Classical Elect & Magnetism
Classical Elect & Magnetism PHYS 4610
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Notes for Phys 4610 Last revised 120104 These notes are just intended to give an overview of the major equations covered in class 1 Vector Analysis 11 Products of Vectors Dot product A B AB cos0 14MBm AyBy AZBZ gt6 Ngt Cross product UCJCB H Udggtltltgt UCJCB N H d N A x B 14sz 7 14sz A1131 7 AmBZ y AmBy 7 Ame 2 A X B1 AB sin0l 12 Position Vectors Thevector rm y zi is reserved usually for the point at which we want to calculate the electromagnetic field The vector rx y zi is reserved for a point often integrated over at which we find the electric charges or electric currents The separation vector is the difference of the two 1quot1quot 9090 3ylty 2 52 13 Triple Products etc of Vectors Scalar triple product be H d H d N Q53 ans 353 ABXCBCXACAXB Vector triple product Ax BXCBAC7CAB 14 The Del Operator 0 Gradient o Divergence 4 lt i 8m 8y 82 o Curl 2 SI 2 3 3 3 V x v m g 3 11 W 121 Alt gt 8 8 Alt 8 7 X 834782 y aziam Z 8x781 15 Product Rules Involving V Gradients Vfg nggi VABAXVXBBX VXAAVBBVA Divergences VfAfVAAVf VAXBBVXA7AVXB Curls 16 Second Derivatives 1 V VT Divergence of curl 2 2 2 VVT7 7 7 VZT 2 V x VT Curl of gradient V x VT O 3 VV v Gradient of divergence Nothing interesting about this does not occur often 4 V V x v V V x v 0 5 V x V x v Curl of a curl VX vaVVv7V2v Figure 1 Spherical coordinates 17 The Main Theorems of Vector Calculus AbVTd1TbiTa VVVd39rj vda Avada vdl 18 Spherical Coordinates Spherical coordinates of a point are illustrated in Fig 1 Relation with Cartesian coordinates mrsin0cos yrsin0sin zrcos0 Unit vectors r sin0cos sin0sin cos02 t9 c0s0cos cos0sin isin02 g isin cos Line element d1 m dle 1W m 1029 rsin0d lt2gt Volume element dT r2 sin0dr d0 d Vector differential operators Gradient ET A 1 ET A 1 ET A VTE quotHsmo 3 Figure 2 Cylindrical coordinates Divergence V V 712r2vr sin0v9 Curl V x V sin0v 7 271 r i 7 g vwl 9 06120 Laplacian 2 VZT 712ltr2gt m lt8in0gt 273 19 Cylindrical Coordinates Cylindrical coordinates of a point are illustrated in Fig 2 zscos yssin 22 Unit vectors s cos 2 sin SI 45 Line element Volume element Vector differential operators Ngt d1 isin cos l 2 dlssdl lt2gtdlzi ds sd d22 d r sdsd dz 7 311T 80 l 45 13 15 18 19 Gradient 7 8T 1 ET A ET A VT 7 as E87 2 20 Divergence VV svs 21 curl 18121 8 A avg auz A 1 8 avg A VXVlt8 7Egt5ltE788gt gsv 78 z 22 Laplacian 2 2 WT 38873 8ng 75 23 110 The Dirac Delta Function In one dimension the delta function77 6z 7 c has the property 17 i 6 7 d fa if a lt c lt b a Cm W C m 0 otherwise that is7 the delta function 6z 7 0 picks out77 the value of x at c if c is contained within the interval of integration It has the value of zero for all points except z c If there aren t any mathematicians in the room we can say 6W 7 C 0 m 7 c 00 13 C so that the delta function is a big spike located at z c The spike has area77 1 00 6m 7 6 dm 1 700 A property of the 6 function that we use from time to time is 6kz The threedimensional Dirac delta function treated as a product of three onedimensional delta functions The function 631quot 7 a with a am 1 ayy azi is 63r7a 6z7am6y7 ay6z7az 24 which has the property f1quot531quot 7 a dT a lt25 all space The 3 d delta function is related to derivatives of the functions 1r and frz With m r 7 r 7 we have v 4w63 m and v2 747r63 m 26 111 General Features of Vector Fields ln electromagnetism two vector elds will be of primary interest the electric eld E and the magnetic eld B In the theorems about these elds we will see their divergences and curls To this end the Helmholtz theorem is of great importance If the magnitude of a vector eld goes to zero at in nity then the vector eld is uniquely speci ed by it divergence and curl One can also show that for a vector eld F the following are equivalent a V x F 0 everywhere b 3 F dl is independent of the path from a to b c F dl 0 for any closed loop d F is the gradient of some scalar eld FVV Likewise the following are equivalent for a vector eld F a V F 0 everywhere b f F da is independent of the surface bounded by a given boundary line c F da 0 for any closed surface d F is the curl of some vector F V x A 2 Electrostatics 21 Introduction We rst deal with charges which are at rest We rst deal with the problem of nding the net force on a charge Q the test charge from the presence of other point charges ql qg qg We may think of moving our test charge Q or even changing its value but the other charges will remain at rest Knowing the basic law of electric force we can apply the principle of superposition for the forces ie the total force is the vector sum of the forces from the individual charges 22 Coulomb s Law The force between two point charges in given by Coulomb s law For a point charge q located at r the force on charge Q located at r is 1 6162 A 7 27 4713960 12 m where a r 7 r and 7 2 60 8854 x 10 12 28 In writing Coulomb s law in this way we are making the choice of the SI system of units where charge is measured in Coulombs and force newtons and distance in meters Exercise caution with the units in EampM Most textbooks nowadays use the SI system but a few of them use the Gaussian unit system where Coulomb s law looks like W 2772 where the charge is measured in some other way But the system used by most particle physicists is the HeavisideiLorentz system7 where Coulomb s law looks like W F 4713972 Confusion over the fact that different systems are in use can be deadly 23 The Electric Field The force F on a charge Q due to a set of charges q1 q2 is proportional to the charge Q then FQ depends only on the positions and values of the other charges This vector quantity is the electric field E at the position of Q We have F QE 29 where M 7 739 w 30gt T 4713960 1 For a continuous charge distribution the appropriate expression is 1 Er 7 p0quot NH 31 74713960 v 12 24 Divergence and Curl 0f the E Field A field line diagram is a nice way to represent the direction of the E field at each point If a diagram is drawn with the lines radiating outward from or to the charges in proportion to the size of the charge then the density of field lines in the diagram indicates the magnitude of the eld One should keep in mind though these diagrams just give a representative sampling of the eld You really can t count field lines77 as books so often say because the number you draw is arbitrary Through any surface S the ux of E through the surface is ltIgtESEda 32 One can show that for any closed surface 57 the electric ux is related to the total enclosed charge 1 f E da foam 33 5 60 which is known as Gauss s law It is equivalent to 1 V E 7p 34 60 Some interesting consequences of Gauss7 law 0 Outside of a spherically symmetric distribution of charge7 the electric field is the same as if all the charge were concentrated at the center 0 Inside a spherically symmetric shell of charge the electric field is zero 0 The electric field from an in nite sheet of charge of charge density a is E i 35 where the unit vector n is the unit normal to the sheet pointing toward the observation point One example of an application of Gauss s law to nd the E eld is that of a uniformly charged sphere of radius R Symmetry says that the E eld is radial everywhere E Err Using Gauss s law7 we nd ALKIHltR 7 47mg R3 ETT ll 25 The Electric Potential Vr 7OrEdl 36 The point 9 is the reference point 7 a point chosen where we want the potential to be zero It is arbitrary but the usual choice is to place 9 at infinity77 so that the potential gets small as we go far away from the charges From this it follows that E 7VV 37 For many electrostatics problems it is easier to calculate V at all points in space and then take the gradient to get the electric field E Since the electric field obeys the superposition principle7 the electric potential does also The consequence of Eq 34 and Eq 37 is WV7 Ba 60 The potential of a set of 71 point charges is 1 n 2 lt39 1 Von 4713960 i1 and that of a localized charge distribution pr is 1 M 60 m 40 47139 m Vr These expressions assume that the reference point for V is at in nity In some idealized problems where the charge distribution extends to infinity you can t do this 26 Electrostatic Boundary Conditions One can use Gauss s law to derive some facts about the possible discontinuities in the electric field and potential as we step across a bounding surface First7 the perpendicular component of the E field will have a discontinuity if there is a thin layer of charge a Eiabove 7 Eibelow g Secondly7 the component of E parallel to the surface is continuous EHaboVe EHbelow The potential itself is continuous Vabove Vbelow And the condition on the normal component of the E eld can be written in terms of the potential as avabove 7 avbelow 7 an an 60 27 Work and Energy in Electrostatics Since the electric force on a charge Q is just F QE the electric eld is related to work done the charge as we move it from one place to another The work required of some outside agency to move a charge Q from location a to location b in and electrostatic field is given by W Qle 7 Val 45 The energy associated with a collection of charges or a charge distribution arises from the repulsions and attractions between all the charges In calculating this energyi by which we mean the work required of an outside agency to bring the charges together from in nityi one must be careful about counting For a set of point charges this energy can be expressed as 1 iiqkiiiq i lt46 47quot60 171 71 7 87quot60 11 71 47 i and also as n W Zquri 47 i1 For a continuous charge distribution the wor to assemble it generalizes Eq 47 to get W deT 48 which can also be written is W 0 EZdT 49 2 all space For the moment it s worth stopping to ponder these expressions Do they tell where the energy is located Eq 47 seems to say that the energy is all in the charges while Eq 49 seems to say that the energy is in the space all around the charges contained in the fields For now it makes no difference since these equations should be regarded as prescriptions for calculating the work necessary to assemble the charge density and the question falls in the class of a Zen koan However much later when we associate physical entities with the elds it will make sense to ask where the energy is located It s also interesting to try to calculate the electric energy contained in the simplest electrical system imaginable A single point charge q The field is E 1 72 and Eq 49 gives 47reo T 2 co 1 2 co 1 60 q 2 q Wi7 4 i d id 00 2 47TEO2 700 72gt T T 87139600 72 T Whoal An infinite answer Yes and for that reason we have to ignore the energy required to pack a nite amount of charge into a point 7 When the charge distributions are continuous this problem does not arise The problem with point charges gives the answer to another odd feature of Eqs 47 and 49 The former can clearly be positive or negative while the latter can only be positive That is because the step in going from a point charge distribution to a continuous one ignored the poorlyibehaved energy involved in assembling a point charge In the rst equation the point charges were fully formed when they were out at infinity7 and we must use that one if we are talking about point charges 28 Properties of Conductors In a conductor some of the electrons from the atoms are completely free to roam around The consequences of this are c E 0 inside the conductor 0 p 0 inside the conductor so all charges reside on the surfaces 0 All points of the conductor are at the same value of V the electric potential 0 Just outside the conductors surface7 E is perpendicular to the surface If the conductor has a cavity inside of it and the cavity is free of charge then the field in the cavity is zero regardless of the elds outside If the cavity does contain a charge then the outer surface of the conductor just knows about the net charge inside but not the shape or placement of the cavity The electric field just outside a conductor is E in 50 E0 Equivalently7 the charge density at some spot on the conductors outer surface is 8V 7 7 51 a 60 an The force per unit area on the surface of a conductor due to the other charges on the surface 1 f 7 2 A 52 2600 H or in terms of the field just outside the surface7 P i lfl i 72 i 60E2 53 T T 260 T 2 29 Capacitors A capacitor is a system of two isolated conductors for which we intend to place a charge 62 on one of them and a charge 762 on the other When we do this it turns out that the potential difference the conductors is proportional to Q and the constant of proportionality C is de ned by Q E 7 54 V lt gt Capacitance is measured in farads7 which is just a coulomb per volt Practical capacitors are usually in the microfarad or picofarad range For two parallel plates of area A separated by a small distance d the capacitance is given approximately by 7 A60 0 d 55 10 To charge up a capacitor C to its final charge Q one must do an amount of work Q2 W 7 20 CV2 56 3 Mathematical Techniques in Electrostatics The main problem of electrostatics reduces to solving the differential equation for V 2 1 V V 77p 57 60 subject to the particular boundary conditions of the situation And generally we con ne our attention to the places where there is no charge so that we are solving the Laplace equation 82V 82V 82V 7 VZV7 8x2 8342 W 7 58 The mathematical techniques that we pick up in solving this equation are useful throughout physics so there is a point to teaching impressionable young persons about solving the Laplace equation Trust me 31 Laplace Equation Introduction Consider the Laplace equation in one dimension dZV 7 0 59 d2 with boundary conditions of some sort The solution is easy A line But we gain some insight by looking at the first numerical approximation to the derivative dV Vm h 7 Vm 7 h dZV Vm 2h Vm 7 2h 7 2Vm 7 7 60 dz 2h dz 2h so that from this the Laplace equation implies with e 2h was mm e m a egtgt lt61 that is the value of V at any point is the average of the values of V at the neighboring points As Griffiths points out this is a prescription for making a function as boring as possible In two dimensions the Laplace equation is a partial differential equation 82V 82V 3352 872 0 62 and the boundary condition is a functi0nie the value of V given on a bounding curve If we make a similar first approximation to the derivatives as above we get the condition Wan y iVm7 y 6 Wm y 7 6 V 67 y Wag 7 67 y 63 that is that value of V anyplace is the average of the 4 nearest neighbors 11 N N Q i lt o gt Q I Q V G Heme 4 A Q v A U v Figure 3 a Original electrostatics problem b A di erent electrostatics problem 32 Uniqueness of Solutions Occasionally we need to listen to the mathematicians when they tell us to sharpen up our ideas about the solutions to differential equations It is important to know if the solution to a given differential equation with boundary conditions is unique a study of this question will tell how much in the way of boundary conditions we need to specify for a physical problem before we know we have given all the relevant physics Two theorems are of some importance in the solutions of our electrostatics problems First Uniqueness Theorem The solution to Laplace s equation in some volume V is uniquely determined if V is speci ed on the boundary surface S We note that the outer surface of V can be in nity as long as we specify the behavior of V there and V could have islands in its interior as long as V is speci ed on their surfaces Next Second Uniqueness Theorem In a volume V surrounded by conductors and containing a speci ed charge density p the electric field is uniquely determined if the total charge on each conductor is given 33 The Method of Images Before diving into the more general methods of function expansions for partial differential equations we cover a method of solving problems involving conducting surfaces and simple geometries Now that we have the uniqueness theorems we can be assured that this simple procedure gives the correct answers The method is best introduced with the problem of a point charge q help at a distance 1 above an in nite conducting plane as shown in Fig 3a The points of the plane taken to be the my plane are all at the same potential which we can choose to be 0 The problem is to find the potential in the region above the plane 2 gt 0 On further thought we realize that this could be a complicated problem because the plane will acquire an induced negative charge which contributes to the field and this charge is not uniform Now we consider the solution to a completely different problem With no conducting plane consider a charge q at z H1 and a charge iq at z 7d as shown in Fig 3b The solution v I b Figure 4 a Original electrostatics problem b A di erent electrostatics problem for this problem is easy 1 q q V 7 7 7 7 r 4713960 la 1 71 q 7 q lt64 4713960 2 y2 z 7 12 m2 y2 z 12 where we notice that we have V 0 in the my plane Now compare some features of the original problem and the different one o For both the conditions on the boundary of the z gt 0 region are the same ie V 0 for z 0 andVHOaeroo o In the interior of the region VZV 0 except at the location of the charge both regions contain the same charge distribution By the rst uniqueness theorem the solution to both problems for z gt 0 is the same So Eq 64 also solves the original problem with the point charge and the conducting plane We can nd the force on the charge it must be toward the plate since an opposite charge is induced on the plate it is the same as the force of attraction between q and the image charge g 2 F LL 4713960 2d2 We can nd the charge density a on the plate and the total induced charge on the plate iq However it takes some care to compute the work done in bringing charge q from in nity and setting it at z d It is not the same as the work in forming the ctitious system since the image charge was never really brought in from in nity Next we consider the problem of a point charge q placed at a distance a from the center of a grounded conducting sphere as shown in Fig 4a We need to say that the sphere is grounded ie it is attached by an inconspicuous wire to a large conductor far away so that it was pull in any induced charges that it needs to be at zero potential The eld outside the sphere will be due to q and to the induced negative charge on the sphere so that the solution for Vr for r gt R is not so trivial Now consider a different problem which is little contrived With no conductors present consider a charge q on the z axis at z a and another charge q qua located on the z axis at z b RZa as shown in Fig 4b One can show that with these two point charges in place Vr R 0 Figure 5 Find Vx7 y z within the slot Once again the real problem and the fake problem have the same boundary conditions for the region of interest7 which here is r gt R and so the solution to the fake problem which is fairly easy to write down is the solution to the real problem 34 Separation of Variables We introduce the subject with a problem which contains enough of the elements of the technique that we can make generalizations later on Consider 0 Two infinite grounded plates parallel to the 2 plane7 one at y 0 and the other at y a We consider the space between the plates for z gt O The end of the region at z 0 is insulated from the grounded plates and maintained at a speci c potential given by V0y See Fig 5 We want to nd the potential everywhere within the slot7 ie for z gt 07 0 lt y lt a and all 2 And then we realize that this a hard problem So we see if we can solve an easier one Can we nd a solution to VZV 0 which is a product of functions of z and y individually the solution won t depend on 2 That is7 what kind of solution has the form Wm 24 XYy One can show that with the solution we must have 1d2X7 1 121 YT 2 7 01 and 7 1 Y dm2 7 02 where 01 and Cg are constants with 01 Cg O For reasons to be seen later we make the choice 2 2 dX 2 dY 2 WkX and WikY which have general solutions for X and Y Xz A5 By Yy Csin ky D cos kg The boundary of our particular problem would restrict these solutions to 717139 Xz 05 sinky with k 7 65 But in the end this solution cannot solve our example since the function V0y is arbitrary But a sum of solutions of the form 65 will still satisfy the conditions at y 0 and y a and may satisfy 14 the one onm0 00 Z Cne mma sinmrya 66 n1 The condition at z 0 requires 00 V0y E On sinmrya 67 n1 From the theory of functions it is known that we can nd a set of Cn s that will satisfy 67 and the method for nding them relies on the orthogonality property of the sine functions a 1 1 7 0 n 31 71 A s1nmrya s1nn wyady 7 a2 71 n 68 Using this relation with 67 and specializing to the case V0 constant one can show 2 a 1 0 71 even On gO V0ys1nmryady n Odd Scraping everything together the solution to the slot problem is 4V0 1 7 75 n135 n TLWEa Vxy sinmrya 69 7139 35 Separation of Variables in Spherical Coords The Laplace equation in spherical coordinates is 1 8 av 1 8 av 1 82V 2 7 if 27 i 7 7 V V 7287 T 8rgt 1 712sin 819ltsmt9819gt mum 63 0 70 Throughout this course we will assume azimuthal symmetry ie the problem have a symmetry in so that the solution V will be independent of 0 Again we look for solutions which are products of r and 19 Vr 19 Rr 0 Putting all of this into the Laplace equation gives 1oilt2dRgt 1 dlti0d gt70 Rdr T dr sin0d0 sm d0 and as before this implies that each term is equal to a constant With hindsight we know what the form of the constant should be and so we write 1 d ZdR 1 d i 19 g lt7 a 71z1 fesmo lts1n0wgt 7 711 1 71 where l is a constant and as we will see is an integer The rst of the equations in 71 has solution B RU Arl W while the second requires more work after doing all that work isee your local math physics courseli we nd that the solution is 90 Plcos 0 where Plz is known as the Legendre polynomial of order 1 is given by the Rodriques formula ami ywew m and the rst few Plm s are These polynomials have the property Hm1 MAPGW and they are orthogonal speci cally 1 0 if 1 l HMHMMLL Mil m 71 211 1 i The general solution to a spherical problem is a linear combination of such separated solutions so that 00 WT 0 2 AM gt Plcos0 74 7 10 The general solution to Laplace s equation in cylindrical coordinates is treated in Problem 323 One assumes that the boundary conditions and hence the solutions are independent of The result is Vs A0 log 3 Bo Z Anrn cosn an Bnr cosn Ba n1 36 The Multipole Expansion We want to consider how a localized charge distribution looks when we are at large distances from it We assume the charge distribution is centered around the origin We are interested in approximate simple forms for the potential V are large distances For this it will be best to work in spherical coordinates because the distance from the origin r is fundamental in that system Before taking off on the general theory it is worth reViewing the electric dipole In the dipole problem we have a charge q on the z axis at z d2 and a charge iq at z idQ We use the binomial approximation to nd an expression for the potential at large distances from the origin where large means 7 gtgt 1 One nds 1 qd cos0 Vr4n60 T2 75 16 The result is interesting because it show that for this system where the net charge is zero there is still a potential but it gets weaker in inverse proportion to r2 and not r Now we look for a general procedure for nding the potential at large distances from a nite charge distribution We start with the usual 1 1 V 7 7 d 1quot 47r60mp1quot T and use an expansion for 1L which is valid at large T 1 i Pncos0 76 L where we are assuming the vector r points in the z direction in our coordinate system so its direction is xedl This gives Vr F160 2 M711 T Pncos0 pr dT 77 which called the multipole expansion of the charge distribution pr One word of caution7 though it is note completely general since the direction of the field point r is xed Explicitly7 1 1 1 3 1 pI39d r 72 r cos 9pI39d r 73 r 2 cos2 9 7 pr d7 78 1 Vltrgt 4713960 In 78 if the total charge is nonzero7 then the dominant term is 1 1 c2 1 d 7 4713960 TpOI T1 47110 but if the total charge is zero7 the second term dominates unless it too is zerol One can show that the second term can be written in a more general way as 1 A p 2r where p r pr d7 V A dip 47139 60 r The vector p is called the dipole moment of the charge distribution The dipole moment of a set of point charges is given by p Zqir 79gt For the dipole made of two point charges iq we have p qd where d is the vector pointing from iq to q The electric field of a dipole is given by p A i A 1 A A Edip MO cos0r s1n06 m p rr7 p 80 4 Electric Fields in Matter It is an unavoidable fact of life that we have to deal normal matter7 which is made of atoms with their constituent charges and which do not respond to electric fields in the simple way that conductors do In particular7 for insulators the electrons are bound to the individual atoms and do not move around in the material Nevertheless the charge distributions of the atoms will distort under the in uence of external fields and as a result will induce will produce a new electric field from this property of polarizability When a neutral atom is placed in an electric field E the charges are shifted in average po sition so as to give a dipole moment p proportional to the field7 p 04E7 where 04 is a constant characteristic of the type of atom This is a simplification the induced dipole is really given by some matrix 04 which multiplies the E eld Now is a good time to point out a couple mechanical properties of electric dipoles If a dipole p is in an electric field E7 it experiences a torque N p x E 81 and the force on the dipole is given by F p VE 82 Here we note that the force on a dipole arises because of the spatial variation of E if E were uniform the V operator would give zero everywhere The potential energy of a dipole in an E field is U7pE 83 When a macroscopic sample is placed in a magnetic field7 the field will induce a dipole moment in all of the atoms we can then discuss the dipole moment induced per unit volume7 P7 which is called the polarization of the material7 which7 generally speaking is parallel to the direction of the original E field 41 Electric Field Due to a Polarized Object We can use the potential due to a dipole p located at r 1 i p 2 V if r 4713960 z and our definition of P as the dipole moment per unit volume of the source material p P 17quot to show that the potential due to the polarization of a sample is i 1 1 1 1 V74moj wpuai4mmw Pdr 1 Each term on the rhs has the same form as V f qu where there is a surface charge density 47mg ab P n 84 and a volume charge density pb 7V P 85 18 so the equivalent charge distributions ab and 0 give the eld due to polarization of the material These are called the distributions of bound charge7 the crude physical picture being that this charge produces an E field but stays stuck to the atoms of the material it only became separated detectable because of the distortions of the electron clouds in the atoms Griffiths has a couple sections in the book here which are well worth reading but don t add much to the overall theory First he discusses what to make of the bound charge distributions which were formally found in 84 and 85 Views within the community of physics educators are not unanimous Griffiths emphasizes that they are real distributions in the sense that it is charge that cannot be removed77 l but which sticks out in the open uncancelled by others when the material is polarized Surely the solution as always is to get a firm understanding of the subject and then you can view the formal quantities like ab in any way that you want Less controversial and more subtle is the discussion of the field inside a dielectric The difficulty comes from the fact that while we always want to talk about average charge densities7 if you are inside the material the process involves averaging charge densities which are very close to the observation point 42 The Electric Displacement if we identify the bound charge density by Eq 85 then the total charge density at any point can be written 0 pb Pf 86 where pf is the density of free charge Then if we define the field D as D E 60E P 87 the Gauss s law can be written as V 39 D Pf 88 Which as we now know can immediately be written in an integral form7 f D da th 89 You are cautioned against the simplistic idea of completely ditching the quantities E and p and using D and pf instead in essence this is because the free charge77 distribution pf does not determine D in the way that p determines E And there is no potential like V which corresponds to the field D 43 Boundary Conditions One can show that when dielectrics are involved7 the conditions on the discontinuities in field components have to be expressed differently in order to again have a a simple form Dibove 7 Dlielow Uf Elbove 7 E39L elow 0 44 Linear Dielectrics For many substances under a wide set of conditions the polarization eld P is proportional to the electric eld we write this as P 63er 92 where E is the total E eld ie that due to outside charges and the polarized material itself Usually we work with the displacement D for which we have D 60E P EOlt1X5E which is expressed as D 6E where e EOlt1 X5 93 Finally we extract the unitless number 6 as ErElX5i 94 E0 and it is on this number that we focus for our studies of the properties of dielectrics Of course we must remember that the original linear relation was just empirical under conditions of very strong elds it doesn t hold 45 Energy and Force in Dielectric Systems With dielectrics around we need to be even more careful with framing questions about energy in electrical systems A calculation of the energy change in the system caused by dragging a set of free charges into place when there is polarizable material around gives WDEdr 95 This expression would seem to contradict the one given earlier Eq 49 but it s a matter of knowing what energy is included Eq 95 includes the energy change of the dielectric material as the charges in the material get pulled apart this energy in the system analogous to many little springs is present even if the polarization is uniform and thus there is no pb But the earlier expression for the energy just counts up the energy due to the non zero charge densities that is pf 0f pb 05 One can think of all kinds of situations where there might be a force on a dielectric due to an external E field in which we ve placed it but one of the simplest cases is to consider the force on a dielectric slab which is partially inserted between two narrowly separated parallel plates This is what Griffiths does in the final section and he nd that a force tends to pull the dielectric into the capacitor The derivation is interesting because it points out an error one could make by not considering all the energy changes in a physics problem in this case the energy changes in the voltage source if if it attached to the capacitor to keep V constant Also even though the electric elds near the edge where the field drops off from the uniform value to zero the fringing elds are complicated and important in giving rise to the force the derivation does not need their explicit form 5 Magnetostatics We now move on to the study of the magnetic eld how it is produced by electric currents charges in motion and how it gives rise to the magnetic force on a moving charge or electric current While the subject involves electric current charge in motion is still referred to as magnetostatics because we will begin with currents which themselves don t change in time and as a result the magnetic elds will depend on position but not on time 51 Magnetic Fields and Forces Something is going to be tricky with the mathematics of the magnetic interactionl Wires carrying parallel currents attract and those with opposite currents repel Also a compass help near a currenti carrying wire gives the direction of the magnetic eld the field lines go in circles around the wire But these two facts taken together imply that the force on an element of current is perpendicular to both the magnetic field and the current So we re in for some complicated vector mathematics Or maybe it s interesting mathematics take your pick 52 Magnetic Forces The magnetic force on a charge Q moving with speed V in a magnetic field B is F1mg QV x B 96 known as the Lorentz force law The magnetic force is perpendicular to both the velocity and the B eld it is proportional to the charge of the particle With both electric and magnetic elds present the net force on Q is F QE v x 13 97 Recall from Phys 2120 that when a particle has planar motion in a uniform magnetic eld the path is a circle where the radius is given by m1 QB f where Q B and v are the magnitudes of the charge magnetic field and velocity With E and B field both present and mutually perpendicular the motion has the shape of a cycloid We note that since the magnetic force is perpendicular to the velocity a magnetic field can do no work a fact which is confusing when we think of magnetic forces being used to lift things As Griffiths points out using some simple illustrations later on the work comes from the other forces that are present eg from the electric fields the magnetic field serves to redirect the directions of those other forces and they are the ones which do the work In case you forgot the magnetic field is measured in Tesla in the SI system but one often sees a field strength expressed in gauss 1 Tesla equals 104 gauss 53 Currents We begin with the idea of a current owing through a thin wire The current I measures the amount of electric charge passing a given point of the wire per unit time Current is measured in amperes with 1 A 1 Cs ln counting the electric charge which moves a positive charge moving say to the right is the same as a negative charge moving to the left If a line charge were moving along a path with speed 1 then the current would be I Av Actually current is a vector where the direction is given by the orientation of the wire and the sense in which the charge is moving so the real relation is I V The force on a currenticarrying wire is then found to be Fmg dl x B 98 where the integral goes over the section of the wire of interest It s usually the case that the current I is same all through the wire and then we have Fmag Ifdl x B When charge ows over a surface it is described by the surface current density K de ned by 11 K a 99 where d1 is a length measured perpendicular to the ow and 11 is the current contained in that length If the current is made of a surface charge density a in motion then K is given by K av When charge ow is distributed through a threedimensional region it is described by the volume current density J de ned by 11 E 7 100 da where day is a small area taken perpendicular to the ow of charge and 11 is the current contained in that area If the current comes from a charge density p in motion then J is given by J pV The expressions for the magnetic force on surface and volume currents are Fmg K x Bda Fmg J x Bdr 101 Conservation of charge the idea that the loss of charge contained within a surface comes from the ow charge through that surface gives the continuity equation 90 V J 7 75 102 54 The Bi0tSavart Law The magnetic field of a steady line current is given by M0 Igtlt Ii IMO dl gtlt Ii B 7 7d i 7 103 r 47139 12 47139 12 where uo is the permeability of free space 00 47139 x 104 104 The unit of B the Tesla is related to the other SI units by 1 T 1 Expressions for the field arising from surface and volume currents are K r x A J r x A Br da and Br d 105 but we note that in all cases we are evaluating the B field due to a steady current not from a moving point charge or set of point charges A moving point charge does indeed give rise to a magnetic field but its form is too complicated to consider just now Also the consideration of currents is much more practical so we don t have much to complain about 22 55 Divergence and Curl of B Starting with the most general expression for the B eld arising from currents do J Iquot gtlt 3 BrE 2 dT one can show two important facts about the magnetic eld arising from steady currents VB0 and VXBu0J 106 The rst expresses the fact that there are no isolated magnetic charges The second of these is called Ampere s law and can be expressed in integral form as f B d1 uOIem 107 where em is the total current which passes through the loop it can be found by taking 5 J da for any surface S bounded by the loop Also the direction the loop is traversed in the integral and the direction of the area vector da are related by the rightihand rule Ampere s law can be used to calculate the magnetic field for simple geometries with the proviso that we know or can reason out the direction of the B field everywhere because things are generally more complicated than electrostatics Some famous results are the B field for a very long solenoid If the solenoid runs along the z axis and carries a current I 56 Magnetostatics and Electrostatics It is worth taking time to ponder the fundamental equations of electricity and magnetism both static we have acquired up to now We have VE i VXE0 108 E0 VB 0 VXBu0J 109 along with the force law FQEVXB 110 These laws show how the sources of the elds relate to the elds themselves the E field diverges outward from point charges Gauss s law the B field wraps curls around a current Ampere s law But there are no point sources for the B eld The B field must be produced by moving charges and the B field can only be felt by a moving electric charge 57 The Magnetic Vector Potential Since V B 0 always we know that there must be a vector field of which B is the curl Here we need to think back to the electric field which because of V x E O was required to be the gradient of some function which was 7V But there was an ambiguity in the choice of V which was resolved by the choice of a 77reference point Now we introduce the uector potential for the magnetic eld for which we have BVxA 111 23 From the fact that this only gives a requirement on the derivatives of A is leaves some freedom in the choice of A which will be exploited later on From this de nition and Ampere s law one can show VV A 7 VZA qu A useful thing to do from the start is to use the freedom in the choice of A to demand that A have zero divergence V A 0 112 With this choice we get a simpler relation between A and J VZA WOJ 113 and since this is Poisson s equation the assumption that all the currents die off at in nity allows us write the solution for A which is similar in form to Eq 40 H0 JIJ A 7 7 d 114 r 47139 m T which for line and surface currents is written as Aldz LIld1 AEda 47139 m 47139 m 47139 1 What s the point of introducing A Just because we can Being a vector it is trickier to compute than our electrostatic potential V and doesn t provide as useful a tool for nding B Also at this stage of the course it s not clear where it can be useful whereas the electric potential V has a direct connection to the work and energy in systems of electric charge lts importance will come up later one must be patient 58 Magnetostatic Boundary Conditions When there is surface current K there can be a discontinuity in the B field above and below this surface current One can show that at such a boundary the perpendicular component of B will be continuous Balbove BBelow But the part of B which points along the surface has a discontinuity Babove 7 Bbelow HOltK gtlt where n is the unit vector pointing normal to the surface from below to above The vector potential is continuous across a boundary Aabove l Abelow and the derivative of A has a discontinuity from the surface current 8A 8A above 7 below iMOK an an 59 Multipole Expansion of the Vector Potential Starting with Eq 114 and the expansion for 11 given way back in Eq 76 we arrive at an expansion for Ar in inverse powers of r A i 0 00 1 P 0 dl 119 r 7 E 0W r n cos n le if H ifze 2lgt 7 47quot T dlr2 rcos0dlT3 r 2cos19 2 dl 120 where the successive terms here are called magnetic monopole magnetic dipole magnetic quadrupole etc But the rst term in 120 is always zero so the lowest order term is the magnetic dipole For this the vector potential is Adjpr il f r d1 121 47139 which can be rewritten as Adipr Zimjr where rnEIdaa 122 7r 7 The vector In is the magnetic dipole moment of the current loop a is the vector area77 of the loop which is the usual area if the loop is planar We get a pure magnetic dipole by considering a current loop whose size vanishes and current increases such that Ia is constant at a value of m In spherical coordinates the vector potential and magnetic eld of a pure magnetic dipole are uo m sin0 A Adijquot E 2 45 123 ijpr V x A 07 2 cos0f sina 124 3 WT The B eld of a magnetic dipole has the same form as the E eld from an electric dipole 6 Magnetic Fields in Matter 61 Introduction Just as matter contains positive and negative charges which can be separated under the in uence of an E eld matter contains tiny microscopic currents which can be in uenced by a B eld and thus produced magnetic dipoles in the material In treating the magnetization of matter we will draw on our previous formalism for electric elds in matter Magnetic polarization has some differences with electric polarization The magnetization of the material can be opposite to the applied eld These are diamagnets for paramagnets the direction of the polarization is along the direction of the applied eld The treatment of microscopic current has the additional feature that the particles in nature have an intrinsic magnetic moment which does not come from point charges in motion Finally some substances retain their magnetization after the external eld is removed These are ferromagnets and of course are quite common Magnetism in matter is complicated Figure 6 a Electron is in orbit with radius R and speed 2 It gives a magnetic dipole m b Magnetic eld B perpendicular to the orbit plane is turned on R will be held constant As a result 2 and m change 62 Torques and Forces on Magnetic Dipoles Like the electric dipole the magnetic dipole does not experience a net force when placed in a uniform magnetic eld But it does experience a torque given by N m x B 125 It does experience a force in a non uniform eld given by F VmB 126 With all of the similarities between magnetic dipoles and electric dipoles one might wonder if on the atomic scale the dipole might not be possibly or better described by two magnetic monopoles separted by a very small distance with the proviso that for some reason the individual magnetic charges can never be separated In fact one can get away with a model of this form if one wants but in the end it is no good it is known that magnetic moments in matter are not generated this way they really are made by the orbital motion of elementary charges or else by the intrinsic moments of certain particles 63 A Simple Toy Calculation We consider a simplistic but revealing model of how an external magnetic eld can induce a magnetic dipole moment for an orbiting electron The model is shown in Fig 6a Electron orbits a center of force at radius R and speed 1 Initially the dipole moment is Now we introduce a magnetic eld B Bi We assume that the central force on the electron came from a charge 5 at the center and that the radius R stays constant One can show that there will be an increase in speed of AU eRB 2mg and as a result there will be a change in the magnetic moment of 2 2 ieRB 4mg Am which might seem odd because it in the opposite direction from the applied eld B But that s what happens and this example is a very simple model of what happens in diamagnetism 26
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