ELECTROSTATICS I ELECTRIC FIELD AND SCALAR POTENTIAL

Transcription

1 Chapter ELECTROSTATICS I ELECTRIC FIELD AND SCALAR POTENTIAL. Introduction Atoms and molecules under normal circumstances contain eual number of protons and electrons to maintain macroscopic charge neutrality. However, charge neutrality can be disturbed rather easily as we often experience in daily life. "Static electricity" induced when walking on a carpet and caressing a cat is a familiar phenomenon, and is known as frictional (or tribo) electricity. Before the invention of chemical batteries (Volta, 786), electricity generation had been largely done by frictional electricity generators. In nature, lightnings are caused by electrical discharges of electric charges accumulated through friction among cloud particles. In frictional electricity, mechanical disturbance given to otherwise charge neutral molecules either splits electrons o a material body, or adds them. For example, if an ebony rod is rubbed with a piece of fur, electrons are transfered from the fur to the ebony, and the ebony rod becomes negatively charged, while the fur becomes positively charged. Charge neutrality can be disturbed by various other means, such as thermal, chemical, optical and electromagnetic disturbances. It is well known that even a candle ame is weakly ionized (cuased by thermal ionization) and responds to an electric eld. Chemical batteries are capable, through chemical reactions, of separating charges. Some metals are known to emit electrons when exposed to ultraviolet light (photoelectric e ect discovered by Hertz in 897 and later given uantum mechanical explanation by Einstein). Finally, as an example of electromagnetic means of charge separation, the plasma (ionized gas) state in uorescent lamps and plasma TV may be added to the list. Charged bodies exert electric forces (Coulomb force) on each other. Systematic experiments on electric forces had been carried out by Cavendish and Coulomb in the 8th century and led

2 to the establishment of the Coulomb s law. Like charges repel each other, while unlike charges attract each other. Therefore, in any attempt to separate charges out of an initially charge neutral body, energy must be added. For example, in frictional electricity, mechanical energy is expended to separate charges. In chemical batteries, chemical energy is converted into electric energy, and in electric generators, either gravitational (as in hydropower generation) or thermal (as in steam power plants) energy is converted into electric energy. The Coulomb force to act between two charges is inversely proportional to r 2 where r is the separation distance between the charges. A hydrogen atom consists of one electron revolving around a proton. The centripetal Coulomb force (attracting) is counterbalanced by the mechanical centrifugal force, and the electron stays on a circular orbit. (This is a classical electrodynamic model. For a more satisfactory description of a hydrogen atom, uantum mechanical analysis is reuired. However, the concept of force balance is basically correct.) A nucleus of heavier atoms contains more than one proton. The size of a nucleus is of order 5 m (compare this with the atomic size m). Therefore, among the protons packed in a nucleus, a tremendously large repelling Coulomb force should act, and some other force, which is not of electric nature, must keep protons together. This is provided by the so-called strong nuclear force which acts among hadrons such as protons and neutrons. Obviously, the nature of nuclear force is beyond the realm of classical electrodynamics. However, it should be realized that nuclear energy that can be released when a heavy nucleus (such as U 235 ) splits (nuclear ssion process) is nothing but electric energy stored in a nucleus. Electrostatics is one branch of electrodynamics in which electric charges are either stationary or moving su ciently slowly so that magnetic eld induction and electromagnetic radiation can be ignored entirely. All basic laws in electrostatics (Gauss law, Maxwell s euations) follow from the Coulomb s law. Therefore, formulating the electric eld and scalar potential in electrostatics will be deduced from this fundamental law. It should be pointed out that electrostatics (together with magnetostatics) provides us with preparation for more general electrodynamics, electromagnetic wave phenomena in particular. For example, electromagnetic wave propagation in a medium (including vacuum) reuires that the medium be able to store both electric and magnetic energy. Obviously, electric energy storage leads us to the concept of capacitance. Calculation of the capacitance of a given electrode system is one important application of electrostatics, but the concept of capacitance (and inductance) will also play fundamental roles in dynamic electromagnetic phenomena..2 Coulomb s Law In the late 7 s, Cavendish and, independently, Coulomb carried out extensive research on static electricity. (At that time, no batteries were available, and electricity generation was mainly done with frictional electricity machines.) Apparently, Cavendish s discovery of the inverse suare law, now known as Coulomb s law, was made before Coulomb. However, Cavendish did not publish his large amount of work on electricity, while Coulomb wrote several papers on electricity and magnetism. Of course, Cavendish is best known for his gravitational torsion balance experiments 2

3 which experimentally veri ed the inverse suare law of gravitational force originally postulated by Newton. Both Cavendish and Coulomb used similar torsion balance apparatus to measure electric forces acting between charged bodies. Coulomb s law is stated as follows. When two charges and 2 are at a distance r from each other, the electric force exerted on each other is proportional to 2 r 2, F const. 2 r 2 : (.) In the MKS-Ampere unit system (SI unit system), the charge is measured in Coulombs (C), the distance in meters (m), and the force in Newtons (N). In these units, the constant experimentally determined is N m constant 8: C 2 Figure -: Repelling Coulomb force between like charges. The force is a vector uantity. In the case of the Coulomb force, the force is directed along the separation distance vector r, and a more formal expression is given by F const. 2 r 2 e r (.2) where e r r r (.3) is the unit vector along the position vector r. Of course, both charges and 2 experience a force of the same magnitude, but oppositely directed. The sign of the product 2 can be either positive or negative. When 2 >, the force is repelling, and when 2 <, the force is attractive. In the 8th century when Cavendish and Coulomb conducted experiments, the presence 3

4 of two kinds of charges, positive and negative, was known although their origin, namely charged elementary particles, was clari ed much later. The electron was discovered by Thomson in 897. The electronic charge presently established is :6 9 Coulomb. The proportional constant in the Coulomb s law corresponds to the force to act between two eual charges, C each, separated by a distance of m. By arranging such an experimental situation, the constant could be measured as done by Cavendish and Coulomb. (In more modern methods, the velocity of light in vacuum provides an indirect measurement of the constant. This will become clear later in Chapter 7.) In the MKS-Ampere unit system, the connection between mechanical force (Newtons) and basic electromagnetic unit is actually made in terms of magnetic force, rather than the Coulomb force, as will be explained in Chapter 7. When two long parallel currents of eual magnitude separated m exert a force per unit length of 2 7 N/m, the magnitude of the current is de ned to be Ampere. The electric charge, C, is then deduced from, C Ampere sec In other unit systems, the Coulomb s law itself is employed to de ne the unit of electric charges. For example, in the CGS-ESU (ESU for ElectroStatic Unit) system, when two eual charges separated by cm exert a force of dyne on each other, the charge is de ned to be ESU. Since N 7 dyne, and m 2 cm, we can readily see that C For example, the electronic charge in ESU is ESU p 8:99 8 ' ESU 3: 9 :6 9 3: 9 4:8 ESU Although in engineering, the MKS-Ampere (SI) unit system is universally accepted, the CGS-ESU and associated Gaussian unit system is still popular in physics. Both have merits and demerits, and it is di cult to judge one unit system superior to the other. In the MKS-Ampere unit system, the Coulomb s law is rewritten as where F 8: r 2 e r (N) (.4) F 2 r 2 e r (N) (.5) 8:85 2 C 2 N m 2 (.6) is called the vacuum permittivity. The appearance of the numerical factor 4 may be uncomfortable, but cannot be entirely avoided. If we do not introduce the factor 4 in the Coulomb s law, it will pop up in the corresponding Maxwell s euation. In fact, the Maxwell s euation for the static 4

5 Figure -2: In MKS unit system, I Ampere current is de ned if the force per unit length between in nite parallel currents m apart is 2 7 N/m. The magnetic permeability 4 7 H/m is an assigned constant to de ne Ampere current. electric eld in the CGS-ESU system has to be written as re 4 (.7) because the factor 4 is avoided in the corresponding Coulomb s law, F 2 r 2 e r (dynes) The newly introduced constant is one of the fundamental constants in electrodynamics. Note that is a measured constant, since it is derived from the original measured constant, 8:99 9 Nm 2 /C in the Coulomb s law. The permittivity of air (room temperature, atmospheric pressure) is about :4 due to the polarizability of air molecules. For practical applications, the air permittivity can be very well approximated by the vacuum permittivity..3 Electric Field E Interpretation of Coulomb s law was a matter of debate before the concept of electric eld was well established by Faraday and Maxwell. Before Faraday and Maxwell, the so-called theory of action at distance once prevailed. According to this theory, electric e ects (such as Coulomb force) appear through some sort of direct interaction between charges, and space (or vacuum) has nothing to do with the interaction. In the eld theory, a single charge disturbs space surrounding it by creating an electric eld. The Coulomb force to act between two charges is due 5

6 to interaction between one charge and the electric eld produced by the other. The eld theory is now well accepted, and modern electrodynamics is almost entirely described by eld uantities such as electric and magnetic elds. Let us write down the Coulomb force again, This can be written either or F 2 r 2 e r F r 2 e r 2 (.8) F 2 r 2 e r (.9) We may interpret E. (.8) as the force experienced by a charge 2 placed in an electric eld, E r 2 e r (.) produced by a charge while E. (.9) as the force experienced by a charge placed in an electric eld, produced by a charge 2. electric eld given by regardless of the presence of a second charge. E 2 2 r 2 e r (.) A single charge in the space with a permittivity thus produces an E r 2 e r (N/C) (.2) Numerically, the electric eld is euivalent to a force to act on a unit change. Since the force is a vector, so is the electric eld, and complete determination of an electric eld reuires three spatial components. by In general, if a charge is placed in an electric eld E, the force to act on the charge is given F E (N) (.3) This may alternatively be used for de nition of an electric eld, namely, if a stationary charge experiences a force proportional to the amount of the charge, then we de ne that an electric eld is present. (Elementary particles such as protons and electrons do have masses, and they experience gravitational forces as well. However, gravitational forces are orders of magnitude smaller than electromagnetic forces, and in most practical applications, gravitational forces can be ignored.) Although E. (.2) has been deduced from Coulomb s law, the inverse is not always true, that is, electric elds are not necessarily due directly to charges. A time varying magnetic eld produces an electric eld (Faraday s law). Also, an object travelling across a magnetic eld experiences an electric eld (motional electromotive force), as we will study in Chapter 9. However, regardless of the origin of the electric eld, the force euation, E. (.2), holds. 6

7 E 2 E t P E Figure -3: Vectorial superposition of electric eld. 2.4 Principle of Superposition and Electric Field due to a Distributed Charge The expression for the electric eld due to a point charge, E r 2 e r (.4) can be applied repeatedly to nd an electric eld due to a system of charges. For example, if there are two charges at r and 2 at r 2, the total eld can be found from, Note that the uantities, E r r jr r j 3 + r r 2 jr r 2 j 3 2 (.5) r r jr r j ; r r 2 jr r 2 j are the unit vectors in the directions r r and r r 2, respectively. For larger number of charges, the total electric eld can still be found as vector sum of electric elds due to individual charges. This is known as the principle of superposition. For N discrete charges, the electric eld can be written down as, E NX r r i jr r i j 3 i (.6) i As the number of charges increases, the summation becomes rather awkward. In most practical applications, the charge distribution can be regarded continuous, rather than discrete. As one 7

8 would expect, the summation in the case of discrete charges can be replaced by an integral in the case of a continuous charge distribution., A continuous charge distribution can be described by a local charge density, (r) (C/m 3 ) which may vary as a function of position r. Since the size of elementary particles is so small, a collection of those particles (e.g. electrons) can be well approximated by a continuous function (r), just as we treat water as a uid although, microscopically, water consists of discrete water molecules. Let us pick up a small volume dv located at a distance r from a reference point O as shown in Fig..4. The charge contained in the volume dv located at r is d (r )dv By choosing the volume element su ciently small, we may regard the charge d a point charge. We already know how to express the electric eld due to a point charge, de d(r r ) jr r j 3 r r jr r j 3 (r )dv (.7) Therefore, the electric eld at r can be calculated from the following integral, Z E(r) de Z r r jr r j 3 (r )dv (.8) This is a general formula for calculating an electric eld due to an arbitrary distribution of charges. Remember that we have derived it from the Coulomb s law. Although the formula we just derived is a complete solution for static electric elds due to a charge distribution, it is seldom used in practical applications except for simple (or often trivial) cases. There are several reasons for this. First, E. (.8) is a vector euation. In the cartesian coordinates, for example, three components, E x ; E y and E z will have to be evaluated separately. That is, we have to carry out integrations three times for a complete vector solution. Second, in many potential boundary value problems, the charge distribution, (r), is not known a priori. For example, in evaluation of a capacitance of a given electrode system, the problem is reversed, that is, we calculate the electric eld (from the scalar potential) rst, and then evaluate the charge distribution on the surface of electrodes. As we will study in Section 2, the method based on the scalar potential is more convenient in practical applications than evaluating the electric eld using E. (.8). However, if the charge distribution is known, we can certainly use E. (.8) to evaluate the electric eld at an arbitrary point. Let us work on some simple examples. 8

9 de r r' d ρdv' r r' O Figure -4: Di erential charge d dv produces a di erential electric eld de: Charge Sheet Let a large, thin insulating sheet carry a uniform surface charge density (C/m 2 ) ( constant). If the sheet is large enough, or the point at which we wish to evaluate the electric eld is close enough to the sheet, the electric eld should be perpendicular to the sheet because of cancellation by an element located opposite with respect to the origin O. Choosing the cartesian coordinates x; y; z, we then have to evaluate only the z component of the electric eld at a distance z from the sheet. Let us pick up a surface element dx dy located at (x ; y ; z ) on the sheet. The element carries a point charge d dx dy. Then, the magnitude of the electric eld on the z axis is, and its z component is, de dx dy x 2 + y 2 + z 2 (.9) de z z (x 2 + y 2 + z 2 ) 32 dx dy (.2) Therefore, the electric eld on the axis is given by the following integral, E z z Z Z dx dy (x 2 + y 2 + z 2 ) 32 (.2) However, the double surface integral can be replaced by a single integral over the radius r, where r 2 x 2 + y 2 so that E z z Z 2r (r 2 + z 2 ) 32 dr (.22) 9

10 The integral is elementary, and we nd E z z 2 2 z jzj r p r 2 + z 2 r (.23) where p z 2 jzj has been substituted. The solution indicates that the magnitude of the electric eld is independent of the distance from the sheet. The direction of the electric eld is negative in the region z <, and positive for z >. When there are two oppositely charged sheets, and (C/m 2 ), the electric eld outside the sheets if zero, but the eld in between the sheets is given by, E z (.24) as can be readily seen from the principle of superposition. This con guration corresponds to a parallel plate capacitor provided the edge e ects are ignored. Line Charge We assume a long line charge with a line charge density (C/m). At a distance r from the line charge, the di erential electric eld due to a charge d dz located at z is de dz r 2 + z 2 (cos e r + sin e z ) (.25) where, cos r p r 2 + z ; sin z p 2 r 2 + z 2 The z component is an odd function of z. Therefore, the z component of the electric eld vanishes after integration from z to +. The radial (r) component remains nite, and is given by, E r Z r (r 2 + z 2 dz ) 32 2 r (.26) Remember that this result is valid only if the line charge is long, or the point of observation is su ciently close to a line charge of nite length..5 Gauss Law for Static Electric Field The electric eld at a distance r from a point charge is, E r 2 e r;

11 charge sheet σ(c/m 2 ) E σ/2ε E σ/2ε σ σ E E E σ/ε Figure -5: Upper gure: Single charge sheet. The electric eld on both sides is E z 2" : Lower gure: Eual, opposite charge sheets. The eld in-between is E z " : Outside, E z : de r de dez dλdz z r line charge z Figure -6: Electric eld due to a long line charge (line charge density C/m). Note that E z and the radial electric eld E r can also be found using Gauss law. 2" r

12 de n r θ γ a b Figure -7: Gauss law applied to a sphere that is not concentric with the charge. Note r cos + b cos a: On a spherical surface with radius r, the magnitude of the electric eld is constant, and the uantity, 4r 2 E r is eual to. This strongly suggests that the closed surface integral of the electric eld, I EdS (.27) is eual to the amount of charge enclosed by the closed surface S divided by, I EdS S " S Noting and I S Z EdS Z r EdV (r) dv we nd the Maxwell s euation r E " Let us see if this is the case for a spherical surface which is not concentric with the point charge. We denote the distance between the charge and the spherical center by b, and the radius of the sphere by a. The electric eld on the surface is no more uniform. It is not normal to the surface, either. Noting 2

13 the relationship, r 2 a 2 + b 2 2ab cos in Fig.??, we nd the magnitude of the electric eld at angle, E a 2 + b 2 2ab cos (.28) The component normal to the spherical surface is E cos where the angle is related to through r cos + b cos a: Therefore, the entire surface integral reduces to, I S E ds Z a b cos (a 2 + b 2 2ab cos ) 32 2a2 sin d (.29) where ds 2a 2 sin d is the area of the ring having radius asin and width ad. The relevant integral is Z 8 (a b cos ) sin < (a 2 + b 2 d 2ab cos ) 32 : 2 a 2 ; a > b ; a < b (.3) (In integration, it is convenient to change the variable from to through cos. Then the integral reduces to b p a 2 + b 2 R a b (a 2 + b 2 d 2ab) 32 a 2 + b 2 2ab a 2 p b a 2 + b 2 ab 2ab Evaluation of the de nite integral is left for a mathematical exercise. Note that p (a b) 2 ja bj.) Therefore, the surface integral of the electric eld becomes, I 8 < ; a > b EdS : ; a < b (.3) Obviously, the case a > b corresponds to a sphere enclosing the charge, and a < b corresponds to the case in which the charge is outside the closed spherical surface. The result may be generalized to a closed surface of an arbitrary shape. The following formula is called Gauss law, I S EdS Total charge enclosed by S! (.32) 3

14 Again, remember that Gauss law is euivalent to Coulomb s law, since we have deduced the former from the latter. In fact, Gauss law can be mathematically proven if we adopt Coulomb s law. A formal proof will be given after the Maxwell s euation is introduced. Gauss law can be convenient for simple cases in which a system has a high degree of symmetry, such as spherical, cylindrical, and planar symmetries. Let us work on a few simple examples. Uniformly-Charged Insulating Sphere Let an insulating sphere of radius a carry a uniform charge density (C/m 3 ) and a total charge 4 3 a3 (C). The system has complete spherical symmetry, and the only nonvanishing component of the electric eld is the radial component, E r. Outside the sphere (r > a), Gauss law yields, 4r 2 E r or E r a3 3 r 2 (.33) which is identical to the eld due to a point charge concentrated at the center. Inside the sphere (r < a), Gauss law reads 4r 2 E r 4 3 r3 (.34) since in the RHS, only the amount of charge enclosed by the spherical surface (called Gaussian surface) having a radius r (< a) enters. Solving for E r, we obtain, E r 3 r (.35) Therefore, in the sphere, the eld linearly increases with the radius r up to the surface, r a, where the interior eld connects to the exterior eld without dicontinuity. The electric eld is maximum at the surface as shown in Fig. -8. The fact that the electric eld should vanish at the center of a charged (uniformly!) sphere is understandable from spherical symmetry. At the center, the contribution from a charge d to the electric eld can always be cancelled by another located opposite with respect to the center. By analogy, the gravitational eld due to the earth mass itself at the earth s center should be zero. Charged Conducting Sphere A charge given to a conductor must reside entirely on the conductor surface so that the electric eld inside a conductor body should be identically zero in static condition. If an electric eld were not zero in a conductor, a large electric current would ow according to Ohm s law, J E (.36) where (Siemens/m) is the conductivity. of metals is large. (For example, copper has ' 5:9 7 S/m). Therefore, unless E in a conductor, a large current should ow, and this violates the assumption of static electricity. In other words, an electric eld can exist in a conductor 4

15 Er E max a 2a 3a 4a r ρ a Figure -8: Electric eld of a unifrmly charged sphere. The eld is at the center. The maximum eld, E max a 3" ; occurs at the surface. only in dynamic conditions in which current ow is allowed. If an excess charge is given to a conducting sphere, the charge uniformly resides on the surface as a surface charge, with a surface charge density s 4a 2 (C/m 2 ) where a is the sphere radius. The electric eld inside (r < a) the sphere is zero, while outside (r > a), it is given by E r r 2 (r > a) (.37) Note that there is a sudden jump in the eld at the surface (r a) where the surface charge density exists. In general, wherever an in nitesimaly thin surface charge layer exists, the electric eld there becomes discontinuous. We will come back to this subject in Chapter 3 where the concept of the displacement vector D is introduced. Experimental Veri cation of Coulomb s Inverse Suare Law As stated earlier, Gauss law and Coulomb s law are physically identical, in the sense that the former is derivable from the latter. Therefore, if Gauss law can be veri ed experimentally, then Coulomb s law is indirectly veri ed also. Here, we are particularly concerned with a uestion: How valid is the inverse suare law? Or put alternatively, when the Coulomb s law is written as F const: r n e r how close is the power n to 2.? Is it exactly 2.? Experiments in the past have shown that n is 5

16 Er E max a 2a 3a 4a r σ a conducting sphere carrying a surface charge Figure -9: Electric eld in a charged conducting sphere is. The charge given to the sphere must reside on the surface as a surface charge 4a 2 C/m 2 : indeed very close to 2. with an uncertainty of order 6. The power n is probably exactly eual to 2. The experiment performed by Plimpton and Lawton in 936 was to measure the intensity of electric eld inside a charged conducting shell. As we have just seen in the preceding example, an excess charge given to a conductor must reside entirely on its outer surface, and the electric eld in the conductor must be zero. This holds for a conducting shell, too, and no charge can exist on the inner surface of the shell. Plimpton and Lawton established n Further improvement n was achieved later in 97 by Williams et al..6 Di erential Euations for Static Electric Fields-Maxwell s Euations For a given charge distribution (r), the electric eld can be uniuely evaluated from E(r) Z r r jr r j 3 (r )dv (.38) as we have seen in Sec..4. The electric eld also satis es Gauss law, I S EdS Z V (r)dv (.39) 6

17 We may use either formulation in evaluating the electric eld due to a prescribed charge distribution. The electric eld is a vector uantity. A vector can be uniuely de ned if its divergence and curl are both speci ed. This mathematical theorem is known as Kirchho s theorem. (Kirchho is a familiar name in electric circuit theory. However, his contributions are not limited to circuit theory, but uite diversi ed. Kirchho s scalar di raction formula for electromagnetic waves is another important contribution, and had been used extensively until it was replaced by a more accurate vector di raction formula rather recently. See Chapter 3.) A derivation of vector di erential euations for electric and magnetic elds was formulated by Maxwell in the celebrated book Treatise on Electricity and Magnetism published in 873. Each eld (electric and magnetic) reuires both divergence and curl. Therefore, for complete description of electric and magnetic elds, four Maxwell s euations emerge. In electrostatics, the magnetic eld is absent (or ignored). Let us start with the divergence of static electric elds. In the example of a charged sphere in Sec. 2.5, we have seen that the electric eld inside a uniformly charged spherical body can be found from Gauss law, with the result, 4r 2 E r 4 3 r3 (.4) E r 3 r (.4) E. (.4) holds no matter how small r is chosen. The divergence of the electric eld is de ned by div E r E lim V! H S E ds V (.42) In E. (.4), the LHS is a special case of symmetric surface integration, and the uantity 4r 3 3 in the RHS is of course the volume. Therefore, if we take the limit of r! in the ratio, 4r 2 E r lim r! 4 3 r3 it reduces to the de nition of divergence. Conseuently, the divergence of the electric eld satis es, r E (.43) This is one of Maxwell s four euations., Outside the charged sphere, the surface integration, I E ds S vanishes unless S intersects the sphere. Therefore, outside the sphere, r E, as reuired because outside the sphere. The divergence euation alternatively follows from Gauss law, although physical meaning is 7

18 less clear. The surface integral can be rewritten in terms of a volume integral as follows, I Z EdS S V r E dv (Gauss theorem) (.44) This is a mathematical theorem, and has nothing to do with the physical Gauss law, although they are intimately related. Using this transformation, we can rewrite Gauss law as, Z r E dv Z dv (.45) Therefore, r E immediately follows. The third method to derive the divergence euation for the electric eld is to directly take the divergence of E. (.8), r r E(r) Z r r r r jr r j 3 (r )dv (.46) where the subscript r indicates di erentiation with respect to the observer s coordinates r. The function, r r r r jr r j 3 (.47) has a peculiar property. It vanishes wherever r 6 r but is not properly de ned where r r. (In fact, it diverges at r r.) Introducing jr r j R, we indeed see that, r r r r jr r j 3 d R 2 R R 2 dr 2 (R 6 ) However, its volume integral is well de ned and remains nite since R r r rr jr r j 3 dv I r r jr r j 3 ds I I R 2 R2 d d 4 (.48) where d is the di erential solid angle, and 4 is the total solid angle. In other words, the function, has the property of the delta function, r r r r jr r j 3 (.49) r r r r jr r j 3 4(r r ) (.5) 8

19 where (r r ) (x x )(y y )(z z ) (.5) is the three-dimensional delta function having the dimensions of m 3. Therefore, the integral in E. (.46) can be readily performed as Z 4 (r )(r r )dv (r) (.52) and E. (.46) is euivalent to r E (.53) We now evaluate the curl of a static electric eld given in E. (.8). Taking the curl of both sides, we nd, However, in the RHS, the vector, r r E(r) Z r r r r jr r j 3 r r identically vanishes, because the vector, jr r j 3 r r r r jr r j 3 (r )dv (.54) (r r ) + jr r j 3 r r (r r ) (.55) r r jr r j 3 3 r r jr r j 4 (.56) is parallel to the vector r r, and r r (r r ) (.57) identically. In contrast to the divergence in E. (.46), r r r r jr r j 3 (.58) holds even at r r. Therefore, for static electric elds due to charge distributions, we conclude, r E (.59) We thus have speci ed both divergence and curl of static electric elds in Es. (.53) and (.59), respectively, which determine static electric elds (vector) uniuely. The divergence euation holds for any electric elds (electrostatic and non-electrostatic), but the curl euation is the special case of the more general Maxwell s euation, (.6) In static and r E holds only for static electric elds. 9

20 What are the physical meanings of the Maxwell s euations? The nonvanishing divergence of the static electric eld indicates that the charge density is the source (or sink) of the eld. This may be illustrated by the electric lines of force. A positive charge emits electric eld lines, while a negative charge absorbs electric lines of force, just like stream lines in water ow. If two eual, but opposite charges are near by, the eld lines emitted by the positive charge are all absorbed by the negative charge. If the magnitude of the charges are not eual, some eld lines are not absorbed by the negative charge. The vanishing curl of static electric elds indicates that an electric eld line does not close on itself, that is, a eld line does not form a loop. (This is in contrast to non-electrostatic eld for which the curl is nonvanishing.) In other words, static electric eld lines always have heads or tails as clearly seen in the above examples. Static eld lines start at a positive charge and end at a negative charge..7 Scalar Potential (r) The vanishing curl of static electric elds has an important implication. Since the curl of a gradient of an arbitrary scalar function is identically zero, r rf (.6) a static electric eld (vector!) can be deduced from a gradient of some scalar function. We denote this scalar by (r) and call it a scalar potential. The electric eld is given by the gradient of the scalar potential, E(r) r(r) (.62) The negative sign is introduced so that the electric eld is directed from higher to lower potential regions, just like in the familiar gravitational eld and gravitational potential. The dimensions of the scalar potential are Newton meter Coulomb Joule Coulomb This is rede ned as Volts (after Volta). Therefore, an alternative unit for the electric eld is Volts/meter. The physical meaning of the electric eld is (as discussed before) the force to act on a unit charge. Then, the scalar potential can be interpreted as the work reuired to move a unit charge from one point to another, since (r) (r ) Z r r E dr (.63) where (r ) is the potential at r. Clearly, the potential is a relative uantity and is referred to the potential at a speci ed reference point. (Again, the analogy to the gravitational potential 2

21 V E d Φ(x) V d x Figure -: Potential (x) V d x and electric eld E x Vd in a parallel plate capacitor. should be recalled. When we measure the height of a mountain, usually the sea level is chosen as the reference point.) As an example, let us consider a point charge placed at the origin r. The electric eld is given by, E r (r) r 2 ; V/m If the reference potential is chosen on a surface with a radius r, the potential at an arbitrary point r becomes (r) Z r r r 2 dr r r (.64) For a charge system of a nite spatial extent, it is convenient to choose at r, so that relative to the zero potential at r. (r) ; (V) (.65) r Similarly, the potential due to a charged conducting sphere with a total charge and radius a is given by, (r) r (.66) 2

22 relative to at r. The sphere potential can be found by euating the radius to r a, sphere a (.67) Note that this result is independent of whether the conducting sphere is solid or shell, since the electric eld in the conductor must vanish identically. In general, the potentials at any points on and in a conductor must be eual because E in a conductor. In particular, the surface of a conductor, no matter how complicated is its shape, is an euipotential surface. This fact will play an important role in potential boundary problems in Chapter III. In the following, we will work on a few examples in which the potentials can be calculated from known electric elds. Potential of a Charged Insulating Sphere The electric eld for this problem has been worked out in Sec..5, and given by 8 >< E r >: a 3 ; (r > a) 3 r2 3 r; (r < a) (.68) where is the charge density and a is the sphere radius. We choose at r. Then, the exterior potential becomes Z r a 3 (r) 3 r 2 dr a 3 ; (r > a) (.69) 3 r On the surface of the sphere, the potential takes the value, (a) 3 a2 (.7) Therefore, the interior potential becomes, (r) (a) a2 3 a2 2 Z r rdr a 3 r 2 a 2 6 r 2 ; (r < a) (.7) 6 The potential at the sphere center is, (r ) a2 2 (.72) which is the maximum (if > ) potential. Note that the integration should be carried out starting 22

23 at the reference point (r in this case), and radially inward. Potential due to a Long Line Charge The electric eld due to a long line charge has been found in Section.5, and given by E 2 (.73) where (C/m) is the line charge density. Let us choose a reference, zero potential surface at a radius a. (a cannot be in nity, in contrast to the case of spheres, because the potential due to an in nitely long line charge does not vanish at in nity, but diverges. Such a line charge reuires in nitely large amount of energy, and thus is of mathematical interest only.) Then, the potential at an arbitrary radial position can be found as () Z a E d Z a 2 d a 2 (.74) The zero potential surface can be chosen arbitrarily, since the potential is a relative uantity. However, the potential di erence between two radial positions is independent of the choice of the reference. Indeed, the potential di erence between two points at and 2 becomes ( ) ( 2 ) in which the particular radius a has disappeared. a a 2 (.75) In practical applications, the potential we just found can be used to calculate the capacitance of a coaxial cable. Let us consider a coaxial cable having inner and outer conductor radii a and b, respectively, as shown in Fig. -. We connect a dc power supply of a voltage V between the two conductors. If we can calculate the charges to appear on respective conductors, the capacitance can be calculated by de nition from, The charge per unit length is, C V l (F) (.76) (.77) Therefore, the potential di erence between the two conductors becomes, V (a) (b) b 2 a 23

24 l V 2a b Figure -: Coaxial cable. or Hence, V or the capacitance per unit length is given by, b 2 l a (.78) C V 2 l (F) (.79) b a C l 2 (F/m) (.8) b a If the space between the conductors is lled with a dielectric having a permittivity, this should be modi ed as C l 2 b (.8) a.8 Potential due to a Prescribed Charge Distribution In the preceding Section, we have (brie y) learned how to calculate the potential (r) from a known electric eld. However, in most applications, it is more common to reverse the procedure, namely, we rst nd the potential, and then calculate the electric eld from, E r (.82) There are several reasons for inverting the procedure. The basic Maxwell s euations for static electric elds are, re (.83) re (.84) 24

25 Obviously, these are vector di erential euations, and for a complete solution, we must solve three di erential euations for each component of the eld E. However, if we substitute E r into re, we nd a single scalar di erential euation for, r 2 (.85) This is known as Poisson s euation, and mathematically speaking, it is an inhomogeneous version of the Laplace euation, r 2 (.86) for which exhaustive studies have been made since the 8th century. The Laplace and Poisson s euations most freuently appear in physical science and engineering. Analytic solutions to those euations can be found in a relatively limited number of cases. However, with powerful computers becoming more easily accessible these days, numerical solutions to Poisson and Laplace euations are no more prohibitively expensive even for complicated geometries for which analytic solutions are extremely di cult, if not impossible. We will return to the problem of solving Laplace euations in the following Chapter. Here, we derive a formal solution to the Poisson s euation, which enables us to calculate the potential for a prescribed charge distribution. There are several methods to nd the solution to r 2 and we start with a method that is physically most transparent. We have already seen that the potential due to a point charge is given by (r r ) jr r j (.87) where r is the location of the point charge. For a distributed charge, we pick up a small volume dv in the region where the charge density,, exists. The charge contained in the volume dv is d dv (.88) By choosing the volume dv su ciently small, the di erential charge d approaches a point charge. Therefore, the potential due to the charge d located at r becomes d (r ) jr r j dv (.89) 25

26 dφ r r' d r r' O provided the reference potential is at r. By integrating E. (??) over the dummy variable r, we nd (r) Z which is the desired solution to the Poisson s euation. (r ) jr r j dv (.9) The same result can be deduced from the electric eld due to a prescribed charge distribution found earlier, Since E(r) Z r r jr r j 3 (r )dv : r r jr r j 3 r r jr r j where the subscript r indicates di erentiation with respect to the observing point r, we nd E(r) Z r r (r ) jr r j dv (.9) Note that the di erential operator r r can be taken out of the integral, because it operates on r only. Comparing with the basic relationship, E r we readily obtain (r) Z (r ) jr r j dv which is identical to E. (.9). Below, we will apply this formula to a few simple problems. Potential due to a Line Charge of Finite Length Consider a line charge of length 2a carrying a uniform line charge density (Cm). A di erential 26

27 dφ a d z' ρ z a Figure -2: Line charge ( C/m)of nite length 2a: d dz : charge d dz located at z creates a potential at the coordinates (; z), d dz p 2 + (z z ) 2 (.92) Integrating this over z from z (; z) a to +a, we nd Z a dz p a (z z) h p i a (z z) z z p! (z a) a z p (z + a) a z a (.93) Let us examine special cases of the potential. At a point very close to the line charge, so that z!, and a, the potential reduces to The potential of the form () ' () 2a 2 ( a) (.94) 2 + const. (.95) has been worked out in Sec. 2.6 for a long coaxial cable, and the result we just obtained is therefore of an expected form. Far away from the line charge so that ; z a, we expect to recover the potential due to a 27

28 point charge, 2a. On the plane z, the logarithmic function approaches p! 2 + a 2 + a p 2 + a 2 a ' + 2a ' 2a ( a) (.96) Therefore, the potential indeed approaches () ' ( a) (.97) where 2a is the total charge carried by the line. On the z axis,!, and at jzj a, the potential also approaches, (z) ' lim z! + 2a z Capacitance of a Thin Linear Conductor p! (z a) a z p (z + a) a z (.98) The expression for the potential in the vicinity of a line charge, E. (.94), can be directly used to nd the capacitance of a thin linear conductor having a radius b and length 2a with a b. On the surface of the conductor, b. Therefore, the potential of the conductor can be approximated by ( b) 2a 2 b 2a ' a b and the approximate capacitance is given by C a 2" L (Lb) 2a b (.99) where L 2a is the total length of the conductor. Capacitance of Parallel Wire Transmission Line As an application of the potential due to a line charge, we consider a long, two-parallel-wire transmission line, which is often encountered in high voltage power transmission and old-fashioned telephone lines. The geometry is shown in Fig.??. We assume two thin parallel conductor wires 28

29 of an eual radius a separated a distance d. By thin conductor wires, we mean d a. Since the capacitance we are concerned here is the mutual capacitance between the two wires, we let the two wires carry eual, but opposite line charge densities, + and (C/m). Then, the potential at arbitrary point can be written down as (r + ; r ) [ r + + r ] 2 r 2 r + where r + and r are the distance from the observing point to the positive and negative wires, respectively. Note that the reference, zero potential surface is chosen at the midplane on which r + r. When the wire radius a is small compared with the separation distance d, the potential at the surface of the positive wire can be found by letting r + a and r d, + d 2 a (.) Similarly, the potential at the negative wire is, a 2 d d 2 a (.) Therefore, the potential di erence between the wires is V + d a (.2) and the capacitance per unit length of the transmission line is given by, C l V d a (F/m) (.3) The capacitance (per unit length) of a single wire transmission line placed above the ground at a height h can be found in a similar manner. The ground potential can be chosen to be zero. Therefore, the potential di erence between the wire and the ground is given by V 2h 2 a (.4) where 2h is the distance between the wire and an image line charge located at the mirror point, z h, in the ground. Then, the capacitance becomes C l 2 2h a (.5) 29

30 +λ r r + λ d Figure -3: Parallel wire transmission line. Separation distance d and wire radius a: Note that this is twice as large compared with the capacitance of the two-wire transmission line. The latter can be considered to be a series connection of two capacitors, one between the positive wire and the midplane, and the other between the negative wire and the midplane. The midplane, which is chosen at zero potential, can be replaced by a grounded large conducting plate without a ecting the potential and electric eld. The method of images will be discussed more fully in Chapter 5. Both formulae in Es. (.3) and (.5) are subject to the assumption of thin wire radius, a d; h. As the wire radius becomes large, they become inaccurate, but only logarithmically. For example, the exact capacitance of the parallel wire transmission line is given by C l " d + p # (.6) d 2 4a 2 2a Even when a :2d (thick conductors indeed), the error caused by using the approximate formula in E. (.3) is less than 3%. Electric Dipoles Two charges of opposite signs, but eual magnitude, + and separated by a small distance a constitute an electric dipole. (A single charge is called a monopole. Higher order multipoles, uadrupole, octapoles, etc., can be similarly constructed.) Dipoles are the basic elements in dielectric materials as we will study in detail in Chapter 4. jsin j The potential due to two charges, + and (r + ; r ), can be written down as r + r (.7) 3

31 Φ r+ + a θ r Figure -4: Electric dipole consists of two charges and separated by a small distance a: where r + and r related through are the distances to the respective charges from the observation point. They are r 2 + r 2 + a 2 2ar cos (.8) where is the angle between the z axis and the vector r. So far, we have not made any approximations, and E. (??) is exact. At a point far away from the dipole such that r + ; r a, r + may be approximated by r + ' r a cos (.9) Then, the potential at r a becomes (r; ) ' r a cos r a cos r 2 (.) where we have replaced r (' r + ) by r. In contrast to the monopole potential, (r) r the dipole potential is proportional to r 2, and thus of higher order in the series expansion in powers of r. The potentials due to the eual but opposite charges cancel each other in the lowest (monopole) order. Also, the dipole potential has the angular dependence, cos. It is convenient to introduce the dipole moment (vector) p a (.) 3

32 y x Figure -5: Euipotential surfaces of electric dipole p z : where a is the vector directed from the negative charge the angle between a and r, we may rewrite E. (.) as to the positive charge +. Since is p r r 3 (.2) where p r pr cos ar cos The euipotential surface of the dipole is shown in Fig..7. p jsin j The electric eld associated with the dipole can be calculated by taking the gradient of the potential, E r(r; e + e r a r 2 cos r 3 cos e r + sin r 3 e (.3) 32

33 The electric eld lines are described by the following di erential euation, dr rd (.4) E r E since the electric eld is tangent to the eld lines. The components, E r and E are E r a 2 cos (.5) r3 E a r 3 sin Substituting these into E. (??), we nd the following di erential euation to describe the eld line, dr 2r cot d or Integrating both sides, we obtain or dr r 2 cot d (.6) r (c sin 2 ) (.7) r c sin 2 where c is a constant. The electric eld lines of the diople are shown in Fig..8. Note that the electric eld lines are normal to the eui potential surfaces. cos 2.9 Linear Quadrupole Charges +; 2; and + placed along the z axis at z a; ; a constitute a linear uadrupole. The potential at r a can be found as superposition of 3 potemtials, (r; ) ' p 4" r 2 + a 2 2ar cos a2 3 cos 2 4" r 3 where use is made of binomial expansion 2 r + p r 2 + a 2 + 2ar cos p + x 2 x x2 33

34 y x Figure -6: Electrif eld lines of the diople p z a: Note that the dipole potemtial vanishes. This is because the uadrupole consists of two dipoles oppositely directed. The euipotebtial surfaces of the linear uadrupole is shown in Fig.. 3 sin 2 y x 2 The function P 2 (x) 2 3x2 is the Legendre polymonial of order l 2: The binomial expansion of the potential due to the 34

35 charge at z a is p 4" r 2 + a 2 2ar cos 4" r + a a2 3 cos 2 cos + r2 r 3 2 X a l 4" r l+ P l (cos ) l + Likewise 2 p 4" r 2 + a 2 + 2ar cos X ( ) l a l 4" r l+ P l (cos ) l 35