the ﬁeld. 8—6 The energy of a point charge Our new relation, Eq. (8.35), says that even a single point charge q will have
some electrostatic energy. In this case, the electric ﬁeld is given by _ q ,
E — 41r€072 So the energy density at the distance r from the charge is €0E2 q2 2 = 321r2eor4' We can take for an element of volume a Spherical shell of thickness dr and area
47rr2. The total energy is no 2 2 r
_ q _ _ q 1 ,
U ‘ f 81reor2 d’ ‘ 8“,, r .=,, (8.36) T=0 Now the limit at r = on gives no difﬁculty. But for a point charge we are
supposed to integrate down to r = 0, which gives an inﬁnite integral. Equation
(8.35) says that there is an inﬁnite amount of energy in the ﬁeld of a point charge,
although we began with the idea that there was energy only between point charges.
In our original energy formula for a collection of point charges (Eq. 8.3), we did
not include any interaction energy of a charge with itself. What has happened is
that when we went over to a continuous distribution of charge in Eq. (8.27), we
counted the energy of interaction of every inﬁnitesimal charge with all other
inﬁnitesimal charges. The same account is included in Eq. (8.35), so when we
apply it to a ﬁnite point charge, we are including the energy it would take to
assemble that charge from inﬁnitesimal parts. You will notice, in fact, that we
would also get the result in Eq. (8.36) if we used our expression (8.11) for the energy
of a charged sphere and let the radius tend toward zero. We must conclude that the idea of locating the energy in the ﬁeld is incon
sistent with the assumption of the existence of point charges. One way out of the
difﬁculty would be to say that elementary charges, such as an electron, are not
points but are really small distributions of charge. Alternatively, we could say
that there is something wrong in our theory of electricity at very small distances,
or with the idea of the local conservation of energy. There are difﬁculties with
either point of view. These difﬁculties have never been overcome; they exist to this
day. Sometime later, when we have discussed some additional ideas, such as the
momentum in an electromagnetic ﬁeld, we will give a more complete account of
these fundamental difﬁculties in our understanding of nature. 8—12 ...
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 Spring '09
 LeeKinohara
 Physics, Energy

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