Lista de Exercicios de Eletromagnetismo

Static Electric Fields 31 Introduction In Section 12 we mentioned that three essential steps are involved in constructing a deductive theory for the study of a scientific subject They are the definition of basic quantities the development of rules of operation and the postulation of funda mental relations We have defined the source and field quantities for the electromag netic model in Chapter 1 and developed the fundamentals of vector algebra and vector calculus in Chapter 2 We are now ready to introduce the fundamental postu lates for the study of sourcefield relationships in electrostatics A field is a spatial distribution of a scalar or vector quantity which may or may not be a function of time An example of a scalar is the altitude of a location on a mountain relative to the sea level It is a scalar which is not a function of time if longterm erosion and earthquake effects are neglected Various locations on the mountain have different altitudes constituting an altitude field The gradient of altitude is a vector that gives both the direction and the magnitude of the maximum rate of increase the upward slope of altitude On a flat mountaintop or flat land the altitude is constant and its gradient vanishes The gravitational field of the earth representing the force of gravity on a unit mass is a vector field directed toward the center of the earth having a magnitude depending on the altitude of the mass Electric and mag netic field intensities are vector fields In electrostatics electric charges the sources are at rest and electric fields do not change with time There are no magnetic fields hence we deal with a relatively simple situation After we have studied the behavior of static electric fields and mastered the techniques for solving electrostatic boundaryvalue problems we will go on to the subject of magnetic fields and timevarying electromagnetic fields Although electrostatics is relatively simple in the electromagnetics scheme of things its mastery is fundamental to the understanding of more complicated electromagnetic models Moreover the explanation of many natural phenomena such as lightning corona St Elmos fire and grain explosion and the principles of some important industrial 31 Introduction 73 applications such as oscilloscope inkjet printer xerography and electret micro phone are based on electrostatics Many articles on special applications of electro statics appear in the literature and a number of books on this subject have also been publishedf The development of electrostatics in elementary physics usually begins with the experimental Coulombs law formulated in 1785 for the force between two point charges This law states that the force between two charged bodies qt and q2 that are very small in comparison with the distance of separation R12 is proportional to the product of the charges and inversely proportional to the square of the distance the direction of the force being along the line connecting the charges In addition Coulomb found that unlike charges attract and like charges repel each other Using vector notation Coulombs law can be written mathematically as K12 where F12 is the vector force exerted by q1 on q2 aRl2 is a unit vector in the direction from qx to q2 and k is a proportionality constant depending on the medium and the system of units Note that if qt and q2 are of the same sign both positive or both negative F12 is positive repulsive and if q1 and q2 are of opposite signs F12 is negative attractive Electrostatics can proceed from Coulombs law to define electric field intensity E electric scalar potential V and electric flux density D and then lead to Gausss law and other relations This approach has been accepted as logical perhaps because it begins with an experimental law observed in a laboratory and not with some abstract postulates We maintain however that Coulombs law though based on experimental evi dence is in fact also a postulate Consider the two stipulations of Coulombs law that the charged bodies be very small in comparison with the distance of separation and that the force be inversely proportional to the square of the distance The ques tion arises regarding the first stipulation How small must the charged bodies be in order to be considered very small in comparison with the distance In practice the charged bodies cannot be of vanishing sizes ideal point charges and there is diffi culty in determining the true distance between two bodies of finite dimensions For given body sizes the relative accuracy in distance measurements is better when the separation is larger However practical considerations weakness of force existence of extraneous charged bodies etc restrict the usable distance of separation in the laboratory and experimental inaccuracies cannot be entirely avoided This leads to a more important question concerning the inversesquare relation of the second t A Klinkenberg and J L van der Minne Electrostatics in the Petroleum Industry Elsevier Amsterdam 1958 J H Dessauer and H E Clark Xerography and Related Processes Focal Press London 1965 A D Moore Ed Electrostatics and Its Applications John Wiley New York 1973 C E Jewett Electrostatics in the Electronics Environment John Wiley New York 1976 JC Crowley Fundamentals of Applied Electrostatics John Wiley New York 1986 74 3 Static Electric Fields stipulation Even if the charged bodies are of vanishing sizes experimental measure ments cannot be of infinite accuracy no matter how skillful and careful an experi mentor is How then was it possible for Coulomb to know that the force was exactly inversely proportional to the square not the 2000001th or the 1999999th power of the distance of separation This question cannot be answered from an experimental viewpoint because it is not likely that experiments could have been accurate to the seventh place during Coulombs time1 We must therefore conclude that Coulombs law is itself a postulate and that the exact relation stipulated by Eq 31 is a law of nature discovered and assumed by Coulomb on the basis of his experiments of limited accuracy Instead of following the historical development of electrostatics we introduce the subject by postulating both the divergence and the curl of the electric field intensity in free space From Helmholtzs theorem in Section 212 we know that a vector field is determined if its divergence and curl are specified We derive Gausss law and Coulombs law from the divergence and curl relations and we do not present them as separate postulates The concept of scalar potential follows naturally from a vector identity Field behaviors in material media will be studied and expressions for elec trostatic energy and forces will be developed 32 Fundamental Postulates of Electrostatics in Free Space We start the study of electromagnetism with the consideration of electric fields due to stationary static electric charges in free space Electrostatics in free space is the simplest special case of electromagnetics We need to consider only one of the four fundamental vector field quantities of the electromagnetic model discussed in Section 1 2 namely the electric field intensity E Furthermore only the permittivity of free space e0 of the three universal constants mentioned in Section 13 enters into our formulation Electric field intensity is defined as the force per unit charge that a very small stationary test charge experiences when it is placed in a region where an electric field exists That is I F I 32 F E lim o q Vm The electric field intensity E is then proportional to and in the direction of the force F If F is measured in newtons N and charge q in coulombs C then E is in new tons per coulomb NC which is the same as volts per meter Vm The test charge T The exponent on the distance in Coulombs law has been verified by an indirect experiment to be 2 to within one part in 1015 See E R Williams J E Faller and H A Hall Phys Rev Letters vol 26 1971 p 721 32 Fundamental Postulates of Electrostatics in Free Space 75 q of course cannot be zero in practice as a matter of fact it cannot be less than the charge on an electron However the finiteness of the test charge would not make the measured E differ appreciably from its calculated value if the test charge is small enough not to disturb the charge distribution of the source An inverse relation of Eq 32 gives the force F on a stationary charge q in an electric field E 33 The two fundamental postulates of electrostatics in free space specify the diver gence and curl of E They are 34 and F qE N tes of electrosta VE V x E 0 tic 35 In Eq 34 p is the volume charge density of free charges Cm3 and e0 is the permittivity of free space a universal constant1 Equation 35 asserts that static electric fields are irrotational whereas Eq 34 implies that a static electric field is not solenoidal unless p 0 These two postulates are concise simple and independent of any coordinate system and they can be used to derive all other relations laws and theorems in electrostatics Such is the beauty of the deductive axiomatic ap proach Equations 34 and 35 are point relations that is they hold at every point in space They are referred to as the differential form of the postulates of electro statics since both divergence and curl operations involve spatial derivatives In prac tical applications we are usually interested in the total field of an aggregate or a distribution of charges This is more conveniently obtained by an integral form of Eq 34 Taking the volume integral of both sides of Eq 34 over an arbitrary volume V we have jy Edv jypdv 36 In view of the divergence theorem in Eq 2115 Eq 36 becomes 37 f The permittivity of free space e0 x 1T9 Fm See Eq 111 3 67c 76 3 Static Electric Fields where Q is the total charge contained in volume V bounded by surface S Equation 37 is a form of Gausss law which states that the total outward flux of the elec tric field intensity over any closed surface in free space is equal to the total charge enclosed in the surface divided by e0 Gausss law is one of the most important re lations in electrostatics We will discuss it further in Section 34 along with illustrative examples An integral form can also be obtained for the curl relation in Eq 35 by inte grating V x E over an open surface and invoking Stokess theorem as expressed in Eq 2143 We have 38 The line integral is performed over a closed contour C bounding an arbitrary surface hence C is itself arbitrary As a matter of fact the surface does not even enter into Eq 38 which asserts that the scalar line integral of the static electric field intensity around any closed path vanishes The scalar product E d integrated over any path is the voltage along that path Thus Eq 38 is an expression of Kirchhoffs voltage law in circuit theory that the algebraic sum of voltage drops around any closed circuit is zero This will be discussed again in Section 53 Equation 38 is another way of saying that E is irrotational conservative Referring to Fig 31 we see that if the scalar line integral of E over the arbitrary closed contour CXC2 is zero then or or f Ed f Edt 0 JCi JC2 JPl 2 Along Cj Along C2 r Ede r Ede JPl JPl JPl Along Cj 39 310 311 Along C2 Pi CTT FIGURE 31 An arbitrary contour 33 Coulombs Law 77 Equation 311 says that the scalar line integral of the irrotational E field is inde pendent of the path it depends only on the end points As we shall see in Section 35 the integrals in Eq 311 represent the work done by the electric field in moving a unit charge from point P1 to point P2 hence Eqs 38 and 39 imply a statement of conservation of work or energy in an electrostatic field The two fundamental postulates of electrostatics in free space are repeated below because they form the foundation upon which we build the structure of electrostatics Postulates of Electrostatics in Free Space Differential Form Integral Form VE Vx E 0 E d 0 We consider these postulates like the principle of conservation of charge to be repre sentations of laws of nature In the following section we shall derive Coulombs law 33 Coulombs Law We consider the simplest possible electrostatic problem of a single point charge q at rest in a boundless free space In order to find the electric field intensity due to q we draw a hypothetical spherical surface of a radius R centered at q Since a point charge has no preferred directions its electric field must be everywhere radial and has the same intensity at all points on the spherical surface Applying Eq 37 to Fig 32a we have or jsEds jsREJdRds ER ds ER4nR2 Therefore 312 Equation 312 tells us that the electric field intensity of a positive point charge is in the outward radial direction and has a magnitude proportional to the charge and inversely proportional to the square of the distance from the charge This is a very important basic formula in electrostatics Using Eq 2139 we can verify that 78 3 Static Electric Fields 7 a Point charge at the origin b Point charge not at the origin FIGURE 32 Electric field intensity due to a point charge V x E 0 for the E given in Eq 312 A fluxline graph for the electric field intensity of a positive point charge q will look like Fig 225b If the charge q is not located at the origin of a chosen coordinate system suitable changes should be made to the unit vector aR and the distance R to reflect the locations of the charge and of the point at which E is to be determined Let the position vector of q be R and that of a field point P be R as shown in Fig 32b Then from Eq 312 q EP a qP 4n0R R 12 where aqP is the unit vector drawn from q to P Since R R V I R R we have 313 314 315 EXAMPLE 31 Determine the electric field intensity at P 02 0 23 due to a point charge of 5 nC at g02 01 25 in air All dimensions are in meters Solution The position vector for the field point P R O P a x02a z23 The position vector for the point charge Q is R OQ ax02 a01 az25 The difference is R R ax04 av01 az02 33 Coulombs Law 79 which has a magnitude R R 042 012 02212 0458 m Substituting in Eq 315 we obtain EP 1 qR R 4ne0J R R 5 x 10 9 9 x 1Q9 04 583 M 4 a a i az2 2145ax0873 a218 az0437 Vm The quantity within the parentheses is the unit vector aQP R RR R and EP has a magnitude of 2145 Vm B Note The permittivity of air is essentially the same as that of the free space The factor l47ie0 appears very frequently in electrostatics From Eq 111 we know that e0 lc2fi0 But 0 An x 10 7 Hm in SI units so 1 c 2 i n 7 4ne0 An 10 7c 2 mF 316 exactly If we use the approximate value c 3 x 108 ms then l47ie0 9 x 109 mF When a point charge q2 is placed in the field of another point charge q1 at the origin a force F 1 2 is experienced by q2 due to electric field intensity E12 of q1 at q2 Combining Eqs 33 and 312 we have F 1 2 2 E 1 2 a N 317 Equation 317 is a mathematical form of Coulombs law already stated in Section 31 in conjunction with Eq 31 Note that the exponent on R is exactly 2 which is a consequence of the fundamental postulate Eq 34 In SI units the propor tionality constant k equals l47ie0 and the force is in newtons N EXAMPLE 32 A total charge Q is put on a thin spherical shell of radius b Determine the electric field intensity at an arbitrary point inside the shell Solution We shall solve this problem in two ways a At any point such as P inside the hollow shell shown in Fig 33 an arbitrary hypothetical closed surface a Gaussian surface may be drawn over which we apply Gausss law Eq 37 Since no charge exists inside the shell and the surface is arbitrary we conclude readily that E 0 everywhere inside the shell 80 3 Static Electric Fields FIGURE 33 A charged shell Example 32 b Let us now examine the problem in more detail Draw a pair of elementary cones of solid angle dQ with vertex at an arbitrary point P The cones extend in both directions intersecting the shell in areas ds1 and ds2 at distances rx and r2 re spectively from the point P Since charge Q distributes uniformly over the spherical shell there is a uniform surface charge density 318 The magnitude of the electric field intensity at P due to charges on the ele mentary surfaces ds and ds2 is from Eq 312 ps fdsx ds2 4ne0 I r r dZ r d r Y 319 But the solid angle dQ equals ds dsy dQ cos a cos a 320 r r Combining the expressions of dE and dQ we find that ps dQ dQ n dE P 0 321 47re0 cos a cos ay Since the above result applies to every pair of elementary cones we conclude that E 0 everywhere inside the conducting shell as before s It will be noted that if Coulombs law as expressed in Eq 312 and used in Eq 319 was slightly different from an inversesquare relation the substitution of Eq 320 which is a geometrical relation in Eq 319 would not yield the result dE 0 Consequently the electric field intensity inside the shell would not vanish indeed it would vary with the location of the point P Coulomb originally used a torsion balance to conduct his experiments which were necessarily of limited accuracy Nevertheless he was brilliant enough to postulate the inversesquare law Many 33 Coulombs Law 81 scientists subsequently made use of the vanishing field inside a spherical shell illus trated in this example to verify the inversesquare law The field inside a charged shell if it existed could be detected to a very high accuracy by a probe through a small hole in the shell EXAMPLE 33 The electrostatic deflection system of a cathoderay oscilloscope is depicted in Fig 34 Electrons from a heated cathode are given an initial velocity u0 azw0 by a positively charged anode not shown The electrons enter at z 0 into a region of deflection plates where a uniform electric field Ed nyEd is main tained over a width w Ignoring gravitational effects find the vertical deflection of the electrons on the fluorescent screen at z L Solution Since there is no force in the zdirection in the z 0 region the horizontal velocity u0 is maintained The field Ed exerts a force on the electrons each carrying a charge e causing a deflection in the ydirection F eEd ayeEd From Newtons second law of motion in the vertical direction we have m du eEd where m is the mass of an electron Integrating both sides we obtain dy e dt m uy n EdU where the constant of integration is set to zero because uy 0 at t 0 Integrating again we have Screen FIGURE 34 Electrostatic deflection system of a cathoderay oscilloscope Example 33 82 3 Static Electric Fields The constant of integration is again zero because y 0 at t 0 Note that the elec trons have a parabolic trajectory between the deflection plates At the exit from the deflection plates t wu0 1m u0j and w y Uylt eEd iw m u When the electrons reach the screen they have traveled a further horizontal distance of L w which takes L wu0 seconds During that time there is an additional vertical deflection h w eEd wL w d2 uyl m Hence the deflection at the screen is d0 dL d2 eEd i w wU2 mu Inkjet printers used in computer output like cathoderay oscilloscopes are de vices based on the principle of electrostatic deflection of a stream of charged particles Minute droplets of ink are forced through a vibrating nozzle controlled by a piezo electric transducer The output of the computer imparts variable amounts of charges on the ink droplets which then pass through a pair of deflection plates where a uniform static electric field exists The amount of droplet deflection depends on the charge it carries causing the ink jet to strike the print surface and form an image as the print head moves in a horizontal direction 331 ELECTRIC FIELD DUE TO A SYSTEM OF DISCRETE CHARGES Suppose an electrostatic field is created by a group of n discrete point charges qu q2 q located at different positions Since electric field intensity is a linear func tion of proportional to aRqR2 the principle of superposition applies and the total E field at a point is the vector sum of the fields caused by all the individual charges From Eq 315 we can write the electric intensity at a field point whose position vector is R as 322 Although Eq 322 is a succinct expression it is somewhat inconvenient to use be cause of the need to add vectors of different magnitudes and directions 33 Coulombs Law 83 Let us consider the simple case of an electric dipole that consists of a pair of equal and opposite charges q and q separated by a small distance d as shown in Fig 35 Let the center of the dipole coincide with the origin of a spherical coor dinate system Then the E field at the point P is the sum of the contributions due to q and q Thus E q 4ne0 j M 3 Rl r j 323 The first term on the right side of Eq 323 can be simplified if d R We write i 3 J 5 32 4 2132 324 n J 3 R d where the binomial expansion has been used and all terms containing the second and higher powers of dR have been neglected Similarly for the second term on the right side of Eq 323 we have 2 R Substitution of Eqs 324 and 325 in Eq 323 leads to q I K d E 4ne0R 325 326 d q R d2 RJ R d2 FIGURE 35 Electric field of a dipole 3 Static Electric Fields The derivation and interpretation of Eq 326 require the manipulation of vec tor quantities We can appreciate that determining the electric field caused by three or more discrete charges will be even more tedious In Section 35 we will introduce the concept of a scalar electric potential with which the electric field intensity caused by a distribution of charges can be found more easily The electric dipole is an important entity in the study of the electric field in di electric media We define the product of the charge q and the vector d going from q to q as the electric dipole moment p p d Equation 326 can then be rewritten as E 1 Ae0R R2 R p 327 328 where the approximate sign over the equal sign has been left out for simplicity If the dipole lies along the zaxis as in Fig 35 then see Eq 277 p azp paR cos 6 ae sin 0 R p Rp cos 6 and Eq 328 becomes E 4ne0R aK2 cos 6 ae sin 0 Vm 329 330 331 Equation 331 gives the electric field intensity of an electric dipole in spherical co ordinates We see that E of a dipole is inversely proportional to the cube of the dis tance R This is reasonable because as R increases the fields due to the closely spaced q and q tend to cancel each other more completely thus decreasing more rapidly than that of a single point charge 332 ELECTRIC FIELD DUE TO A CONTINUOUS DISTRIBUTION OF CHARGE The electric field caused by a continuous distribution of charge can be obtained by integrating superposing the contribution of an element of charge over the charge distribution Refer to Fig 36 where a volume charge distribution is shown The volume charge density p Cm3 is a function of the coordinates Since a differential element of charge behaves like a point charge the contribution of the charge p dv in a differential volume element dv to the electric field intensity at the field point P is dE aR pdv 4ne0R 332 p 85 FIGURE 36 Electric field due to a continuous charge distribution We have or since a RR E 4kJyaWdv Vm E ii Vm 333 334 Except for some especially simple cases the vector triple integral in Eq 333 or Eq 334 is difficult to carry out because in general all three quantities in the inte grand aR p and R change with the location of the differential volume dv If the charge is distributed on a surface with a surface charge density ps Cm2 then the integration is to be carried out over the surface not necessarily flat Thus For a line charge we have E 1 f aRds 47T0 JS R R2 Vm ve 4ne0 h R R2 Vm 335 336 where p Cm is the line charge density and L the line not necessarily straight along which the charge is distributed EXAMPLE 34 Determine the electric field intensity of an infinitely long straight line charge of a uniform density p in air 86 3 Static Electric Fields dEr dEr dE FIGURE 37 An infinitely long straight line charge Solution Let us assume that the line charge lies along the zaxis as shown in Fig 37 We are perfectly free to do this because the field obviously does not depend on how we designate the line It is an accepted convention to use primed coordinates for source points and unprimed coordinates for field points when there is a possibility of confusion The problem asks us to find the electric field intensity at a point P which is at a distance r from the line Since the problem has a cylindrical symmetry that is the electric field is independent of the azimuth angle it would be most convenient to work with cylindrical coordinates We rewrite Eq 336 as 337 For the problem at hand pe is constant and a line element d dz is chosen to be at an arbitrary distance z from the origin It is most important to remember that R is the distance vector directed from the source to the field point not the other way around We have R arr azz 338 The electric field dE due to the differential line charge element p M pe dz is pez a rra zz 339 dE 4n0 r2 z232 ardEr azdEz where dEr per dz 4n0r2 z2fl2 339a 34 Gausss Law and Applications 87 and dE pez dz 4ne0r2 z232 339b In Eq 339 we have decomposed dE into its components in the ar and az directions It is easy to see that for every pedz at z1 there is a charge element pdz at z which will produce a dE with components dEr and dEz Hence the az components will cancel in the integration process and we only need to integrate the dEr in Eq 339a pfr foo dz 4ne0Jr2 z2f12 or 340 Equation 340 is an important result for an infinite line charge Of course no phys ical line charge is infinitely long nevertheless Eq 340 gives the approximate E field of a long straight line charge at a point close to the line charge 34 Gausss Law and Applications Gausss law follows directly from the divergence postulate of electrostatics Eq 34 by the application of the divergence theorem It was derived in Section 32 as Eq 37 and is repeated here on account of its importance 341 Gausss law asserts that the total outward flux of the Efield over any closed surface in free space is equal to the total charge enclosed in the surface divided by 0 We note that the surface S can be any hypothetical mathematical closed surface chosen for convenience it does not have to be and usually is not a physical surface Gausss law is particularly useful in determining the Efield of charge distributions with some symmetry conditions such that the normal component of the electric field intensity is constant over an enclosed surface In such cases the surface integral on the left side of Eq 341 would be very easy to evaluate and Gausss law would be a much more efficient way for finding the electric field intensity than Eqs 333 through 337 On the other hand when symmetry conditions do not exist Gausss law would not be of much help The essence of applying Gausss law lies first in the rec ognition of symmetry conditions and second in the suitable choice of a surface over which the normal component of E resulting from a given charge distribution is a 3 Static Electric Fields constant Such a surface is referred to as a Gaussian surface This basic principle was used to obtain Eq 312 for a point charge that possesses spherical symmetry con sequently a proper Gaussian surface is the surface of a sphere centered at the point charge Gausss law could not help in the derivation of Eq 326 or 331 for an electric dipole since a surface about a separated pair of equal and opposite charges over which the normal component of E remains constant was not known EXAMPLE 35 Use Gausss law to determine the electric field intensity of an infi nitely long straight line charge of a uniform density p in air Solution This problem was solved in Example 34 by using Eq 336 Since the line charge is infinitely long the resultant E field must be radial and perpendicular to the line charge E arr and a component of E along the line cannot exist With the obvious cylindrical symmetry we construct a cylindrical Gaussian surface of a radius r and an arbitrary length L with the line charge as its axis as shown in Fig 38 On this surface Er is constant and ds 2irrddz from Eq 253a We have js E ds JQ L Err dj dz 2rLEr There is no contribution from the top or the bottom face of the cylinder because on the top face ds azr dr dcj but E has no zcomponent there making E ds 0 Sim ilarly for the bottom face The total charge enclosed in the cylinder is Q peL Sub stitution into Eq 341 gives us immediately 2rLEr PeL Cylindrical Gaussian surface Infinitely long uniform line charge pp FIGURE 38 Applying Gausss law to an infinitely long line charge Example 35 34 Gausss Law and Applications 89 or E arr a r 2n0r This result is of course the same as that given in Eq 340 but it is obtained here in a much simpler way We note that the length L of the cylindrical Gaussian surface does not appear in the final expression hence we could have chosen a cylinder of a unit length mm EXAMPLE 36 Determine the electric field intensity of an infinite planar charge with a uniform surface charge density ps Solution It is clear that the E field caused by a charged sheet of an infinite extent is normal to the sheet Equation 335 could be used to find E but this would in volve a double integration between infinite limits of a general expression of lR2 Gausss law can be used to much advantage here We choose as the Gaussian surface a rectangular box with top and bottom faces of an arbitrary area A equidistant from the planar charge as shown in Fig 39 The sides of the box are perpendicular to the charged sheet If the charged sheet coincides with the xyplane then on the top face Ed zEgagds Egds On the bottom face Eds azEzazds Ezds Since there is no contribution from the side faces we have jj E ds 1EZ j A ds 2EZA The total charge enclosed in the box is Q psA Therefore 2EZA PA surface j Area 4 a Infinite uniform surface charge ps FIGURE 39 Applying Gausss law to an infinite planar charge Example 36 3 Static Electric Fields from which we obtain and 342a 342b Of course the charged sheet may not coincide with the xyplane in which case we do not speak in terms of above and below the plane but the E field always points away from the sheet if ps is positive It is obvious that the Gaussian surface could have been a pillbox of any shape not necessarily rectangular ma The lighting scheme of an office or a classroom may consist of incandescent bulbs long fluorescent tubes or ceiling panel lights These correspond roughly to point sources line sources and planar sources respectively From Eqs 312 340 and 342 we can estimate that light intensity will fall off rapidly as the square of the distance from the source in the case of incandescent bulbs less rapidly as the first power of the distance for long fluorescent tubes and not at all for ceiling panel lights EXAMPLE 37 Determine the E field caused by a spherical cloud of electrons with a volume charge density p p0 for 0 R b both p0 and b are positive and p 0 for R b Solution First we recognize that the given source condition has spherical symmetry The proper Gaussian surfaces must therefore be concentric spherical surfaces We must find the E field in two regions Refer to Fig 310 a 0 R b A hypothetical spherical Gaussian surface St with R b is constructed within the electron cloud On this surface E is radial and has a constant magnitude E 2LRER ds aR ds The total outward E flux is E ds ER j s ds ER4nR2 The total charge enclosed within the Gaussian surface is Q jrfdv An P J 91 Electron cloud FIGURE 310 Electric field intensity of a spherical electron cloud Example 37 Substitution into Eq 37 yields 0 R b We see that within the uniform electron cloud the E field is directed toward the center and has a magnitude proportional to the distance from the center b Rb For this case we construct a spherical Gaussian surface S0 with R b outside the electron cloud We obtain the same expression for jSo E ds as in case a The total charge enclosed is Consequently E a Pob3 3e0i2 Rb which follows the inverse square law and could have been obtained directly from Eq 312 We observe that outside the charged cloud the E field is exactly the same as though the total charge is concentrated on a single point charge at the center This is true in general for a spherically symmetrical charged region even though p is a function of R n 92 3 Static Electric Fields The variation of ER versus R is plotted in Fig 310 Note that the formal solution of this problem requires only a few lines If Gausss law is not used it is necessary 1 to choose a differential volume element arbitrarily located in the electron cloud 2 to express its vector distance R to a field point in a chosen coordinate system and 3 to perform a triple integration as indicated in Eq 333 This is a hopelessly involved process The moral is Try to apply Gausss law if symmetry conditions exist for the given charge distribution 35 Electric Potential In connection with the null identity in Eq 2145 we noted that a curlfree vector field could always be expressed as the gradient of a scalar field This induces us to define a scalar electric potential V such that E V 343 because scalar quantities are easier to handle than vector quantities If we can deter mine V more easily then E can be found by a gradient operation which is a straight forward process in an orthogonal coordinate system The reason for the inclusion of a negative sign in Eq 343 will be explained presently Electric potential does have physical significance and it is related to the work done in carrying a charge from one point to another In Section 32 we defined the electric field intensity as the force acting on a unit test charge Therefore in moving a unit charge from point P1 to point P2 in an electric field work must be done against the field and is equal to L Vd JCorV a Jpi 344 Many paths may be followed in going from P1 to P2 Two such paths are drawn in Fig 311 Since the path between Px and P2 is not specified in Eq 344 the FIGURE 311 Two paths leading from P1 to P2 in an electric field 35 Electric Potential 93 question naturally arises how does the work depend on the path taken A little thought will lead us to conclude that Wq in Eq 344 should not depend on the path if it did one would be able to go from P1 to P2 along a path for which W is smaller and then to come back to Px along another path achieving a net gain in work or energy This would be contrary to the principle of conservation of energy We have already alluded to the pathindependent nature of the scalar line integral of the irrotational conservative E field when we discussed Eq 38 Analogous to the concept of potential energy in mechanics Eq 344 represents the difference in electric potential energy of a unit charge between point P2 and point Pv Denoting the electric potential energy per unit charge by V the electric potential we have 345 Mathematically Eq 345 can be obtained by substituting Eq 343 in Eq 344 Thus in view of Eq 288 rErFHa j Vv What we have defined in Eq 345 is a potential difference electrostatic voltage between points P2 and P It makes no more sense to talk about the absolute potential of a point than about the absolute phase of a phasor or the absolute altitude of a geographical location a reference zeropotential point a reference zero phase usually at t 0 or a reference zero altitude usually at sea level must first be specified In most but not all cases the zeropotential point is taken at infinity When the reference zeropotential point is not at infinity it should be specifically stated We want to make two more points about Eq 343 First the inclusion of the negative sign is necessary in order to conform with the convention that in going against the E field the electric potential V increases For instance when a dc battery of a voltage V0 is connected between two parallel conducting plates as in Fig 312 positive and negative charges cumulate on the top and bottom plates respectively The E field is directed from positive to negative charges while the potential increases in the opposite direction Second we know from Section 26 when we defined the gradient of a scalar field that the direction of V is normal to the surfaces of constant H N I VZ3 Direction of increasing V FIGURE 312 Relative directions of E and increasing V 94 3 Static Electric Fields V Hence if we use directed field lines or streamlines to indicate the direction of the E field they are everywhere perpendicular to equipotential lines and equipotential surfaces 351 ELECTRIC POTENTIAL DUE TO A CHARGE DISTRIBUTION The electric potential of a point at a distance R from a point charge q referred to that at infinity can be obtained readily from Eq 345 which gives 4n0R nRdR 346 347 This is a scalar quantity and depends on besides q only the distance R The potential difference between any two points P2 and Px at distances R2 and Ru respectively from q is dt 348 This result may appear a little surprising at first since P2 and Px may not lie on the same radial line through q as illustrated in Fig 313 However the concentric circles spheres passing through P2 and P1 are equipotential lines surfaces and Vp2 VPl is the same as VPl VPy From the point of view of Eq 345 we can choose the path of integration from Px to P 3 and then from P 3 to P2 No work is done from PA to P 3 because F is perpendicular to d a dj along the circular path Edt 0 The electric potential at R due to a system of n discrete point charges qlt q2 qn located at Rl9 R 2 RJ is by superposition the sum of the potentials due to FIGURE 313 Path of integration about a point charge 35 Electric Potential 95 the individual charges 349 Since this is a scalar sum it is in general easier to determine E by taking the negative gradient of V than from the vector sum in Eq 322 directly As an example let us again consider an electric dipole consisting of charges q and q with a small separation d The distances from the charges to a field point P are designated R and as shown in Fig 314 The potential at P can be written down directly q 1 1 V If d R we have 4n0 R R 350 and 1 7 i RI R cos 2 R cos 1 1 d R 2RCS l s n i 2 cos Substitution of Eqs 351 and 352 in Eq 350 gives 351 352 or V T qd cos 6 4ne0R2 Pa 4ne0R2 V 353a 353b where p fd The approximate sign has been dropped for simplicity d q FIGURE 314 An electric dipole 3 Static Electric Fields The E field can be obtained from V V In spherical coordinates we have dR Rd9 354 P 3 aR2 cos 9 ae sin 9 4ne0R Equation 354 is the same as Eq 331 but has been obtained by a simpler proce dure without manipulating position vectors EXAMPLE 38 Make a twodimensional sketch of the equipotential lines and the electric field lines for an electric dipole Solution The equation of an equipotential surface of a charge distribution is ob tained by setting the expression for V to equal a constant Since q d and e0 in Eq 353a for an electric dipole are fixed quantities a constant V requires a constant ratio cos 9R2 Hence the equation for an equipotential surface is R Cyjcos 9 355 where cv is a constant By plotting R versus 9 for various values of cv we draw the solid equipotential lines in Fig 315 In the range 0 9 n2 V is positive R is maximum at 9 0 and zero at 9 90 A mirror image is obtained in the range n2 9 n where V is negative The electric field lines or streamlines represent the direction of the E field in space We set d cE 356 where k is a constant In spherical coordinates Eq 356 becomes see Eq 266 a dR agR d9 aR sin 9 dj kaRER aeEe a 357 which can be written dR Rd9 Rsin9dl E 1T E For the electric dipole in Fig 315 there is no E component and dR Rd9 2 cos 6 sin 9 or dR 2 dsin 9 R sin 9 Integrating Eq 359 we obtain R cE sin2 9 360 359 35 Electric Potential 97 FIGURE 315 Equipotential and electric field lines of an electric dipole Example 38 where cE is a constant The electric field lines are drawn as dashed lines in Fig 315 They are rotationally symmetrical about the zaxis independent of and are everywhere normal to the equipotential lines mm The electric potential due to a continuous distribution of charge confined in a given region is obtained by integrating the contribution of an element of charge over the charged region We have for a volume charge distribution V J f L dv V 4ne0 Jv R v 361 98 3 Static Electric Fields For a surface charge distribution and for a line charge 4ne0 Js R V 4ne0 JL1 R V 362 363 We note here again that the integrals in Eqs 361 and 362 represent integrations in three and two dimensions respectively EXAMPLE 39 Obtain a formula for the electric field intensity on the axis of a circular disk of radius b that carries a uniform surface charge density ps Solution Although the disk has circular symmetry we cannot visualize a surface around it over which the normal component of E has a constant magnitude hence Gausss law is not useful for the solution of this problem We use Eq 362 Working with cylindrical coordinates indicated in Fig 316 we have ds rdrdt and R Jz2 r2 The electric potential at the point P0 0 z referring to the point at infinity is Ps V AneQ Jo Jo z2 r212 drdy TLfc2 b2ll2z 364 p0 0 z FIGURE 316 A uniformly charged disk Example 39 35 Electric Potential 99 Therefore 8V E V K a z oz a z l z z 2 b 2 n z 0 a z l z z 2 0 2 r 1 2 z 0 ze0 365a 365b The determination of E field at an offaxis point would be a much more difficult problem Do you know why For very large z it is convenient to expand the second term in Eqs 365a and 365b into a binomial series and neglect the second and all higher powers of the ratio b2z2 We have b2Ym b2 i s i 2 Substituting this into Eqs 365a and 365b we obtain inb2 Ps E a 4ne0zz Q a 4ne0z2 Q 4ne0z 2 z 0 z 0 366a 366b where Q is the total charge on the disk Hence when the point of observation is very far away from the charged disk the E field approximately follows the inverse square law as if the total charge were concentrated at a point EXAMPLE 310 Obtain a formula for the electric field intensity along the axis of a uniform line charge of length L The uniform linecharge density is pe Solution For an infinitely long line charge the E field can be determined readily by applying Gausss law as in the solution to Example 35 However for a line charge of finite length as shown in Fig 317 we cannot construct a Gaussian surface over which E ds is constant Gausss law is therefore not useful here Instead we use Eq 363 by taking an element of charge M dz at z The distance R from the charge element to the point P0 0 z along the axis of the line charge is R zz z Here it is extremely important to distinguish the position of the field point un primed coordinates from the position of the source point primed coordinates We 100 3 Static Electric Fields z z O P0 0 z dz L2 L t L2 I y FIGURE 317 A finite line charge of a uniform line density pe Example 310 integrate over the source region V 4ne0 Pe 47ln In L2 dz W zz z L2 z L2 L ZY 367 The E field at P is the negative gradient of V with respect to the unprimed field coordinates For this problem dV pL L E a z a z 4 7 r e 0 z 2 L 2 2 ZY 368 The preceding two examples illustrate the procedure for determining E by first finding V when Gausss law cannot be conveniently applied However we emphasize that if symmetry conditions exist such that a Gaussian surface can be constructed over which E ds is constant it is always easier to determine E directly The potential V if desired may be obtained from E by integration 36 Conductors in Static Electric Field So far we have discussed only the electric field of stationary charge distributions in free space or air We now examine the field behavior in material media In general we classify materials according to their electrical properties into three types conduc tors semiconductors and insulators or dielectrics In terms of the crude atomic model of an atom consisting of a positively charged nucleus with orbiting electrons the electrons in the outermost shells of the atoms of conductors are very loosely held 36 Conductors in Static Electric Field 101 and migrate easily from one atom to another Most metals belong to this group The electrons in the atoms of insulators or dielectrics however are confined to their orbits they cannot be liberated in normal circumstances even by the application of an external electric field The electrical properties of semiconductors fall between those of conductors and insulators in that they possess a relatively small number of freely movable charges In terms of the band theory of solids we find that there are allowed energy bands for electrons each band consisting of many closely spaced discrete energy states Be tween these energy bands there may be forbidden regions or gaps where no electrons of the solids atom can reside Conductors have an upper energy band partially filled with electrons or an upper pair of overlapping bands that are partially filled so that the electrons in these bands can move from one to another with only a small change in energy Insulators or dielectrics are materials with a completely filled upper band so conduction could not normally occur because of the existence of a large energy gap to the next higher band If the energy gap of the forbidden region is relatively small small amounts of external energy may be sufficient to excite the electrons in the filled upper band to jump into the next band causing conduction Such materials are semiconductors The macroscopic electrical property of a material medium is characterized by a constitutive parameter called conductivity which we will define in Chapter 5 The definition of conductivity is not important in this chapter because we are not dealing with current flow and are now interested only in the behavior of static electric fields in material media In this section we examine the electric field and charge distri bution both inside the bulk and on the surface of a conductor Assume for the present that some positive or negative charges are introduced in the interior of a conductor An electric field will be set up in the conductor the field exerting a force on the charges and making them move away from one another This movement will continue until all the charges reach the conductor surface and redistribute themselves in such a way that both the charge and the field inside vanish Hence Inside a Conductor Under Static Conditions P 0 E 0 369 370 When there is no charge in the interior of a conductor p 0 E must be zero be cause according to Gausss law the total outward electric flux through any closed surface constructed inside the conductor must vanish The charge distribution on the surface of a conductor depends on the shape of the surface Obviously the charges would not be in a state of equilibrium if there were a tangential component of the electric field intensity that produces a tangential 102 3 Static Electric Fields 0 FIGURE 318 A conductorfree space interface force and moves the charges Therefore under static conditions the E field on a conductor surface is everywhere normal to the surface In other words the surface of a conductor is an equipotential surface under static conditions As a matter of fact since E 0 everywhere inside a conductor the whole conductor has the same elec trostatic potential A finite time is required for the charges to redistribute on a con ductor surface and reach the equilibrium state This time depends on the conductivity of the material For a good conductor such as copper this time is of the order of 1019 s a very brief transient This point will be elaborated in Section 54 Figure 318 shows an interface between a conductor and free space Consider the contour abcda which has width ab cd Aw and height be da Ah Sides ab and cd are parallel to the interface Applying Eq 381 letting Ah 0 and noting that E in a conductor is zero we obtain immediately Ede EtAw 0 Jabcda or Et 0 371 which says that the tangential component of the E field on a conductor surface is zero In order to find En the normal component of E at the surface of the conductor we construct a Gaussian surface in the form of a thin pillbox with the top face in free space and the bottom face in the conductor where E 0 Using Eq 37 we obtain Eds EHAS Js e0 or En f 372 We assume that Eqs 37 and 38 are valid for regions containing discontinuous media Free space V Conductor E 36 Conductors in Static Electric Field 103 Hence the normal component of the E field at a conductorfree space boundary is equal to the surface charge density on the conductor divided by the permittivity of free space Summarizing the boundary conditions at the conductor surface we have Boundary Conditions at a ConductorFree Space Interface Et 0 E 371 372 When an uncharged conductor is placed in a static electric field the external field will cause loosely held electrons inside the conductor to move in a direction opposite to that of the field and cause net positive charges to move in the direction of the field These induced free charges will distribute on the conductor surface and create an induced field in such a way that they cancel the external field both inside the conductor and tangent to its surface When the surface charge distribution reaches an equilibrium all four relations Eqs 369 through 372 will hold and the conductor is again an equipotential body EXAMPLE 311 A positive point charge Q is at the center of a spherical conducting shell of an inner radius Rt and an outer radius R0 Determine E and V as functions of the radial distance R Solution The geometry of the problem is shown in Fig 3 19a Since there is spheri cal symmetry it is simplest to use Gausss law to determine E and then find V by in tegration There are three distinct regions a R R0 b Rt R R0 and c R R Suitable spherical Gaussian surfaces will be constructed in these regions Obviously E RER in all three regions a R R0 Gaussian surface Sx Eds ERAnR2 s K1 e Q or ERI Q 373 4ne0R2 The E field is the same as that of a point charge Q without the presence of the shell The potential referring to the point at infinity is Q VifRJER1dR 374 4ne0R b RiRR0 Gaussian surface S2 Because of Eq 370 we know that ER2 0 375 104 3 Static Electric Fields Conducting shell ER 0 Ri R0 b FIGURE 319 Electric field intensity and potential variations of a point charge Q at the center of a conducting shell Example 311 Since p 0 in the conducting shell and since the total charge enclosed in surface S2 must be zero an amount of negative charge equal to Q must be induced on the inner shell surface at R Rt This also means that an amount of positive charge equal to Q is induced on the outer shell surface at R R0 The con ducting shell is an equipotential body Hence Q K K 376 c R Rt Gaussian surface S3 Application of Gausss law yields the same formula for ER3 as ERl in Eq 373 for the first region Q i JR3 4ne0R 377 The potential in this region is V3JER3dR C Q c 4n0R where the integration constant C is determined by requiring V3 at R Rt to equal V2 in Eq 376 We have 4JK 0 R Rh 37 Dielectrics in Static Electric Field 105 and nfil lY 378 The variations of ER and V versus R in all three regions are plotted in Figs 319b and 319c Note that while the electric intensity has discontinuous jumps the potential remains continuous A discontinuous jump in potential would mean an infinite electric field intensity MM 3 7 Dielectrics in Static Electric Field Ideal dielectrics do not contain free charges When a dielectric body is placed in an external electric field there are no induced free charges that move to the surface and make the interior charge density and electric field vanish as with conductors How ever since dielectrics contain bound charges we cannot conclude that they have no effect on the electric field in which they are placed All material media are composed of atoms with a positively charged nucleus surrounded by negatively charged electrons Although the molecules of dielectrics are macroscopically neutral the presence of an external electric field causes a force to be exerted on each charged particle and results in small displacements of positive and negative charges in opposite directions These displacements though small in comparison to atomic dimensions nevertheless polarize a dielectric material and create electric dipoles The situation is depicted in Fig 320 Inasmuch as electric dipoles do have nonvanishing electric potential and electric field intensity we expect that the induced electric dipoles will modify the electric field both inside and outside the dielectric material The molecules of some dielectrics possess permanent dipole moments even in the absence of an external polarizing field Such molecules usually consist of two or tti FIGURE 320 External E A cross section of a polarized dielectric medium 3 Static Electric Fields more dissimilar atoms and are called polar molecules in contrast to nonpolar mole cules which do not have permanent dipole moments An example is the water molecule H20 which consists of two hydrogen atoms and one oxygen atom The atoms do not arrange themselves in a manner that makes the molecule have a zero dipole mo ment that is the hydrogen atoms do not lie exactly on diametrically opposite sides of the oxygen atom The dipole moments of polar molecules are of the order of 1030 Cm When there is no external field the individual dipoles in a polar dielectric are randomly oriented producing no net dipole moment macroscopically An applied electric field will exert a torque on the individual dipoles and tend to align them with the field in a manner similar to that shown in Fig 320 Some dielectric materials can exhibit a permanent dipole moment even in the absence of an externally applied electric field Such materials are called electrets Electrets can be made by heating softening certain waxes or plastics and placing them in an electric field The polarized molecules in these materials tend to align with the applied field and to be frozen in their new positions after they return to normal temperatures Permanent polarization remains without an external electric field Electrets are the electrical equivalents of permanent magnets they have found important applications in high fidelity electret microphones1 371 EQUIVALENT CHARGE DISTRIBUTIONS OF POLARIZED DIELECTRICS To analyze the macroscopic effect of induced dipoles we define a polarization vector P as nAv I Pfc P lim Cm2 379 Au0 Av where n is the number of molecules per unit volume and the numerator represents the vector sum of the induced dipole moments contained in a very small volume Av The vector P a smoothed point function is the volume density of electric dipole moment The dipole moment dp of an elemental volume dv is dp P dv which produces an electrostatic potential see Eq 353b dV dv 380 Integrating over the volume V of the dielectric we obtain the potential due to the polarized dielectric 1 See for instance J M Crowley Fundamentals of Applied Electrostatics Section 83 Wiley New York 1986 37 Dielectrics in Static Electric Field 107 47Tn JV R2 381t where R is the distance from the elemental volume dv to a fixed field point In Cartesian coordinates R2 x x2 y y2 z z2 382 and it is readily verified that the gradient of R with respect to the primed coordi nates is vi R2 Hence Eq 381 can be written as 4ne0 RJ Recalling the vector identity Problem 228 VA VA AV and letting A P and 1R we can rewrite Eq 384 as V 4ne0 V P Jv R Jv R dv 383 384 385 386 The first volume integral on the right side of Eq 386 can be converted into a closed surface integral by the divergence theorem We have y d t J t m 387 4ne0 Js R 4ne0 Jv R where aj is the outward normal from the surface element ds of the dielectric Com parison of the two integrals on the right side of Eq 387 with Eqs 362 and 361 respectively reveals that the electric potential and therefore the electric field intensity also due to a polarized dielectric may be calculated from the contributions of surface and volume charge distributions having respectively densities PPs P a KSM and P P V P 388 389 f We note here that V on the left side of Eq 381 represents the electric potential at a field point and V on the right side is the volume of the polarized dielectric 1 The prime sign On a and V has been dropped for simplicity since Eqs 388 and 389 involve only source coordinates and no confusion will result 108 3 Static Electric Fields These are referred to as polarization charge densities or boundcharge densities In other words a polarized dielectric may be replaced by an equivalent polarization surface charge density pps and an equivalent polarization volume charge density pp for field calculations v TPirds T f w 390 47re0 Js R 4ne0 Jv R y Although Eqs 388 and 389 were derived mathematically with the aid of a vector identity a physical interpretation can be provided for the charge distributions The sketch in Fig 320 clearly indicates that charges from the ends of similarly oriented dipoles exist on surfaces not parallel to the direction of polarization Con sider an imaginary elemental surface As of a nonpolar dielectric The application of an external electric field normal to As causes a separation d of the bound charges positive charges q move a distance d2 in the direction of the field and negative charges q move an equal distance against the direction of the field The net total charge AQ that crosses the surface As in the direction of the field is nq dAs where n is the number of molecules per unit volume If the external field is not normal to As the separation of the bound charges in the direction of a will be d a and AQ nq aAs 391 But nqd the dipole moment per unit volume is by definition the polarization vector P We have AQ P aAs 392 and A s P a as given in Eq 388 Remember that a is always the outward normal This relation correctly gives a positive surface charge on the righthand surface in Fig 320 and a negative surface charge on the lefthand surface For a surface S bounding a volume V the net total charge flowing out of V as a result of polarization is obtained by integrating Eq 392 The net charge remaining within the volume V is the negative of this integral 2 dPMs Js 393 jvFdv jvppdv which leads to the expression for the volume charge density in Eq 389 Hence when the divergence of P does not vanish the bulk of the polarized dielectric appears to be charged However since we started with an electrically neutral dielectric body the total charge of the body after polarization must remain zero This can be readily 38 Electric Flux Density and Dielectric Constant 109 verified by noting that Total charge b ppsds ppdv jsPads jyFdv 0 where the divergence theorem has again been applied 394 38 Electric Flux Density and Dielectric Constant Because a polarized dielectric gives rise to an equivalent volume charge density pp we expect the electric field intensity due to a given source distribution in a dielectric to be different from that in free space In particular the divergence postulated in Eq 34 must be modified to include the effect of pp that is Using Eq 389 we have VE p pJ V0E P p 395 396 We now define a new fundamental field quantity the electric flux density or electric displacement D such that D e0E P Cm2 397 The use of the vector D enables us to write a divergence relation between the electric field and the distribution of free charges in any medium without the necessity of dealing explicitly with the polarization vector P or the polarization charge density pp Combining Eqs 396 and 397 we obtain the new equation V D p Cm3 398 where p is the volume density of free charges Equations 398 and 35 are the two fundamental governing differential equations for electrostatics in any medium Note that the permittivity of free space e0 does not appear explicitly in these two equations The corresponding integral form of Eq 398 is obtained by taking the volume integral of both sides We have j v V D dv f p dv or D ds Q C 399 3100 110 3 Static Electric Fields Equation 3100 another form of Gausss law states that the total outward flux of the electric displacement or simply the total outward electric flux over any closed surface is equal to the total free charge enclosed in the surface As was indicated in Section 3 4 Gausss law is most useful in determining the electric field due to charge distributions under symmetry conditions When the dielectric properties of the medium are linear and isotropic the polar ization is directly proportional to the electric field intensity and the proportionality constant is independent of the direction of the field We write P e0XeE 3101 where xe is a dimensionless quantity called electric susceptibility A dielectric medium is linear if xe is independent of E and homogeneous if e is independent of space coordinates Substitution of Eq 3101 in Eq 397 yields D e0l XeE e0erE eE Cm2 where r 1 Xe 3102 3103 is a dimensionless quantity known as the relative permittivity or the dielectric constant of the medium The coefficient e e0er is the absolute permittivity often called simply permittivity of the medium and is measured in farads per meter Fm Air has a dielectric constant of 100059 hence its permittivity is usually taken as that of free space The dielectric constants of some common materials are included in Table 31 on p 114 and Appendix B3 Note that er can be a function of space coordinates If er is independent of posi tion the medium is said to be homogenous A linear homogeneous and isotropic medium is called a simple medium The relative permittivity of a simple medium is a constant Later in the book we will learn that the effects of a lossy medium can be rep resented by a complex dielectric constant whose imaginary part provides a mea sure of power loss in the medium and is in general frequencydependent For anisotropic materials the dielectric constant is different for different directions of the electric field and D and E vectors generally have different directions permittivity is a tensor In matrix form we may write 3 2 3104 For crystals the reference coordinates can be chosen to be along the principal axes of the crystal so that the offdiagonal terms of the permittivity matrix in Eq 3104 38 Electric Flux Density and Dielectric Constant 111 are zero We have Dx Dv Dz l 0 0 0 e2 0 0 0 esJ E E E 3105 Media having the property represented by Eq 3105 are said to be biaxial We may write Dx lE 3106a Dy e2Ey 3106b Dz e3Ez 3106c If further e1 e2 then the medium is said to be uniaxial Of course if e1 e2 e3 we have an isotropic medium We shall deal only with isotropic media in this book EXAMPLE 312 A positive point charge Q is at the center of a spherical dielectric shell of an inner radius R and an outer radius R0 The dielectric constant of the shell is er Determine E V D and P as functions of the radial distance R Solution The geometry of this problem is the same as that of Example 311 The conducting shell has now been replaced by a dielectric shell but the procedure of solution is similar Because of the spherical symmetry we apply Gausss law to find E and D in three regions a R R0 b RR R0 and c R R Potential V is found from the negative line integral of E and polarization P is determined by the relation P D e0E 0er 1E 3107 The E D and P vectors have only radial components Refer to Fig 321a where the Gaussian surfaces are not shown in order to avoid cluttering up the figure a R R0 The situation in this region is exactly the same as that in Example 311 We have from Eqs 373 and 374 Q 4TL0R2 Q 4ne0R From Eqs 3102 and 3107 we obtain and DRI CQERI PRI 0 Q 4TLR 3108 3109 112 3 Static Electric Fields Dielectric shell a PR 0 c v R R0 R FIGURE 321 Field variations of a point charge Q at the center of a dielectric shell Example 312 b Rt R R0 The application of Gausss law in this region gives us directly Q Q ER2 DR2 R24e0erR24nzR2 Q AnR 2 JiS 3110 3111 3112 Note that DR2 has the same expression as DR1 and that both ER and PR have a discontinuity at R R0 In this region 1R R0 4neJRoR2 Q 47l0 4TL JRo R 3113 c i K Since the medium in this region is the same as that in the region R R0 the application of Gausss law yields the same expressions for ER DR and PR in 38 Electric Flux Density and Dielectric Constant 113 both regions Q ER347Z0R2 D Q R3 4nR 2 PR3 0 To find V3 we must add to V2 at R R the negative line integral of ER3 Vi VlRiPRi EdR Q 4u0 1 JR0 v1 C J R R 3114 The variations of 0ER and DR versus R are plotted in Fig 321b The difference DR 0ER is PR and is shown in Fig 321c The plot for V in Fig 321d is a composite graph for Vu V2 and V3 in the three regions We note that DR is a con tinuous curve exhibiting no sudden changes in going from one medium to another and that PR exists only in the dielectric region am It is instructive to compare Figs 321b and 321d with Figs 319b and 319c respectively of Example 311 From Eqs 388 and 389 we find 1 P l H l Ki PR2R Rt on the inner shell surface 3115 PARRO P RRRO PR2RRC Y ri 3116 on the outer shell surface and V P WmR2pR2 o 3117 Equations 3115 3116 and 3117 indicate that there is no net polarization volume charge inside the dielectric shell However negative polarization surface charges exist on the inner surface and positive polarization surface charges on the outer surface These surface charges produce an electric field intensity that is directed radially inward thus reducing the E field in region 2 due to the point charge Q at the center 114 3 Static Electric Fields TABLE 31 Dielectric Constants and Dielectric Strengths of Some Common Materials Material Air atmospheric pressure Mineral oil Paper Polystyrene Rubber Glass Mica Dielectric Constant 10 23 24 26 2340 410 60 Dielectric Strength Vm 3 x 106 15 x 106 15 x 106 20 x 106 25 x 106 30 x 106 200 x 106 381 DIELECTRIC STRENGTH We have explained that an electric field causes small displacements of the bound charges in a dielectric material resulting in polarization If the electric field is very strong it will pull electrons completely out of the molecules The electrons will accelerate under the influence of the electric field collide violently with the molecular lattice structure and cause permanent dislocations and damage in the material Avalanche effect of ionization due to collisions may occur The material will become conducting and large currents may result This phenomenon is called a dielectric breakdown The maximum electric field intensity that a dielectric material can with stand without breakdown is the dielectric strength of the material The approxi mate dielectric strengths of some common substances are given in Table 31 The dielectric strength of a material must not be confused with its dielectric constant A convenient number to remember is that the dielectric strength of air at the atmospheric pressure is 3 kVmm When the electric field intensity exceeds this value air breaks down Massive ionization takes place and sparking corona discharge follows Charge tends to concentrate at sharp points In view of Eq 372 the electric field intensity in the immediate vicinity of sharp points is much higher than that at points on a relatively flat surface with a small curvature This is the principle upon which a lightning arrester with a sharp metal lightning rod on top of tall buildings works When a cloud containing an abundance of electric charges ap proaches a tall building equipped with a lightning rod connected to the ground charges of an opposite sign are attracted from the ground to the tip of the rod where the electric field intensity is the strongest As the electric field intensity ex ceeds the dielectric strength of the wet air breakdown occurs and the air near the tip is ionized and becomes conducting The electric charges in the cloud are then discharged safely to the ground through the conducting path The fact that the electric field intensity tends to be higher at a point near the surface of a charged conductor with a larger curvature is illustrated quantitatively in the following example 38 Electric Flux Density and Dielectric Constant 115 EXAMPLE 313 Consider two spherical conductors with radii bx and b2 b2 i that are connected by a conducting wire The distance of separation between the conductors is assumed to be very large in comparison to b2 so that the charges on the spherical conductors may be considered as uniformly distributed A total charge Q is deposited on the spheres Find a the charges on the two spheres and b the electric field intensities at the sphere surfaces Solution a Refer to Fig 322 Since the spherical conductors are at the same potential we have 2i Q2 4ne0b1 4n0b2 or Q2 b2 Hence the charges on the spheres are directly proportional to their radii But since fii Qi Q we find that b The electric field intensities at the surfaces of the two conducting spheres are SO E2n bj Q2 bx The electric field intensities are therefore inversely proportional to the radii being higher at the surface of the smaller sphere which has a larger curvature h FIGURE 322 Two connected conducting spheres Example 313 116 3 Static Electric Fields 39 Boundary Conditions for Electrostatic Fields Electromagnetic problems often involve media with different physical properties and require the knowledge of the relations of the field quantities at an interface between two media For instance we may wish to determine how the E and D vectors change in crossing an interface We already know the boundary conditions that must be satisfied at a conductorfree space interface These conditions have been given in Eqs 371 and 372 We now consider an interface between two general media shown in Fig 323 Let us construct a small path abcda with sides ab and cd in media 1 and 2 respectively both being parallel to the interface and equal to Aw Equation 38 is applied to this path If we let sides be da Ah approach zero their contribu tions to the line integral of E around the path can be neglected We have abcda E d E1 Aw E2 Aw EuAw E2tAw 0 Therefore EU E it Vm 3118 which states that the tangential component of an E field is continuous across an inter face Eq 3118 simplifies to Eq 371 if one of the media is a conductor When media 1 and 2 are dielectrics with permittivities ex and e2 respectively we have Du D it 3119 In order to find a relation between the normal components of the fields at a boundary we construct a small pillbox with its top face in medium 1 and bottom FIGURE 323 An interface between two media 39 Boundary Conditions for Electrostatic Fields 117 face in medium 2 as illustrated in Fig 323 The faces have an area AS and the height of the pillbox Ah is vanishingly small Applying Gausss law Eq 3100 to the pillbox1 we have Dds D1all2 D2aII1AS a 2D 1D 2AS pAS 3120 where we have used the relation a2 a n l Unit vectors aHl and a2 are respec tively outward unit normals from media 1 and 2 From Eq 3120 we obtain a 2 D 1 D 2 pJ or Dln D2n Ps Cm2 3121a 3121b where the reference unit normal is outward from medium 2 Eq 3121b states that the normal component ofD field is discontinuous across an interface where a surface charge existsthe amount of discontinuity being equal to the surface charge density If medium 2 is a conductor D 2 0 and Eq 3121b becomes Dm ii Ps 3122 which simplifies to Eq 372 when medium 1 is free space When two dielectrics are in contact with no free charges at the interface ps 0 we have Dln D2n 3123 or iEln e2E2n 3124 Recapitulating we find that the boundary conditions that must be satisfied for static electric fields are as follows Tangential components Eu E2t Normal components a2 Dt D2 ps 3125 3126 EXAMPLE 314 A lucite sheet er 32 is introduced perpendicularly in a uniform electric field E0 axE0 in free space Determine E D and P inside the lucite f Equations 38 and 3100 are assumed to hold for regions containing discontinuous media See C T Tai On the presentation of Maxwells theory Proceedings of the IEEE vol 60 pp 936945 August 1972 118 3 Static Electric Fields E0 XE0 D0 axeoEx Free space EL Lucite cr 32 E0 Free space FIGURE 324 A lucite sheet in a uniform electric field Example 314 Solution We assume that the introduction of the lucite sheet does not disturb the original uniform electric field E0 The situation is depicted in Fig 324 Since the interfaces are perpendicular to the electric field only the normal field components need be considered No free charges exist Boundary condition Eq 3123 at the left interface gives or Bt axe0E0 There is no change in electric flux density across the interface The electric field intensity inside the lucite sheet is The polarization vector is zero outside the lucite sheet P0 0 Inside the sheet 1 Pi Bi0Ei axn 0E0 ax06S15eoEo Cm2 mm Clearly a similar application of the boundary condition Eq 3123 on the right interface will yield the original E0 and D0 in the free space on the right of the lucite sheet Does the solution of this problem change if the original electric field is not uniform that is if E0 axj EXAMPLE 315 Two dielectric media with permittivities ex and e2 are separated by a chargefree boundary as shown in Fig 325 The electric field intensity in medium 1 at the point Px has a magnitude Et and makes an angle ocx with the normal Determine the magnitude and direction of the electric field intensity at point P2 in medium 2 119 FIGURE 325 Boundary conditions at the interface between two dielectric media Example 315 Solution Two equations are needed to solve for two unknowns E2t and E2n After E2t and E2 have been found E2 and a2 will follow directly Using Eqs 3118 and 3123 we have E2 sin a2 E1 sin at 3127 and e2E2 cos a2 e1E1 cos a1 3128 Division of Eq 3127 by Eq 3128 gives 3129 The magnitude of E2 is E2 yElt Ej JE2 sin a22 E2 cos a22 E1 sin a j 2 I Et cos at I 2l2 or 3130 By examining Fig 325 can you tell whether e1 is larger or smaller than e2 EXAMPLE 316 When a coaxial cable is used to carry electric power the radius of the inner conductor is determined by the load current and the overall size by the voltage and the type of insulating material used Assume that the radius of the inner conductor is 04 cm and that concentric layers of rubber err 32 and polystyrene erp 26 are used as insulating materials Design a cable that is to work at a voltage 120 3 Static Electric Fields rating of 20 kV In order to avoid breakdown due to voltage surges caused by lightning and other abnormal external conditions the maximum electric field inten sities in the insulating materials are not to exceed 25 of their dielectric strengths Solution From Table 31 p 114 we find the dielectric strengths of rubber and polystyrene to be 25 x 106 Vm and 20 x 106 Vm respectively Using Eq 340 for specified 25 of dielectric strengths we have the following In rubber Max Er 025 x 25 x 106 In polystyrene Max E 025 x 20 x 106 1 Pe 2e0 32rt Pt 1 3131a 3131b 27re0 26rp Combination of Eqs 313la and 313lb yields rp 154r 0616 cm 3132 Equation 3132 indicates that the insulating layer of polystyrene should be placed outside of that of rubber as shown in Fig 326a It would be interesting to deter mine what would happen if the polystyrene layer were placed inside the rubber layer v 625 S 500 2 406 o n r P ro r b FIGURE 326 Cross section of coaxial cable with two different kinds of insulating material Example 316 310 Capacitance and Capacitors 121 The cable is to work at a potential difference of 20000 V between the inner and outer conductors We set rp Epdr n Erdr 20000 where both Ep and Er have the form given in Eq 340 The above relation leads to Pi f1 l n l n U 20000 or 2TL0 rp rp err r 1 In T i In 154 1 20000 3133 2ne0 26 154rf 32 Since rt 04 cm is given r0 can be determined by finding the factor pJ2n0 from Eq 313la and then using it in Eq 3133 We obtain p2ne0 8 x 104 and ro 208r 0832 cm In Figs 326b and 326c are plotted the variations of the radial electric field intensity E and the potential V referred to that of the outer sheath Note that E has discontinuous jumps while the V curve is continuous The reader should verify all the indicated numerical values 310 Capacitance and Capacitors From Section 36 we understand that a conductor in a static electric field is an equipotential body and that charges deposited on a conductor will distribute them selves on its surface in such a way that the electric field inside vanishes Suppose the potential due to a charge Q is V Obviously increasing the total charge by some factor k would merely increase the surface charge density ps everywhere by the same factor without affecting the charge distribution because the conductor remains an equipo tential body in a static situation We may conclude from Eq 362 that the potential of an isolated conductor is directly proportional to the total charge on it This may also be seen from the fact that increasing V by a factor of k increases E V by a factor of k But from Eq 372 E apse0 it follows that ps and consequently the total charge Q will also increase by a factor of k The ratio QV therefore remains unchanged We write Q cv 3134 where the constant of proportionality C is called the capacitance of the isolated con ducting body The capacitance is the electric charge that must be added to the body per unit increase in its electric potential Its SI unit is coulomb per volt or farad F Of considerable importance in practice is the capacitor which consists of two conductors separated by free space or a dielectric medium The conductors may be of arbitrary shapes as in Fig 327 When a dc voltage source is connected between the conductors a charge transfer occurs resulting in a charge Q on one conductor 122 3 Static Electric Fields FIGURE 327 A twoconductor capacitor and Q on the other Several electric field lines originating from positive charges and terminating on negative charges are shown in Fig 327 Note that the field lines are perpendicular to the conductor surfaces which are equipotential surfaces Equa tion 3134 applies here if V is taken to mean the potential difference between the two conductors V12 That is 3135 The capacitance of a capacitor is a physical property of the twoconductor system It depends on the geometry of the conductors and on the permittivity of the medium between them it does not depend on either the charge Q or the potential difference V12 A capacitor has a capacitance even when no voltage is applied to it and no free charges exist on its conductors Capacitance C can be determined from Eq 3135 by either 1 assuming aF 1 2 and determining Q in terms of V12 or 2 assuming a Q and determining V12 in terms of Q At this stage since we have not yet studied the methods for solving boundaryvalue problems which will be taken up in Chapter 4 we find C by the second method The procedure is as follows 1 Choose an appropriate coordinate system for the given geometry 2 Assume charges Q and Q on the conductors 3 Find E from Q by Eq 3122 Gausss law or other relations 4 Find Vl2 by evaluating V12fEd from the conductor carrying Q to the other carrying Q 5 Find C by taking the ratio QV12 310 Capacitance and Capacitors 123 EXAMPLE 317 A parallelplate capacitor consists of two parallel conducting plates of area S separated by a uniform distance d The space between the plates is filled with a dielectric of a constant permittivity e Determine the capacitance Solution A cross section of the capacitor is shown in Fig 328 It is obvious that the appropriate coordinate system to use is the Cartesian coordinate system Follow ing the procedure outlined above we put charges Q and Q on the upper and lower conducting plates respectively The charges are assumed to be uniformly dis tributed over the conducting plates with surface densities ps and ps where From Eq 3122 we have E Q which is constant within the dielectric if the fringing of the electric field at the edges of the plates is neglected Now njrSMaS Therefore for a parallelplate capacitor C Q S 3136 which is independent of Q or V12 For this problem we could have started by assuming a potential difference V12 between the upper and lower plates The electric field intensity between the plates is uniform and equals E a V 12 d Dielectric permittivity e zzz Area S FIGURE 328 Cross section of a parallelplate capacitor Example 317 124 3 Static Electric Fields The surface charge densities at the upper and lower conducting plates are ps and ps respectively where in view of Eq 372 Ps eEy eVf Therefore Q psS eSdVl2 and C QVl2 eSd as before EXAMPLE 318 A cylindrical capacitor consists of an inner conductor of radius a and an outer conductor whose inner radius is b The space between the conductors is filled with a dielectric of permittivity e and the length of the capacitor is L Deter mine the capacitance of this capacitor Solution We use cylindrical coordinates for this problem First we assume charges Q and Q on the surface of the inner conductor and the inner surface of the outer conductor respectively The E field in the dielectric can be obtained by applying Gausss law to a cylindrical Gaussian surface within the dielectric a r b Note that Eq 3122 gives only the normal component of the E field at a conductor surface Since the conductor surfaces are not planes here the E field is not constant in the dielectric and Eq 3122 cannot be used to find E in the a r b region Referring to Fig 329 and applying Gausss law we have E arr a r 2 3137 zneLr Again we neglect the fringing effect of the field near the edges of the conductors The potential difference between the inner and outer conductors is 3138 e jr 2neL Dielectric e FIGURE 329 A cylindrical capacitor Example 318 310 Capacitance and Capacitors Therefore for a cylindrical capacitor 125 3139 We could not solve this problem from an assumed Vab because the electric field is not uniform between the inner and outer conductors Thus we would not know how to express E and Q in terms of Vab until we learned how to solve such a boundary value problem EXAMPLE 319 A spherical capacitor consists of an inner conducting sphere of radius R and an outer conductor with a spherical inner wall of radius R0 The space in between is filled with a dielectric of permittivity e Determine the capacitance Solution Assume charges Q and Q on the inner and outer conductors respec tively of the spherical capacitor in Fig 330 Applying Gausss law to a spherical Gaussian surface with radius RRt R R0 we have E aRR aR Q 4neR Therefore for a spherical capacitor Q dRQ f1 4neR2 4TT i v j Ane 1 3140 For an isolated conducting sphere of a radius Ru R0 oo C AneRt FIGURE 330 A spherical capacitor Example 319 126 3 Static Electric Fields d Q c2 Q QQ Q gitgl e Q FIGURE 331 Series connection of capacitors 3101 SERIES AND PARALLEL CONNECTIONS OF CAPACITORS Capacitors are often combined in various ways in electric circuits The two basic ways are series and parallel connections In the series or headtotail connection shown in Fig 331T the external terminals are from the first and last capacitors only When a potential difference or electrostatic voltage V is applied charge cumulations on the conductors connected to the external terminals are Q and Q Charges will be induced on the internally connected conductors such that Q and Q will appear on each capacitor independently of its capacitance The potential differences across the individual capacitors are QCl5 QC2 QC and Q Q Q Q F sr 1 2 Cw where Csr is the equivalent capacitance of the seriesconnected capacitors We have 3141 In the parallel connection of capacitors the external terminals are connected to the conductors of all the capacitors as in Fig 332 When a potential difference V is applied to the terminals the charge cumulated on a capacitor depends on its capacitance The total charge is the sum of all the charges Q 2i Qi Qn c1v c2v cnvcllv Therefore the equivalent capacitance of the parallelconnected capacitors is 1 sr 1 Ci 1 c2 1 cn CM Cx C2 C 3142 f Capacitors whatever their actual shape are conventionally represented in circuits by pairs of parallel bars 127 FIGURE 332 Parallel connection of capacitors We note that the formula for the equivalent capacitance of seriesconnected capacitors is similar to that for the equivalent resistance of parallelconnected resistors and that the formula for the equivalent capacitance of parallelconnected capacitors is similar to that for the equivalent resistance of seriesconnected resistors Can you explain this EXAMPLE 320 Four capacitors Cx 1 F C2 2 F C3 3 F and C4 4 JAF are connected as in Fig 333 A dc voltage of 100 V is applied to the external terminals ab Determine the following a the total equivalent capacitance between terminals ab b the charge on each capacitor and c the potential difference across each capacitor Q C2 I I C VV v2 c3 Vi 100F c4 V4 FIGURE 333 A combination of capacitors Example 320 3 Static Electric Fields Qi Q2 Ci C2 Q1 Q C3 C4 Q2 Qi c3 100 24 Solution a The equivalent capacitance C12 of Cx and C2 in series is C 1 2 2 12 1CJ 1C2 C1 C2 3 The combination of C12 in parallel with C3 gives C123 C12 C 3 OxF The total equivalent capacitance Cab is then C 7 r 1913 F b Since the capacitances are given the voltages can be found as soon as the charges have been determined We have four unknowns Qx Q2 Q3 and 24 Four equa tions are needed for their determination Series connection of Cx and C2 Qx Q2 Kirchhoffs voltage law V1 V2 V3 Kirchhoffs voltage law V3 V4 100 Series connection at d Using the given values of Cl5 C2 C3 and C4 and solving the equations we obtain 81 62 348 o n 63 1565 W 4400 64 231913 QiQ c Dividing the charges by the capacitances we find V 348 V F2 1 7 4 V F3 522 V F4 478 V c 4 These results can be checked by verifying that V1 V2 V3 and that V3 V4 100 V n 310 Capacitance and Capacitors 129 3102 CAPACITANCES IN MULTICONDUCTOR SYSTEMS We now consider the situation of more than two conducting bodies in an isolated system such as that shown in Fig 334 The positions of the conductors are arbi trary and one of the conductors may represent the ground Obviously the presence of a charge on any one of the conductors will affect the potential of all the others Since the relation between potential and charge is linear we may write the following set of N equations relating the potentials Vx V2 VN of the N conductors to the charges Qu Q2 QN i PnQi P12Q2 PINQN Vl P2lQl PllQl P2NQN VN PmQi PN2Q2 PNNQN In Eqs 3143 the pys are called the coefficients of potential which are constants whose values depend on the shape and position of the conductors as well as the permittivity of the surrounding medium We note that in an isolated system 6i Q2 23 QN 0 3144 The N linear equations in 3143 can be inverted to express the charges as functions of potentials as follows 61 c11V1 c12V2 c1NVN Q2 c217i c22V2 c2NVN QN cN1V cN2V2 cNNVN where the cys are constants whose values depend only on the pys in Eqs 3143 The coefficients cHs are called the coefficients of capacitance which equal the ratios of the charge Qt on and the potential Vt of the ith conductor i 1 2 N with all other conductors grounded The cys i j are called the coefficients of induction If a positive Qt exists on the ith conductor Vt will be positive but the charge Qj 0 2 I FIGURE 334 A multiconductor system 130 3 Static Electric Fields induced on theth i conductor will be negative Hence the coefficients of capac itance cu are positive and the coefficients of induction ctJ are negative The condition of reciprocity guarantees that ptJ pfi and ctJ cn To establish a physical meaning to the coefficients of capacitance and the coef ficients of induction let us consider a fourconductor system as depicted in Fig 334 with the stipulation that the conductor labeled N is now the conducting earth at zero potential and is designated by the number 0 A schematic diagram of the four conductor system is shown in Fig 335 in which the conductors 1 2 and 3 have been drawn as simple dots nodes Coupling capacitances have been shown between pairs of nodes and between the three nodes and the ground If Qu Q2 Q3 and VXV2 V3 denote the charges and the potentials respectively of conductors 1 2 and 3 the first three equations in 3145 become Qi c11V1 c12V2 c13V3 Q2 c12V1c22V2 c23V3 Qi c137i c23V2 c33V3 3146a 3146b 3146c where we have used the symmetry relation ctj cn On the other hand we can write another set of three Q V relations based on the schematic diagram in Fig 335 Q C10V CX2VX V2 C13VX V3 Qi c20v2 c12v2 vx c23v2 v3 Q3 C30V3 C13V3 Vx C23V3 V2 3147a 3147b 3147c where C10C20 and C30 are selfpartial capacitances and CyOy are mutual partial capacitances Equations 3147a 3147b and 3147c can be rearranged as Gi c10 c12 c 1 3m c12v2 c13v3 Qi c12v c20 c12 c23v2 c23v3 Qi cv c23v2 c30 c13 c23v3 3148a 3148b 3148c FIGURE 335 Schematic diagram of three conductors and the ground 310 Capacitance and Capacitors 131 Comparing Eqs 3148 with Eqs 3146 we obtain C l l 10 12 1 3 C22 20 12 2 3 C33 30 Cl3 2 3 and C12 1 2 C23 C 2 3 C13 C 1 3 3149a 3149b 3149c 3150a 3150b 3150c On the basis of Eq 3149a we can interpret the coefficient of capacitance c n as the total capacitance between conductor 1 and all the other conductors connected together to ground similarly for c22 and c33 Equations 3150 indicate that the coefficients of inductances are the negative of the mutual partial capacitances In verting Eqs 3149 we can express the conductortoground capacitances in terms of the coefficients of capacitance and coefficients of induction ClO CU C12 C13 C20 C22 C12 C23 C30 c33 4 c13 c23 3151a 3151b 3151c EXAMPLE 321 Three horizontal parallel conducting wires each of radius a and isolated from the ground are separated from one another as shown in Fig 336 As suming d a determine the partial capacitances per unit length between the wires Solution We designate the wires as conductors 01 and 2 as indicated in Fig 336 Choosing conductor 0 as the reference and using Eq 3138 we can write two equa tions for the potential differences V10 and V20 due to the three wires as follows or 10 2ne0 d 2ne0 a 2ne0 2d T 1 a 1 d 3 2ii0V10 p0 In pn In pe2 In d a 2 3152a FIGURE 336 Three parallel wires Example 321 1 3 2 3 Static Electric Fields where peo pn and pf2 denote the charges per unit length on wires 0 1 and 2 respectively Similarly 1 a d 3d 2n0V20 p0 In Pn In pe2 In 3l52b 3d 2d a For the isolated system of three conductors we have peo pn pe2 0 or Peo iPn Pn 3153 Combination of Eqs 3l52a 3l52b and 3153 yields 2ne0V10 pn2 In pe2 In 3154a a la 2ne0V20 pn In pe22 In 3154b Equations 3154a and 3154b can be used to solve for pn and pe2 as functions of V10 and V20 Pn A0 F102 In V20 In j 3155a where Pei A0 F10 In F202 In j 3155b A o T 7 r V w 3156 4 In In In a a 2a Comparing Eqs 3155 with Eqs 3146 3148 and 3151 we obtain the fol lowing partial capacitances per unit length for the given threewire system C i 2 c i 2 A 0ln 3157a 2a 3d 3d Cio en c12 A0f 2 In Inj 3157b C20 c22 c12 A 0 2 I n I n j 3157c 3103 ELECTROSTATIC SHIELDING Electrostatic shielding a technique for reducing capacitive coupling between con ducting bodies is important in some practical applications Let us consider the situation shown in Fig 337 in which a grounded conducting shell 2 completely 133 FIGURE 337 Illustrating electrostatic shielding encloses conducting body 1 Setting V2 0 in Eq 3147a we have 2i C107 C12V C13V V3 3158 When Qx 0 there is no field inside shell 2 hence body 1 and shell 2 have the same potential V1 V2 0 From Eq 3158 we see that the coupling capacitance C13 must vanish since V3 is arbitrary This means that a change in V3 will not affect Qlt and vice versa We then have electrostatic shielding between conducting bodies 1 and 3 Obviously the same shielding effectiveness is obtained if the grounded con ducting shell 2 encloses body 3 instead of body 1 311 Electrostatic Energy and Forces In Section 35 we indicated that electric potential at a point in an electric field is the work required to bring a unit positive charge from infinity at reference zero potential to that point To bring a charge Q2 slowly so that kinetic energy and radiation effects may be neglected from infinity against the field of a charge Q1 in free space to a distance R12 the amount of work required is w Q v QJki 3159 Because electrostatic fields are conservative W2 is independent of the path followed by Q2 Another form of Eq 3159 is This work is stored in the assembly of the two charges as potential energy Combining Eqs 3159 and 3160 we can write W2QiV1Q2V2 3161 Now suppose another charge Q3 is brought from infinity to a point that is R13 from Q1 and R23 from Q2 an additional amount of work is required that equals W Q3V3 QJQ J 3162 47ce0i13 4n0R23J 134 3 Static Electric Fields The sum of AW in Eq 3162 and W2 in Eq 3159 is the potential energy W3 stored in the assembly of the three charges Qu Q2 and Q3 That is W3 W2 AW 1 4ne0 V R Q1Q2 2i23 Q2Q3 L12 i R 23 3163 We can rewrite W3 in the following form W Qi Q Q 4TZ0R12 61 Qi 4013 Q2 Q 61 Qi 4ne0Rl2 4ne0R23 K4ne0Rl3 4TZ0R23 HQiV1 Q2V2 Q3V3 3164 In Eq 3164 Vlt the potential at the position of Qu is caused by charges Q2 and Q3 it is different from the Vt in Eq 3160 in the twocharge case Similarly V2 and V3 are the potentials at Q2 and 23 respectively in the threecharge assembly Extending this procedure of bringing in additional charges we arrive at the following general expression for the potential energy of a group of N discrete point charges at rest The purpose of the subscript e on We is to denote that the energy is of an electric nature We have 3165 where Vk the electric potential at Qk is caused by all the other charges and has the following expression 47T60 li Rv vk 3166 Vk U Two remarks are in order here First We can be negative For instance W2 in Eq 3159 will be negative if Qt and Q2 are of opposite signs In that case work is done by the field not against the field established by Qt in moving Q2 from infinity Second We in Eq 3165 represents only the interaction energy mutual energy and does not include the work required to assemble the individual point charges them selves selfenergy The SI unit for energy joule J is too large a unit for work in physics of elemen tary particles where energy is more conveniently measured in terms of a much smaller unit called electronvolt eV An electronvolt is the energy or work required to move an electron against a potential difference of one volt 1 eV 160 x 1019 x 1 160 x 10 119 J 3167 Energy in eV is essentially that in J per unit electronic charge The proton beams of the worlds most powerful highenergy particle accelerator collide with a kinetic 311 Electrostatic Energy and Forces 135 energy of two trillion electronvolts 2 TeV or 2 x 1012 x 160 x 1019 320 x 107J A binding energy of 5x 1019J in an ionic crystal is equal to We 5 x 1019160 x 1019 3125 eV which is a more convenient number to use than the one in terms of joules EXAMPLE 322 Find the energy required to assemble a uniform sphere of charge of radius b and volume charge density p Solution Because of symmetry it is simplest to assume that the sphere of charge is assembled by bringing up a succession of spherical layers of thickness dR At a radius R shown in Fig 338 the potential is 4ne0R where QR is the total charge contained in a sphere of radius R QR 4nR3 The differential charge in a spherical layer of thickness dR is dQR p4nR2 dR and the work or energy in bringing up dQR is 4n dWVRdQR p2R4dR 3e0 Hence the total work or energy required to assemble a uniform sphere of charge of radius b and charge density p is W lm 4Rgi J 3168 In terms of the total charge 3e0 J o 15e0 epy3 FIGURE 338 Assembling a uniform sphere of charge Example 322 3 Static Electric Fields we have W 32 2 20neob J 3169 Equation 3169 shows that the energy is directly proportional to the square of the total charge and inversely proportional to the radius The sphere of charge in Fig 338 could be a cloud of electrons for instance mm For a continuous charge distribution of density p the formula for We in Eq 3165 for discrete charges must be modified Without going through a separate proof we replace Qk by p dv and the summation by an integration and obtain vpVdv J W 3170 In Eq 3170 V is the potential at the point where the volume charge density is p and V is the volume of the region where p exists EXAMPLE 323 Solve the problem in Example 322 by using Eq 3170 Solution In Example 322 we solved the problem of assembling a sphere of charge by bringing up a succession of spherical layers of a differential thickness Now we assume that the sphere of charge is already in place Since p is a constant it can be taken out of the integral sign For a spherically symmetrical problem W 2 Jv 7 Jo 3171 where V is the potential at a point R from the center To find V at R we must find the negative of the line integral of E in two regions 1 E aRER1 from R oo to R b and 2 E2 aRER2 from R b to R R We have and Ejti R Eja aR Q 4n0R2 QR a Rb 4ne0R2 a R 3e0 Pb3 3e 0 2 0Rb Consequently we obtain V J E dR M ER1 dR jb R ER2 dR pb J 3e0R2 Jt RdR b 3e0 3172 311 Electrostatic Energy and Forces 137 Substituting Eq 3172 in Eq 3171 we get e 2 Jo 3e0 V2 2 15e0 which is the same as the result in Eq 3168 m Note that We in Eq 3170 includes the work selfenergy required to assemble the distribution of macroscopic charges because it is the energy of interaction of every infinitesimal charge element with all other infinitesimal charge elements As a matter of fact we have used Eq 3170 in Example 323 to find the selfenergy of a uniform spherical charge As the radius b approaches zero the selfenergy of a mathematical point charge of a given Q is infinite see Eq 3169 The selfenergies of point charges Qk are not included in Eq 3165 Of course there are strictly no point charges inasmuch as the smallest charge unit the electron is itself a distribution of charge 3111 ELECTROSTATIC ENERGY IN TERMS OF FIELD QUANTITIES In Eq 3170 the expression of electrostatic energy of a charge distribution contains the source charge density p and the potential function V We frequently find it more convenient to have an expression of We in terms of field quantities E andor D without knowing p explicitly To this end we substitute V D for p in Eq 3170 We jvBVdv 3173 Now using the vector identity from Problem P228 V VD V D D V 3174 we can write Eq 3173 as We iLVDdviLTVdv I r 3 175 ijsiVDndsivBEdv where the divergence theorem has been used to change the first volume integral into a closed surface integral and E has been substituted for W in the second volume integral Since V can be any volume that includes all the charges we may choose it to be a very large sphere with radius R As we let R oo electric potential V and the magnitude of electric displacement D fall off at least as fast as lR and lR2 respectively1 The area of the bounding surface S increases as R2 Hence the surface integral in Eq 3175 decreases at least as fast as lR and will vanish as R oo We are then left with only the second integral on the right side of Eq 3175 For point charges V oc lR and D oc lR2 for dipoles V oc lR2 and D oc lR3 138 3 Static Electric Fields W jvDEdv J 3176a Using the relation D eE for a linear medium Eq 3176a can be written in two other forms W jveE2dv J and 3176b 3176c We can always define an electrostatic energy density we mathematically such that its volume integral equals the total electrostatic energy We can therefore write or or We vwedv we Y E Jm3 we eE2 Jm2 D2 we Jm3 3177 3178a 3178b 3178c However this definition of energy density is artificial because a physical justification has not been found to localize energy with an electric field all we know is that the volume integrals in Eqs 3176a b c give the correct total electrostatic energy EXAMPLE 324 In Fig 339 a parallelplate capacitor of area S and separation d is charged to a voltage V The permittivity of the dielectric is e Find the stored electrostatic energy Solution With the dc source batteries connected as shown the upper and lower plates are charged positive and negative respectively If the fringing of the field at Area5 K r 1 FIGURE 339 A charged parallelplate capacitor Example 324 311 Electrostatic Energy and Forces 139 the edges is neglected the electric field in the dielectric is uniform over the plate and constant across the dielectric and has a magnitude E Using Eq 3176b we have iGHrK5 3179 The quantity in the parentheses of the last expression eSd is the capacitance of the parallelplate capacitor see Eq 3136 So We CV2 J Since Q CV Eq 3180a can be put in two other forms We iQV J and Q2 w lc J 3180a 3180b 3180c It so happens that Eqs 3180a b c hold true for any twoconductor capacitor see Problem P343 EXAMPLE 325 Use energy formulas 3176 and 3180 to find the capacitance of a cylindrical capacitor having a length L an inner conductor of radius a an outer conductor of inner radius b and a dielectric of permittivity e as shown in Fig 329 Solution By applying Gausss law we know that Q E zrEr ar IneLr a r b The electrostatic energy stored in the dielectric region is from Eq 3176b 4f edy 2 L 2 Q2 cb dr Q2 i b 4mL Jo r 4neL a 3181 140 3 Static Electric Fields On the other hand We can also be expressed in the form of Eq 3180c Equating 3180c and 3181 we obtain jQL b 2C AneL a or 2TTL a which is the same as that given in Eq 3139 3112 ELECTROSTATIC FORCES Coulombs law governs the force between two point charges In a more complex sys tem of charged bodies using Coulombs law to determine the force on one of the bodies that is caused by the charges on other bodies would be very tedious This would be so even in the simple case of finding the force between the plates of a charged parallelplate capacitor We will now discuss a method for calculating the force on an object in a charged system from the electrostatic energy of the system This method is based on the principle of virtual displacement We will consider two cases 1 that of an isolated system of bodies with fixed charges and 2 that of a system of conduct ing bodies with fixed potentials System of Bodies with Fixed Charges We consider an isolated system of charged conducting as well as dielectric bodies separated from one another with no connec tion to the outside world The charges on the bodies are constant Imagine that the electric forces have displaced one of the bodies by a differential distance d a virtual displacement The mechanical work done by the system would be dW FQ d 3182 where FQ is the total electric force acting on the body under the condition of constant charges Since we have an isolated system with no external supply of energy this mechanical work must be done at the expense of the stored electrostatic energy that is dW dWe Qd 3183 Noting from Eq 288 in Section 26 that the differential change of a scalar resulting from a position change d is the dot product of the gradient of the scalar and d we write dWe VWQ d 3184 Since d is arbitrary comparison of Eqs 3183 and 3184 leads to QWe N 3185 311 Electrostatic Energy and Forces 141 Equation 3185 is a very simple formula for the calculation of FQ from the electro static energy of the system In Cartesian coordinates the component forces are dWe dx dWe dy FQX T 3186a Foy 1TT 3186b FQI TT 3186c If the body under consideration is constrained to rotate about an axis say the zaxis the mechanical work done by the system for a virtual angular displacement df would be dW TQzdj 3187 where TQZ is the zcomponent of the torque acting on the body under the condition of constant charges The foregoing procedure will lead to 3188 System of Conducting Bodies with Fixed Potentials Now consider a system in which conducting bodies are held at fixed potentials through connections to such external sources as batteries Uncharged dielectric bodies may also be present A displacement d6 by a conducting body would result in a change in total electrostatic energy and would require the sources to transfer charges to the conductors in order to keep them at their fixed potentials If a charge dQk which may be positive or negative is added to the feth conductor that is maintained at potential Vk the work done or energy sup plied by the sources is VkdQk The total energy supplied by the sources to the system is dWJVdQ 3189 k The mechanical work done by the system as a consequence of the virtual displace ment is dW v d 3190 where v is the electric force on the conducting body under the condition of constant potentials The charge transfers also change the electrostatic energy of the system by an amount dWe which in view of Eq 3165 is dWe 1YjVkdQk ldWs 3191 3 Static Electric Fields Conservation of energy demands that dW dWe dWs Substitution of Eqs 3189 3190 and 3191 in Eq 3192 gives Fvd dWe Wedt or 3192 F Wa N 3193 Comparison of Eqs 3193 and 3185 reveals that the only difference between the formulas for the electric forces in the two cases is in the sign It is clear that if the con ducting body is constrained to rotate about the zaxis the zcomponent of the electric torque will be 3194 which differs from Eq 3188 also only by a sign change EXAMPLE 326 Determine the force on the conducting plates of a charged parallel plate capacitor The plates have an area S and are separated in air by a distance x Solution We solve the problem in two ways a by assuming fixed charges and then b by assuming fixed potentials The fringing of field around the edges of the plates will be neglected a Fixed charges With fixed charges Q on the plates an electric field intensity Ex Qe0S Vx exists in the air between the plates regardless of their separa tion unchanged by a virtual displacement From Eq 3180b where Q and Ex are constants Using Eq 3186a we obtain 8x ie 4 e 2 3195 where the negative signs indicate that the force is opposite to the direction of increasing x It is an attractive force b Fixed potentials With fixed potentials it is more convenient to use the expression in Eq 3180a for We Capacitance C for the parallelplate air capacitor is e0Sx We have from Eq 3193 dI L l cvA v fas e SV2 dx 8x Fyx 2 2 dx 2x 3196 Review Questions 143 How different are FQX in Eq 3195 and Fvx in Eq 3196 Recalling the relation x we find W Fvx 3197 The force is the same in both cases in spite of the apparent sign difference in the for mulas as expressed by Eqs 3185 and 3193 A little reflection on the physical problem will convince us that this must be true Since the charged capacitor has fixed dimensions a given Q will result in a fixed V and vice versa Therefore there is a unique force between the plates regardless of whether Q or V is given and the force certainly does not depend on virtual displacements A change in the conceptual con straint fixed Q or fixed V cannot change the unique force between the plates The preceding discussion holds true for a general charged twoconductor capaci tor with capacitance C The electrostatic force e in the direction of a virtual displace ment A for fixed charges is dWe d Q2 Q2 dC Fwvyhw 3198 For fixed potentials 2 2 d 2C2 Vv C V 1 W 3 3199 It is clear that the forces calculated from the two procedures which assumed different constraints imposed on the same charged capacitor are equal Review Questions R31 Write the differential form of the fundamental postulates of electrostatics in free space R32 Under what conditions will the electric field intensity be both solenoidal and irrotational R33 Write the integral form of the fundamental postulates of electrostatics in free space and state their meaning in words R34 When the formula for the electric field intensity of a point charge Eq 312 was derived a why was it necessary to stipulate that q is in a boundless free space b why did we not construct a cubic or a cylindrical surface around ql R35 In what ways does the electric field intensity vary with distance for a a point charge b an electric dipole 3 Static Electric Fields R36 State Coulombs law R37 Explain the principle of operation of inkjet printers R38 State Gausss law Under what conditions is Gausss law especially useful in determining the electric field intensity of a charge distribution R39 Describe the ways in which the electric field intensity of an infinitely long straight line charge of uniform density varies with distance R310 Is Gausss law useful in finding the E field of a finite line charge Explain R311 See Example 36 Fig 39 Could a cylindrical pillbox with circular top and bottom faces be chosen as a Gaussian surface Explain R312 Make a twodimensional sketch of the electric field lines and the equipotential lines of a point charge R313 At what value of 9 is the E field of a zdirected electric dipole pointed in the negative zdirection R314 Refer to Eq 364 Explain why the absolute sign around z is required R315 If the electric potential at a point is zero does it follow that the electrical field intensity is also zero at that point Explain R316 If the electric field intensity at a point is zero does it follow that the electric potential is also zero at that point Explain R317 If an uncharged spherical conducting shell of a finite thickness is placed in an external electric field E what is the electric field intensity at the center of the shell Describe the charge distributions on both the outer and the inner surfaces of the shell R318 What are electrets How can they be made R319 Can VlR in Eq 384 be replaced by Rl Explain R320 Define polarization vector What is its SI unit R321 What are polarization charge densities What are the SI units for P a and V P R322 What do we mean by simple medium R323 What properties do anisotropic materials have R324 What characterizes a uniaxial medium R325 Define electric displacement vector What is its SI unit R326 Define electric susceptibility What is its unit R327 What is the difference between the permittivity and the dielectric constant of a medium R328 Does the electric flux density due to a given charge distribution depend on the properties of the medium Does the electric field intensity Explain R329 What is the difference between the dielectric constant and the dielectric strength of a dielectric material R330 Explain the principle of operation of lightning arresters R331 What are the general boundary conditions for electrostatic fields at an interface between two different dielectric media R332 What are the boundary conditions for electrostatic fields at an interface between a conductor and a dielectric with permittivity e Problems 145 R333 What is the boundary condition for electrostatic potential at an interface between two different dielectric media R334 Does a force exist between a point charge and a dielectric body Explain R335 Define capacitance and capacitor R336 Assume that the permittivity of the dielectric in a parallelplate capacitor is not constant Will Eq 3136 hold if the average value of permittivity is used for e in the formula Explain R337 Given three 1F capacitors explain how they should be connected in order to obtain a total capacitance of aiF bfF cfF d3F R338 What are coefficients of potential coefficients of capacitance and coefficients of induction R339 What are partial capacitances How are they different from coefficients of capacitance R340 Explain the principle of electrostatic shielding R341 What is the definition of an electronvoltl How does it compare with a joule R342 What is the expression for the electrostatic energy of an assembly of four discrete point charges R343 What is the expression for the electrostatic energy of a continuous distribution of charge in a volume on a surface along a line R344 Provide a mathematical expression for electrostatic energy in terms of E andor D R345 Discuss the meaning and use of the principle of virtual displacement R346 What is the relation between the force and the stored energy in a system of stationary charged objects under the condition of constant charges Under the condition of fixed potentials Problems P31 Refer to Fig 34 a Find the relation between the angle of arrival a of the electron beam at the screen and the deflecting electric field intensity d b Find the relation between w and L such that d d020 P32 The cathoderay oscilloscope CRO shown in Fig 34 is used to measure the voltage applied to the parallel deflection plates a Assuming no breakdown in insulation what is the maximum voltage that can be measured if the distance of separation between the plates is K b What is the restriction on L if the diameter of the screen is D c What can be done with a fixed geometry to double the CROs maximum measurable voltage P33 The deflection system of a cathoderay oscilloscope usually consists of two pairs of parallel plates producing orthogonal electric fields Assume the presence of another set of plates in Fig 34 that establishes a uniform electric field Ex axEx in the deflection region Deflection voltages vxt and vyt are applied to produce Ex and E respectively Determine 3 Static Electric Fields the types of waveforms that vxt and vyt should have if the electrons are to trace the following graphs on the fluorescent screen a a horizontal line b a straight line having a negative unity slope c a circle d two cycles of a sine wave P34 Write a short article explaining the principle of operation of xerography Use library resources if needed P35 Two point charges Qt and Q2 are located at 1 2 0 and 2 0 0 respectively Find the relation between Qt and Q2 such that the total force on a test charge at the point P110 will have a no xcomponent b no ycomponent P36 Two very small conducting spheres each of a mass 10 x 104 kg are suspended at a common point by very thin nonconducting threads of a length 02 m A charge Q is placed on each sphere The electric force of repulsion separates the spheres and an equilibrium is reached when the suspending threads make an angle of 10 Assuming a gravitational force of 980 Nkg and a negligible mass for the threads find Q P37 Find the force between a charged circular loop of radius b and uniform charge density pe and a point charge Q located on the loop axis at a distance h from the plane of the loop What is the force when h b and when h 01 Plot the force as a function of h P38 A line charge of uniform density pe in free space forms a semicircle of radius b Determine the magnitude and direction of the electric field intensity at the center of the semicircle P39 Three uniform line chargespn pn and pn each of length Lform an equilateral triangle Assuming that pn 2p2 2pn determine the electric field intensity at the center of the triangle P310 Assuming that the electric field intensity is E ax100x Vm find the total electric charge contained inside a a cubical volume 100 mm on a side centered symmetrically at the origin b a cylindrical volume around the zaxis having a radius 50 mm and a height 100 mm centered at the origin P311 A spherical distribution of charge p p0 R2b2 exists in the region 0 R b This charge distribution is concentrically surrounded by a conducting shell with inner radius Rtb and outer radius R0 Determine E everywhere P312 Two infinitely long coaxial cylindrical surfaces r a and r b b a carry surface charge densities psa and psb respectively a Determine E everywhere b What must be the relation between a and b in order that E vanishes for r bl P313 Determine the work done in carrying a 2 jiC charge from Pj2 1 1 to JP28 2 1 in the field E uxy uyx a along the parabola x 2y2 b along the straight line joining Px and P2 P314 At what values of 9 does the electric field intensity of a zdirected dipole have no zcomponent P315 Three charges q 2q and q are arranged along the zaxis at z d2 z 0 and z d2 respectively Problems 147 a Determine V and E at a distant point PR 9 4 b Find the equations for equipotential surfaces and streamlines c Sketch a family of equipotential lines and streamlines Such an arrangement of three charges is called a linear electrostatic quadrupole P316 A finite line charge of length L carrying uniform line charge density pe is coincident with the xaxis a Determine V in the plane bisecting the line charge b Determine E from pe directly by applying Coulombs law c Check the answer in part b with V P317 In Example 35 we obtained the electric field intensity around an infinitely long line charge of a uniform charge density in a very simple manner by applying Gausss law Since E is a function of r only any coaxial cylinder around the infinite line charge is an equipotential surface In practice all conductors are of finite length A finite line charge carrying a constant charge density pe along the axis however does not produce a constant potential on a concentric cylindrical surface Given the finite line charge pe of length L in Fig 340 find the potential on the cylindrical surface of radius b as a function of x and plot it A Oi dx f4 L x x FIGURE 340 A finite line charge Problem P317 Hint Find dV at P due to charge pgdx and integrate P318 A charge Q is distributed uniformly over an L x L square plate Determine V and E at a point on the axis perpendicular to the plate and through its center P319 A charge Q is distributed uniformly over the wall of a circular tube of radius b and height h Determine V and E on its axis a at a point outside the tube then b at a point inside the tube P320 An early model of the atomic structure of a chemical element was that the atom was a spherical cloud of uniformly distributed positive charge Ne where N is the atomic number and e is the magnitude of electronic charge Electrons each carrying a negative charge e were considered to be imbedded in the cloud Assuming the spherical charge cloud to have a radius R0 and neglecting collision effects a find the force experienced by an imbedded electron at a distance r from the center b describe the motion of the electron c explain why this atomic model is unsatisfactory 148 3 Static Electric Fields P321 A simple classical model of an atom consists of a nucleus of a positive charge Ne surrounded by a spherical electron cloud of the same total negative charge N is the atomic number and e is the magnitude of electronic charge An external electric field E0 will cause the nucleus to be displaced a distance r0 from the center of the electron cloud thus polarizing the atom Assuming a uniform charge distribution within the electron cloud of radius b find P322 The polarization in a dielectric cube of side L centered at the origin is given by P P0axx ayy azz a Determine the surface and volume boundcharge densities b Show that the total bound charge is zero P323 Determine the electric field intensity at the center of a small spherical cavity cut out of a large block of dielectric in which a polarization P exists P324 Solve the following problems a Find the breakdown voltage of a parallelplate capacitor assuming that conducting plates are 50 mm apart and the medium between them is air b Find the breakdown voltage if the entire space between the conducting plates is filled with plexiglass which has a dielectric constant 3 and a dielectric strength 20 kVmm c If a 10mm thick plexiglass is inserted between the plates what is the maximum voltage that can be applied to the plates without a breakdown P325 Assume that the z 0 plane separates two lossless dielectric regions with erl 2 and er2 3 If we know that E1 in region 1 is ax2y ay3x az5 z what do we also know about E2 and D 2 in region 2 Can we determine E2 and D 2 at any point in region 2 Explain P326 Determine the boundary conditions for the tangential and the normal components of P at an interface between two perfect dielectric media with dielectric constants erl and er2 P327 What are the boundary conditions that must be satisfied by the electric potential at an interface between two perfect dielectrics with dielectric constants erl and r2 P328 Dielectric lenses can be used to collimate electromagnetic fields In Fig 341 the left surface of the lens is that of a circular cylinder and the right surface is a plane If E1 at point Pr0 45 z in region 1 is ar5 a3 what must be the dielectric constant of the lens in order that E3 in region 3 is parallel to the xaxis O 45c x D FIGURE 341 A dielectric lens Problem P328 Problems 149 P329 Refer to Example 316 Assuming the same r and r0 and requiring the maximum electric field intensities in the insulating materials not to exceed 25 of their dielectric strengths determine the voltage rating of the coaxial cable a ifrp175r b ifrp135rf c Plot the variations of Er and V versus r for both part a and part b P330 The space between a parallelplate capacitor of area S is filled with a dielectric whose permittivity varies linearly from e at one plate y 0 to e2 at the other plate y d Neglecting fringing effect find the capacitance P331 Assume that the outer conductor of the cylindrical capacitor in Example 318 is grounded and that the inner conductor is maintained at a potential V0 a Find the electric field intensity Ea at the surface of the inner conductor b With the inner radius b of the outer conductor fixed find a so that Ea is minimized c Find this minimum Ea d Determine the capacitance under the conditions of part b P332 The radius of the core and the inner radius of the outer conductor of a very long coaxial transmission line are ri and r0 respectively The space between the conductors is filled with two coaxial layers of dielectrics The dielectric constants of the dielectrics are erl for ri r b and er2 for b r r0 Determine its capacitance per unit length P333 A cylindrical capacitor of length L consists of coaxial conducting surfaces of radii rt and r0 Two dielectric media of different dielectric constants erl and er2 fill the space between the conducting surfaces as shown in Fig 342 Determine its capacitance r Problem P333 P334 A capacitor consists of two coaxial metallic cylindrical surfaces of a length 30 mm and radii 5 mm and 7 mm The dielectric material between the surfaces has a relative permittivity er 2 4r where r is measured in mm Determine the capacitance of the capacitor P335 Assuming the earth to be a large conducting sphere radius 637 x 103 km surrounded by air find a the capacitance of the earth b the maximum charge that can exist on the earth before the air breaks down P336 Determine the capacitance of an isolated conducting sphere of radius b that is coated with a dielectric layer of uniform thickness d The dielectric has an electric susceptibility xe 150 3 Static Electric Fields P337 A capacitor consists of two concentric spherical shells of radii Rt and R0 The space between them is filled with a dielectric of relative permittivity er from Rt to bRi b R0 and another dielectric of relative permittivity 2er from b to R0 a Determine E and D everywhere in terms of an applied voltage V b Determine the capacitance P338 The two parallel conducting wires of a power transmission line have a radius a and are spaced at a distance d apart The wires are at a height h above the ground Assuming the ground to be perfectly conducting and both d and h to be much larger than a find the expressions for the mutual and selfpartial capacitances per unit length P339 An isolated system consists of three very long parallel conducting wires The axes of all three wires lie in a plane The two outside wires are of a radius b and both are at a distance d 500b from a center wire of a radius 2b Determine the partial capacitances per unit length P340 Calculate the amount of electrostatic energy of a uniform sphere of charge with radius b and volume charge density p stored in the following regions a inside the sphere b outside the sphere Check your results with those in Example 322 P341 Einsteins theory of relativity stipulates that the work required to assemble a charge is stored as energy in the mass and is equal to mc2 where m is the mass and c 3 x 108 ms is the velocity of light Assuming the electron to be a perfect sphere find its radius from its charge and mass 91 x 1031 kg P342 Find the electrostatic energy stored in the region of space R b around an electric dipole of moment p P343 Prove that Eqs 3180 for stored electrostatic energy hold true for any twoconductor capacitor P344 A parallelplate capacitor of width w length L and separation d is partially filled with a dielectric medium of dielectric constant er as shown in Fig 343 A battery of V0 volts is connected between the plates a Find D E and ps in each region b Find distance x such that the electrostatic energy stored in each region is the same T d I I V Q FIGURE 343 A parallelplate capacitor Problem P344 P345 Using the principle of virtual displacement derive an expression for the force between two point charges Q and Q separated by a distance x in free space P346 A constant voltage V0 is applied to a partially filled parallelplate capacitor shown in Fig 344 The permittivity of the dielectric is e and the area of the plates is S Find the force on the upper plate P347 The conductors of an isolated twowire transmission line each of radius b are spaced at a distance D apart Assuming D b and a voltage V0 between the lines find the force per unit length on the lines i I T x L A Problems 151 FIGURE 344 A parallelplate capacitor Problem P346 x P348 A parallelplate capacitor of width w length L and separation d has a solid dielectric slab of permittivity e in the space between the plates The capacitor is charged to a voltage V0 by a battery as indicated in Fig 345 Assuming that the dielectric slab is withdrawn to the position shown determine the force acting on the slab a with the switch closed b after the switch is first opened Switch T d r VQ FIGURE 345 A partially filled parallelplate capacitor Problem P348 P315 Three charges q 2q and q are arranged along the zaxis at z d2 z 0 and z d2 respectively a Determine V and E at a distant point PR θ ϕ b Find the equations for equipotential surfaces and streamlines c Sketch a family of equipotential lines and streamlines P319 A charge Q is distributed uniformly over the wall of a circular tube of radius b and height h Determine V and E on its axis a at a point outside the tube then b at a point inside the tube P322 The polarization in a dielectric cube of side L centered at the origin is given by P P0axx ayy azz a Determine the surface and volume boundcharge densities b Show that the total bound charge is zero P325 Assume that the z 0 plane separates two lossless dielectric regions with εr1 2 and εr2 3 If we know that E1 in region 1 is ax2y ay3x az5 z what do we also know about E2 and D2 in region 2 Can we determine E2 and D2 at any point in region 2 Explain P330 The space between a parallelplate capacitor of area S is filled with a dielectric whose permittivity varies linearly from ε1 at one plate y 0 to ε2 at the other plate y d Neglecting fringing effect find the capacitance P344 A parallelplate capacitor of width w length L and separation d is partially filled with a dielectric medium of dielectric constant εr as shown in Fig 343 A battery of V0 volts is connected between the plates a Find D E and ρs in each region b Find distance x such that the electrostatic energy stored in each region is the same 315 Exercício referente à seção 35 que trata sobre potencial elétrico a Temos V 2q4πε0π q4πε0π ππ1 q4πε0π ππ2 Logo V q4πε0π ππ1 ππ2 2 Mas π12 π2 d24 πd cosθ π12π2 1 d24π2 d cosθπ Luego sqrtπ2π12 ππ1 1 dπ d4π cosθ12 if π d ππ1 1 d2π d4π cosθ 38 dπ d4π cosθ2 ππ1 1 d28π2 d cosθ2π 38 d416π4 d2 cos2θπ2 2d3 cosθ4π3 onde desprezamos termos de ordem superior a 1π2 Logo ππ1 1 d cosθ2π d28π2 3 cos2θ 1 De modo análogo ππ2 1 d cosθ2π d28π2 3 cos2θ 1 Daí substituindo na expressão do potencial V q4πε0π 1 d cosθ2π d28π2 3 cos2θ 1 1 d cosθ2π d28π2 3 cos2θ 1 2 V q4πε0π d24π2 3 cos2θ 1 V q d216πε0π3 3 cos2θ 1 Sabemos que E V logo E Vr an 1r Vθ aθ Daí E 3 q d² 3 cos² θ 116πε0 r⁴ an 6 q d² sinθ cosθ16πε0 r⁴ aθ b Para as equipotenciais V constante Logo r³ q d² 3 cos² θ 116πε0 V Cv r³ Cv 3 cos² θ 1 Para as linhas de fluxo d rEr r dθ Eθ Logo d r 3 cos² θ 1 r dθ 2 sinθ cosθ d rr 3 cos² θ 1 2 sinθ cosθ dθ ln r ln sinθ 19 ln cosθ ln CE Daí ln r ln sinθ cosθ12 CE r sinθ cosθ12 CE Daí r² CE sin² θ cosθ c Em particular temos as equipotenciais 310 Exercício referente à seção 35 que trata sobre potencial elétrico novamente Vamos considerar que o cilindro está com sua base no plano xy a Temos V ρs 4πε0 σ 0h b dθ d z z z² b² ρs b 2ε0 0h d z z z² b² Com ρs Q 2 π b h A integral acima é tabulada e temos V ρs b 2 ε0 ln b² z₀² z b² hz² zh e E dVdz az ρs b 2 ε0 b² zh² zh b² z² z ddz b² hz² zh b² z² z az E ρs b 2 ε0 1b² hz² 1b² z² az b Agora z h e temos V fracps bvarepsilon0 left int0z fracdzsqrtzz2b2 intzh fracdzsqrtzz2b2 right Logo o potencial se reduz ao mesmo do item anterior e o campo também consequentemente b hetab L3 cdot rhob 3 rho0 L3 hetas 6 L2 cdot ps 6 L2 cdot fracL2 3 rho0 L3 Logo hetas hetab 0 como esperado 325 Essa questão requer os conceitos das seções 38 e 39 que tratam do vetor D e das condições de contorno Sabemos que vecE1 2y hatax 3x hatay 5z hataz Em z0 vecE1 2y hatax 3x hatay quad vecE1N 5 hataz Mas ali vecE1 vecE2 e quad vecD1N vecD2N Logo quad varepsilon1 vecE1N varepsilon2 vecE2N Com isso se varepsilon1 2 varepsilon0 e varepsilon2 3 varepsilon0 vecE2z0 2y hatax 3x hatay quad vecE2Nz0 frac103 hataz Além disso vecD2N z0 10 varepsilon0 hataz vecD2 z0 varepsilon2 vecE2 6 y varepsilon0 hatax 9 x varepsilon0 hatay Logo em z0 vecE2 2 y hatax 3 x hatay frac103 hataz vecD2 varepsilon0 left6 y hatax 9 x hatay 10 hataz right O campo em pontos da região 2 pode ser tal que tais condições de contorno sejam satisfeitas 330 Essa questão utiliza os conceitos da seção 310 que discute sobre capacitores Temos Ey varepsilon1 varepsilon2 varepsilon1 fracyd Se a placa em yd está positivamente carregada logo Ē q Aε y ây q 5εr εg εI y d ây Daí V 0d Ē dℓ q 5 0d dy εI εg εI y d Logo V q d ln εg εI 5 εg εI Com isso se C q V C 5 εg εI d ln εg εI 344 Seções 310 e 311 capacitores e energia eletrostática a Podemos pensar nesse sistema como dois capacitores em paralelo ar dielétrico Ceq C1 C2 ε0 L x m d εr ε0 x m d Carga em C1 q1 C1 V0 ε0 L x m d V0 ε0 A1 V0 d Daí ρS1 A1 ε0 A1 V0 d ρS1 ε0 V0 d Carga em C2 q2 C2 V0 ρS2 A2 εr ε0 A2 V0 d logo ρS2 εr ε0 V0 d Com isso Ē1 ρS1 ây ε0 V0 ây d Ē2 ρS2 ây εr ε0 V0 ây d e Ē0 ρS0 ây εr ε0 V0 ây d Daí D1 ε0 Ē1 ε0 V0 ây d D2 εr ε0 Ē2 εr ε0 V0 ây d b Energia em C1 U1 C1 V02 2 ε0 L x m V02 2d energia em C2 é U2 C2 V02 2 εr ε0 m x V02 2d De U1 U2 εr x L x x L 1 εr