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Electricity and Magnetism I

by: Quinn Larkin

Electricity and Magnetism I PHY 481

Marketplace > Michigan State University > Physics 2 > PHY 481 > Electricity and Magnetism I
Quinn Larkin
GPA 3.67

Phillip Duxbury

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Phillip Duxbury
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This 80 page Class Notes was uploaded by Quinn Larkin on Saturday September 19, 2015. The Class Notes belongs to PHY 481 at Michigan State University taught by Phillip Duxbury in Fall. Since its upload, it has received 6 views. For similar materials see /class/207646/phy-481-michigan-state-university in Physics 2 at Michigan State University.

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Date Created: 09/19/15
PHY481 Lecture 20 Chapter 6 of PS A Planar capacitor containing a dielectric Examples 345 of PS Consider a parallel plate capacitor of area A and plate separation d that has dielectric material with dielectric permittivity 6 between the places Find the electric eld between the plates when a charge Q is placed on the capacitor Find the voltage across the capacitor and the energy stored in the capacitor ii Now consider connecting a battery to the capacitor7 with voltage V on the top plate and the lower plate at ground Find the charge on the capacitor plates and the energy stored in the capacitor In both cases compare your results to the case where there is no dielectric between the plates That is7 in the two cases of xed charge and xed voltage7 does the stored energy go up or down when the dielectric is added Also7 if the dielectric slab is free to move7 is it drawn into the space between the capacitor plates or is it pushed out Find the force in both cases and ii This is a typical electro mechanical actuator7 used in doorbells amongst other things Solution to When a charge Q is placed on the capacitor7 we can use Gauss7s law or superposition7 to nd the diplacement eld7 a A a o A D 70k so that E 7k 1 The voltage across the plates is given by V E d g The capacitance is found from Q CV7 and using7 od Qd EA Vii sothat Ci 2 and the energy stored in the capacitor is U QZQC If there is no dielectric between the plates7 the capacitance is 00 eoAd7 and hence7 U 2 2 O E 60 1H for isolated capacitor 3 so that the energy is reduced when the dielectric is between the plates Note that from this calculation we deduce that the energy density by writing7 i Q 7 mm 20 26A 1 uV with u EOI E2 4 where u is the energy density7 V Ad is volume and we used oe Now note that7 f E 5 The result u E7 is a general result for linear isotropic dielectrics Solution to ii Now consider the case where the voltage across the plates is xed to V In that case 1 2 50V UU447 0 cbV2 CCO H for fixed voltage across capacitor 6 The case of a xed voltage applies to energy storage7 as capacitors are usually charged at xed voltage7 eg from a generator or solar cell High energy storage materials require high values of H highly polarizable material and high V very good insulator with low leakage B An electric eld applied to a uniform dielectric sphere We consider an uncharged uniform dielectric sphere of radius a and dielectric constant 6 in a constant applied eld V E0 E07 so that V iEoz rcos 7 Since the dielectric sphere is uniform and there is no free charge7 we have7 VD0V6E6VEO uniforme 8 Using E7 i V we then nd that V still obeys Laplace7s equation7 so we try the solutions7 O 3 9 VW 701T60567 Vm iEorcos 20 05 7 We impose continuity of V7 and the condition eoEff Xa7 t9 eEfLma7 t97 to nd7 7E0 02 7E0 7 202 7601 0 which lead to7 07E 3 oiEK an 1 0K27 2 0 2 where H 660 Using the fact that Vim 7012 we nd that the electric eld inside the sphere is uniform quot 3 A Em 7 E k 12 t 32 0H 2 and that the dipole of the sphere is a C a3 H 7 1 psphe39re 2 4WEOElOaSK 2 PHY481 Lectures 7 and part of 8 Sections 3536 of PS A Finding the electric eld continued m Uniform shell of charge shell theorems The shell theorems state that the electric eld inside a uniform shell of charge is zero and ii that the eld outside the uniform shell of charge is the same as that of a point charge with the same total charge as the shell These results are easy to derive from Gauss7s law in the same way as for the cylindrical case However now we use spherical surfaces of radius r lt R and r gt R where R is now the radius of the spherical shell of charge The shell has charge density 039 and the total charge on the shell is Q 47TR2039 We rst consider a Gaussian surface consisting of a spherical shell which is concentric with the charged shell with radius r lt R Noting that by symmetry the electric eld is directed radially and only depends on r then since there is no enclosed charge the electric eld for r lt R is zero in agreement with the rst shell theorem Considering a spherical shell of radius r gt R we have a a a 1 f SE dA 47TR2ET 3 so that E R37 1 which proves the second shell theorem Now lets prove the rst shell theorem directly Consider a general point P inside the sphere If we place a spherical co ordinate system at that point then we can draw a cone of in nitesimal solid angle d9 sm d drb extending from this origin to the spherical surface The cone has two intersections with the spherical shell on opposite sides of the sphere The areas of the spherical shell subtended by this solid angle are dA1 r d and dA2 rgd The magnitude of the electric eld at que to these two areas of charge is given by dA dA dE mi e 72 0 lt2 71 T2 This construction applies to all parts of the spherical surface proving the rst shell theorem Proof of the second shell theorem is more involved It is also relatively straightforward to demonstrate the second shell theorem by direct integration using polar co ordinates it may be shown that the electric eld in the radial direction for r gt R where R is the radius of the shell is given by E r 039 riRcos wRZstd 0 4m lt3 1 which reduces to the point charge result in deriving this expresion we used dA 27TR2 sin0d0 vii Uniformly charged sphere of radius R Again we carry out the ux integral to nd for a surface of radius r E Er47rr2 4 The charge enclosed by this surface changes with 7 For 7 lt R the enclosed charge is given by 4 W lt R 3 lt5 while for r gt R we have 4 W gt R gm lt6 Using Gauss7s law E qEO we thus have rp Q r E lt R 7 if 7 T 360 47TEO R3 where Q 47TpR33 is the total charge on the sphere The electric eld outside the sphere of charge is like that of a point charge shell theorem 1 Er gt R mg 8 In both cases the direction of the electric eld is f see problem 319 B Electric potential energy and electric potential Physical de nition The electric potential energy U is the potential energy due to the electrostatic force As always only di ferences in potential energy correspond to physical observables However we de ne a reference potential energy and calculate all differences in potential energy with respect to this reference In electrostatics the potential energy is de ned to be zero when the charges are an in nite distance apart The difference in potential energy in moving a charge between two positions a and b is de ned in terms of the work done in moving the charge between these two positions so that b a a b a a AUabFeztd8iFd8 9 1 1 2 PHY481 Lecture 30 Chapter 9 of PS Sections 723 724 of Gri iths A Linear magnetic materials Linear magnetic materials are characterized by a linear relation between the magnetiza tion and the magnetic eld intensity7 M XmH7 which is similar to the de nition of linear dielectrics7 13 eoxe but not completely analogous We also have7 g Mo t o M01Xm 1 which is analogous to D 6E in dielectrics a is the permeability This should not be confused with the magnetic moment for a current ring7 which is also sometimes called This is horrible notation7 but it is entrenched in the area We shall have to live with it Most of the time I use 773 for the magnetic moment and A71 for the magnetization B Paramagnets linear materials with Xm gt 0 Paramagnets do not exhibit spontaneous magnetic order7 nevertheless they can have large magnetic susceptibilities The magnetic moment of paramagnetic materials tries to align in the direction of the applied magnetic eld Actually all materials will magnetically order at suf ciently low temperatures7 but when the ordering temperature is very low7 materials are called paramagnetic The susceptibility of paramagnetic materials obeys the Curie Law7 amp Xm T 2 Paramagaetz39e materials are attracted to magnets C Diamagnets linear materials with 71 lt Xm lt 0 If elementary particles did not have an intrinsic magnetic moment7 then all materials would be diamagnetic That is7 the magnetic moment of materials would be opposite the direction of the applied eld This is due to Lenz7s law Superconductors are the best diamagnets7 but many pure normal conductors are too eg Cu At low enough values7 magnetic elds are completely excluded from the interior of a superconductors The phase in which this occurs is called the Meissner phase of a superconductor From the expression7 g Molt1 XMH 3 1 V it is evident that in order for ux to be completely expelled so that B 0 inside the superconductor we must have7 Xm 71 A measurement of Xm is one of the rst measurements that people do to determine if a material is in the superconducting state Diamagnetle materials are repelled from magnets This enables the possibility of magnetic levitation Since superconductors are the best diamagnets7 they are the primary candidates for possible magnetic levitation applications D Ferromagnets nonlinear magnetic materials hysteresis ln ferromagnetic materials7 the magnetic moments of the atoms in the material seek to align in the same direction Examples are Fe and Permalloy 55 Fe7 45 Ni It is actually quite dif cult to nd good ferromagnetic materials There is a continuing search for ferromagnetic materials which have large local magnetic moments A group at GM research in Detroit made a major breakthrough in this area about a decade ago They helped develop the Niodymium7 lron7 Boron magnets The production of these magnets is now a multibillion dollar industry Calculation of the elds around magnetics is carried out in a similar manner to the xed magnetization case discussed above7 eg for a uniformly magnetized sphere A more general calculation uses a non linear consitutive law Sometimes ferromagnets are treated as a linear dielectric with a large positive value of Xm this is not completely correct7 but it gives an indication of the expected behavior Ferromagnetic materials are very important in technology For example the hard drives in most computers are made using small domains on ferromagnetic materials A small sensor or read head scans the surface of the hard drive On the hard drive surface are small domains of ferromagnetic material These domains are oriented in the plane of the surface and they have a prefered direction The read head measures a resistivity which is sensitive to the local magnetic eld The technology of magnetic storage eg hard drives relies on a particular property of ferromagnetic materials This property is called hysteresis Hysteresis is a property which occurs when a magnetic eld is applied to a ferromagnet which is below its Curie temperature In order to describe hysteresis we must describe the way in which we vary the temperature and the magnetic eld Let us start at high temperatures and quench to a temperature well below the Curie temperature The magnetic material is frozen in a domain structure by this process Now we apply a positive external eld The domains now begin to align with the magnetic eld At suf ciently high magnetic eld the atomic magnetic moments are all aligned with the applied eld This is called the saturation magnetization Now consider reducing the applied eld until it is oriented in the opposite direction to the direction of the magnetic moments However7 the magnetic moments in a ferromagnetic material prefer to have the same orientation so they do not want to follow the direction of the magnetic eld at rst The magnetic moment then remains oriented opposite the applied eld until a suf ciently large opposite magnetic eld is applied At this point a sudden switching of the orientation of the magnetic moment occurs This is the switching eld He ln magnetic storage7 when we write information onto the hard drive7 we are switching the orientation of the magnetic domains The read operation does not do this7 instead it just senses the direction of the local eld This magnetic memory is non volatile as it is not necessary to have a power source continually applied to the material in order to maintain the orientation of the spins Magnetic materials with very large reversal elds Hcare called magnetically hard materials7 while those with small hysteresis loops are called soft magnetic materials Antiferromagnetics and complex magnets nonlinear materials Antiferromagnetic materials have atomic magnetic moments which prefer to have their nearest neighbors in an antiparallel alignment In the simplest case7 the magnetic moments alternate between one orientation and another This is easily possible in material structures which are bipartite eg the square lattice or the cubic lattice7 however in other lattice structures7 the magnetic order can be extremely complex These complex ordered states are called spin glasses or frustrated magnets Antiferromagnetic materials loose their order at a critical temperature called the Neel temperature Complex magnets loose their order at a glass temperature or ordering temperature Often complex materials exhibit hysteresis and time dependent e fects that make reproducible measurements more dif cult There are many antiferromagnets and complex magnets This is the usual behavior of compounds and some elements Examples are NiO7 07quot E Calculations involving linear magnetic materials Example Magnetic eld enhancement in a solenoid containing iron For a solenoid with 71 turns per unit length and carrying current I we found B0 Mon 4 This is the result for a solenoid in air Now if we place a material inside the solenoid we use Ampere7s law for the eld intensity and g pH to nd H m and B pH Mo1 Xmm39 5 From this expression it is evident that the magnetic eld inside the solenoid is greatly enhanced if the center the core of the solenoid is composed of a magnetic material which has large magnetic susceptibility Xm for example permalloy Note that this seems di erent to dielectrics where the electric eld is reduced when a dielectric is placed between the plates of an isolated capacitor However if a capacitor is connected to a battery the electric eld is unaltered by the addition of the dielectric however the charge stored increases by a large amount If we associate the charge stored on the capacitor with the ux in the solenoid then the two devices appear more similar This is the analogy that is usually used The energy stored in the solenoid is still Liz2 but we need to calculated L again L is de ned through L N so that LN2MAl 6 This is just the formula that we have for vacuum but with no a u The energy stored in the inductor thus increases dramatically when a large permeability material is used for the core of the inductor As an example consider an inductor containing permalloy with M 10000 and with N 10000 A 017712 l 1m carrying a current of 20A then U NZMAz39Zl 4 gtlt 1013 This is a very large number and looks attractive for energy storage applications However there are a number of critical limitations ranging from hysteresis to resistive losses and the e ects of large magnetic elds on materials and people Materials with positive Xm are drawn toward regions of high eld ie the solenoid while those with negative Xm are repelled from regions of high eld A co acm39al cable Consider a co aXial cable with an inner conductor of radius a and a material with permeability M in the region a lt r lt b A conducting cylinder completes the cable and has inner radius b and outer radius 0 Find the magnetic eld in each of the four regions 4 PHY481 Lecture 26 Sections 83 1021 115 of PS Section 73 92 of Gri iths A The effect of time varying electric and magnetic elds Electrostatics is the study of time independent static electric elds Magnetostatics is the study of time independent static magnetic elds The source of static electric elds is the fundamental charge q of elementary particles for example the electron charge 6 The sources of magnetic elds are The fundamental or intrinsic magnetic moment of elementary particles the magnitude of the magnetic moment of elementary particles gig where g is the g factor m is the mass and s is the spin of the particle eg for the electron s 712 and q 6 ii Steady state currents DC Note that the intrinsic magnetic moment formula g s applies to elementary particles ie quarks leptons photons gluons etc For example the neutron is neutral but it has a magnetic moment because it is a composite particle made up of quarks In fact the magnetic moment of the neutron is negative so that the neutron spin aligns in the opposite direction to the magnetic eld The fact the neutron has a nite magnetic moment and yet is not charged is very important to its use in scattering studies of materials particularly polymeric materials biological materials and magnetic materials The e fects of time varying electric and magnetic elds are included through two quite simple physical observations 1 A time varying magnetic ux leads to an induced electric eld Faraday 2 A time varying electric ux leads to an induced magnetic eld Maxwell7s displacement current term Faraday7s law takes the forms i i 6m a a 6453 a a 63 mducedemfiei W or Edli W or VAEi E 1 This equation states that a time varying magnetic ux induces a voltage or electromotive force The minus sign on the RHS is understood through Lenz7s law which states that the induced emf acts to oppose the change in ux Example of Faraday 8 law Consider the simplest time varying magnetic eld B 0L 020 that is uniform in space ie does not dependent on yz Place a circular conducting ring in this magnetic eld with its unit normal along the Z direction ie the vector area of the ring is 6 WRZIQ Faraday7s law states that there is an induced emf in the ring given by 7 31537 3 a 7 60102t7TR27 emf 6t 7 atB39da 6t 7 CZWRZ 2 Some questions where is this emf does the emf exist if there is no conducting current ring what direction is the emf which direction does the current ow Maxwell7s displacement current term is an additional source term added to Ampere7s law of magnetostatics namely 6E E Ampere 7 Maxwell law 3 Maxwell7s term describes the fact that a time varying electric eld induces a magnetic a a 63 a a a fBdlMozMoeon VABM0JM060 eld Maxwell noticed that when a capacitor is charging there is a logical inconsistency in Ampere7s law To understand this inconsistency consider an initially uncharged capacitor connected to a voltage source at t 0 For simplicity we consider a parallel plate capacitor Current begins to ow in the circuit at t 0 charging up the capacitor Now we can construct a loop around the wire in the circuit However this loop does not really enclose the current in the wire The loop can pass through the capacitor without cutting the wire Therefore when the capacitor is charging Amperes law would state that there is no enclosed current and hence the magnetic eld is zero This is wrong There is a magnetic eld produced by the current Maxwell resolved this dif culty by adding a new term which includes the effect of the electric eld which builds up between the capacitor plates His idea was to related this electric eld to the current owing the circuit From Gauss7 law we have m i 4 dt 60 or the 77Maxwell displacement current77 is d M 60 5 This relates the current owing into the capacitor to the electric eld between the plates Maxwell realized that if Amperes law is modi ed to a a d fB dl u0z u 02 060 6 2 then the logical inconsistency in the case of a charging capacitor is removed This extra term is called the displacement current as it has the same dimensions as the true current in the circuit His insight was brilliant as this equation is correct in general Example 30 7 A parallel plate capacitor is being charged at 239 105 If the plates are circular with radius7 R 01m7 and are separated by d 10m7 nd the magnetic eld as a function of distance from the central axis of the capacitor Consider a circular loop or radius r centered on the axis of the capacitor and lying parallel to the plates Erom Ampere7s law we have7 f5 39 df Moi M0 odg 7 The electric ux through the loop is given by7 E 7TT2E 8 so the rate of change of the electric ux is7 d E ZdE 7T7 2 dV 7T7 2 dQ 7T7 2 dQ 7 7T 7 i i 7 dt dt VE WE ETAE 9 Here we have used the relation for a capacitor Q CV and the relation between electric eld and voltage for a parallel plate capacitor E Vd7 and the expression for the capacitance of a parallel plate capacitor C eoAd Evaluating the path integral for the magnetic eld and equating it to the displacement current term7 we then have7 7T7 2 dQ 27TTBltTgt 60ij 10 01 BO E W E W 2A dt 27TR2 dt 27TR2 For 7 gt R7 the magnetic eld is rltR 11 B Electromagnetic waves Optics7 Electricity and Magnetism were considered to be unrelated subjects prior to 1820 In 1820 Ampere and Oersted demonstrated that electric current in uences magnets and founded the eld of magnetostatics Faraday extended this to include the effect of time 3 varying magnetic elds However it was not until 1864 that optics and other electromag netic waves were uni ed and their relation to electricity and magnetism was made clear by James Clerk Maxwell though addition of his displacement current term and then solving the equations to show that they predict wave motion that describes EM waves across all frequencies This prediction was subsequently con rmed by Hertz The demonstration of EM waves is actually quite straightforward ln free space there are no wires so the term he is not needed and there are no charges so we remove the term pEO from Gauss7s law so that 7 7 V a a a 6B a 6E VE0 VB0 VAE7E VABMOEOE freespace 12 If we take a time derivative of Ampere7s law and use Faraday7s law we nd a 6217 V V E 7mg 13 If we take a time derivative of Faraday7s law then use Ampere7s law we nd a 621 B 77 14 v v gt a lt gt An identity that is easy to prove eg using Mathematical is VAVEVVE7V2E 15 Now note that in free space V E V g 0 Using these expressions in Eqs 13 and 14 we nd a mi VZE MOEOW a 621 VZB M060 These are both wave equations which just means that they have solutions which are of the form Emxyz E0005kz 7 wt b Ey 0 E1 0 18 The sin function also works and solutions like this apply to the y direction and to the z direction We have to choose the solutions to t the equations and the initial conditions Eq 18 describes an EM wave that travels in the z 7 direction and whose electric eld oscillates in the z direction Similar solutions exist for waves travelling in the z 7 direction 4 PHY481 Lecture 2 Sections 21 and 22 of Pollack and Stump PS A Co ordinate systems we will use We shall be using three orthogonal co ordinate systems7 cartesion7 cylindrical and spherical polar We need the transformations between these systems We have7 1 Cartesian co ordinates f9 y29 y52 1 2 Cylindrical co ordinates 2 7362 rerzAr 2 z r0056 y 7192716 2 y2 r2 3 3 Spherical polar co ordinates 9 73945 W 4 z rcosrbsz n y rsz nrbsm g z r0050 2 y2 22 r2 5 Following convention and as used in PS 7 and 6 have di erent meanings for cylindrical as opposed to the spherical polar case B Rotation matrices Rotation matrices are used to rotate a co ordinate system about an axis We may rotate about the X7 y7 or Z axes and there is a di erent matrix for each case However the matrices are very similar so we only need to consider one in detail Lets consider rotation about the Z axis Consider that in the original co ordinate system the unit vectors are in hatk We then rotate the co ordinate system through an angle 6 about the Z aXis In this new rotatedprimed co ordinate system7 the new unit vectors along the z y z directions are 33319 A vector i may be written in either of these co ordinate systems7 ie7 f d 21 x y 2 6 Now it is easy to show that the relationships between the unit vectors in the original and rotated co ordinate systems are7 2 00502 52710 3quot 7527102 00503 I I 7 Substituting these expressions in the last of Eq 67 we nd that7 f 20050 7 y527102 052710 0050 2 2 2 8 so that7 z 20050 7 y 52710 y 052710 20050 2 2 9 or x 20050 y52710 y 7252710 20050 2 z 10 The latter equation may be written in matrix form7 x 0050 52710 0 z y 752710 0050 0 y or i Fi 11 2 0 0 1 z where F is the rotation matrix in the middle equation It is easy to show that the inverse of F is the transpose of F7 so that7 RTE I 12 where I is the identity matrix Rotation matrices for co ordinate rotations by angle 0 around the x7y7z axis are respectively7 1 0 0 0050 0 752710 0050 52710 0 0 0050 52710 0 1 0 752710 0050 0 13 0 752710 0050 52710 0 0050 0 0 1 C What is a vector What is an invariant A vector is a quantity that behaves like the position vector 2 under co ordinate rotations An invariant is unaltered under rotations of the co ordinate system For example7 we expect the dot product of two vectors and the cross product of two vectors to be unaltered by rotations of the co ordinate system as they depend on the angle between the two vectors and not on the angle itself they should be invariant Eg to check this for the dot product7 we need to prove that E a FE E 7 which is easily seen as follows Ming XTFTEE A E 14 D Some help with proving vector identities LeviCivita tensor The completely antisymmetric Levi Civita tensor7 Sigk seems obscure7 however it helps a lot in proving vector identities and is worth learning The Levi Civita tensor has 27 components since Lj k can each take on three values which refer to co ordinates in 3 space However all of its entries are zero except the ones where z jk are all different There are thus 6 non zero entries which correspond to the 3 permutations of 1 23 The values of the 6 non zero entries are either 1 or 1 depending on the number of transpositions required to take the permutation from the natural ording 123 to the desired permutation7 thus7 8123 1 8132 1i8312 1 5213 1 5231 1 321 1 15 The cross product may then be written7 EA Z ZeijkAJBk or in suf x notation eijkAJBk 16 j k In su m 07quot Einstein notation7 if an index is repeated it implies that a sum over that index should be carried out Example 1 Prove the vector identity EA Proof eijkAiBJCk Also EA eijkAJBkCi The latter expression would be equivalent to the former if we were to make the replacement Eijk a 8M This is true as it involves two transpositions of the indices and hence preserves all entries in the Levi Civita tensor Note that this is considerably simpler than writing out the two expressions in the identity explicitly Example 2 Prove the vector identity EA 7 Proof The LHS may be written as Using the expansion of the cross product again7 this becomes7 A EijkEklmAjBlOm 3 PHY481 Lecture 9 Sections 3738 of PS A The electric potential energy of a charge distribution Potential energy The potential energy of a charge q at position Fis U qVf In the case of a continuum this becomes fpFVFdF However in using this continuum equation it is assumed that the potential is unaltered by the added charge fpFdF Potential energy stored in a charge distribution A di erent question is What is the potential energy of a distribution of charges7 that is7 what is the potential energy stored in a distribution of charges In this case we muts take into account the way in which the electrostatic potential changes as charge is added to the system The potential energy stored in a distribution of charges is equal to the work done in setting up the distribution of charges7 provided there is no dissipation and no kinetic energy is generated To set up a distribution of charges Qi at positions 771 we need to bring each of the charges in from in nity and place it at its allocated position The work required to place the rst charge is zero no other charges are there yet The work required to place the second charge is QgVn7 where V21 lergl is the electric potential at position 7 due to charge Q1 Note that r21 r12 1772 7 F11 The work required to place charge 3 at its position is equal to Qngl legg7 and so on7 once all of the it charges are in position7 we have7 1 M inQ39 M inQ39 Un i Z Z 1 i7 j ii iltj ii In these expressions each pair interaction is counted once and the total potential energy is the sum of the potential energies of all pairs Note in writing this energy we have ignored the self energy of each charge The self energy is n gtk Eself and is the same regardless of where the charges are places Un is the interaction energy between the charges Taking the continuum limit the electric potential due to a charge distribution is7 1 pf pF dFdF39 1 U 7 7 V d 2 2 WWW 2 m o r ltgt Note that this is 12 the value which would be true if the potential were xed and when the charge as added This factor of two is thus quite fundamental it is also a source of considerable confusion The energy stored in the electric eld Using the relations7 PV 760V2VW 3 Using the vector identity VEnL and E 76V this may be rewritten as fawnw 7606 NV EONV 4 Using Gauss7s theorem the rst term on the RHS becomes V V dfT which goes to zero at r a infinity The only surviving term is the last term on the RHS7 so that the energy density in the electric eld may then be written as7 1 2 ma mE a where we used the fact that E7 76V This is the energy density in the electric eld and is the energy required to set up the charge distribution Problem 328 can be solved by integrating this energy density over the volume of interest Note that if we integrate the eld due to an isolated charge we get in nityl However we are interested in changes in potential energy due to changing the charge con guration The intinite self energy of each charge is there no matter what the charge arrangement is7 so it plays no role in the physics of the problem However it is important in trying to formulate a quantum version of EM B The multipole expansion The multipole expansion is a systematic perturbation theory of the general expressions for the potential due to a charge distribution7 kql kp7jd7j 7 6 lFe l lFe l vm The perturbation expansion requires that the locations of the charges are all signi cantly less than the position at which we plan to nd the electrostatic potential and the electric eld7 ie Fgtgt 771 In the case of the continuum integral7 the charge density pF39 must have a maximum extent which is much less than F The result of the expansion for the potential is7 A B C V77727mu 7 where ABC in general depend on f leading to an important angle dependence for the second two terms Finding ABC is where the hard work in the multipole expansion resides The rst term is called the monopole term because it looks like a point charge term while the second is the dipole term and the third the quadupole term All of them have very important applications in physics chemistry and medicine eg MRI NMR etc When the electric eld is found by taking a gradient of the terms in this expansion the rst term which results is again the monopole term while the second and third terms are again the dipole and quadrupole terms respectively The dependence on r for the electric eld multipole expansion is 172 for the monopole 173 for the dipole and 17 4 for the quadrupole The multipole expansion is carried out by expanding the denomenator of Eq 6 by using the cosine rule to write 777 r2 r 7 2rncost912 8 Therefore 1 1 1 777 r2 r 7 2rncost912 r1 7 212 9 Now we use the expansion 1 1 32 W17 6 6 10 with 6 727139 We then have 1 1 1 Ff r 3 F f r 77i72l 12 if 1 22 11 TTTT T 2T1 T Tl 8T1 T Tl ltgt Which may be written in the form 1 1 77 37742 1772 FETT T2 T3 7 m 12 This may also be written in terms of the angle between F and F T T was 2 30052071 13 r2 r3 This is the basis of the multipole expansion to quadrupole order and shows the angular dependence characteristic of the dipole and quadrupole terms To complete the expansion we substitute this expansion in Eq 6 to nd kqiricos i kqlr in in 2 1 E E 7 E E 73 071 0 7 14 i i 7 139 T2 139 2T3 COS T4 PHY481 Lecture 27 Sections 101103 PS 72 of Grif ths In the last lecture we went through the most remarkable consequence of including dynamics in the study of electricity and magnetism EM waves It is important to emphasize that the although the relations between time varying magnetic ux and induced emf Faraday7s law and between time varying electric us and induced magnetic eld Maxwell displacement term7 both involve loops7 these terms exist regardless of whether there is a material loop in the experiment Today we look at some of the remarkable effects of Faraday7s law that occur when there are materials in the experiment A Faraday7s law and the Lorentz force law A moving conducting rod in a constant eld Faraday7s law and the Lorentz force law are closely related as can be seen by considering a conducting wire moving through a constant magnetic eld Consider a uniform and constant magnetic eld7 B7 directed along the z axis Now consider moving a conducting rod7 of length l7 which is directed along the y direction at constant speed 1 along the x direction First we use Faraday7s law to show that a motional emf is developed between the ends of this wire To nd this emf7 consider a rectangular loop which is composed of a side of length L lying on the y axis centered at the origin7 the moving piece of wire7 and the two sides which join them to form the rectangle These two joining sides have length L mt where we assume that the conducting rod starts at the origin at time t 0 The rate of change of the ux is given by7 7 d Bi dLi 5 dt 7 Bzdt B11 1 This emf is induced around the loop7 however the only conducting part of the loop is the piece of wire The charges in the piece of wire move until the voltage drop between the ends of the wire just balance the induced emf7 ie Vwm Blv7 so that the voltage inside the rod is zero If the rod was made of an insulating material7 the charges would not move and there would be a voltage across the rod The behavior described above can also be understood from the Lorentz force law Con sider the wire to be composed of negatively charge carriers The motion of these carriers in the magnetic eld leads to the Lorentz force law FB 7611B The charges build up at the ends of the wire until the induced electric eld produces a force on the conductors which just balances the magnetic force We then have7 6E 6 611B so that Vwm Eli 2 as found using Faraday7s law A more general derivation of from the Lorentz force law as is given in PS 1012 An important deduction from this example is that when an observer moves through a constant magnetic eld7 the observer sees an electric eld The effect of transformation to a moving co ordinate system is then very interesting and is believed to be the way in which Einstein rst started thinking about relativity B A conducting loop and magnetic drag Now consider a square loop of wire which lies in the X y plane7 and where each side has length l Consider that the half space z lt 0 constains a constant and uniform magnetic7 B7 directed along the positive Z axis Now consider the situation in which the square loop is initially within the B eld and it is drawn out of the B eld at velocity 1 along the X axis If the loop has resistance R nd the induced current The rate of change of ux is given by7 61453 57731 3 dt v The induced emf is then Blv The current in the loop is thus7 Blv Z f 4 The direction of the current is to oppose the changing ux Since the ux is decreasing7 the current ows counterclockwise in the X y plane This produces an induced ux in the Z direction which opposes the changing ux induced by the motion of the loop out of the uniform B eld Note that the induced current is small if the resistance of the loop is large7 while the induced current is large if the resistance is small In cases where an induced current ows7 eg a conducting loop7 there is dissipation in the loop This energy loss must be equal to the work done by an external force7 but what is the origin of the force it is the Lorentz force and the most convenient form in this problem is d 239de In the simple example above the rod direction is perpendicular to the eld so we nd that the magnitude of the magnetic drag force is BZZZ deg z lB Y direction opposite 7 5 In general a moving magnet near a conductor leads to induced eddy77 currents that oppose the motion This e ect can be strong leading to strong braking e ects eg when a magnet is dropped down a copper tube The external force provides power given by a BZLZWZ 82 Pewteran Femte39rmzl 39 Z7 R 7 R The mechanical power that is supplied is dissipated as resistive losses due to current ow in the loop Surprisingly the smaller the resistance the larger the resistive losses and the greater the drag force A more samples erample Another popular type of problem is to change the way in which the loop is removed from the eld for example consider the case where the loop above has mass m and falls under gravity out of the eld In that case taking the X direction to be the direction downward in the gravitational eld lezu d1 le2 77 h R or dt 9 ow were a WR 7 d1 mam9degm9 The latter equation is solved by rewriting it as d1 9 70w 71 dt so that Elmg 7 ow t C or lng 7 vt 70 7 040 8 solving this equation for vt and using the initial condition 00 0 gives we 917 er lt9 04 The terminal velocity occurs when 7 ng W 10 g ow or UGO 90 C Rotating loops Electric Generators and motors We consider for illustration the case of a homopolar generatormotor 1831 where a coil is rotated through a constant magnetic eld This generatormotor has the disadvantage 3 that the coils rotate7 requiring the use of communtators7 as in the DC motor Tesla 1880s introduced the idea of rotating the magnet inside the coils which removes the need for the communtator Modern AC motors and generators are almost exclusively induction devices having stationary coils and rotating magnets The principle of operation of these devices is basically the same physics as magnetic braking or drag as described above However in the case of generators7 the current generated is shunted to carry out work in the electrical grid Consider a homopolar coil with N turns and area A rotating about its central axis at constant angular speed w The angle between the magnetic moment of the loop and the applied magnetic eld is 1 This angle increases as wt wt due to the constant angular speed of the coil The rate of change of the magnetic ux is d B d W ampBA0051 iBAw smwt The induced emf is given by Faraday7s which gives7 kw dt NBAw smwt The potential energy is given by7 U in E NIAB com The torque is given by7 T m B NIAB smut The mechanical power that must be supplied to rotate the coil is7 Pmech F AD NIABw swab This must be equal to the electrical power delivered to the load7 this is given by7 82 NBA 2 PeleciM r 2 R R 5271 wt The average power dissipated is 1 27 NBAw2 2 PM A R 5271 6d Using the result that7 1 27r 1 i sz n26d0 7 27139 0 2 4 11 12 13 14 15 16 17 18 PHY481 Lecture 24 Chapter 78 of PS Chapter 5 of Gri iths A Cyclotrons Another Lorentz force law problem One of the most notable features of motion in a constant magnetic eld is that the classical angular frequency we qBm does not depend on radius of the orbit Cyclotron7s take advantage of this fact to accelerate charged particles using a constant frequency electric eld The radius of the orbit is given by7 R vw so as the particles velocity is increased7 the radius increases7 however the frequency does not change This is not true once the particle becomes relativistic in which case there is a relativistic correction7 w IEWm wc1vC212 B Current and current density The relation between current and current density is the same as the relation between any ux variable and the ux density7 so that7 S S S For any 77 0w77 variable7 there is an associated ux density variable The electric eld and magnetic eld are quite strange as in classical physics nothing is owing7 the elds are static These elds act like a real classical ow as in current or mass ux etc Now we want to relate current to velocity7 as when a charge is moving with velocity 17 it clearly produces a current We can nd the relation easily by using the de nition of current as charge per unit time Consider a small element of area dA with its normal along the z 7 axis lmagine we have a number density of particles 71 each with charge q and travelling with velocity Dz along the Z axis The charge per unit time crossing a surface is given by the number of particles per unit time crossing the surface7 times their charge7 ie a dN 62 A A j 7 gm 7 qn k i nquk 2 This expression can be written in vector form as nqz7 which relates the motion of particles to the current density and hence to the current This is important in the theory of electrical resistivity the Drude model Note that in the Drude model the velocity 17 is not the full velocity of the charge carriers Instead it is the drift velocity we will discuss this further later in the course C The force between current carrying wires is just the Lorentz force law We have learned that a moving charge experiences a force which is equal to F3 q A g Since current ow corresponds to moving charges a current carrying wire should also experience a force when it is placed in a perpendicular magnetic eld The force on a straight wire carrying current I is easily derived from the magnetic force The velocity which contributes to current is the drift velocity w The current density is given by nqud and the current is I jA Anqud The force on each charge is then on average a a f a F i7 B 7 B 3 B I D Anq The force on the wire is proportional to the number of charge carriers are travelling through the magnetic eld ie N an 71qu where l is the length of the wire We thus have 137N137 Alf Aging 4 l T B 7 mg Am 7 The direction of the current ow can be kept in either I or in the length Z If we have a small piece of wire in a magnetic eld we may then write dF 1de B 113527109 5 Here we have assigned the direction of current ow to the vector direction of the wire The direction of the force is given by the right hand rule as is the case for the magnetic force F33 First we show that this is consistent with the Ampere7s result for the force between two parallel current carrying wires Quiz27rd D The vector potential We have learned that several of the tools of magnetostatics 6580 vAB toj FqE17 6 Now we learn another method the use of the vector potential First how do we introduce it In magnetostatics the magnetic eld is divergence free and we have the vector identity v 0 for any vector function F therefor if we write g v A then we ensure that the magnetic eld is divergence free This is how we can PHY481 Lecture 10 Sections 38 41 42 of PS A A dipole in an electric eld The simple mew In the simplest case we consider a simple dipole consisting of two charges placed in a uniform electric eld We take the angle between the electric eld and the dipole to be 6 Since F 1E7 there are equal and opposite forces on the two charges of magnitude q in the dipole so the net force is zero and the net center of mass motion is zero However there is a torque on the dipole d NFF2 sm6qE E 1 where d is the separation between the two charges in the dipole and 151s the dipole moment The torque is zero when the dipole aligns with the eld and 6 0 The state of zero energy is taken to be at the angle 6 900 where the torque is maximum The potential energy of the dipole in the eld is then found from 9 a U i MmeAda ipEeos6 716E 2 7r 2 The lowest energy state is when 6 0 and the dipole is aligned with the applied eld The highest energy state is when 6 7139 and at this point the torque is also zero so it is a point of unstable equilibrium Any slight change in 6 away from 7139 makes the dipole have a torque pushing it to towards the lowest energy state General calculation For a general charge distribution we have NFAd FApEdFpFAEdF 3 If the eld is uniform then the electric eld can be taken out of the integral We also use the general expression for the dipole 15 fpf FdF to nd A7 17A E 4 The energy of a dipole in a eld can be found by starting with the general expression for the energy cost of placing a small amount of charge a xed potential U pF VF dW pVF W 7 7 W WW 5 1 Using E 7VV assuming that the electric eld is constant7 and taking the total charge fpF dF 0 to be zero7 we nd that the last expression on the RHS reduces to7 WWW W 151 6 If the total charge is not zero7 or if the electric eld is not constant7 then the dipole will have a center of mass motion in addition to its rotation toward alignment with the eld B Basic properties of conductors Conductors require care as we cannot simply assume that the charges on the surface of a conductor have xed locations Instead the charges move in response to an electric eld leading to a coupled problem that in general seems hard to solve In this chapter we develop some ideas and tools to gure out how a metal responds in the presence of charges andor an applied electric eld Things we already know about conductors i In electrostatic equilibrium Electric eld inside a conductor is zero This implies that the potential inside and on the surface of a conductor is constant and that the net charge inside a conductor is zero However the charge density on the surface of the conductor can and often is non zero In fact this charge density arranges itself to ensure that the electric eld inside the conductor is zero ii The electric eld at the surface of a conductor is of 60 and is perpendicular to the surface of the conductor This is true even if the the surface has a complex shape and if the charge density is di ferent in di ferent parts of the conductor One further property that is very useful in solving problems in electrostatics and in many other areas of science and engineering is as follows iii Uniqueness If we nd a potential function which satis es the voltage or charge boundary conditions then it is the unique correct solution This is a general property of linear partial differential equations with appropriate boundary conditions This means that if you guess a solution which works7 it is the correct solution This is remarkably useful especially when using superposition to solve problems7 as we shall see below iv When a conductor is grounded7 it means that the potential of the conductor is xed at zero In general we may specify either the total charge or the potential of a conductor The total charge may be distributed in complex ways on the surface of the conductor7 but PHY481 Lecture 3 Sections 2225 of Pollack and Stump PS The Divergence The divergence is the dot product of the gradient operator and a vector function7 Fm Fy FZ7 so that a a 31 BE an V39Fia tw l az 1 We want to nd a co ordinate independent representation ofthe divergence7 which we achieve by considering a small cube of dimension 6 Now consider an integral of the ux through this surface7 that is7 3 s dA E5eei2 71227 eel2 2 where 616263 3 are introduced to simplify the expression Eq 22 reduces to7 6m 63 v 13w 3 F dA is 11 x The co ordinate independent representation of the divergence of a vector function is then7 a a 1 a a V F l39 a i F dA 4 mm 0V is The divergence is then proportional to the ux of the function7 though the surface of the volume V The Laplaez39ah The Laplacian V2 is a scalar operator found by taking the dot product of the gradient operator with itself7 ie7 2 a a 2 62 32 32 V V V in cartesion co 7 ordinates V w 672 Q A U V The Curl The curl of a vector function7 V F is de ned in the same way as the cross product a a 3F 3F 3F 3F 3F 3F AF 717yi717y4 6 v 6m ay 61 g 32 7 32 h h 3y FmFy F1 Using Levi Civita and suf x notation this is a a 51 To nd a co ordinate independent representation consider a square loop placed in the X y plane with edge length 6 Consider a path integral around the loop the circulation f 13 df emf 7 eej2 7 61 7 em 7 6mm eej2 7 emf 7 6522 8 loop f 39 loop The general co ordinate independent form of v F is then This reduces to 6F 7 6F 3 3x7 L H gt7 Woke lt9 a a 1 a a A AFl39 n7fFdz 10 71V ZmA 0A 0 F Derivative operator identities Table 22 of PS gives a list of indentities for derivative operators We will look at a couple of interesting examples and in the homework you will need to use both these identities and the vector identities in Table 21 of the text Identity v 0 Proof eijk Because there is a sum over jk pairs of terms appear with opposite signs due to the assymmetry of the Levi Civita tensor Summing these pairs of terms gives zero a210 Identity 6 0 Proof Using Levi Civita notation the LHS is eijkamzam Again the terms on the RHS can be collected into pairs with opposite signs due to the antisymmetric nature of the Levi Civita tensor The identity is then proved G Summary of key geometrical concepts This has been a busy lecture with many mathematical details If you have not seen these mathematical tools before it will take some time for you to become familiar with them However the most important lessons to take from these tools is the key concepts that enable insight into the physics behind the math Here are four key concepts 1 The cross product EA E is perpendicular to both X and 2 The de nition of the gradient through df di f demonstrates that f is perpen diclar to equipotentials of the function f PHY481 Lecture 19 Chapter 6 of PS Chapter 6 of PS covers the use of Maxwell7s equations in dielectric media First we go through some general relations that apply in dielectric media7 before specializing to the case of linear dielectrics Most of the problems in PS involve linear dielectrics A General Relations Polarization is the primary new effect and leads to reduction screening of electric eld7 as compared to vacuum Screening is produced by opposing electric elds due to bound charge which in term originates from the polarization Bound charge may occur at the surface of a dielectric material in which case it is denoted by ab the surface bound charge density or in the bulk where it is the bulk bound charge density7 pb The relations between these quantities and the polarization are7 Ub 3 pb 13 1 where 13 is the polarization density7 ie the polarization per unit volume The bound charge acts as a source of the electric eld so we have7 a a 1 V E0PfPb 2 where pf is the so called free charge and is really the excess charge Now we de ne the displacement eld through 6 D pf so that7 emiw iv scum 6054713 3 60 The displacement eld can be found by using Gauss7s law7 that is7 pf implies f dgqf 4 There are thus several ways to approach problems in dielectrics The boundary conditions at interfaces also follow from the equations above7 for example continuity of the electrostatic potential Of course we can also still use E7 76V In order to solve dielectric problems we need to be given either 13 pb and pf andor7 a a we can be given a relation between P and E7 ie some function or constitutive relation We solve a few problems eg 65 where we are given the polarization7 in which case direction application of the formula above provides a solution Below we derive the result for a uniformly polarized sphere However usually we dont know 13 and we need more information than that contained in the results above in order to solve EM problems in dielectric media The key additional information is usually in the form of a constitutive law giving a relation between the polarization and the local electric eld7 ie some function At the molecular level we already noted two relations between the degree of alignment of dipoles as a function of applied electric eld These laws can be extended to the polarization as we discuss later However at this point we focus on one simple constitutive law7 that for linear isotropic dielectric materials where 13 60X5E B Linear isotropic dielectrics Where 13 60X5E D 6D n 660 Most of the problems that we solve will take the simplest7 but still very important7 case of a linear isotropic dielectric where7 13 emf 5 Using the general relation7 60D D 7 13 with the above equation7 we nd that7 13 D 7 Egg 60X5D so that D 6D with 6 601 X5 6 where X5 is the dielectric susceptibility and E is the permittivity Dielectric materials are usually characterized by the ratio n 6607 the relative permittivity The relative permittivity of air is close to 17 while that of engineering polymers is around 3 High values of n are found in perovskite materials such as Strontium Titanate and Barium Titanate Solution strategies for problems involving dielectrics The goal is again to nd the electric eld and the electrostatic potential from which all of the physics follows The following two strategies are generally useful Use Gauss7s law for D7 then nd D ii In regions where there are no free charges7 and where E is uniform7 then7 V VD06VD76V2V0 7 That is7 Laplace7s equation holds We can then use the methods developed in Chapters 4 and 5 to solve a variety of problems The key thing to get right is the boundary conditions PHY481 Lecture 16 Chapter 5153 of PS We will not cover 54 55 A Separation of variables in circular co or39dinates Circular co ordinates are the 73 part of cylindrical polar co ordinates The Laplacian in cylindrical polar co ordinates is given by see Table 23 page 34 of PS 1 6 6V 1 62V 62V VZV if 7 if 7 1 r 67f 67quot r2 6 2 622 If there is no dependence on 2 the third term is zero so 6 6V 62V 7 7 7 2 Ter 6 W Assuming that V RrltIgt we nd r 6 6R 1 62 77 7 77 0 3 R6rltr6r ltrgte 2 We now separate the equation using 6 6R 71 62 3 r7 if 2 lt4 R67quot 67 ltIgt 6 2 The b equation has solutions cosk gt smk when k is nite When k 0 it has solution C Dab Noting that ltIgt gt ltIgt 27139 is required to ensure that ltIgt is single valued we set k n where n 01 We could also use 71 negative but that would just change the sign of the constants Dn in a general solution of the form Oncosn Dnsmn The 7 equation has the form dZR dB 2 2 7 7 7 R 0 5 T d7quot Tdr n which for n 31 0 has solutions R AnrnBnr while for n 0 R A Elm With these observations we nd that the general solution to Laplace7s equation in circular co ordinates ie cylindrical co ordinates with no dependence on 2 is The general solution is Vr A Elm C Dab EXAM 77LCncosn Dnsmn 6 n1 A simple emample Consider an uncharged conducting cylinder of radius R has its axis along the Z aXis A constant electric eld E0 is applied to the cylinder with the eld oriented along the X aXis Assume that the cylinder is at potential V 0 and has radius R Find the electrostatic potential for r gt R Solution The potential as r 7 00 is 7E0x7 which in cylindrical co ordinates is iEorcoso The rst thing to do is try the simplest solution which only contains this term at in nity7 that is7 Vr 7 Ar 57pm 7 To satisfy the boundary condition at in nity7 we can take A 7E07 giving7 B Vr b iEor 7cos 8 The boundary condition V 0 at r R implies that7 VR b 0 7E0R cos 9 This is satis ed provided B EORZ7 so we have7 R2 Vr b 77quot 7E0005 10 The electric eld should be normal to the surface of the conductor7 so we also need to check that E R 0 We have7 1 6V R2 767 71 Eosm 11 This is clearly zero at the surface of the cylinder where r R7 as required So that7s it for this example again a simple case where only one term in the superposition is required B Separation of variables in spherical polar co ordinates The Laplacian in spherical polar co ordinates is the most important case and is an es sential part of quantum mechanics as well as EM Laplace7s equation in spherical polar co ordinates is7 iazav 166V 162V 2V 7 7 7 67 7 v r2 67 T 67 7352716 668m 66 73527126 6 2 0 12 Again lets start with the case where there is no dependence on 157 so V0 6 b 7 V036 Rr6 Making this substitution into Laplace7s equation and dividing through by R9 gives7 1 6 6R 1 6 6 2 7 2 r V V 7 Bar 7 52716 9 a ym E O 13 The reason why we choose the separation constant C ll 1 will become clear later With this choice7 we have the two equations7 6 26B E0 ill1R0 14 and 1 3 39 m em ll1 i 0 15 The R equation is solved by Rr Alrl Blrl1 The 9 equation is more interesting and requires a series solution First we make the substitution7 u 0050 and Pu 967 which imply that 6 6 ism g 16 so that7 d 2 6Pu 7 1iu 6U lll1Pu70 17 This equation is call Legendre7s equation and is solved using a series solution7 Pu En Gnu Substituting this series into Legendre7s equation we nd that7 a 0 0 7 217 u20nnu 1 2 11 new 0 18 au n1 n0 and7 Z Cnnn 71u 2 7 Z Cnnn 1u Zll10nu 0 19 n2 n1 n0 The coef cient of u in this equation must be zero7 implying that7 nn 1 7ll 1 Cn2n2n10nll17nn1 0 hence On On n 1n 2 20 For a given value of l7 we get two series of solutions7 one starting with a value of CO and producing 027 C4 and the other starting with 01 and producing 037 C5 The key physical observation is that ifl is an integer7 the series terminates at n l7 leading to a nite polynomial solution In contrast if l is not an integer7 the series does not terminate and the coef cients remain nite at in nity7 a solution that does not have physicial meaning The constants CO and 01 are xed by requiring that the polynomials be normalized on the interval 7117 which corresponds to 77T in the original variable 0 The normalization condition is 2 inudu T 1 21 PHY481 Lecture 15 Chapter 5153 of PS We will not cover 54 55 Solutions to Laplace7s equations with variations in two directions In two dimensional cases where there is no dependence on one of the directions eg Z but there is dependence on the other two directions eg Ly In this kind of problem we can choose to use either Cartesian co ordinates or cylindrical co ordinates The domain on which the solution is required usually determines our decision of which co ordinate system to use If we are asked to solve a problem in a square or rectangular domain then it is usually better to go with Cartesian while on a circular domain it better to use cylindrical co ordinates In practice the domain is usually irregular so numerical computations are required It is relatively simple to do simulations in Cartesian co ordinates As we will see later A Separation of variables in Cartesian co ordinates in two dimensions Laplace7s equation is then 62V 62V 7 7 lt1 3x2 y This can no longer be directly integrated so we try to reduce it to a simpler form by assuming a solution of the form Vzy Subsituting this expression into Eq 0 we nd that dZX dZY 1 dZX 1 dZY iY X70 or 7777700n5tant7kz 2 dz dyz X dz Y dyz The requirement that the last expressions have to equal a constant occurs as if we have f gy then both f and 9 must be constant It is convenient to call the constant k2 as we shall see Also the sign di erence between the X equation and the Y equation and whether k is real or imaginary is very important and determines whether the solutions to X or Y are oscillatory or decaying functions From this expression we have Xz Army 31051 Yy O keky may 3 A more convenient form of these solutions is to write them as combinations of odd and even functions namely Xz Akcoskz Bksmkz Yy Ckcoshky Dksmhky 4 PHY481 Lecture 23 Chapter 8 of PS Chapter 5 of Gri iths A A circular current ring and magnetic dipoles Consider a circular loop of radius r centered at the origin and lying in the x y plane A steady current 239 ows in the positive a direction Find the magnetic eld on the z axis We use the Biot Savart where dT rd 3 and 13 is along the vector from a position on the circle to a point on the Z axis If we take the angle to the z axis to be a from the geometry we nd that dT rd 3 is perpendicular to The vector dTA 1 makes an angle of 90 7 a to the Z axis and its projection onto the x y plane is at angle b to the x axis On the Z axis by symmetry the magnetic eld points in the Z direction and we nd r r r 2 EM 1 Expanding the expression above at large distances 2 gives N MOZ39TZ 3r2 N MOZ39A 17 W N 223 2713923 BAZ 2 22 2 The leading order behavior is thus like that of a electrostatic dipole there is no monopole term like 172 for a localized current loop However if we consider a point magnetic charge which we call N and an opposite point magnetic charge that we call S Further imagine that this magnetic dipole where centered at the origin with N and S separated by distance d What would the magnetic eld look like First we have to decide what the eld due to a magnetic monopole looks like Though the magnetic monopole has never been found it is assumed that if it existed its magnetic eld would be just like that of an electric charge as we would have f g dfT uoqm so that a 7 0 gm B 7 E7 3 In that case the magnetic eld of a dipole would be 7 EM 4 T 47139 7 Now note that on the z 7 ans this reduces to 7 M0 27 747139 r3 W Z 7 axis 5 Comparing this expression with that for the current loop7 we see that they are the same7 provide we take the magnetic moment of the dipole to be7 773 2A This is a general result for current loops and provides a general concept that connects current loops to magnetic dipoles Notice that the origin of the difference between magnetic and electric elds is purely due to the difference in the sources of the elds charges for the E7 eld7 current for the B eld The elds themselves are perfectly analogous Since the magnetic and electric elds are completely analogous7 we dipole formulae also apply7 so that Uim1i F A o The Lorentz Force law Lorentz derived a general expression for the force on a charge7 q7 moving with velocity 17 in electric and magnetic elds E and E F E AE m A charge moving in a magnetic eld A charge q moving in a magnetic eld E with velocity 17 experiences a force7 TB 117 g 8 Or equivalently7 FB qusz39n0 9 where 6 is the angle between the velocity vector and the magnetic eld vector The direction of F3 is given by the right hand rule Note the following i The force on the charged particle is always perpendicular to both the velocity vector and to the magnetic eld vector ii If the particle moves in the direction of the magnetic eld7 it experiences no magnetic force iii If the particle moves perpendicular to the magnetic eld it experiences the maximum force iv Since the force is perpendicular to the magnetic eld lines and to the velocity vector7 the particle spirals77 around the magnetic eld lines The larger the magnetic eld the larger the magnetic force and the tighter the spiral v No work is done by the magnetic eld on the charged particle The kinetic energy of the particle is therefore a constant7 if no other forces act on the charged particle Constant B eld perpendicular to 17 A useful notation for drawing vectors in three dimensions A cross indicates a vector into the page A dot indicates a vector coming out of the page Consider the simplest case in which a particle of charge q has velocity 17 which is perpendic ular to the direction of the magnetic eld7 then7 FB qu 10 The direction of this force is perpendicular to both the direction of the magnetic eld and the direction of the velocity It also has constant magnitude The charged particle then undergoes circular motion7 with7 my qu 11 This is an incredibly important phenomenon A vast array of devices use the ability of magnetic elds to bend charges Note however that a more precise theory nds that power is dissipated by an accelerating charge7 including charges in uniform circular motion This is why high energy particle accelerators have very large circumference The radiation from an accelerated charged particle is proportional to the acceleration squared An important use of this effect is for producing high intensity X rays for a variety of purposes Now look at various quantities that are important in the circular motion of a charge in a magnetic eld i Radius of the orbit R milQB ii Period of the orbit 739 27TR1 27TmqB iii Frequency Cylotron frequency of the orbit f qB27Tm Notice that the period and frequency of the orbit do not depend on the radius They only depend on the charge to mass ratio qm and the applied magnetic eld B In addition7 if we know the speed of a charged particle and the magnetic eld7 we can nd the charge to mass ratio This is used in mass spectrometers Often the frequency of this orbit is called the cyclotron frequency The circular motion described about is the basis of understanding more complex motion For example PHY481 Lecture 28 Section 103105 of PS Sections 723 724 of Gri iths A Faraday7s law for many loops Faraday7s law is de ned for one one that is the emf generated around a closed loops is E ida Bdt However if we coil a wire so that the ux 15 passes through many loops of the wire then the total emf is multiplied by N the number of loops in the wire This is very important in a variety of applications ranging from magnetic eld detection to transformers and antenna If the ux through each of N loops is the same then we have 6 7N for N loops 1 B Self Inductance Faraday7s law relates the induced emf to the change in magnetic ux however the mag netic ux in circuits is often generated by currents so we would like to relate ux to the current and to write Faraday7s law in terms of the current This is easy to do as the ux comes from the magnetic eld and the magnetic eld is proportional to the current as is seen from the Biot Savant law or from Ampere7s law therefore b olt 239 Using this proportionality we write Faraday7s law as e 7N7 7L 2 This relation de nes the self inductance L that is dependent on the geometry and on the materials used but not on the current or the voltage Sometimes this relation is stated a different way through the ux linkage A Lz39 N B which leads to the same result as above Calculation of L is important as every circuit has a self inductance and its in uence can be dominant in some frequency ranges In general inductance is important at high frequency as there the current is changing rapidly so that the ux changes rapidly leading to high voltage through Faraday7s law In most cases calculation of L has to be carried out numerically or at least is quite complicated However there are a few cases where it can be done analytically and these cases are tabulated We now go through three simple cases Solenoid We nd L by considering a solenoid with N turns and length l and cross sectional area A in air A current 239 passes through the solenoid and from Faraday7s law we have N B Lz39 PHY481 Lecture 14 End of Chapter 4 and Start of Chapter 5 of PS Solutions to Laplace s equations By now we have gone through several methods for nd the electric eld and electrostatic potential of charge distributions The basic tools where Gauss7s law and superposition integration eg for rods7 discs etc In the last few lectures we have extended the superposition method to the case of metals where we found the induced charge on various metal shapes using the method of images Now we start using the di erential equation method7 ie Poisson7s equation VZV pE Actually in many cases we are interested in the electric eld and electrostatic potential in places where there is no charge density7 so we need to solve the simpler equation VZV 07 which is Laplace7s equation First we solve a simple problem using this approach A Conducting sphere or cylinder in a constant electric eld Now consider the case of a grounded conducting sphere placed in an electric eld In this case it is not clear how to use the method of images as there are no charges de ned in the problem Nevertheless we can use what we know about the solutions to Laplace7s equation to solve the problem We know the form of the multipole expansion and taking the monopole and dipole terms we have V030 Ar B 000597 2 where the angle 6 is with respect the the electric eld direction In addition a constant electric eld7 E07 in the 1 direction leads to a potential iEoz iEorcos Collecting these terms by superposition we argue that the potential can have the general form7 0050 T2 A V030 i B C Drcos l r We know that each term in this expression satis es Laplaces equation7 so the sum does too that is superposition again Note that this is not the general form of the solution in spherical co ordinates7 but it will turn out to be suf cient for the problem of a conducting sphere in an electric eld We don7t expect the rst two terms to contribute as the rst corresponds to a net charge on the sphere and the second to a nite potential By symmetry we expect the rst to be zero and the second is zero as the sphere is grounded To ensure that the solution is correct as r a 007 we need D 7E0 Now notice that the last two terms have the same angular dependence and cancel on the surface of the sphere if we choose 0 NEG7 therefore the solution is7 3 V036 iEocos r 7 in 2 T It is then easy to show that 2a3 a3 ET E01 730050 E9 7E01 7 73an 3 r r Notice that E9a0 0 as it must be to ensure that the electric eld at the surface of the conductor is perpendicular to the surface The charge density at the surface is 00 360E00056 lntegrating over the sphere surface gives the net charge on the sphere7 Q 27Ta25m0360E00050d0 0 4 0 which shows that no charge is transferred from ground to the sphere In that case this result also applies when the sphere is isolated However if we integrate over the top hernisphere7 7r2 Q 27Ta25m0360E00050d0 37Ta260E0 5 0 we see that a positive charge is induced in the top hernisphere There is an equal amount of negative charge on the lower hernisphere In a rnultipole expansion this system has no rnonopole terrn7 but it has a dipole term The dipole strength is proportional to the applied electric eld If the sphere is charged7 with an additional charge Q07 this additional charge is distributed uniformly on the sphere surface and its effect on the potential is found by superposition see eg Grif ths problem 3207 ie we add kQOr to the potential If we raise the potential of the sphere to a value V57 this is equivalent to placing a charge Q0 aVOk on the sphere Exactly the same procedure works for conducting cylindrical in a constant applied electric eld In this case7 the form of the potential is7 2 608 Drcosg a iEorcosrb Egg 6 Vr AlnrBC T by following the same reasoning as for the spherical case above The electric eld compo nents are ET E0005 1a2r2 E9 7E05 n 1 7 1272 and the surface charge density is 260E0005 lntegration shows that there is no net charge transfer to the conducting cylinder so this solution also applies for an isolated conduting cylinder in a constant applied eld Addition of a charge to an isolated cylinder leads to a uniform charge density on the surface of the cylinder and an additional term Alnr in the potential NOTE this calculation does not allow the cylinder to move What if we allow the cylinder to rotate What orientation do you think it will adopt B Proof of uniqueness Dirichelet V is xed on the boundaries Neurnan 3V6n is xed on the boundaries Uniqueness theorern states that if we nd two solutions to Laplace7s equation V1 and V2 then V1 and V2 are the same for Dirichelet boundary conditions while V1 and V2 differ at most by a constant for Neurnann boundary conditions Proof We have V2V1 VZVZ 0 so that V2V1 7 V2 V22 0 7 where we de ned 1 V1 7 V2 For Dirichelet conditions on the boundary we have V1 7 V2 110 while for Neurnann conditions on the boundary 3V1 7 V26 aoan 0 Apply the divergence theory to the funtion ovo so that a a a a o V oVodgr f oVo dA f oldA 0 8 vol 57M 57m 371 Now note that 671 UVZU W But V21 0 so we have WWW 7 f taldA 7 0 9 vol 7 Su39r 3n 7 Now the RHS is zero while the LHS is an integral over a quantity that is zero everywhere therefore 11 0 Therefore 1 constant For Dirichelet boundary conditions we know that o 0 on the boundary where the constant is speci d to be zero therefore V1 V2 For Neurnann boundary conditions the constant cannot be speci ed so V1 and V2 can differ by a constant C Separation of variables Separation of variables is a standard tool in all branches of physics as well as all other branches of science and engineering We will go through separation of variables for the three co ordinate systems of interest narnely cartesian polar and cylindrical co ordinates Alternative methods include Fourier transformation Laplace transforrnation You will see that the monopole dipole etc solutions occur in Laplace7s equation and that the separation 3 PHY481 Lecture 8 Sections 3537 of PS A The electric potential continued Mathematical Derivation The electrostatic potential can also be deduced on purely mathematical grounds using the relation TAE 0 This equation is satis ed when E7 i V due to the vector identity v 0 which holds for any scalar function V In terms of the electrostatic potential the di erential form of Gauss7s law then becomes VZV ipeo 1 This is Poisson7s equation and is the most commonly solved form of Gauss7s law The special case p 0 is also very important and is called Laplace7s equation In fact the whole of Chapter 5 of PS is devoted to Laplace7s equation Laplace7s equation is most non trivial when boundary conditions corresponding to nite charge or electostatic potential are used We then need to solve boundary value problems of partial di erential equations Methods for calculating the electrostatic potential 1 Direct method Find the electric eld by using eg Gauss7s law and then carry out the integral Vr7TEdf 2 If this integral diverges we cannot use a reference potential at in nity but instead can only talk of voltages across nite distances For example if the electric eld is a constant and directed in the f direction then we have Val7 iTrb E7 df 7Erb 7 Ta iEd if E is constant 3 where d Tb 7 Ta 2 Superposition Do a scalar addition of the contributions from each charge kg we V d 4 F wm View where the last expression applies to a continuous charge distribution with charge density p 3 Dz erentz39alform Here we use Poisson7s equation7 VZV 73 5 60 To complete our mathematical skills lets introduce the delta function7 which allows us to treat discrete charge distributions using Poisson7s equation The delta function has the properties7 607772 1 if 777 It is zero otherwise 6 In addition the integral of the delta function is one 5Lad 71 7 For a point charge at position 771 we then write7 pm 116077 2 lt8 Since the potential for a point charge at position is kQW m then we already know the solution to the differential equation7 VZGOTi 77 76077 77 9 It is G077 F39 147Tl777 where G is the Greens function That is7 ql gtk GEO is the potential for a point charge qi It is then clear that for a general charge density we have7 1 W i WWW m lt10 60 These rnanipulations seem pretty redundant in this problern7 however the concept of a Green7s function is central to higher level developments in all branches of physics ln electrostatics7 the Greens function is the solution to the smallest element of charge and superposition then gives the general solution for any charge distribution There are many ways to construct a function which obeys these relations and this is great for analytic work In fact a lot of theoretical physics requires working with delta functions Problem 319 of the third hornework uses one form of a one dimensional delta function Another one which is used a lot is the Gaussian Note that the delta function is de ned by a limiting process in both cases There are other forms which use integrals to achieve the same effect Choosing the right form of the delta function in complex problems is quite an art that takes some e ort to master i Electric potential due to a point charge The electric eld due to a point charge is a ka E 7 11 The electric potential due to a point charge is then vierfdiTkadrfikQT 12 00 W 00 7 00 r ii Electric potential due to an electrostatic dipole The electric potential and electric eld of an electrostatic dipole are given by V 7 5quot d E 7k3 39ff d 13 7 i 7 an F 7 T 7 gtgt ln problem 337 you will calculate the electric eld from the potential Actually you already did this in Assignment 1 so it is just a reminder Below we explicitly show how the expression for the dipole potential is calculated and why it is only valid for 7 gtgt d You can solve problem 338 of the homework using a similar expansion except in that case the expansion has to be taken to order dr2 Consider a dipole lying on the Z axis with the positive charge at z d2 and the negative charge at z 7d2 The charges have magnitude q A point in the x y plane is de ned by its radial distance to the origin 7 and by an angle 6 between the direction f and the Z axis We want to nd the electric potential due to the dipole as a function V030 The exact expression for the potential of a dipole is then 1 1 V039 k4 I 14 where 2 d 2 2 d d 2 2 i r 7 2 rc05180 7 l9 r 7 drcos0 15 and 2 d 2 2 d d 2 2 hr r 7 2 rcos0 r 7 drcosl9 16 Using these expressions we rewrite Eq 8 as kg 1 1 V o W 12 7 12 17 r gt 42 T2 7 m lt1 m 1 lt gt 2 52r2 2r2 3 PHY481 Lecture 29 Chapter 9 of PS Chapters 67 of Gri iths A Energy stored in inductors and in magnetic elds An external voltage source must be used to set up a magnetic eld in an inductor The rate at which work is done by the external source is idW P775m1 1 d ltgt We have shown that the voltage across an inductor is iLdIdt so we have dW d W L1E 2 The total energy stored in an inductor is the integral of the work so that dW 139 d 1 U7dtL17dteL392 3 dt 0 dt 2 Z This energy is stored in the magnetic eld of the inductor By considering a solenoid we can nd the energy density in the magnetic eld through 7121 2 21 2 U72Lz 72071 Al 72MB Al 4 Using Al volume the energy density in the magnetic eld is then to be 1 u E2 27m 5 We have introduced the concept of self inductance which characterizes the ability of a wire con guration to store energy in the form of magnetic elds This energy storage can be large and has been proposed as an alternative to batteries provided superconducting wires are used in order to minimize the resistive losses As you will see in the assignment very large amounts of energy are stored in the earth7s magnetic eld and in the magnetic elds of galaxies B Mutual inductance Self inductance is de ned through Nab Lz39 A where A is the ux linkage Self inductance is a measure of the ability of a conducting system to react to a current ow in the system However from consideration of Faraday7s law it is clear that the magnetic eld of one current carring wire con guraiton may in uence a second one that is not physically connected to it This coupling is described by the mutual inductance We write d d39 N2 2 Mlgll SO lZ iNgg lling This coupling is the basis of antennas transformers and a variety of other essential elements of technology Calculations of mutual inductance are similar to that of self inductance We consider two examples A rectangular loop near a wire carrying current i1 and N2 N1 1 We take the loop to have its long sides of length l2 parallel to a long straight wire carrying current i1 The side which is closest to the wire is distance a from the wire The short sides have length b1 and the normal to the loop is in the i5 direction with respect to the axis of the wire In that case we need to nd oz Mlgil so we have a by now quite familiar calculation i2 0l dy AW gsznaa ba so that M12 gznaa ba 7 Two solenoids one inside the other The two solenoids are characterized by cross sectional areas A1A2 turns N1N2 and lengths l1 l2 with the former having A1 gt A2 We also take the lengths of both solenoids to be much larger than their radii so the formula B nine applies with n Nl If a current i1 ows in the larger solenoid the ux in the smaller solenoid is then 152 NlilAgnoll and the so using N2 2 Mlgll M12 NlNZloAgll Now lets consider the reverse situation where a current i2 ows in the smaller solenoid In that case there is no magnetic eld outside the smaller solenoid so the ux in the larger solenoid is 151 NZiZMOlg then using Ni 1 M1222 M21 NiNZloAglg In this expression we use A2 because we assume that there is no eld outside the smaller solenoid N is the number of loops in length l2 so Ni lznl lgNlll With this substitution it is evident that M21 M12 so the mutual inductance is symmetric Now suppose we want to nd the relation between current and voltage in the rst solenoid We have I V1 7L7 7 M7 10 That is both the directly coupled loop and the remotely coupled loop oppose a change in ux We will return to this later after we have studied magnetic materials as it is most naturally de ned in terms of transformers where magnetic materials are used to control the direction that magnetic ux takes C Maxwell7s equations in magnetic materials Diamagnetic materials expel magnetic ux while paramagnetic and ferromagnetic ma terials provide favorable pathways for magnetic ux Strongly diamagnetic materials such as type I superconductors expel ux so strongly that they can levitate when placed on top of a magnetic pole Strongly ferromagnetic materials provide such good pathways for ux that they are used in magnetic circuits such as in transformers where ux is directed from the primary coil to the secondary coil Understanding of these responses starts with an understanding of magnetization and of currents that lead to magnetization Recall that in dielectric materials we de ned the polarization 13 as the electric dipole moment per unit volume In magnetic materials we de ne magnetization M to be the magnetic moment per unit volume As in the discussion of dielectric materials we shall discuss the general formulation and then concentrate on linear isotropic magnetic materials where M mef Recall that linear dielectric materials obey 13 eoxe Recall that bound charges produce polarization and that at surfaces the bound charge is related to the polarization through 0 13 while the bulk bound charge density is m 7v Since the sources of magnetic eld in Maxwell7s equations of magnetostatics are steady currents it is evident that the bound currents can produce magnetization The relations between bound currents and the magnetization are KbM jbvAM 11 The reasoning behind these expression is similar to that used to understand bound charges Basically if the magnetization is a constant only surface currents are required to produce the magnetization while if the magnetization varies in space a bulk bound current density is needed With these de nitions Ampere7s law becomes vA Mo Mo vMooWAHMvw 12 where H is the magnetic intensity and obeys TAHf sothat fHdfif and u0 l17I 13 From this equation it is seen that Amperian contours can be used to relate if to the magnetic intensity or Auxilliary eld The magnetic intensity and magnetization have units Amp per meter as can be seen from the de nition of Two types of magnetic material problems are typically posed Given a magnetisation nd the magnetic eld eg inside a uniformaly magnetized cylinder ii Given a linear magnetic material nd the magnetic eld inside and outside the material eg a linear magnetic sphere in the presence of a uniform applied magnetic eld Later we will look again at solenoids and transformers to see the effect of using magnetic materials in these systems particularly the effect of magnetic materials on the energy stored in an inductor A uniformly magnetized cylinder with length gtgt radius We take the magnetization to be uniform A71 Md and to be oriented along the axis of the cylinder There are no free currents so there is no gain in using the magnetic intensity To nd the magnetic eld note that a uniform magnetisation is like the uniform eld inside a solenoid so surface currents owing around the cylinder provide a solution The system is linear so if we nd a solution it is unique Alternatively we can nd the suface currents through Rb M r Moog so that Elm uoKbh 14 Where the magnetic eld inside the cylinder is then found using an Amperian loop and assuming that the eld outside the cylinder is zero This result applies in central regions of the magnetized cylinder If we now consider a point P which is well removed from the cylinder ie r gtgt l then we can treat the cylinder as a dipole with dipole moment 771w Volume gtlt A71 and then use the dipole formula a no 307102 7 77h B 7 15 T 47139 TS A Uniformly magnetized sphere of radius R PHY481 Lecture 25 Chapter 78 of PS Chapter 5 of Gri iths A Basic equations of magnetostatics The analogy with electrostatics is very close7 with the major di erence that the sources of magnetic elds in magnetostatics are current rings and the intrinsic magnetic moment of elementary particles a N7 7 a S enables us to draw magnetic eld lines in analogy with electrostatics Basic equations7 f d60 v 0 noname 1 which implies that g ln magnetostatics we choose the Coulomb gauge7 mamafl 0 fg df not 6 no Amperes law 2 Boundary conditions on E Bi 7 Bi 0 from 1 Bf 7 Br M from 2 lt3 Superposition laws a town a Mo r Fi dgr dB 7 B 7 4 47139 r2 or 47139 777 W3 and a a Mo jWd3r c A E W 01 VZA qu 5 This last equation looks like 3 Poisson7s equations In parts of space where there are no current sources7 we have 3 Laplace7s equations We can then solve these equations with various boundary conditions To do this we need the boundary conditions on the vector potential7 that are7 9A gen 7 AVnt i 7 7 an ext 7 an lemt 7 MOR 6 Another useful relation7 frdrv da da g lt7 In some cases this makes it easy to nd the vector potential from the magnetic eld7 eg a solenoid7 where B pom so that 3 Ba7 and Ar emu2m The direction is the 1 same as the direction of the current from supperposition law above B Magnetic dipoles There are two sources of magnetic dipoles currents and intrinsics magnet moments of elementary particles In either case7 we have a magnetic dipole moment 77 that gives the magnitude and direction of the magnetic dipole This magnetic dipole moment is directly analagous to the electric dipole moment 15 and we have7 a 3M7 r Few T277149 UseB 8 747139 7 Two key new things here are that the magnetic moment of a current ring is given by7 ml a 7712 7 9 where 6 is the area of the current loop and the direction is normal to the current loop ii In addition we have a new relation for the vector potential of a magnetic dipole7 a AE7 10 It is a useful excersize to show that 6 fT gives the dipole eld formula this was an assigned problem 28 C Lorentz force law and force on current carrying Wire A charged particle of charge q with velocity 17 moving in a magnetic eld E and electric eld E experiences a force given by7 qE 17A Lorentz force law 11 A wire segment carrying current 239 and with vector length dfin a magnetic eld E experiences a force7 dF 239de E 12 D Additional equations related to current The current and current density are related through7 d4 13 PHY481 Lecture 12 Sections 42 43 of PS A More on problems solved using image charge methods First lets solve the simplest image charge problem completely using polar co ordinates The geometry is a grounded semiin nite metal with normal along positive I and metal for all z lt 0 A charge q is placed on the Z aXis at position 20 Find the potential for z gt 0 From this expression show that the potential at z 0 is zero Find the electric eld and show that at z 0 the electric eld is directed along the Z aXis Using the image charge method we have 1 1 V 9 k r i W T a r2 23 7 2r20005012 r2 23 2722000519l2J 1 The surface 2 0 corresponds to 6 7r2 At this value of 6 it is easy to show that the potential The electric eld in polar co ordinates is given by 3V A 1 3V A 1 3V A 77 7 77 lt2 37 r 30 7192716 6 The a component is zero as there is no b dependence in the potential Doing the derivative for the other two components yields 2052710 2052710 E k f 1 3 9 a r2 23 7 2r20005032 r2 23 2r20005032J and E 7 L r 7 200050 7 7quot 200056 4 T 7 WWquot 28 7 2r20005032 r2 28 27quot200051932J Now we need to show that these equations satisfy the required boundary conditions that Vz 0 0 and that 0 7E9z 0 that is the potential at the surface is zero and the electric eld is normal to the surface In polar co ordinates the surface is at Vr7r2 Evaluating this demonstrates that Vr7r2 0 as required To show that the electric eld is normal to the surface we need to show that Er7r2 0 and E90 7T2 is nite These results follow from Eq 3 and 4 moreover we nd that 2kq20 E90quot 7T2 W 5 Finally we would like to show that the total induced charge at the surface of the metal is 7g to do that we use the fact that at a metal surface E U eo which follows from Gauss7s 1 law The induced charge at the metal surface is given by 00 00 2quo dA7 2 d E7 2 d i 6 Q 0 0 7139 T60 9 0 7139 T60 T2Zg32 q The minus sign in the second expression arises as the 6 direction is 77 for this surface Some problems can be solved using many image charges one example where four image charges works is the case of a charge q at aa0 when the regions y lt 0 and z lt 0 are conducting and grounded In that case three image charges iq at a 7a 0 q at 7a 7a 0 and g at 7aa 0 ensure that the potential on the surfaces of the metal regions are all zero In a similar way the method of images can be used for a wedge of angle 7rn the case above has wedge angle 7T2 when a charge is placed on the central axis of the wedge In that case alternating image charges placed at symmetric positions at the centers of all wedges inside the metal sum perfectly to ensure that the potential is zero on the wedge surfaces In some cases an in nite set of image charges can be used with one example being a charge lying between two grounded metal sheets at locations 2 d and z 7d The solution is found by adding image charges iteratively to nd 00 k W ml lt7 This can be generalized to the non symmetric case by using a similar procedure This approach can also be used for some problems involving two spheres or two cylinders Another case which can be solved using one image charge is the case of a charge outside a grounded spherical conductor Now we can try to solve the problem in the same way as for the at surface but now using spherical co ordinates centered at the sphere center There is azimuthal symmetry so we can restrict consideration to a function V030 and we have the boundary condition VR0 0 as the sphere is grounded The point charge is at 20 gt R In this case we need to allow the image charge to have a general location and a general charge so that q 1 V 9 kl i T L0 28 7 2r20005012 r2 262 7 2r26005012J 8 Clearly there are two unknowns which can be determined by choosing two convenient loca tions for example 6 0 7T 7 B These two cases imply that kq kq kq W i 0 i 0 9 R720 R720 y RZO RZO y Taking the minus sign of the rst equation with the positive sign on the second equation and subtracting the two equations yields7 qR R2 q 77 20 7 20 20 10 Using these expressions in Eq 2 and simplifying yields7 1 R V 9 k i i i T a r2 23 7 2r20005012 r223 R4 7 27quotz312009t9 lL2J 11 Now lets check that this is zero for all points on the surface of the sphere7 by evaluating at r R7 where7 1 R V R 0 k l 7 l 12 7 a R2 23 7 2R20005612 R223 R4 7 220R3005012J It is clear that this is zero take R2 out of the square root in the second term The electric eld in E7 7vV ET E9 E457 where E 0 as there is no b dependence in the potential For the other two components7 we have7 7r20 52710 r20 R3 52710 1617 kq 7 f E 77 7 9 r 30 7 L0 28 7 2r20005032 l 13 r228 R4 7 2r20R2005632J Evaluating this at r B it is again easy to see that it reduces to zero take R2 out of denominator of the second term leads to a factor of RC 7 which cancels with the R3 in the numerator This is necessary to ensure that there is no parallel or tangential component to the electric eld at the surface of the sphere ie the eld must be normal to the surface and this normal component is given by the electric eld in the radial direction7 3E r 7 200050 rzgR 7 20R30050 T 7 77 kql l 14 37quot r2 23 7 2r20005032 r223 R4 7 2r20R2005632 Evaluating at r R gives7 R 7 200050 R222 7 20R36080 ET B k l 0 1 15 z 7 20005 2 7 20 005 l l R2 3 2R 932 R23 R4 2B R2 9321 l which reduces to7 kq R2 7 23 E R 7 16 R R2 23 7 2R20005632 so the surface charge on the sphere 06 b 60E0 b is given by7 W32 i 23 6 17 U 47TRltR2 23 7 2R20005032 PHY481 Lecture 13 A line charge near a grounded conducting cylinder A line charge A at position 20 on the X aXis is near a grounded conducting cylinder of radius R which has its center at zy 00 with its central axis along the Z aXis The line charge lies parallel to the central axis of the cylinder Find the electrostatic potential for r gt R in plane cylindrical co ordinates r 15 Note that the potential does not depend on 2 Using Gauss7s law it is easy to show that the electric eld near a uniform line charge is E70 Af27T60r The potential is then of the form Vr Alnconstantr27T60 The constant is chosen to t the boundary conditions Let7s assume that the problem of a line charge near a grounded conducting cylinder is solved by using an image charge which is located at position 6 and with charge per unit length 7A Now we need to nd 6 and we need to check that Vr 0 and that ER E773 The potential is given by superposition so that V0 b lnclr1 7 lnclr2 02 1 27TEO where 01 and 02 are constants Using the cosine rule we have 2 7 2 2 2 7 2 2 r1 7 r 0 7 2rxocos r2 7 r 0 7 2rz0005 2 Lets also assume that the reciprocal relation holds why not ie 26 Rzzo We then have A r2 R4z2 7 2TR2xocos V 71 0 3 T b 47TEO n r2 3 7 2rz0005 CZ At the surface of the cylinder we have A R2 R2 2 7 2Rx0005 V R 71 7 0 4 lt 4 460 n3R2H372Rx0008 62 ltgt This must be zero for our solution to be correct which implies that 02 72EolnRxo 5 The solution to our problem is then A l71 T2 BAX3 T 2TRZgcoww 7 AanRyco 6 V0345 260 47TEO r2 3 7 2rz0005 1 or Vo jw i A lnltxgr2R472r0chos 7 A 47TEO WP lt7 Now we need to check that the electric eld is given correctly We nd that r2 3 7 2rz0005 27TEO 61 7 7A 2rx0R25m i 2rzos n r 6 7 47m 37quot R4 7 2rz0R2005 r2 3 7 2rzocos E i 8 From Eq 7 it is evident that VR b 0 as required and from Eq 8 we nd E R b 0 We have therefore found a solution which satis es the boundary condition so it is correct For completeness the electric eld in the radial direction is given by 3V 7 7A 271 7 20R2005 2r 7 2zocos E if L 7 L 37 47TEO 37 R4 7 2rz0R2005 r2 x3 7 2rzocos Closing remarks on generalizing image charge problems We have solved three basic image charge problerns a point charge near a grounded at conducting surface Lecture 12 ii a point charge near a grounded conducting sphere Lecture 12 iii a line charge near a grounded conducting cylinder This Lecture Lets call these solutions VOW The extension to problems where the conductor is at some nite voltage instead of zero requires adding charges to produce that voltage The charges have to be placed syrnrnetrically to ensure that no electric eld is generated in the metal Eg if we want a sphere of radius R at potential Vb then we place an image charge Q0 at the center of the sphere so that Vb kQOR This corresponds to distributing the charge Q0 uniformly on the surface of the sphere The electrostatic potential for r gt R of this problem is found by superposition ie VF V00 kQOr In the case of a conducting slab a sheet of image charge is placed at the center of the slab while in the case of a conducting cylinder a line charge is placed at the center of the cylinder In a similar way if we are given a problem where a point charge is near an isolated conducting sphere cylinder or slab which has total charge Q then we again have to place an image charge at the center of the sphere However now the magnitude of the image charge at the center of the metal sphere has to be the sum of the total charge on the sphere plus the value of the image charge of the grounded system For example an isolated conducting sphere of charge Q requires that an image charge of Q 7 1 be placed at its center so the total potential for r gt R becornes V007 kQ 7 q r where q 7qRzo is the image charge of the grounded sphere PHY481 Lecture 5 Sections 3135 of PS All of electrostatics follows from Coulomb7s law superposition A Coulomb7s law Force between two charges The starting point in electrostatics is Coulomb7s law which gives the force between two stationary charges a qu Qq Fk rk r 1 Qq are stationary charges Their units are coulombs C f is a unit vector along the line between the two charges F is the vector distance between the two charges k 9 X109Nm2CZ 9 X109kgm30252 k 14713960 60 is the permittivity of free space E Force between many charges superposition Force on a charge q due to many other charges Q1 Q2 Q is just the sum of the forces due to each of these charges ie a quot Qiq A Eat k 2 Ti 1 Ti 1 This looks simple but of course it is a vector sum so the math can get messy r is the distance between charge 239 and the charge q f is a unit vector along the vector which goes from charge q to charge Q The principle of superposition also applies when there is a continuous distribution of charge For example charge distributions on rods discs spheres etc However when treating these distibutions the sum in Eq 5 becomes an integral ln treating these problems we de ne a small element of charge dQ This is the amount of charge in a small part of the continuous charge distribution We shall consider three cases Lines Then dQ Adz where A is the linear charge density Surfaces Then dQ UdA where 039 is the surface charge density Volumes Then dQ pdV where p is the volume charge density Ear ample uniform ring of charge Consider a thin perfectly circular ring centered at the origin with radius R lying in the X y plane and being uniformly charged with linear charge density A Find the force on a charge q placed at position F 002 Solution By symmetry the X and y components of the force on the charge are zero We are thus left with the task of calculating the Z component The distance r from the charge to any point on the ring is given by r2 R2 22 3 If we de ne 6 to be the angle that the vector F makes with the Z axis then 0050 27quot 4 The total force on the charge is then Q2 7 27 16qu kg 2 kg Fi 0 7 6 R22232 cos if r2 R2 22 r 27r ARdgs 0 where we used Q 27TRA C The electric eld De nition Electric eld force on a unit charge The electric eld at a point que to a charge distribution is de ned in terms of the force on a positive test charge q placed at position F The precise de nition is F F mameo e We can think of the electric eld as the force per unit positive charge at position F Notice that once we have found we can nd the force on any charge q by using Since the Electric eld is basically a force superposition applies to the electric eld All of the properties of the electric eld can be derived from Eq and superposition as we now show First from the de nition of the electric eld we write Coulomb7s law as kf a which is the electric eld of a point charge In a similar way7 the electric eld due to a distribution of discrete charges is given by7 EYF Z 8 l 1 Below we show how the integral and di erential forms of Gauss7s law follow from this expression An extremely useful concept in developing these results is the concept of an electric eld line An electric eld line is a series of vectors where at each point the vector points in the direction of the force on a unit charge at that point and it has a length equal to the magnitude of the force ie we plot the vector function The proporties of electric eld lines constructed in this way are as follows Erample discrete charge distribution Consider three charges located and xed at X7y positions F1 0 0 F2 2 0 F3 0 2 in meters and having charges ql qz 2710 qs 73710 Find the force acting on charge 13 due to the other two charges What is the electric eld at position F3 due to charges Q1 and 12 Solution We need to treat the force as a vector In this problem the best way to treat the force is to resolve the Coulomb force between each pair of charges into its X cornponents and it y cornponents The charges form an isosceles right angle triangle7 with charge ql lying at the right angle corner The other angles in the triangle are 450 The X cornponent of the force on charge 13 is given by7 12st f1 chos45 0 9 T23x The y cornponent of the force on charge 13 is given by7 12 k smxm lt10 723 T13 The electric eld at position F3 due to charges Q1 and q2 is7 Em fwQ3 kqiicosOlS Ey fyqs kqiismMB qul 11 723 723 713 Example uniform ring of charge Find the electric eld for the thin charge ring discussed above Solution sz Ez0 Ey0 EzFzQW 12 D Electric eld lines Faraday7s ideas 1 At each point along an electric eld line7 the force on a positive test charge is in a direction tangent to the eld line at that point 2 The density of lines at any point in space is proportional to the magnitude of the electric eld at that point 3 Electric eld lines begin andor end at charges7 or they continue off to in nity ie they do not begin or end in free space 4 Electric eld lines do not cross Very important special case conductors 1 If there is no current owing7 then the electric eld is zero7 E 07 inside a conductor 2 If there is no current owing7 then at the surface of a conductor7 the electric eld is normal to the surface of the conductor E Gauss7s Law The integral form of Gauss7s law in free space is7 f E all W 13 S 60 where gem is the total charge inside the closed surface S This law follows from Coulornb7s law and superposition in combination with the properties of electric eld lines The proof of Gauss7s law in general follows from the following statements Property 1 For a charge q with a spherical shell at radius r it is easy to prove that Gauss7s law is correct ii For a charge that is outside of a spherical shell it is also easy to prove that the total ux through the shell is zero This is proven by noting all ux lines originate or terminate at a charge7 or go to in nity Therefore a ux line originating from a charge outside the spherical shell and which enters the spherical shell must also leave the 4 PHY481 Lecture 18 Chapter 6 of PS Chapters 6 of PS covers the use of electrostatic theory to understand the effect of electric elds on dielectric materials and the generalization of Maxwell7s equations to the calcula tion of electric elds and electrostatic potential inside and outside of these materials The treatment of dielectric materials begins with observation that the behavior of dielectric ma terials can often be understood based on the concept of polarization Since this concept is so central to the discussion it is worth de ning it formally Polarization at the atomicmolecular scale requires a treatment of the response of atoms and molecules to the dielectric properties of their neighbors and to an applied eld These calculations are a subject of intense research as dielectric materials underlie many aspects of integrated circuits insulators capacitors energy storage devices capacitors batteries and insulators in the electrical power industry Two key materials properties are the dielec tric constant and the dielectric breakdown eld Power insulation requires high dielectric breakdown eld while energy storage needs both high dielectric constant and high break down eld lC applications always requires high breakdown eld and sometimes either high or low dielectric constant High dielectric constant implies a tendency to high polarization while a low dielectric constant implies a lower tendency to polarization Currently there is an intensive worldwide search for improved dielectric materials De nition of Static Polarization Polarization is the average density of aligned electric dipoles inside a material Polarization can be modeled at the atomicmolecular level where the polarization is indicated by 15 However Maxwell7s equations are a continuum theory found by averaging over the local behavior The continuum or macroscopic polarization 13 is the fundamental quantity in Maxwell7s equations in dielectric media The macroscopic polarization is found by averaging the atomicmolecular polarization over a small region that is nevertheless much larger than the atomic scale 13 2 lt1 igav A Polarization at the atomic molecular scale To understand the broad aspects of dielectric materials and polarization we rst look at two simple models of atomic level polarization These models apply to dilute gases where the effect of the neighboring atomsmolecules is ignored The rst model applies to both polar atomsmolecules that have a permanent dipole moment and to non polar atomsmoledules that do not have a permanent dipole moment The second model is an additional polarization arising in materials with a permanent dipole moment Induced polarizationdipole moment due to an electric eld Atoms and symmetric andor covalently bonded molecules do not have a permanent dipole moment Nevertheless these molecules do respond to an electric eld In the simplest case the polarization oriented dipole induced by an electric eld is linear in the magnitude of the electric eld so that 5 LE 2 where 04 called the polarizeability In the case of a metal sphere we found that p 47TEOR3E so that 04 47TEOR3 It is not too surprising that the polarizeability of inert atoms is about the same as this value and that bigger atoms have larger polarizeability Of course this can also be calculated quantum mechanically Taking the radius of Helium to be about 1 Angstrom we nd 044713960 10 30m3 while a full quantum calculation gives 23 gtlt 10 30m3 The polarizeabilities of atoms and molecules are tabulated in many places and are one of their fundamental properties The induced dipole dipole interaction is dependent on the polarizeability and is the origin of the attactive term in dispersion forces also known and the van der Waal7s attraction or the attractive part of the Lennard Jones interaction Quantum mechanically the dipole moment is lt wl 7 eflw gt where 7 is the wave function in the presence of an electric eld For the Hydrogen atom the result is 9 15 mg Eo while a conducting sphere of radius aB has 3 while a conducting sphere of radius 13 have an induced dipole moment given by 1 3 p gaBEo lt4 The reason that the conducting sphere of radius 13 has a lower polarization as compared to the H atom is that the application of the electric eld leads to mixing with higher orbitals of the H atom and their orbitals have signi cantly larger radii than 13 Eg if we use 213 instead of 13 we get a factor of 8 in the conducting sphere case The choice of the radius used for the sphere is thus critical Nevertheless the trend is captured correctly Polarization of permanent dipoles due to temperature and electric eld Molecules that have ionic bonding have permanent dipole moments for example H20 The dipole moment is larger for more ionic bonding and for larger molecules The unit of molecular dipole moments is 1D where D stands for Debye 1D 334 gtlt 10 300771 10 gtlt 10 210m2s The permanent dipole moment of a water molecule is 184D If we consider that in H20 the H atoms donate their electron completely to 0 then we would nd p 2ed actually the charge transfer is smaller than this where the bond length is about 1Angstrom and the bond angle is about 1045quot If we take the effective distance between the positive and negative charges to be 05Angstrom we get p 10490771 x 3D This is in the right ballpark but is too large primarily because we assumed perfect charge transfer in the ionic bond When an electric eld is applied to a polar molecule two things can happen there is an induced polarization as occurs for non polar molecules and there is tendency toward alignment of the permanent dipoles in the molecules with the applied eld due to the energy gain U 715 ln small polar molecules the permanent dipole term dominates and the polarization effect of an applied electric eld is due to the alignment of the permanent dipoles with the electric eld rather than the induced polarization In this case we need to use statistical physics to calculate the polarization effect This calculation proceeds by assuming that initially we have a set of dipoles of magnitude p0 but that are randomly oriented due to thermal disorder and there is no macroscopic polarization When an electric eld is applied the dipoles can reduce their energy by aligning with the electric eld We calculate the average polarization by using statistical physics to average over all orientations of the molecules so that for an electric eld in the z direction lt pz gt 02quot f3quot e UkBTpocos clobsinQdQ 0739 f0 e UkBTdobstdQ 5 or 02739 f0 ePOEmQkBTpocos do sinQdQ W Doing the o integral and using 5 0050 this reduces to ltpzgt E1 duepoEukBTudu 7 p0 6ault71 au 1 i 1 filduemEuWdu a win 1 a Cot1a lt7 ltpz gtpu where we used f e mudu 6 71aua2 and a pOEkBT This is called the Langevin formula and is a non linear relation between the applied electric eld and the polarization In the high temperature limit it becomes linear and we have lt pz gt ngBkBT in that limit with polarizablility Oz 040 pgBkBT 8 where 040 is the polarizability of the atomic orbitals and pgZ kaT is the average alignment of the permanent dipoles Most of our discussion will involve linear dielectrics where the response at the atomic level is of the form 5 LE 9 As we shall demonstrate below at the macroscopic level there is often a linear response which is written in the form V 13 60X5E 10 where X5 is the dielectric susceptibility and is the quantity that is frequently measured It is important to nd connections between the atomic level polarizabilities oz and the macroscopic behavior characterized by the dielectric susceptibility X5 We shall discuss some simple models for this relation below But rst we need to develop some of the macroscopic concepts further particularly the connection between 13 and the concept of bound charge B Polarization and bound charge Polarization and bound charge are the two most fundamental quantities in extending Maxwell7s equations to insulating dielectric materials The macroscopic polarization used in Maxwell7s equation is related to the dipole moments at the atomic level We call this variable 13 which can be thought of as an average over the atomic level dipole moments in a very small region of the material we can think of the material being made up of a vast number of small electric dipoles so that W i 5 at mew lt11 6V 6V m We want to generalize Maxwell7s equations to take account of the effect of these dipoles in addition to the 77free77 charges that we treated before when we studied metals Note that the term 77free charge77 is a misnomer7 the free charge is really excess charge or net charge that appears for example in Gauss7s law In contast7 the charge in the atomic level dipoles is always balanced7 ie there are as many negative charges as positive charges For this reason the charges in the dipoles are called the bound charges Therefore we introduce two charge densities the free charge pf and the bound charge pb Many of the methods we developed before can be extended to nd the electric eld and electrostatic potential generated by the free charge pf The new aspect we have to understand is the bound charge pb To understand the origin of bound charge consider a sheet of dielectric material where the polarization is aligned in the direction normal to the sheet surface We also assume that there is no applied electric eld7 so this materials is an electret or ferroelectric which can have nite polarization in the absence of an applied eld At the top of the sheet7 there is an excess of positive charge7 while at the bottom of the sheet7 there is an excess of negative charge If the polarization is uniform7 the interior of the material is charge neutral The excess charge density at the two surfaces which must be equal and opposite is called the bound charge From this discussion it is clear that bound charge occurs due to the fact that the dielectric is made up of little dipoles Moreover it is evident that the bound charge density is proportional to the polarization The relation between the density of the polarization 13 and the bound charge is7 ab 13 12 Due to the bound charge at the surfaces of the material7 there is an electric eld inside the material E7 01760 Another important relation is that the density of bound charge is related to the polar ization through7 Pb iv 39 13 13 This is a generalization of the bound charge at a surface and can be understood in a similar way First note that if there is no gradient in the dipole density7 there is no gradient in the bound charge remember that is averaged over regions much larger that the dipole size d However if there is a gradient in the polarization then there is a constant bound charge density in the interior of the material Once the bound charge density has been determined7 we can calculate the electric eld inside the material from the bound charge7 assuming that there are no other charges in the 5 PHY481 Lecture 11 Sections 42 43 of PS A Electric elds near two charged metal sheets The two sheets have the same area L2 and have total charge Q on the top sheet and q on the bottom sheet The two metal sheets have thickness 1 and are separated by a distance d ltlt L so we can treat the sheets as in nite The sheets have normals oriented along the Z aXis Let the charge on the top of the top sheet be Q2 and the charge on the bottom of the top sheet be Q21 The charge at the top of the bottom sheet is QM and the charge at the bottom of the lower sheet is Q11 We need to nd these charges the electric eld between the sheets and the voltage difference between the two sheets The sheets are isolated so charge conservation requires that Qlu Q11 9 and Q2u f Q21 Q 1 In addition we require that the electric eld inside the upper metal sheet and inside the lower metal sheet are zero Using superposition to calculate these electric elds we nd note that Gauss7s law does not help us here Q21 7 271 Q11 Qlu 260A 260A 260A 260A and 271 Q21 Qlu Q11 260A 260A 260A 260A Combining the second of equations 1 with Eq 2 we nd Q2 Q q2 Similarly we nd that Q21 Q iq2 and QM 7Q iq2 and Q11 Qq2 The outer surfaces of the two sheets have the same charge and inner surfaces have charge of the same magnitude by opposite sign This is consistent with the results found for the case Q Qq 7Q where there is no charge on the outer surfaces Notice that if the charges of the two plates are the same then all of the charge is on the outer surfaces of the metal sheets The electric eld is found by superposition or by using Gauss7s law Using Gauss7s law is easier but it is good to check it using superposition We nd 7 Q 7 a U Ebetween z 7 y Euppe39r iEloue39r 0 2A60 Q a 2A60 Finally the voltage between the plates is Ed V with the positively charged sheet at the higher potential B De nition of capacitance Capacitance measures the ability of a system to store charge It is assumed that if the applied voltage is zero no net charge is stored and that when a voltage is applied charge starts to be stored The basic geometry is set up by attaching one piece of metal to the positive lead of a battery while another piece of metal is attached to the other electrode of a battery The fundamental relation is then Q C V 5 where V is the voltage of the battery and Q is the charge stored on the pieces of metal Q on the piece of metal attached to the positive electrode of the battery and 7Q on the piece of metal attached to the negative electrode From the calculation above for two metal sheets we have Qd a QUA Eg and VEdE 6 Which can be written as QVOV 7 so that the capacitance of a parallel plate capacitor is Cplate eoAd as you know However the de nition of capacitance of Eq 11 is true for any linear dielectric which applies to most materials in the low voltage limit Example 1 A coamz39al cable Consider two coaxial cylindrical metal shells of radii a and b where b gt a and d b 7 a A voltage V is applied to the outer cylinder and the inner cylinder is grounded so the total charge on the two shells is Q on the outer and 7Q on the inner shell The electric eld for r gt b and for r lt a is zero by Gauss7s theorem The electric eld for a lt r lt b is found from Gauss7s law E27TTL iQeo so that Er i QLEO 8 Now we nd the potential di erence by integration b a a b Q Q V7Edl7dil b 9 b a a QWTLEO T 27TL60 09 a The voltage di erence is V V as Va 0 because the inner cylinder is grounded Note also that we are not using the usual rule that the potential at in nity is zero in this calculation Anyway all that matters is the potential di erence and from Eq 11 27TL60 Q lowa 10 so that Cw 27TL60logba Example 2 Concentric spherical shells Consider two concentric rnetal shells of radii a and b7 where b gt a and d b 7 a A voltage V is applied to the outer cylinder and the inner cylinder is grounded so the total charge on the two shells is Q on the outer and 7Q on the inner shell The electric eld for r gt b and for r lt a is zero by Gauss7s theorern The electric eld for a lt r lt b is found from Gauss7s law7 Q E4 27 tht E 77 11 7139 Qeo so a r 4W60T2 Now we nd the potential di erence by integration7 bee 1 Q Q11 Qba V7Edl7d777777 12 b a a 47mmquot T 47TEO b a 47Tab60 Again Va 0 as the inner cylinder is grounded We then have7 47Tab60 V 13 Q b 7 a lt gt so that Osphere 47Tab60b 7 a For a general geornetry determination of the capacitance requires that we calculated the potential di erence between two pieces of metal for a given charge on the metals or equivalently the charge on the two pieces of metal given the potential di erence The latter is easier as the voltage on all parts of the metal is the same and hence we know the voltage boundary conditions for Laplace7s equation Nevertheless in general this problem can only be solved numerically C The method of images The method of images is a method for reducing some problems where a charge is near a conductor to an equivalent problem involving only point charges with the conductor removed 3 PHY481 Lecture 17 Chapter 5153 of PS We will not cover 54 55 A Legendre Polynomials Series solution 3 6Pu 311 This equation is call Legendre7s equation and is solved using a series solution7 Pu 17112 lz1u0 1 220 Gnu Substituting this series into Legendre7s equation we nd that7 a CO CO 717 112 Z Cnnunil Zll10nu 0 2 6 n1 n0 and7 Z Cnnn 71u 2 7 Z Cnnn 1u Zll10nu 0 3 n2 n1 n0 The coef cient of u in this equation must be zero7 implying that7 nn 1 7 ll 1 Cn2n 2n 1 Cnll 1 7 nn 1 0 hence On On n 1n 2 4 For a given value of l7 we get two series of solutions7 one starting with a value of CO and producing 027 C4 and the other starting with 01 and producing 037 C5 The key physical observation is that if l is an integer7 the series terminate at n l7 leading to a nite polynomial solution In contrast if l is not an integer7 the series does not terminate and the coef cients remain nite at in nity7 a solution that does not have physicial meaning The constants CO and 01 are xed by requiring that the polynomials be normalized on the interval 7117 which corresponds to 77T in the original variable 0 The normalization condition is or 1311 1 5 1 P2udui L 1 l 2l1 With this choice the rst few Legendre polynomials are7 P0u1 P1u u P2u 3112 712 P3u Bu3 7 3u2 6 Higher order functions 131u7 are found by choosing an arbitrary starting constant7 applying the recursion relation till termination7 then normalization using Eq 5 xes the starting constant The solution to Laplace7s equation for cases where there is no dependence on b is then of the form 00 1 Bl V036 2Alr WPlcost9 7 10 If we set A 0 in this expansion we get terms which are of the form of the general multipole expansion with B0 the monopole term B1 the dipole term and B2 the quadrupole term In lecture 9 we carried out the multipole expansion to order l 2 using the expansion 1 ricos 7 7 2 Ti 2 2T3300526 7 1 8 r r Now we recognize the angle dependent terms in this expansion as Legendre polynomials In fact this expansion can be carried out to arbitrary order in terms of Legendre polynomials yielding 1 lFi l oo Tl Z TilP cosQ r gt n 9 10 which of course reproduces the multipole terms we found before Also since this is an expansion to in nite order we can extend it to the regime r lt n where the expansion takes the form 1 00 Tl gl cos r lt n 10 Of course these two terms are just the terms in the general solution to Laplace7s equation in polar co orindates Eq A simple emample Consider a sphere in a constant applied electric eld E7 E0 The potential is then Vz iEoz iEorcos Again we tried the simplest solution rst which corresponds to l 1 so that we Ag 521m 0050 Ag 590059 11 r T To satisfy the boundary condition at in nity we set A1 7E0 and to satisfy the boundary condition at the surface of the sphere we require B VR6 0 iEoR Epcos so that B1 EOR3 12 yielding the solution E0R3 T2 V030 iEor 0050 13 From this expression it is clear that the induced dipole moment can be found by comparing m 9 E R3 161 0720056 so that pmd 47TEOE0R3 14 r r Notice that EORZk has dimensions of charge A more interesting example Example 1 of PS Consider a square region of space7 centered at the origin and with of dimensions a gtlt a The sides of the square are parallel to the x and y axes The sides at y ia2 are held at a xed potential V57 while the sides at z ia2 are held are grounded7 ie V 0 there Find an expression for the potential everywhere on the interior of the square domain Solution The rst observation is that the boundary conditions in the boundary condi tions in the x direction are symmetric about the origin so we choose functions of the form Xz 005kz Similarly the boundary conditions in the y direction are symmetric so we choose Yy coshky Since there is dependence on both x and y directions7 we expect the one dimensional solutions will not be useful7 so we do not include them We then have7 Vzy ZAkcoskxcoshky 15 k At this point we don7t know what values of k are needed We can nd a set of values of k by imposing the boundary conditions in the x direction where Va2y V7a2y 0 These boundary conditions can be satis ed by choosing7 2n 17T a 7 005ka20 or k with n012 16 Notice that we don7t need to include negative values of 71 due to the fact that the cosine function is even The electrostatic potential is then given by Egt Vy An0052n 1 cosh2n 1 17 a Our remaining task is to satisfy the boundary conditions in the y direction7 Vz ia2 V57 so we need7 v5 Ancos2n 1005h2n 1 Ancos2n 1 18 where A n Ancosh2n 17r2 Our problem reduces to nding the Fourier cosine series for a constant function Due to the orthogonality of the basis functions7 we can extract the coef cient A n7 by multiplying both sides by cos2n 17Txa and then integrating both sides over the interval fa2a27 7m 12 7T 0 12 Vbcos2n 17dz Z A 005271 1 ill2 a 0 2 7a a 005271 1dz 19 3 PHY481 Lecture 4 Sections 2425 of Pollack and Stump PS A Curvilinear co ordinates Scale factors h1h2 hg In general a set of curvilinear co ordinates can be orthogonal or non orthogonal We focus on the orthogonal case which includes cartesion cylindrical and spherical co ordinates We denote the unit vectors to be 61 6263 and that a position vector i is written as f 111112112 111 1 11262 11363 For the Cartesion Cylindrical and Spherical Polar cases we have 111112112 x y z r 0 z r 0 0 respectively Now imagine displacing the co ordinates by a small amount dul d112 d113 this leads to a change in the vector i by an amount d In general we can write dg 1111161 112h2 2 1131363 where hl 12 fig are the scale factors They are central to deriving the relations that we need Lets derive them for the three cases of interest done in class The results are Cartesian hm 1hy 1 hz 1 Cylindrical hr 1 his r hz 1 Spherical Polar hr 1 he r h rsin B Unit Vector Transformations ln lecture 2 we noted the transformations between Cartesian and cylindrical co ordinates z r0054 y rsinrb x2 y2 r2 3 and that between Cartesian and polar co ordinates z rcosrbsin g y rsinrbsin g z r0050 2 y2 22 r2 4 Relations between unit vectors in the three co ordinate systems are also useful First consider cylindrical co ordinates where we use either 19 or b for the angle and r for the in plane radius It is easy to see that r cosrbi gimp 1 isimbi 005 5 and f 005 7 Sinai sma f 005mg 6 The transformations in the case of polar co ordinates are7 f sz n cosg i sm smabj 00501 6 cos cosg i cos sma j 7 527101 a 75m 005 7 and f sm cosa f cos cosa 7 Sinai j sm sma f cos sz na cosa ag I 0056 7 527166 8 C Volume element Grad Div Curl and Laplacian in curvilinear co ordinates We write down the expressions for these quantities in terms of the scale factors7 hi7 and explicit expressions are found by A volume element dV at position i is given by7 dV hlhghgduldUZhLLQ Z The gradient is found by using two forms for the expansion of a scalar function 6f 6f 6f d 7d 7d 7d 10 f 6u1 U1 6u2 uz QUE us and dfd fd51 f1d82 f2d536f3 11 Using Eq 7 we also have d81 hldulg d82 h2d u2 d83 h3d d3 Combining Eq 7 yields7 a 1 6 1 6 1 6 vlt777777gt lt13 11 6111 hg 6u2 hg 6113 2 The divergence is found by using the same procedure that we used in Eq 7 but now using curvilinear co ordinates 6 F dv Flhzhglwdul i FlhghglullduZdu3 2 FZhthlu2du2 F2h1h3lu2ldu1dus FShthlu3du3 FShthluglduldUZ 14 From this expression and using dV h1h2h3d d1d d2d d3 we nd7 a a 1 6 6 6 F ith ith ith 15 V hlhzhglaul 1 2 36u2 2 1 36u3 3 1 2 3 The Laplacz39an of a scalar function f is v W which reduces to7 1 6 hhk 6f 2 7 16 Wk 4 The Curl is developed in a similar way yielding7 ihghg jhghl IQhlhz vAF 66u1 66112 66113 17 ME th2 11st These expressions are used7 along with the expressions for the scale factors hi to write down expressions for Grad7 Div7 Curl and Laplacian in cylindrical and in polar co ordinates We shall use these expressions repeatedly later in the course D Helmholtz theorem Any vector eld with the properties 127717va a 0 lz39mrmv F a 0 may be broken up into a divergence free part and an irrotational curl free part7 so that7 1476va 18 where the scalar potential ltIgt and the vector potential ff are related to the vector function F through 1F lt1gt7V1 d 19 V47Tlr 777 a AFf A7 Law 20 V47Tlr 777l As we shall see later7 these expression are derived in physics by superposition of the electro static potential topic expression and vector potential lower expression A consequence of Helrnholtz7s theorem is that if we know the div and curl of a vector function7 then the function is speci ed This is important in electrostatics where we have the two equations 6Ep60 AE0 21 3 PHY481 Lecture 6 Sections 3436 of PS A Finding the electric eld We have three methods for nding the electric eld in electrostatics i Superposition7 a had 7 7339 E r or f A re s Slightly different continuous forms are used for line or surface charge cases ii Gauss7s law in integral form7 f E drA qgncl 2 s 60 iii Gauss7s law in differential form7 Ep60 and AE0 3 Chapter 5 of PS is devoted to the differential approach Here we rst concentrate on the rst two approaches Remember that in all electrostatics problems superposition can be used so if you can break the problem or structure of interest into a sum of structures whose solutions you know7 then the solution is often straightforward to write down i In nite sheet of charge Charge density 0Cm2 Gauss7s law Draw a rectangular box around the sheet Note that by symmetry the electric eld lines are perpindicular to the sheet We then have7 f dh2EAUAEO or E039260 4 ii Just outside the surface of a conductor Charge density 0Cm2 Use an in nitesimal rectangle and note that the electric eld inside the conductor is zero7 therefore7 f dAEdAUdAEO or E03960 5 Note two things The electric eld at the surface is twice that of a charge sheet with the same charge density ii If an electric eld is applied to a conductor7 the charges move to screen out the applied eld the surfaces become charged leading to an induced dipole This is used to trap small particles using focused laser beams or laser tweezers As we shall discuss later many molecules have permanent dipoles as well The size of the induced dipole on a molecule depends on the polarizeability of the electron cloud of the molecule metals are highly polarizeable iii In nite sheet of uniform charge density 039 using integration Place the sheet in the X y plane and note by symmetry that only E1 is nite There is no dependence on the X y co orindates as there is translational symmetry in the X y plane To carry out the integral we use cylindrical co ordinates and use the result for a ring of charge Lecture 5 E1 szTZ 22327 00 WW A E 0 W With dQr 027Trdr 6 so we nd 00 U27Trdrz 702 039 E1 co 7 7 0 47T60r2 2232 260r2 2212 0 260 The simplest charged capacitors consist of two sheets of metal with equal and opposite charges on the surfaces of the sheets What is the electric eld inside and outside this type of capacitor iv In nite cylindrical shell of radius R Assume the rod has radius R with its axis along the Z aXis and we use cylindrical co ordinates To apply Gauss7s law we note that by symmetry the electric eld is in r direction Consider two closed cylindrical Gaussian surfaces S of radius r lt R and r gt B respectively In both cases the length of the cylinder is L By symmetry there is no ux through the top and bottom of the cylinder Moreover if r lt B there is no enclosed charge so the electric eld is zero For the case r gt R we have E7 d21 27TrLEr QWRLUEO 8 S so that E 5E A 9 60 r 27T60r where A 27TRU is the charge per unit length of the cylinder Capacitors are sometimes made using concentric cylinders which hold equal and opposite charge The electric eld is again found by adding the elds due to the two cylinders PHY481 Lecture 21 Chapter 6 of PS A Force pulling a dielectric slab between capacitor plates To nd the force on the slab we have to solve the problem when the slab partially penetrates the slab It is easiest to see the effect in the case of an isolated xed charge systems Consider an l gtlt w slab where the plates are penetrated by an area z gtlt w of the slab while an area l 7 s gtlt w remains lled with air The voltage on the plates are at equipotentials they are metal so the electric eld in both regions is the same However the capacitance in the two regions is different The total capacitance can be found by adding the charge on the two parts of the plate eExw 60El 7 w Q 1 Using V Ed we then nd iQi 60l7xw Oiv7 d d 2 Note that we could have used the law of addition of capacitors to get this result Now we can nd the force on the slab by using F dUx d de dew 6 7 60 a dx 7 dx 2exw 6017 zw 2exw 6017 w2 3 In the case of a xed voltage system we have to take into account the work done by the battery or generator dU dCVZ dQV i Fdz dOVZ i Fdz 4 so that 1dC E 7 60 F 77W V2 7 5 2 dz w 2d B Uniformly polarized sphere pf 0 no applied eld If we have a sphere where there is a uniform density of polarization 13 1301 then at the surface of a sphere the bound charge is ab f P01 P000519 6 We can construct an electrostatic potential corresponding to this charge distribution by trying the l 1 spherical solution in which case 02 Vimrt9 Clrcos g Vmrt9 70050 7 r The potential must be continuous across at r a which implies that 01 0203 Using E idVdr the radial component of the electric eld obeys 20 20 O P 6 Efmat9 7 E ta t9 720050 010050 720050 720050 8 60 a3 a3 a3 60 From this we nd Cg Peas360 01 P0360 so that P P P 3 Vim forcos 702 t 7 0a 0050 9 360 360 3 2 The electric eld inside the sphere is given by a 6V A Poi Em 7 k 10 t 32 360 The electric eld is thus uniform and is in the opposite direction to the polarization The potential outside the sphere is like that of a dipole and the dipole moment is found by equating the general expression for a dipole with the potential Vem 1 1637 Poa3 47TEO r2 360T2 0050 11 which shows that the magnitude of the dipole moment of the uniformly polarized sphere is given by p 413P0 As expected the magnitude of the dipole moment of the sphere is the volume of the sphere times the uniform polarization of the sphere C Relation between atomic dipoles and polarization At the beginning of our discussion we noted that there is often a linear relation between the polarization of an atom or molecule and the electric eld that is applied to it 15 04E Now we want to connect this behavior with the behavior of the polarization density used in Maxwell7s equations P eoxe Since is for one atom or molecule while P is the polarization per unit volume a naive assumption might be to write P n where n is the number of atoms or molecules per unit volume in the dielectric material This assumption is valid if the electric eld is not altered by the dielectric material which from the discussion above is rarely true Nevertheless it is PHY481 Lecture 22 Chapter 8 of PS Chapter 5 of Gri iths A Basics of Magnetostatics Magnetostatics is the study of static magnetic elds and the steady currents DC that generate them The MKS unit of magnetic eld is the Tesla T and often the magnetic eld is given in Gauss G the CGS unit The relation is 10 000G 1T so small elds are often given in Gauss In the early 1800s several researchers noted that there is a force between two parallel wires that carry a steady current If the wires are separated by distance d F 7 iiodz39lz39g attractive for currents in the same direction 1 7r where no 47139 gtlt 10 7NA2 is the permeability of vacuum If the currents are in opposite directions the wires repel This result can be compared with the force between two in nite parallel line charges with charge densities A1 A2 The magnitude of the force between the line charges when separated by d is given by H l Adz 27reod lEl repulsive for like charges 2 where we used the electric eld near a line charge Er A27T60r Other than the direction of the force these two results are similar Clearly excess charge leads to electric elds and a DC leads to magnetic elds The total force is a superposition of these two effects so that electrostatics and magnetostatics don7t affect each other Ampere and Oersted also noticed that magnets are affected by a DC current in a wire and Ampere found that EU 3 Magnetic eld near a wire 3 This is not such a leap as we already know that the electric eld due to a line charge is E70 if Electric eld near a line charge 4 27m The most remarkable difference between these two results is the direction of the elds The electric eld diverges from the line charge and is curl free E7 0 The magnetic eld forms circles around the steady current and is divergence free v g 0 Notice that the magnetic eld is not moving even though the steady current is 1 owing and even though we often talk about the circulation of the magnetic eld about the current The magnetic eld is static The direction of circulation is given by the right hand rule7 applied to either the magnetic eld or to the current B Ampere7s law and related problems We start with Section 4 of PS or Section 533 of Grif ths by writing down Ampere7s law Ampere realised that his measurements for the magnetic eld near a wire may be written in the form of a path integral f5 df not poj dfT Amperes law 5 Here d is a small vector along the direction of the path For example for a circle the unit vector is tangent to the circle Now consider the magnetic eld around a long straight wire7 where the current7 239 is in the I direction7 then we have7 027r Br 3 rd 3 27TTBT not 6 Solving for Br yields7 I Em A lt7 Ampere7s law is correct only for DC currents Later in the course we will add an additional term which enables us to use this equation even when there are time dependent elds the Maxwell displacement current term There is a set of important problems that can be solved analytically using the integral form of Ampere7s law The most basic is the line charge case as illustrated above Another simple case is a sheet of current In that case7 consider a sheet of current lying in the x y place7 with the current directed along the positive x direction7 so that the surface current density is Ki where K is the current per unit length If we draw an L gtlt H rectangular amperian loop around the current sheet7 with L the length of a section parallel to the y axis and H a section parallel to the Z direction a 1 a a 1 a 2LBM0LK so that B M0Kj7 zgt0 B M0Kj zlt0 8 A practically important and closely related case is that of an in nite solenoid of radius R with its axis along the Z axis7 where there are 71 turns per unit length In that case7 the eld outside the solenoid is by symmetry zero The nite eld inside the solenoid is found from LB MOLm39 so that B Mom 7 lt 0 9 where 239 is the current in the circuit Using the RHR ifthe solenoid current is rotating in the a direction the magnetic eld is along the 1 direction Large magnetic elds can be achieved by increasing n or by increasing 239 which is the basic goal in design of electromagnets Copper electromagnets can achieve magnetic elds of up to 10T though values around 1 7 5T are more standard Superconducting electromagnets can achieve higher values but require liquid Helium cooling Large magnets used in physics experiments such as the NSCL LHC etc use superconducting electromagnets Use of magnets in NMR and in medicine MRI also require large magnetic elds and rely on electromagnets Electromagnets provide very nice control of the size of the magnetic eld William Stugeon invented the electromagnet in 1823 High termperature superconductors hold the promise of superconducting magnets operating at liquid Nitrogen instead of Liquid He which would be a very important advance Another class of problems that can be solved with Ampere7s law is the case of a wire with current density jr up to its radius R with one example being a constant current density jo for r lt R It is a useful excersize to do this case and nd the magnetic eld in the two regimes Br Major2 for r lt R and B0 MOjORzQr for r gt R C Biot Savart law and related problems Ampere7s law is convenient for cases with high symmetry but we need a more general law for cases where the current carrying wire is not so symmetric for example in current loops The Bio Savart law is the generalization of Ampere7s law to these cases and is given by dB 4W T2 7 4W T3 Biot 7 Savart law 10 Notice that this looks a lot like Coulomb7s law a 1 d A dE 7Q 11 47TEO r2 The complication is that in the Biot Savart case we now have to do the cross product but in practice that is not so hard Integration of the Biot Savart law is used with to nd the magnetic eld due to wires carrying direct currents


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