R
U 0
2 2
2 2
4 E r dr
E r dr.
(1.67)
2
0
R
Using Eqs. (19) and (22) for, respectively, the external and internal regions, we get
2
2
R
Qr
Q
Q
0
2
2
1
1
2
U
4
r dr
r dr 1
.
(1.68)
2
4
4
2
r
5
4 2 R
0
0
R
0
0
This is (fortunately :-) the same answer as given by Eq. (66), but to some extent, Eq. (68) is more informative because it shows how exactly the electric field’s energy is distributed between the interior and exterior of the charged sphere.26
24 This transformation follows from the divergence theorem MA (12.2) applied to the vector function f = , taking into account the differentiation rule MA Eq. (11.4a): () = ()() + () = ()2 + 2.
25 In the Gaussian units, the standard replacement 0 1/4 turns the last of Eqs. (65) into u(r) = E 2/8.
26 Note that U at R 0. Such divergence appears at the application of Eq. (65) to any point charge. Since it does not affect the force acting on the charge, the divergence does not create any technical difficulty for analysis Chapter 1
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We see that, as we could expect, within the realm of electrostatics, Eqs. (60) and (65) are equivalent. However, when we examine electrodynamics (in Chapter 6 and beyond), we will see that the latter equation is more general and that it is more adequate to associate the electric energy with the field itself rather than its sources – in our current case, the electric charges.
Finally, let us calculate the potential energy of a system of charges in the general case when both the internal interaction of the charges and their interaction with an external field are important. One might fancy that such a calculation should be very hard since, in both ultimate limits, when one of these interactions dominates, we have got different results. However, once again we get help from the almighty linear superposition principle: in the general case, for the total electric field we may write Er E r E
r
r r r
(1.69)
int
ext ,
int ext ,
where the index “int” now marks the field induced by the charge system under analysis, i.e. the variables participating (without indices) in Eqs. (56)-(65). Now let us imagine that our system is being built up in the following way: first, the charges are brought together at Eext = 0, giving the potential energy U int expressed by Eq. (60), and then Eext is slowly increased. Evidently, the energy contribution from the latter process cannot depend on the internal interaction of the charges, and hence may be expressed in the form (55). As the result, the total potential energy27 is the sum of these two components: 1
U U U
(r) (r)
(r) (r)
.
(1.70)
int
ext
d 3 r
int
d 3 r
ext
2
Now making the transition from the potentials to the fields, absolutely similar to that performed in Eqs.
(62)-(65), we may rewrite this expression as
U u r 3
( ) d r,
with u r
0
( )
2
E
r
( ) 2E
.
(1.71)
int
int r E ext r
2
One might think that this result, more general than Eq. (65) and perhaps less familiar to the reader, is something entirely new; however, it is not. Indeed, let us add to, and subtract E 2
ext (r) from the
sum in the brackets, and use Eq. (69) for the total electric field E(r); then Eq. (71) takes the form
U 0 E 2 r d 3 r
0
E 2 r
.
(1.72)
ext
d 3 r
2
2
Hence, in the most important case when we are using the potential energy to analyze the statics and dynamics of a system of charges in a fixed external field, i.e. when the second term on the right-hand side of Eq. (72) may be considered as a constant, we may still use for U an expression similar to the familiar Eq. (65), but with the field E(r) being the sum (69) of the internal and external fields.
Let us see how this works in a very simple situation. A uniform external electric field Eext is applied normally to a very broad, plane layer that contains a very large and equal number of free electric charges of both signs – see Fig. 7. What is the equilibrium distribution of the charges over the layer?
of charge statics or non-relativistic dynamics, but it points to a possible conceptual problem of classical electrodynamics as the whole at describing point charges. This issue will be discussed at the very end of the course (Sec. 10.6).
27 This total U (or rather its part dependent on our system of charges) is sometimes called the Gibbs potential energy of the system. (I will discuss this notion in detail in Sec. 3.5.)
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Eext
Fig. 1.7. A simple model of the electric
field screening in a conductor. Here
Eint
(and in all figures below) the red and
blue colors are used to denote the
opposite charge signs.
Since any area-uniform distribution of the charge inside the layer does not affect the field (and hence its energy) outside it, and the equilibrium distribution has to minimize the total potential energy of the system, Eq. (72) immediately gives the answer: the distribution should provide E Eint + Eext = 0
inside the whole layer – the effect called the electric field screening. The only way to ensure this equality is to have enough free charges of opposite signs residing on the layer’s surfaces to induce a uniform field Eint = –Eext, exactly compensating the external field at each point inside the layer – see Fig. 7. According to Eq. (24), the areal density of these surface charges should equal , with =
E ext/0. This is a rudimentary but reasonable model of conductors’ polarization – to be discussed in detail in the next chapter.
1.4. Exercise problems
1.1. Calculate the electric field of a thin, long, straight filament, electrically charged with a constant linear density , using two approaches:
(i) directly from the Coulomb law, and
(ii) using the Gauss law.
1.2. Two thin, straight parallel filaments, separated by distance , carry
equal and opposite uniformly distributed charges with linear density – see the
figure on the right. Calculate the force (per unit length) of the Coulomb interaction
of the wires. Compare its dependence on with the Coulomb law for the force
between two point charges, and interpret their difference.
1.3. Calculate the electric field of the following spherically-symmetric charge distribution: ( r) =
0exp{– r}.
1.4. A sphere of radius R, whose volume had been charged with a constant density , is split with a very narrow planar gap passing through its center. Calculate the force of the mutual electrostatic repulsion of the resulting two hemispheres.
1.5. A thin spherical shell of radius R, which had been charged with a constant areal density , is split into two equal halves with a very narrow, planar cut passing through the sphere’s center. Calculate the force of electrostatic repulsion between the resulting hemispheric shells, and compare the result with that of the previous problem.
1.6. Calculate the distribution of the electrostatic potential created by a straight, thin filament of a finite length 2 l, charged with a constant linear density , and explore the result in the limits of very small and very large distances from the filament.
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1.7. A thin plane sheet, perhaps of an irregular shape, carries an electric charge with a constant areal density .
(i) Express the electric field’s component normal to the plane, at a certain distance from it, via the solid angle at which the sheet is visible from the observation point.
(ii) Use the result to calculate the field in the center of a cube, with one face charged with a constant density .
1.8. Can one create, in an extended region of space, electrostatic fields with the Cartesian components proportional to the following products of Cartesian coordinates { x, y, z}: i
( ) yz, xz,
xy ,
ii xy, xy, yz?
1.9. Distant sources have been used to create different uniform electric fields in two semi-spaces: z
Er
E ,
at z ,
0
n
r
R
z
E ,
at z ,
0
E
everywhere except for a transitional region of scale R near the origin, where
the field is perturbed but still axially-symmetric. (As will be discussed in the
next chapter, this may be done, for example, using a thin conducting
2 R
membrane with a round hole of radius R in it – see the figure on the right.) E
Prove that such field may serve as an electrostatic lens for charged particles
flying along the z-axis, at distances << R from it, and calculate the focal distance f of the lens. Spell out the conditions of validity of your result.
q
r 3
1.10. Eight equal point charges q are located at the corners of a cube of side a. Calculate all Cartesian components Ej of the electric field, and their spatial derivatives Ej/ rj’, at the cube’s center, where rj are the Cartesian coordinates 0
r 2
oriented along the cube’s sides – see the figure on the right. Are all your results r valid for the center of a plane square, with four equal charges at its corners?
1
a
1.11. By a direct calculation, find the average electric potential of a spherical surface of radius R, created by a point charge q located at a distance r > R from the sphere’s center. Use the result to prove the following general mean value theorem: the electric potential at any point is always equal to its average value on any spherical surface with the center at that point, and containing no electric charges inside it.
1.12. Two similar thin, circular, coaxial disks of radius R, separated
R
by distance 2 d, are uniformly charged with equal and opposite areal
densities – see the figure on the right. Calculate and sketch the 2 d distribution of the electrostatic potential and the electric field of the disks
along their common axis.
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1.13. The electrostatic potential created by some electric charge distribution, is
r
1
1
r
C
exp
,
r 2 r
r
0
0
where C and r 0 are constants, and r r is the distance from the origin. Calculate the charge distribution in space.
1.14. A thin flat sheet cut in the form of a rectangle of size a b, is electrically charged with a constant areal density . Without an explicit calculation of the spatial distribution (r) of the electrostatic potential induced by this charge, find the ratio of its values at the center and at the corners of the rectangle.
Hint: Consider partitioning the rectangle into several similar parts and using the linear superposition principle.
1.15. Calculate the electrostatic energy per unit area of the system of two thin, parallel planes with equal and opposite charges of a constant areal density , separated by distance d.
1.16. The system analyzed in the previous problem (two thin,
Eext
parallel, oppositely charged planes) is now placed into an external,
uniform, normal electric field E ext = /0 – see the figure on the right. Find
d
the force (per unit area) acting on each plane, by two methods:
(i) directly from the electric field distribution, and
(ii) from the potential energy of the system.
1.17. Explore the relationship between the Laplace equation (42) and the condition of the minimum of the electrostatic field energy (65).
1.18. Prove the following reciprocity theorem of electrostatics:28 if two spatially-confined charge distributions 1(r) and 2(r) create respective distributions 1(r) and 2(r) of the electrostatic potential, then
r r
r r
.
1 2 d 3 r
2 1 d 3 r
Hint: Consider the integral E E d 3 r .
1
2
1.19. Calculate the energy of electrostatic interaction of two spheres, of radii R 1 and R 2, each with a spherically-symmetric charge distribution, separated by distance d > R 1 + R 2.
R
R
1.20. Calculate the electrostatic energy U of a (generally, thick) spherical shell, 1
2
Q
with charge Q uniformly distributed through its volume – see the figure on the right.
Analyze and interpret the dependence of U on the inner cavity’s radius R 1, at fixed Q
and R 2.
28 This is only the simplest one of several reciprocity theorems in electromagnetism – see, e.g., Sec. 6.8 below.
Chapter 1
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Chapter 2. Charges and Conductors
This chapter starts our discussion of the very common situations when the electric charge distribution in space is not known a priori, but rather should be calculated in a self-consistent way together with the electric field it creates. The simplest situations of this kind involve conductors and lead to the so-called boundary problems in that the partial differential equations describing the field distribution have to be solved with appropriate boundary conditions. Such problems are also typical for other parts of electrodynamics ( and indeed for other fields of physics as well) , so that following tradition, I will use this chapter’s material as a playground for a discussion of various methods of boundary problem solution, and the special functions most frequently encountered on that way.
2.1. Polarization and screening
The basic principles of electrostatics outlined in Chapter 1 present the conceptually full solution of the problem of finding the electrostatic field (and hence Coulomb forces) induced by electric charges distributed over space with some density (r). However, in most practical situations, this function is not known but should be found self-consistently with the field. For example, if a sample of relatively dense material is placed into an external electric field, it is typically polarized, i.e. acquires some local charges of its own, which contribute to the total electric field E(r) inside, and even outside it – see Fig. 1a.
(a)
(b)
Fig. 2.1. Two typical electrostatic
situations involving conductors:
(a) polarization by an external
field, and (b) re-distribution of
conductor’s own charge over its
surface – schematically. Here and
below, the red and blue points
denote charges of opposite signs.
The full solution of such problems should satisfy not only the fundamental Eq. (1.7) but also the so-called constitutive relations between the macroscopic variables describing the sample’s material.
Due to the atomic character of real materials, such relations may be very involved. In this part of my series, I will have time to address these relations, for various materials, only rather superficially,1
focusing on their simple approximations. Fortunately, in most practical cases such approximations work very well.
In particular, for the polarization of good conductors, a very reasonable approximation is given by the so-called macroscopic model, in which the free charges in the conductor are is treated as a charged continuum that is free to move under the effect of the force F = qE exerted by the macroscopic electric field E, i.e. the field averaged over space on the atomic scale – see also the discussion at the end 1 A more detailed discussion of the electrostatic field screening may be found, e.g., in SM Sec. 6.4. (Alternatively, see either Sec. 13.5 of J. Hook and H. Hall, Solid State Physics, 2nd ed., Wiley, 1991; or Chapter 17 of N.
Ashcroft and N. Mermin, Solid State Physics, Brooks Cole, 1976.)
© K. Likharev
Essential Graduate Physics
EM: Classical Electrodynamics
of Sec. 1.1. In electrostatics (which excludes the case dc currents, to be discussed in Chapter 4 below), there should be no such motion, so that everywhere inside the conductor the macroscopic electric field should vanish:
E 0 .
(2.1a)
This is the electric field screening 2 effect, meaning, in particular, that conductors’ polarization in an Conductor: external electric field has the extreme form shown (rather schematically) in Fig. 1a, with the field of the macroscopic induced surface charges completely compensating the external field in the conductor’s bulk. Note that model Eq. (1a) may be rewritten in another, frequently more convenient form:
const ,
(2.1b)
where is the macroscopic electrostatic potential related to the macroscopic field by Eq. (1.33).3 (If a problem includes several unconnected conductors, the constant in Eq. (1b) may be specific for each of them.)
Now let us examine what we can say about the electric field in free space just outside a conductor, within the same macroscopic model. At close proximity, any smooth surface (in our current case, that of a conductor) looks planar. Let us integrate Eq. (1.28) over a narrow ( d << l) rectangular loop C encircling a part of such plane conductor’s surface (see the dashed line in Fig. 2a), and apply it to the electric field vector E the well-known vector algebra equality – the Stokes theorem 4
( E d 2
)
r E dr ,
(2.2)
n
S
C
where S is any surface limited by the contour C.
(a)
(b)
E
0
const
free space
E
C
c
d
conductor
l
E 0
c
Fig. 2.2. (a) The surface charge layer at a conductor’s surface, and
(b) the electric field lines and equipotential surfaces near it.
In our current case, the contour is dominated by two straight lines of length l, so that if l is much smaller than the characteristic spatial scale of field’s changes but much larger than the interatomic distances, the right-hand side of Eq. (2) may be well approximated as [( E)in – ( E)out] l, where E is the tangential component of the corresponding macroscopic field, parallel to the surface. On the other hand, according to Eq. (1.28), the left-hand side of Eq. (2) equals zero. Hence, the macroscopic field’s 2 This term, used for the electric field, should not be confused with shielding – the term used for the description of magnetic field’s reduction by magnetic materials – see Chapter 5 below.
3 Since averaging of a function over space is a linear operation, any linear relation between genuine (microscopic) variables, including Eq. (1.33), is also valid for the corresponding macroscopic variables.
4 See, e.g., MA Eq. (12.1).
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component E should be continuous at the surface, and to satisfy Eq. (1a) inside the conductor, the component has to vanish immediately outside it: ( E)out = 0. This means that the electrostatic potential immediately outside of a conducting surface cannot change along it. In other words, the equipotential surfaces outside a conductor should “lean” to the conductor’s surface, with their potential values approaching the constant potential of the conductor – see Fig. 2b.
So, the electrostatic field just outside any conductor has to be normal to its surface. To find this normal field, we may apply the universal relation (1.24) to our macroscopic field E. Since in our current case En = 0 inside the conductor, we get
Surface
charge
,
(2.3)
0 E
n out
0
n
0
density
n
where is the macroscopic areal density of the conductor’s surface charge. Note that deriving this universal relation between the normal component of the field and the surface charge density, we have not used any cause-vs-effect arguments, so that Eq. (3) is valid regardless of whether the surface charge is induced by an externally applied field (as in the case of conductor’s polarization, shown in Fig. 1a), or the electric field is induced by the electric charge placed on the conductor and then self-redistributed over its surface (Fig. 1b), or it is some combination of both effects.
Before starting to use the macroscopic model for the solution of particular problems of electrostatics, let me use the balance of this section to briefly discuss its limitations. (The reader in a rush may skip this discussion and proceed to Sec. 2; however, I believe that every educated physicist has to understand when this model works, and when it does not.)
Since the argumentation which has led us to Eq. (1.24) and hence to Eq. (3) is valid for any thickness d of the Gauss pillbox, within the macroscopic model, the whole surface charge is located within an infinitely thin surface layer. This is of course impossible physically: for one, this would require an infinite volumic density of the charge. In reality, the charged layer (and hence the region of the electric field’s crossover from the finite value (3) to zero) has a nonzero thickness . At least three effects contribute to .
(i) Atomic structure of matter. Within each atom, and frequently between the adjacent atoms as well, the genuine (“microscopic”) electric field is highly nonuniform. Thus, as was already stated above, Eq. (1) is valid only for the macroscopic field, i.e. the field averaged over distances of the order of the atomic size scale a 0 ~ 10-10 m,5 and cannot be applied to the field changes on that scale. As a result, the surface layer of charges cannot be much thinner than a 0.
(ii) Thermal excitation. According to Eq. (1.9), in the whole field-free bulk of a conductor, the net charge density, = e( n – n e), 6 has to vanish, so that the numbers of protons in atomic nuclei ( n) and electrons ( n e) per unit volume have to be balanced. However, if an external electric field penetrates a conductor, free electrons can shift in or out of its affected part, depending on the field’s contribution to their potential energy, U = q e = – e. (Here the arbitrary constant in is chosen to give = 0 well inside the conductor.) In classical statistics, this change is described by the Boltzmann distribution:7
5 This scale originates from the quantum-mechanical effects of electron motion, characterized by the Bohr radius r
2
B 2/ m e( e 2/40) 0.5310-10 m – see, e.g., QM Eq. (1.10). It also defines the scale E B = e/40 r B ~ 1012 SI units (V/m) of the microscopic electric fields inside atoms. (Please note how large these fields are.) 6 In this series, e denotes the fundamental charge, e 1.610-19 C > 0, so that the electron’s charge equals (– e).
7 See, e.g., SM Sec. 3.1.
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U (r)
n r n
(2.4)
e
exp
,
k T
B
where T is the absolute temperature in kelvins (K), and k B 1.3810-23 J/K is the Boltzmann constant.
As a result, the net charge density is
e r
r
en1 exp
.
(2.5)
k T
B
The penetrating electric field polarizes the atoms as well. As will be discussed in the next chapter, such polarization results in the reduction of the electric field by a material-specific dimensionless factor
(larger, but typically not too much larger than 1), called the dielectric constant. As a result, the Poisson equation (1.41) takes the so-called Poisson-Boltzmann form,8
2
d
en
e
exp
1 ,
(2.6)
2
dz
k T
0
0
B
where we have taken advantage of the 1D geometry of the system to simplify the Laplace operator, with the z-axis normal to the surface.
Even with this simplification, Eq. (6) is a nonlinear differential equation allowing an analytical but rather bulky solution. Since our current goal is just to estimate the field penetration depth , let us simplify the equation further by considering the low-field limit: e ~ e E << k B T. In this limit, we may extend the exponent into the Taylor series, and keep only two leading terms (of which the first one cancels with the following unity). As a result, Eq. (6) becomes linear,
2
2
d
en
e
d
1
,
i.e.
,
(2.7)
2
2
2
dz
k T
dz
0
B
where the constant , in this case, is called the Debye (or “Debye-Hückel”) screening length D:
k T
Debye
2
0 B
.
(2.8)
screening
D
e 2 n
length
As the reader certainly knows, Eq. (7) describes an exponential decrease of the electric potential, with the characteristic length D: exp{- z/D}, where the z-axis is directed into the conductor.
Plugging in the involved fundamental constants into Eq. (8), we get the following estimate: D[m]
70( T[K]/ n[m-3])1/2. According to this formula, in semiconductors at room temperature, the Debye length may be rather substantial. For example, in silicon ( 12) doped to the free charge carrier concentration n = 31018 cm-3 (the value typical for modern integrated circuits),9 D 2 nm, still well 8 This equation and/or its straightforward generalization to the case of charged particles (ions) of several kinds is also (especially in the theories of electrolytes and plasmas) called the Debye-Hückel equation.
9 There is a good reason for making an estimate of D for this case: the electric field created by the gate electrode of a field-effect transistor, penetrating into doped silicon by a depth ~D, controls the electric current in this most important electronic device – on whose back all our information technology rides. Because of that, D establishes the possible scale of semiconductor circuit shrinking, which is the basis of the well-known Moore’s law.
(Practically, the scale is determined by integrated circuit patterning techniques, and Eq. (8) may be used to find the proper charge carrier density n and hence the necessary level of silicon doping – see, e.g., SM Sec. 6.4.) Chapter 2
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above the atomic size scale a 0, thus justifying the estimate However, for typical good metals ( n ~ 1029
m-3, ~ 10) the same formula gives D ~ 10-11 m, less than a 0. In this case, Eq. (8) should not be taken literally, because it is based on the assumption of a continuous charge distribution.
(iii) Quantum statistics. Actually, the last estimate is not valid for good metals (and highly doped semiconductors) for one more reason: their free electrons obey the quantum ( Fermi-Dirac) statistics rather than the Boltzmann distribution (4).10 As a result, at all realistic temperatures the electrons form a degenerate quantum gas, occupying all available energy states below some energy level EF >> k B T, called the Fermi energy. In these conditions, the screening of relatively low electric field may be described by replacing Eq. (5) with
e n n eg E
U
e g E
(2.9)
e
( )(
)
2
( ) ,
F
F
where g(E) is the density of quantum states (per unit volume per unit energy) at the electron’s energy E.
At the Fermi surface, the density is of the order of n/EF.11 As a result, we again get the second of Eqs.
(7), but with a different characteristic scale , defined by the following relation:
Thomas-
Fermi
2
0
~
0 F
E ,
(2.10)
screening
TF
2
e g(
length
E )
2
e n
F
and called the Thomas-Fermi screening length. Since for most good metals, n is of the order of 1029 m-3, and EF is of the order of 10 eV, Eq. (10) typically gives TF close to a few a 0, and makes the Thomas-Fermi screening theory valid at least semi-quantitatively.
To summarize, the electric field penetration into good conductors is limited to a depth ranging from a fraction of a nanometer to a few nanometers, so that for problems with a characteristic linear size much larger than that scale, the macroscopic model (1) gives a very good accuracy, and we will use them in the rest of this chapter. However, the reader should remember that in many situations involving semiconductors, as well as at some nanoscale experiments with metals, the electric field penetration should be taken into account.
Another important condition of the macroscopic model’s validity is imposed on the electric field’s magnitude, which is especially significant for semiconductors. Indeed, as Eq. (6) shows, Eq. (7) is only valid if e << k B T, so that E ~ /D should be much lower than k B T/ eD. In the example given above (D 2 nm , T = 300 K), this means E << Et ~107V/m 105V/cm – the value readily reachable in the lab. In larger fields, the field penetration becomes nonlinear, leading in particular to the very important effect of carrier depletion; it will be discussed in SM Sec. 6.4. For typical metals, such linearity limit, Et ~ EF/ eTF is much higher, ~1011 V/m, but the model may be violated at lower fields by other effects, such as the impact-ionization leading to electric breakdown, which may start at ~106 V/m.
2.2. Capacitance
Let us start using the macroscopic model from systems consisting of charged conductors only, with no so-called stand-alone charges in the free space outside them.12 Our goal here is to calculate the 10 See, e.g., SM Sec. 2.8. For a more detailed derivation of Eq. (10), see SM Chapter 3.
11 See, e.g., SM Sec. 3.3.
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distributions of the electric field E and potential in space, and the distribution of the surface charge density over the conductor surfaces. However, before doing that for particular situations, let us see if there are any integral measures of these distributions, which should be our primary focus.
The simplest case is of course a single conductor in the otherwise free space. According to Eq.
(1b), all its volume should have the same electrostatic potential , evidently providing one convenient global measure of the situation. Another integral measure is provided by the total charge Q d 3 r d 2 r ,
(2.11)
V
S
where the last integral is extended over the whole surface S of the conductor. In the general case, what can we tell about the relation between Q and ? At Q = 0, there is no electric field in the system, and it is natural (though not absolutely necessary) to select the arbitrary constant in the electrostatic potential to have = 0 everywhere. Then, if the conductor is charged with a non-zero Q, according to the linear Eq.
(1.7), the electric field at any point of space has to be proportional to that charge . Hence the electrostatic potential at all points, including its value inside the conductor, is also proportional to Q:
pQ .
(2.12)
The proportionality coefficient p, which depends on the conductor’s size and shape, but on neither nor Q, is called its reciprocal capacitance (or, not too often, “electric elastance”). Usually, Eq. (12) is rewritten in a different form,
1
Q C, with C
,
(2.13) Self-
p
capacitance
where C is called self-capacitance. (Frequently, C is called just capacitance, but as we will see very soon, for more complex situations the latter term may be ambiguous.)
Before calculating C for particular geometries, let us have a look at the electrostatic energy U of a single conductor. To calculate it, of the several relations discussed in Chapter 1, Eq. (1.61) is most convenient, because all elementary charges qk are now parts of the conductor charge, and hence reside at the same potential – see Eq. (1b) again. As a result, the equality becomes very simple: 1
1
U q Q .
(2.14)
k
2
k
2
Moreover, using the linear relation (13), the same result may be re-written in two more forms: 2
Q
C
Electro-
2
U
.
(2.15)
static
2 C
2
energy
We will discuss several ways to calculate C in the next sections, and right now will have a quick look at just the simplest example for that we have calculated everything necessary in the previous chapter: a conducting sphere of radius R. Indeed, we already know the electric field distribution: according to Eq. (1), E = 0 inside the sphere, while Eq. (1.19), with Q( r) = Q, describes the field distribution outside it, because of the evident spherical symmetry of the surface charge distribution.
12 In some texts, these charges are called “free”. This term is somewhat misleading, because they may well be bound, i.e. unable to move freely.
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Moreover, since the latter formula is exactly the same as for the point charge placed in the sphere’s center, the potential’s distribution in space may be obtained from Eq. (1.35) by replacing q with the sphere’s full charge Q. Hence, on the surface of the sphere (and, according to Eq. (1b), through its interior),
1 Q
.
(2.16)
4 R
0
Comparing this result with the definition (13), for the sphere’s self-capacitance we obtain a very simple formula13
C: shere
C 4 R .
(2.17)
0
This formula, which should be well familiar to the reader, is convenient to get some feeling of how large the SI unit of capacitance (1 farad, abbreviated as F) is: the self-capacitance of Earth ( R E
6.34106 m) is below 1 mF! Another important note is that while Eq. (17) is not exactly valid for a conductor of arbitrary shape, it implies an important general estimate
C ~ 2 a
(2.18)
0
where a is the scale of the linear size of any conductor.14
Now proceeding to a system of two arbitrary conductors, we immediately see why we should be careful with the capacitance definition: one constant C is insufficient to describe all electrostatic properties of such a system. Indeed, here we have two, generally different conductor potentials, 1 and
2, that may depend on both conductor charges, Q 1 and Q 2. Using the same arguments as for the single-conductor case, we may conclude that the dependence is always linear:
p Q p Q ,
1
11
1
12
2
(2.19)
p Q p Q ,
2
21
1
22
2
but now has to be described by more than one coefficient. Actually, it turns out that there are three rather than four different coefficients in these relations, because
p
.
(2.20)
12
p 21
This equality may be proved in several ways, for example, using the general reciprocity theorem of electrostatics (whose proof was the subject of Problem 1.17):
r
r
r r
,
(2.21)
1 2 d 3 r
2 1 d 3 r
13 In the Gaussian units, using the standard replacement 40 1, this relation takes an even simpler form: C =
R, very easy to remember. Generally, in the Gaussian units (but not in the SI system!) the capacitance has the dimensionality of length, i.e. is measured in centimeters. Note also that a fractional SI unit, 1 picofarad (10-12 F), is very close to the Gaussian unit: 1 pF = [(110-12)/(4010-2)] cm 0.8998 cm. So, 1 pF is close to the capacitance of a metallic ball with a 1-cm radius, making this unit very convenient for human-scale systems.
14 These arguments are somewhat insufficient to say which size should be used for a in the case of narrow, extended conductors, e.g., a thin, long wire . Very soon we will see that in such cases the electrostatic energy, and hence C, depends mostly on the larger size of the conductor.
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where (1)(r) and (2)(r) are the potential distributions induced, respectively, by two electric charge distributions, 1(r) and 2(r). In our current case, each of these integrals is limited to the volume (or, more exactly, the surface) of the corresponding conductor, where each potential is constant and may be taken out of the integral. As a result, Eq. (21) is reduced to
2
Q
r Q
r .
(2.22)
1
1
1
2
2
In terms of Eq. (19), (2)(r1) is just p 12 Q 2, while (1)(r2) equals p 21 Q 1. Plugging these expressions into Eq.
(22), and canceling the product Q 1 Q 2, we arrive at Eq. (20).
Hence the 22 matrix of coefficients pjj’ (called the reciprocal capacitance matrix) is always symmetric, and using the natural notation p 11 p 1, p 22 p 2, p 12 = p 21 p, we may rewrite it in a simpler form:
p 1
p .
(2.23)
p
p 2
Plugging the relation (19), in this new notation, into Eq. (1.61), we see that the full electrostatic energy of the system may be expressed as a quadratic form of its charges:
p 1 2
p 2 2
U
Q pQ Q
Q .
(2.24)
1
1
2
2
2
2
It is evident that the middle term on the right-hand side of this equality describes the electrostatic coupling of the conductors. (Without it, the energy would be just a sum of two independent electrostatic energies of conductors 1 and 2.)15 Still, even with this simplification, Eqs. (19) and (20) show that in the general case of arbitrary charges Q 1 and Q 2, the system of two conductors should be characterized by three, rather than just one coefficient (“the capacitance”). This is why we may attribute a single capacitance to the system only in some particular cases.
For practice, the most important of them is when the system as the whole is electrically neutral: Q 1 = – Q 2 Q. In this case, the most important function of Q is the difference between the conductors’
potentials, called the voltage:16
V ,
(2.25) Voltage:
1
2
definition
For that function, the subtraction of two Eqs. (19) gives
Q
1
V
, with C
,
(2.26) Mutual
C
p
capacitance
1
p
2
2
p
where the coefficient C is called the mutual capacitance between the conductors – or, again, just
“capacitance” if the term’s meaning is absolutely clear from the context. The same coefficient describes 15 This is why systems with p << p 1, p 2 are called weakly coupled, and may be analyzed using approximate methods – see, e.g., Fig. 4 and its discussion below.
16 A word of caution: in condensed matter physics and electrical engineering, voltage is most commonly defined as the difference between electrochemical rather than electrostatic potentials. These two notions coincide if the conductors have equal workfunctions – for example if they are made of the same material. In this course, this condition will be implied, and the difference between the two voltages ignored – to be discussed in detail in SM
Sec. 6.3.
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the electrostatic energy of the system. Indeed, plugging Eqs. (19) and (20) into Eq. (24), we see that both forms of Eq. (15) are reproduced if is replaced with V, Q 1 with Q, and with C meaning the mutual capacitance:
Capacitor’s
2
Q
C
energy
2
U
V .
(2.27)
2 C
2
The best-known system for that the mutual capacitance C may be readily calculated is the plane (or “parallel-plate”) capacitor: a system of two conductors separated with a narrow plane gap of a constant thickness d and an area A ~ a 2 >> d 2 – see Fig. 3.
Q
d a
Q
Fig. 2.3. Plane capacitor
A
– schematically.
a
Since the surface charges that contribute to the opposite charges Q of the conductors of this system, attract each other, in the limit d << a they sit entirely on the opposite surfaces limiting the gap, so there is virtually no electric field outside of the gap, while (according to the discussion in Sec. 1) inside the gap it is normal to the surfaces. According to Eq. (3), the magnitude of this field is E = /0.
Integrating this field across thickness d of the narrow gap, we get V 1 – 2 = Ed = d/0, so that =
0 V/ d. But due to the constancy of the potential of each electrode, V should not depend on the position in the gap area. As a result, should be also constant over all the gap area A, regardless of the external geometry of the conductors (see Fig. 3 again), and hence Q = A = 0 V/ d. Thus we may write V = Q/ C, with
A
C: Plane
C
0
.
(2.28)
capacitor
d
Let me offer a few comments on this well-known formula. First, it is valid even if the gap is not quite planar – for example, if it gently curves on a scale much larger than d, but retains its thickness.
Second, Eq. (28), which is valid only if A ~ a 2 is much larger than d 2, ignores the nonuniform electric fields spreading to distances ~ d beyond the gap edges. Such fringe fields result in an additional stray capacitance C’ ~ 0 a << C ~ 0 a( a/ d).17 Finally, the same condition ( A >> d 2) assures that C is much larger than the self-capacitance Cj of each conductor – see Eq. (18).
The opportunities opened by the last fact for electronic engineering and experimental physics practice are rather astonishing. For example, a very realistic 3-nm layer of high-quality aluminum oxide, which may provide a nearly perfect electric insulation between two thin conducting films, with an area of 0.1 m2 (a typical area of silicon wafers used in the semiconductor industry) provides C ~ 1 mF,18
larger than the self-capacitance of the whole planet Earth!
17 The exact value of C’ depends on the shape of the conductors. In a rare case when it has been calculated analytically, two thin round concentric disks of radius R, C’ = 0 R [ln(16 R/ d) – 1].
18 Just as in Sec. 1, for the estimate to be realistic, I took into account the additional factor (for aluminum oxide, close to 10) which should be included in the numerator of Eq. (28) to make it applicable to dielectrics – see Chapter 3 below.
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In a plane capacitor with d << a, the electrostatic coupling of the two conductors is evidently very strong. As an opposite example of a weakly coupled system, let us consider two conducting spheres of the same radius R, separated by a much larger distance d (Fig. 4).
R
R
d R
Fig. 2.4. A system of two far-separated,
similar conducting spheres.
In this case, the diagonal components of the matrix (23) may be approximately found from Eq.
(16), i.e. by neglecting the coupling altogether:
1
p p
.
(2.29)
1
2
4 R
0
Now, if we had just one sphere (say, number 1), the electric potential at distance d from its center would be given by Eq. (16): = Q 1/40 d. If we move to this point a small ( R << d) sphere without its own charge, we may expect that its potential should not be too far from this result, so that 2 Q 1/40 d.
Comparing this expression with the second of Eqs. (19) (taken for Q 2 = 0), we get 1
p
p .
(2.30)
4
,
1 2
d
0
From here and Eq. (26), the mutual capacitance
1
C
2 R .
(2.31)
0
p
1
p 2
We see that (somewhat counter-intuitively), in this limit C does not depend substantially on the distance between the spheres, i.e. does not describe their electrostatic coupling. The off-diagonal coefficients of the reciprocal capacitance matrix (20) play this role much better – see Eq. (30).
Now let us consider the case when only one conductor of the two is charged, for example Q 1
Q, while Q 2 = 0. Then Eqs. (19)-(20) yield
p Q .
(2.32)
1
1
1
Now, we may follow Eq. (13) and define C 1 1/ p 1 (and C 2 1/ p 2), just to see that such partial capacitances of the conductors of the system differ from its mutual capacitance C – cf. Eq. (26). For example, in the case shown in Fig. 4, C 1 = C 2 40 R 2 C.
Finally, let us consider one more frequent case when one of the conductors carries a certain charge (say, Q 1 = Q), but the potential of its counterpart is sustained constant, say 2 = 0.19 (This condition is especially easy to implement if the second conductor is much larger than the first one.
Indeed, as the estimate (18) shows, in this case it would take a much larger charge Q 2 to make the potential 2 comparable with 1.) In this case the second of Eqs. (19), with the account of Eq. (20), yields Q 2 = – ( p/ p 2) Q 1. Plugging this relation into the first of those equations, we get 19 In electrical engineering, such a constant-potential conductor is called the ground. This term stems from the fact that in many cases the electrostatic potential of the (weakly) conducting ground at the Earth’s surface is virtually unaffected by laboratory-scale electric charges.
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1
2
p
p
ef
ef
2
Q C ,
with C
.
(2.33)
1
1
1
1
p
1
2
p 2
p 1 p 2 - p
Thus, this effective capacitance of the first conductor is generally different both from both its partial capacitance C 1 and the mutual capacitance C of the system, emphasizing again how accurate one should be using the term “capacitance” without a qualifier.
Note also that none of these capacitances is equal to any element of the matrix reciprocal to the matrix (23):
p 1
p 1
1
p 2
p
.
(2.34)
2
p
p 2
p p p
p
p
1
2
1
Because of this reason, this physical capacitance matrix, which expresses the vector of conductor charges via the vector of their potentials, is less convenient for most applications than the reciprocal capacitance matrix (23). The same conclusion is valid for multi-conductor systems, which are most conveniently characterized by an evident generalization of Eq. (19). Indeed, in this case, even the mutual capacitance between two selected conductors may depend on the electrostatic conditions of other components of the system.
Logically, at this point I would need to discuss the particular, but practically very important case when the regions where the electric field between each pair of conductors is most significant do not overlap – such as in the example shown in Fig. 5a. In this case, the system’s properties may be discussed using the equivalent-circuit language, representing each such region as a lumped (localized) capacitor, with a certain mutual capacitance C, and the whole system as some connection of these capacitors by conducting “wires”, whose length and geometry are not important – see Fig. 5b.
(a)
(b)
Q Q
Q
1
1
Q
Fig. 2.5. (a) A simple system of
1
1
conductors, with three well-
Q
Q
2
3
Q
Q
localized regions of high electric
2
C
3
1
field (and hence surface charge)
Q
C
C
Q
2
2
3
3
concentration,
and
(b)
its
Q
Q
2
3
representation with an equivalent
circuit of three lumped capacitors.
Since the analysis of such equivalent circuits is covered in typical introductory physics courses, I will save time by skipping their discussion. However, since such circuits are very frequently met in the physical experiment and electrical engineering practice, I would urge the reader to self-test their understanding of this topic by solving a couple of problems offered at the end of this chapter,20 and if their solution presents any difficulty, review the corresponding section in an undergraduate textbook.
20 These problems have been selected to emphasize the fact that not every circuit may be reduced to the simplest connections of the capacitors in parallel and/or in series.
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2.3. The simplest boundary problems
In the general case when the electric field distribution in the free space between the conductors cannot be easily found from the Gauss law or a particular symmetry, the best approach is to try to solve the differential Laplace equation (1.42), with the boundary conditions (1b):
Typical
boundary
2 ,
0
S
,
(2.35)
k
problem
k
where Sk is the surface of the k th conductor of the system. After this boundary problem has been solved, i.e. the spatial distribution (r) has been found at all points outside the conductors, it is straightforward to use Eq. (3) to find the surface charge density, and finally the total charge
Q
d 2
r
(2.36)
k
Sk
of each conductor, and hence any component of the reciprocal capacitance matrix. As an illustration, let us implement this program for three very simple problems.
(i) Plane capacitor (Fig. 3). In this case, the easiest way to solve the Laplace equation is to use the linear (Cartesian) coordinates with one axis (say, z) normal to the conductor surfaces – see Fig. 6.
z
d
Fig. 2.6. The plane capacitor as the system for the
simplest illustration of the boundary problem
(35) and its solution.
0
x
In these coordinates, the Laplace operator is just the sum of three second derivatives.21 It is evident that due to the problem’s translational symmetry within the [ x, y] plane, deep inside the gap (i.e.
at any lateral distance from the edges much larger than d) the electrostatic potential may only depend on the coordinate normal to the gap surfaces: (r) = ( z). For such a function, the derivatives over x and y vanish, and the boundary problem (35) is reduced to a very simple ordinary differential equation 2
d
( z) ,
0
(2.37)
2
dz
with boundary conditions
( )
0 ,
0 ( d) V.
(2.38)
(For the sake of notation simplicity, I have used the discretion of adding a constant to the potential, to make one of the potentials vanish, and also the definition (25) of the voltage V.) The general solution of Eq. (37) is a linear function: ( z) = c 1 z + c 2, whose constant coefficients c 1,2 may be readily found from the boundary conditions (38). The final solution is
z
V .
(2.39)
d
21 See, e.g. MA Eq. (9.1).
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From here the only nonzero component of the electric field is
d
V
E
,
(2.40)
z
dz
d
and the surface charge of the capacitor plates is
V
E E
,
(2.41)
0
n
0
z
0 d
where the upper and lower signs correspond to the upper and lower plate, respectively. Since does not depend on x and y, we can get the full charges Q 1 = – Q 2 Q of the surfaces by its multiplication by the gap area A, giving us again the already obtained result (28) for the mutual capacitance C Q/ V. I believe that this calculation, though very easy, may serve as a good illustration of the boundary problem solution approach, which will be used below for more complex cases.
(ii) Coaxial-cable capacitor. Coaxial cable is a system of two round cylindrical, coaxial conductors, with the cross-section shown in Fig. 7.
b a
0
a
Fig. 2.7. The cross-section of a coaxial cable.
Evidently, in this case the cylindrical coordinates {, , z}, with the z- axis coinciding with the common axis of the cylinders, are most convenient.22 Due to the axial symmetry of the problem, in these coordinates E(r) = n E(), (r) = (), so that in the general expression for the Laplace operator23 we may take / = / z = 0. As a result, only the radial term of the operator survives, and the boundary problem (35) takes the form
1 d d
,
0
( a) V ,
( b) 0
.
(2.42)
d d
The sequential double integration of this ordinary linear differential equation is elementary (and similar to that of the Poisson equation in spherical coordinates, carried out in Sec. 1.3), giving d
d "
c ,
c
c c ln c .
(2.43)
1
1
2
1
2
d
"
a
a
The constants c 1,2 may be found using boundary conditions (42):
22 I am sorry for using for the 2D radius the same letter as for the volumic density of charge. (Both notations are too common to refuse.) I do not believe this may lead to confusion, because the letter will not be used in two different meanings during any particular discussion.
23 See, e.g., MA Eq. (10.3).
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b
c V ,
c ln c 0 ,
(2.44)
2
1
2
a
giving c 1 = – V/ln( b/ a), so that Eq. (43) takes the following form:
ln( / a)
V 1
.
(2.45)
ln( b / a)
Next, for our axial symmetry, the general expression for the gradient of a scalar function is reduced to its radial derivative, so that
E
d
V
.
(2.46)
d
ln b / a
This expression, plugged into Eq. (2), allows us to find the density of conductors’ surface charge. For example, for the inner electrode
V
E
0
,
(2.47)
a
0
a a ln b/ a
so that its full charge (per unit length of the system) is
Q
2 V
2
0
a
.
(2.48)
l
a
ln( b / a)
(It is straightforward to check that the charge of the outer electrode is equal and opposite.) Hence, by the definition of the mutual capacitance, its value per unit length is
C
Q
2
0
.
(2.49) C: Coaxial
l
lV
ln( b / a)
cable
This expression shows that the total capacitance C is proportional to the systems length l (if l >> a, b), while being only logarithmically dependent on is the dimensions of its cross-section. Since the logarithm of a large argument is an extremely slow function (sometimes called a quasi-constant), if the external conductor is made very large ( b >> a), the capacitance diverges, but very weakly. Such logarithmic divergence may be cut by any minuscule additional effect, for example by the finite length l of the system. This fact yields the following very useful estimate of the self-capacitance of a single round wire of radius a:
2 l
C
0
,
l
for a .
(2.50)
ln l
( / a)
On the other hand, if the gap d between the conductors is very narrow: d b – a << a, then ln( b/ a) ln(1 + d/ a) may be approximated as d/ a, and Eq. (49) is reduced to C 20 al/ d, i.e. to Eq.
(28) for the plane capacitor, of the appropriate area A = 2 al.
(iii) Spherical capacitor. This is a system of two conductors, with the central cross-section similar to that of the coaxial cable (Fig. 7), but now with the spherical rather than axial symmetry. This symmetry implies that we may be better off using spherical coordinates, so that the potential depends only on one of them: the distance r from the common center of the conductors: (r) = ( r). As we already know from Sec. 1.3, in this case the general expression for the Laplace operator is reduced to its Chapter 2
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first (radial) term, so that the Laplace equation takes the simple form (1.47). Moreover, we have already found the general solution of this equation – see Eq. (1.50):
c
( r)
1
c ,
(2.51)
2
r
Now acting exactly as above, i.e. determining the (only essential) constant c 1 from the boundary condition ( a) – ( b) = V, we get
1
1
1 1
V 1 1
c V ,
that
so
( r) c .
(2.52)
1
2
a b
r a b
Next, we can use the spherical symmetry to find the electric field, E(r) = n rE( r), with
1
d
V 1 1
E( r)
,
(2.53)
2
dr
r a b
and hence its values on conductors’ surfaces, and then the surface charge density from Eq. (3). For example, for the inner conductor’s surface,
1
V 1 1
E( a)
,
(2.54)
a
0
0
2
a a b
so that, finally, for the full charge of that conductor we get the following result:
1
1
Q 4 a 2
1
4 V .
(2.55)
0 a b
(Again, the charge of the outer conductor is equal and opposite.) Now we can use the definition (26) of the mutual capacitance to get the final result:
C: Spherical
Q
1 1 1
ab
capacitor
C
4
4
.
(2.56)
0
V
a b
0 b a
For b >> a, it coincides with Eq. (17) for the self-capacitance of the inner conductor. On the other hand, if the gap d between two conductors is narrow, d b – a << a, then a( a d)
2
a
C 4
4
,
(2.57)
0
0
d
d
i.e. the capacitance approaches that of the planar capacitor of the area A = 4 a 2 – as it should.
All this seems (and indeed is) very straightforward, but let us contemplate what was the reason for such easy successes. In each of the cases (i)-(iii) we have managed to find such coordinates that both the Laplace equation and the boundary conditions involved only one of them. The necessary condition for the former fact is for the coordinates to be orthogonal. This means that the three vector components of the local differential dr, due to small variations of the new coordinates (say, dr, d , and d for the spherical coordinates), are mutually perpendicular.
2.4. Using other orthogonal coordinates
The cylindrical and spherical coordinates used above are only the simplest examples of the curvilinear orthogonal (or just “orthogonal”) coordinates, and that approach may be extended to other Chapter 2
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coordinate systems of this type. As an example, let us calculate the self-capacitance of a thin, round conducting disk. The cylindrical or spherical coordinates would not give much help here, because while they have the appropriate axial symmetry, they would make the boundary condition on the disk too complicated: involving two coordinates, either and z, or r and . Help comes from noting that the flat disk, i.e. the area with z = 0, r < R, may be viewed as the limiting case of an axially-symmetric ellipsoid (or “degenerate ellipsoid”, or “ellipsoid of rotation”, or “spheroid”) – the surface formed by rotation of the usual ellipse about one of its major axes – which would be also the symmetry axis of the disk – in Fig. 8, the z-axis.
z
2
1
0
1
0
x
–
0
R
Fig. 2.8. Solving the disk’s capacitance problem. (The
cross-section of the system by the vertical plane y = 0.)
Analytically, this ellipsoid may be described by the following equation:
2
2
2
x y
z
1,
(2.58)
2
2
a
b
where a and b are the so-called major semi-axes, whose ratio determines the ellipse’s eccentricity – the degree of its “squeezing”. For our problem, we will only need oblate ellipsoids with a b; according to Eq. (58), they may be represented as surfaces of constant in the oblate spheroidal (also called
“degenerate ellipsoidal”) coordinates { , , } that are related to the Cartesian coordinates as follows:24
x R cosh sin cos,
0
,
y R cosh sin sin, with 0
,
(2.59)
z R sinh cos
,
0 2.
Such spheroidal coordinates are an evident generalization of the spherical coordinates, which correspond to the limit >> 1 (i.e. r >> R). In the opposite limit, the surface of constant = 0 describes our thin disk of radius R, with the coordinate describing the distance ( x 2 + y 2)1/2 = R sin of its point from the z-axis. It is almost evident (and easy to prove) that the curvilinear coordinates (59) are also orthogonal; the Laplace operator in them is:
1
cosh
1
cosh
2
.
(2.60)
2
2
R (cosh
2
sin )
1
1
1
2
sin
2
2
sin
sin cosh 2
Though this expression may look a bit intimidating, let us notice that since in our current problem, the boundary conditions depend only on :25
24 For solution of some problems, it is convenient to use Eqs. (59) with – < < + and 0 /2.
25 I have called the disk’s potential V, to distinguish it from the potential at an arbitrary point of space.
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V ,
0
,
(2.61)
0
there is every reason to assume that the electrostatic potential in all space is a function of alone; in other words, that all ellipsoids = const are the equipotential surfaces. Indeed, acting on such a function
() by the Laplace operator (60), we see that the two last terms in the square brackets vanish, and the Laplace equation (35) is reduced to a simple ordinary differential equation
d
d
cosh
.
0
(2.62)
d
d
Integrating it twice, just as we did in the three previous problems, we get
d
( ) c
.
(2.63)
1 cosh
This integral may be readily worked out using the substitution sinh (which gives d cosh d, i.e. d = d/cos, and cosh2 = 1 + sinh2 1 + 2): sinh d
( ) c
c c tan 1 sinh
(
) c .
(2.64)
1
1
2
2
1
2
0
The integration constants c 1,2 may be simply found from the boundary conditions (61), and we arrive at the following final expression for the electrostatic potential:
2
V
1
2
1
1
( ) V 1
tan
sinh
(
)
tan
.
(2.65)
sinh
This solution satisfies both the Laplace equation and the boundary conditions. Mathematicians tell us that the solution of any boundary problem of the type (35) is unique, so we do not need to look any further.
Now we may use Eq. (3) to find the surface density of electric charge, but in the case of a thin disk, it is more natural to add up such densities on its top and bottom surfaces at the same distance =
( x 2 + y 2)1/2 from the disk’s center. The densities are evidently equal, due to the problem symmetry about the plane z = 0, so that the total density is = 20 En z=+0. According to Eq. (65), and the last of Eqs. (59), the electric field on the upper surface is
()
2
1
2
1
E
V
V
, (2.66)
n z 0
z 0
z
( R sinh cos ) 0
R cos
( 2
2
R )1/ 2
and we see that the charge is distributed over the disk very nonuniformly:
4
1
V
,
(2.67)
0
( 2
2
R )1/ 2
with a singularity at the disk edge. Below we will see that such singularities are very typical for sharp edges of conductors. Fortunately, in our current case the divergence is integrable, giving a finite disk charge:
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R
R
d
2
4
2
1
d
Q
d ()2 d V
4 VR
RV
(2.68)
0
( 2
2
R )1/ 2
0
1/ 2
disk
0
0
0 1
8
.
0
surface
Thus, for the disk’s self-capacitance we get a very simple result,
2
C 8 R
4 R,
(2.69)
0
0
a factor of /2 1.57 lower than that for the conducting sphere of the same radius, but still complying with the general estimate (18).
Can we always find such a “good” system of orthogonal coordinates? Unfortunately, the answer is no, even for highly symmetric geometries. This is why the practical value of this approach is limited, and other, more general methods of boundary problem solution are clearly needed. Before proceeding to their discussion, however, let me note that in the case of 2D problems (i.e. cylindrical geometries26), the orthogonal coordinate method gets much help from the following conformal mapping approach.
Let us consider a pair of Cartesian coordinates { x, y} of the cylinder’s cross-section plane as a complex variable z x + iy,27 where i is the imaginary unit ( i 2 = –1), and let w(z) = u + iv be an analytic complex function of z.28 For our current purposes, the most important property of an analytic function is that its real and imaginary parts obey the following Cauchy-Riemann relations:29
u
v
v
u
,
.
(2.70)
x
y
x
y
For example, for the function
2
w z x iy2 x 2 y 2 ixy
2
,
(2.71)
whose real and imaginary parts are
u Re w x 2 y 2 ,
v Im w 2 xy ,
(2.72)
we immediately see that u/ x = 2 x = v/ y, and v/ x = 2 y = – u/ y, in accordance with Eq. (70).
Let us differentiate the first of Eqs. (70) over x again, then change the order of differentiation, and after that use the latter of those equations:
2
2
u
u
v
v
u
u
,
(2.73)
2
2
x
x
x
x
y
y
x
y
y
y
26 Let me remind the reader that the term cylindrical describes any surface formed by a translation, along a straight line, of an arbitrary curve, and hence more general than the usual circular cylinder. (In this terminology, for example, a prism is also a cylinder of a particular type, formed by a translation of a polygon.) 27 The complex variable z should not be confused with the (real) 3rd spatial coordinate z! We are considering 2D
problems now, with the potential independent of z.
28 An analytic (or “holomorphic”) function may be defined as the one that may be expanded into the Taylor series in its complex argument, i.e. is infinitely differentiable in the given point. (Almost all “regular” functions, such as z n, z1/ n, exp z, ln z, etc., and their linear combinations are analytic at all z, maybe besides certain special points.) If the reader needs to brush up on their background on this subject, I can recommend a popular textbook by M.
Spiegel et al., Complex Variables, 2nd ed., McGraw-Hill, 2009.
29 These relations may be used, in particular, to prove the Cauchy integral formula – see, e.g., MA Eq. (15.1).
Chapter 2
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