Semester II, 2017-18 Department of Physics, IIT Kanpur PHY103A: Lecture # 11 (Text Book: Intro to Electrodynamics by Griffiths, 3rd Ed.) Anand Kumar Jha 24-Jan-2018 Notes • No Class on Friday; No office hour. • HW # 4 has been posted • Solutions to HW # 5 have been posted 2 Summary of Lecture # 10: The potential in the region above the infinite grounded conducting plane? 1 1 V = x + y + (z d) x + y + (z + d) 0 2 2 2 − 2 2 2 4 − The potential outside the grounded conducting sphere? = = 2 ′ − 1 V = + r r′ ′ 0 4 3 Question: The potential in the region above the infinite grounded conducting plane? 1 1 V = x + y + (z d) x + y + (z + d) 0 2 2 2 − 2 2 2 4 − Q: Is kV also a solution, where k is a constant? Ans: No Corollary to First Uniqueness Theorem: The potential in a volume is uniquely determined if (a) the charge desity throughout the region and (b) the value of V at all boundaries, are specified. In this case one has to satisfy the Poisson’s V = equation. But kV does not satisfy it. 2 1 − 0 Exception: Only when = 0 (everywhere), kV can be a solution as well But then since in this case V = , it is a trivial solution. 4 Multipole Expansion (Potentials at large distances) • What is the potential due to a point charge (monopole)? 1 V = (goes like / ) • What is the potential4 due0 to a dipole at large distance? 1 V = r r − 40 + − r = + cos = 1 cos + ± 2 2 4 2 2 2 2 ∓ ∓ 2 1 cos (for ) 2 for , and using binomial≈ expansion ∓ ≫ 1 1 1 1 1 1 ≫= 1 cos − 1 ± cos So, = cos r± 2 r r ∓ 1 ≈ cos − 2 + − V = 2 (goes like / at large ) 5 2 40 Multipole Expansion (Potentials at large distances) • What is the potential due to a quadrupole at large distance? (goes like / at large ) • What is the potential due to a octopole at large distance? (goes like / at large ) Note1: Multipole terms are defined in terms of their dependence, not in terms of the number of charges. Note 2: The dipole potential need not be produced by a two-charge system only . A general -charge system can have any multipole contribution. 6 Multipole Expansion (Potentials at large distances) • What is the potential due to a localized charge distribution? 1 1 ( ) V( ) = = r r ′ ′ � � Using the cosine40 rule, 40 r = + cos 2 2 2 ′ − 2′ r = 1 + 2 2 cos ′ ′ Source coordinates: ( , , ) 2 2 − Observation point coordinates:′ ′ ′ ( , , ) r = 1 + 2cos ′ ′ Angle between and : − r = 1 + ′ 2cos Define: ′ ′ 1 1 So, = 1 + / ≡ − r −1 2 1 1 1 3 5 Or, = 1 + + (using binomial expansion) r 2 8 16 7 2 3 − − ⋯ Multipole Expansion (Potentials at large distances) • What is the potential due to a localized charge distribution? 1 1 ( ) V( ) = = r r ′ ′ � � 40 40 1 1 1 3 5 = 1 + + r 2 8 16 2 3 − − ⋯ 1 1 3 5 = 1 2cos + 2cos 2cos + 2 ′ ′ 8 ′ 2 ′ 2 16 ′ 3 ′ 3 − − − − − ⋯ 1 = 1 + cos + 3cos 1 /2 5cos 3cos /2 + ′ ′ 2 ′ 3 2 3 − − − ⋯ 1 = (cos ) ∞ ′ (cos ) are Legendre polynomials � =0 1 ( ) 1 1 V( ) = = ∞ (cos ) ′ r 8 ′ ′ ′ +1 � 0 � � � 40 4 =0 Multipole Expansion (Potentials at large distances) • What is the potential due to a localized charge distribution? 1 1 ( ) V( ) = = r r ′ ′ � � 40 40 1 1 V( ) = ∞ (cos ) ′ ′ +1 0 � � � 4 =0 1 1 1 3 1 = + (cos ) + cos 2 2 ′ ′ ′ ′ ′ ′ 2 2 ′ ′ � 2 � 3 � − + 40 40 40 ⋯ Monopole potential Dipole potential Quadrupole potential ( 1/ dependence) ( 1/ dependence) ( 1/ dependence) 2 3 Multipole Expansion of ( ) 9 Multipole Expansion (Few comments) V 1 1 1 3 1 = + (cos ) + cos 2 2 ′ ′ ′ ′ ′ ′ 2 2 ′ ′ � 2 � 3 � − + 40 40 40 ⋯ Monopole potential Dipole potential Quadrupole potential ( 1/ dependence) ( 1/ dependence) ( 1/ dependence) 2 3 • It is an exact expression, not an approximation. • A particular term in the expansion is defined by its dependence • At large , the potential can be approximated by the first non-zero term. • More terms can be added if greater accuracy is required 10 Questions 1: Q: In this following configuration, is the “large ” limit valid, since the source dimensions are much smaller than ? Ans: No. The “large ” limit essentially mean | |. In majority of the situations, the charge distribution is centered at the origin and therefore the “large ” limit≫ is′ the same as source dimension being smaller than . 11 Multipole Expansion (Monopole and Dipole terms) Monopole term: • = is the total charge 1 1 1 • If = 0, monopole′ ′ term is zero. V = = ∫ ′ ′ • For a collection of point charges mono � 40 40 = � Dipole term: =1 1 1 V = (cos ) ′ ′ ′ • is called the dipole dip 2 is the angle between0 � and . moment′ of a′ charge′ distribution 4 ≡ ∫ So, cos = ′ • If = 0, dipole term is zero. ′ 1 1 V = � ⋅ ′ • For a collection of point charges. ′ ′ ′ dip 2 � ⋅ � = 410 V = 12 ⋅ � � ′ dip 2 =1 40 Multipole Expansion (Monopole and Dipole terms) Monopole term: 1 1 1 (for point V = = charges) ′ ′ Dipolemono term: � → � 40 40 =1 1 1 1 (for point V = (cos ) = charges) ′ ′ ′ ⋅ � dip 2 2 � → 0 � ′ 40 4 =1 Example: A three-charge system = = � − =1 = = + = � ′ � − − − � � =1 Therefore the system will have both monopole and dipole contributions 13 Multipole Expansion (Monopole and Dipole terms) Monopole term: 1 1 1 (for point V = = charges) ′ ′ Dipolemono term: � → � 40 40 =1 1 1 1 (for point V = (cos ) = charges) ′ ′ ′ ⋅ � dip 2 2 � → 0 � ′ Example: Origin40 of Coordinates 4 =1 1 1 1 V = = V = = r mono mono = 0 V = 0 1 0 ≠ 0 40 = V = 4 4 dip � � dip ⋅ 2 14 40 Questions 2: 1 1 = V = r mono 1 ≠ = V = 40 40 ⋅ � � dip 2 Q: Why not calculate the potential40 directly ?? Ans: Yes, that is what should be done. For a point charge, we don’t need a multipole expansion to find the potential. This is only for illustrating the connection. 15 The electric field of pure dipole ( = ) = 0 And 0 Assume = 1 ≠ 1 �1 cos V = = = ⋅ � � ⋅ � dip 2 2 2 0 0 0 = 4V 4 4 − cos = = = (2cos + sin ) 2 3 − 0 dip 3 � ̂ 1 4sin 40 = = 3 − 0 1 4 = = 0 sin − 16 .