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) 𝑞𝑞 𝐫𝐫 0 2 2 2 − 2 2 2 4𝜋𝜋𝜖𝜖 −
The potential outside the grounded conducting sphere?
= = 2 ′ 𝑅𝑅 𝑅𝑅 𝑞𝑞 − 𝑎𝑎 𝑞𝑞 𝑏𝑏 1 𝑎𝑎 V = + r r′ 𝑞𝑞 𝑞𝑞′ 𝐫𝐫 0 4𝜋𝜋𝜖𝜖 3 Question: The potential in the region above the infinite grounded conducting plane? 1 1 V = x + y + (z d) x + y + (z + d) 𝑞𝑞 𝐫𝐫 0 2 2 2 − 2 2 2 4𝜋𝜋𝜖𝜖 − Q: Is kV also a solution, where k is a constant?
Ans: No 𝐫𝐫
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 𝜌𝜌 𝛁𝛁 1 − 𝐫𝐫 𝜖𝜖0 Exception: Only when = 0 (everywhere), kV can be a solution as well But then since in this case V = , it is a trivial solution. 𝜌𝜌 𝐫𝐫 4 𝐫𝐫 𝟎𝟎 Multipole Expansion (Potentials at large distances)
• What is the potential due to a point charge (monopole)? 1 V = (goes like / ) 𝑞𝑞 𝐫𝐫 𝟏𝟏 𝒓𝒓 • What is the potential4𝜋𝜋 due𝜖𝜖0 to𝑟𝑟 a dipole at large distance? 1 V = r r 𝑞𝑞 𝑞𝑞 𝐫𝐫 − 4𝜋𝜋𝜖𝜖0 + − r = + cos = 1 cos + ± 2 2 4 2 2 2 𝑑𝑑 2 𝑑𝑑 𝑑𝑑 𝑟𝑟 ∓ 𝑟𝑟𝑟𝑟 𝜃𝜃 𝑟𝑟 ∓ 𝜃𝜃 2 1 𝑟𝑟cos (for𝑟𝑟 ) 2 𝑑𝑑 for , and using binomial≈ expansion𝑟𝑟 ∓ 𝜃𝜃 𝑟𝑟 ≫ 𝑑𝑑 𝑟𝑟 1 1 1 1 1 1 𝑟𝑟 ≫= 𝑑𝑑 1 cos − 1 ± cos So, = cos r± 2 r r 𝑑𝑑 𝑑𝑑 𝑑𝑑 ∓ 𝜃𝜃 1 ≈ cos 𝜃𝜃 − 2 𝜃𝜃 𝑟𝑟 𝑟𝑟 + − V = 𝑟𝑟 2𝑟𝑟 (goes like / at large 𝑟𝑟) 5 𝑞𝑞𝑞𝑞 𝜃𝜃 𝟐𝟐 𝐫𝐫 2 𝟏𝟏 𝒓𝒓 𝒓𝒓 4𝜋𝜋𝜖𝜖0 𝑟𝑟 Multipole Expansion (Potentials at large distances)
• What is the potential due to a quadrupole at large distance?
(goes like / at large ) 𝟑𝟑 𝟏𝟏 𝒓𝒓 𝒓𝒓
• What is the potential due to a octopole at large distance? (goes like / at large ) 𝟒𝟒 𝟏𝟏 𝒓𝒓 𝒓𝒓
Note1: Multipole terms are defined in terms of their dependence, not in terms of the number of charges. 𝒓𝒓 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
𝑛𝑛 Multipole Expansion (Potentials at large distances) • What is the potential due to a localized charge distribution? 1 1 ( ) V( ) = = r r ′ 𝑑𝑑𝑑𝑑 𝜌𝜌 𝐫𝐫 ′ 𝐫𝐫 � � 𝑑𝑑𝜏𝜏 Using the cosine4𝜋𝜋𝜖𝜖0 rule, 4𝜋𝜋𝜖𝜖0 r = + cos 2 2 2 𝑟𝑟 𝑟𝑟� − 2𝑟𝑟𝑟𝑟� 𝛼𝛼 r = 1 + 2 2 cos ′ ′ Source coordinates: ( , , ) 2 2 𝑟𝑟 𝑟𝑟 𝑟𝑟 − 𝛼𝛼 ′ ′ ′ ( , , ) 𝑟𝑟 𝑟𝑟 Observation point coordinates: r = 1 + 2cos 𝑟𝑟 𝜃𝜃 𝜙𝜙 ′ ′ Angle between and : 𝑟𝑟 𝑟𝑟 𝑟𝑟 𝜃𝜃 𝜙𝜙 𝑟𝑟 − 𝛼𝛼 r = 1 + 𝑟𝑟 𝑟𝑟 𝐫𝐫 𝐫𝐫′ 𝛼𝛼 2cos Define: ′ ′ 1 𝑟𝑟1 𝜖𝜖 𝑟𝑟 𝑟𝑟 So, = 1 + / 𝜖𝜖 ≡ − 𝛼𝛼 r 𝑟𝑟 𝑟𝑟 −1 2 1 1 1𝜖𝜖 3 5 Or, = 𝑟𝑟 1 + + (using binomial expansion) r 2 8 16 7 2 3 − 𝜖𝜖 𝜖𝜖 − 𝜖𝜖 ⋯ 𝑟𝑟 Multipole Expansion (Potentials at large distances) • What is the potential due to a localized charge distribution? 1 1 ( ) V( ) = = r r ′ 𝑑𝑑𝑑𝑑 𝜌𝜌 𝐫𝐫 ′ 𝐫𝐫 � � 𝑑𝑑𝜏𝜏 4𝜋𝜋𝜖𝜖0 4𝜋𝜋𝜖𝜖0 1 1 1 3 5 = 1 + + r 2 8 16 2 3 − 𝜖𝜖 𝜖𝜖 − 𝜖𝜖 ⋯ 𝑟𝑟 1 1 3 5 = 1 2cos + 2cos 2cos + 2 ′ ′ 8 ′ 2 ′ 2 16 ′ 3 ′ 3 𝑟𝑟 𝑟𝑟 𝑟𝑟 𝑟𝑟 𝑟𝑟 𝑟𝑟 − − 𝛼𝛼 − 𝛼𝛼 − − 𝛼𝛼 ⋯ 𝑟𝑟1 𝑟𝑟 𝑟𝑟 𝑟𝑟 𝑟𝑟 𝑟𝑟 𝑟𝑟 = 1 + cos + 3cos 1 /2 5cos 3cos /2 + ′ ′ 2 ′ 3 𝑟𝑟 𝑟𝑟 2 𝑟𝑟 3 𝛼𝛼 𝛼𝛼 − − 𝛼𝛼 − 𝛼𝛼 ⋯ 1𝑟𝑟 𝑟𝑟 𝑟𝑟 𝑟𝑟 = (cos ) ∞ ′ 𝑛𝑛 (cos ) are Legendre polynomials 𝑟𝑟 � 𝑃𝑃𝑛𝑛 𝛼𝛼 𝑛𝑛 𝑟𝑟 𝑛𝑛=0 𝑟𝑟1 ( ) 1𝑃𝑃 𝛼𝛼 1 V( ) = = ∞ (cos ) ′ r 8 𝜌𝜌 𝐫𝐫 ′ ′ 𝑛𝑛 ′ 𝑛𝑛+1 𝑛𝑛 𝐫𝐫 � 𝑑𝑑𝜏𝜏 0 � � 𝑟𝑟 𝑃𝑃 𝛼𝛼 𝜌𝜌 𝐫𝐫 𝑑𝑑𝑑𝑑� 4𝜋𝜋𝜖𝜖0 4𝜋𝜋𝜖𝜖 𝑛𝑛=0 𝑟𝑟 Multipole Expansion (Potentials at large distances) • What is the potential due to a localized charge distribution? 1 1 ( ) V( ) = = r r ′ 𝑑𝑑𝑑𝑑 𝜌𝜌 𝐫𝐫 ′ 𝐫𝐫 � � 𝑑𝑑𝜏𝜏 4𝜋𝜋𝜖𝜖0 4𝜋𝜋𝜖𝜖0 1 1 V( ) = ∞ (cos ) ′ 𝑛𝑛 ′ 𝑛𝑛+1 𝑛𝑛 𝐫𝐫 0 � � 𝑟𝑟 𝑃𝑃 𝛼𝛼 𝜌𝜌 𝐫𝐫 𝑑𝑑𝑑𝑑� 4𝜋𝜋𝜖𝜖 𝑛𝑛=0 𝑟𝑟 1 1 1 3 1 = + (cos ) + cos 2 2 ′ ′ ′ ′ ′ ′ 2 2 ′ ′ � 𝜌𝜌 𝐫𝐫 𝑑𝑑𝜏𝜏 2 � 𝑟𝑟 𝛼𝛼 𝜌𝜌 𝐫𝐫 𝑑𝑑𝜏𝜏 3 � 𝑟𝑟 𝛼𝛼 − 𝜌𝜌 𝐫𝐫+𝑑𝑑𝜏𝜏 4𝜋𝜋𝜖𝜖0𝑟𝑟 4𝜋𝜋𝜖𝜖0𝑟𝑟 4𝜋𝜋𝜖𝜖0𝑟𝑟 ⋯
Monopole potential Dipole potential Quadrupole potential ( 1/ dependence) ( 1/ dependence) ( 1/ dependence) 2 3 𝑟𝑟 𝑟𝑟 Multipole Expansion𝑟𝑟 of ( ) 9 𝐕𝐕 𝐫𝐫 Multipole Expansion (Few comments)
V 1 1 1 3 1 =𝐫𝐫 + (cos ) + cos 2 2 ′ ′ ′ ′ ′ ′ 2 2 ′ ′ � 𝜌𝜌 𝐫𝐫 𝑑𝑑𝜏𝜏 2 � 𝑟𝑟 𝛼𝛼 𝜌𝜌 𝐫𝐫 𝑑𝑑𝜏𝜏 3 � 𝑟𝑟 𝛼𝛼 − 𝜌𝜌+𝐫𝐫 𝑑𝑑 𝜏𝜏 4𝜋𝜋𝜖𝜖0𝑟𝑟 4𝜋𝜋𝜖𝜖0𝑟𝑟 4𝜋𝜋𝜖𝜖0𝑟𝑟 ⋯
Monopole potential Dipole potential Quadrupole potential ( 1/ dependence) ( 1/ dependence) ( 1/ dependence) 2 3 𝑟𝑟 𝑟𝑟 𝑟𝑟
• 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 the𝒓𝒓 first non-zero term.
• More terms𝒓𝒓 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 ? 𝐫𝐫 Ans: No. The “large ” limit essentially mean | |. 𝐫𝐫 In majority of the situations, the charge distribution is centered at the origin𝐫𝐫 and therefore the “large 𝐫𝐫” limit≫ 𝐫𝐫 is′ the same as source dimension being smaller than . 𝐫𝐫 𝐫𝐫
11 Multipole Expansion (Monopole and Dipole terms)
Monopole term: • = is the total charge 1 1 1 • If = 0, monopole′ ′ term is zero. V = = 𝑄𝑄 ∫ 𝜌𝜌 𝐫𝐫 𝑑𝑑𝜏𝜏 ′ ′ 𝑄𝑄 • For a collection of point charges mono 𝐫𝐫 � 𝜌𝜌 𝐫𝐫 𝑑𝑑𝜏𝜏 𝑄𝑄 4𝜋𝜋𝜖𝜖0 𝑟𝑟 4𝜋𝜋𝜖𝜖0 𝑟𝑟 = 𝑛𝑛
𝑄𝑄 � 𝑞𝑞𝑖𝑖 Dipole term: 𝑖𝑖=1 1 1 V = (cos ) ′ ′ ′ • is called the dipole dip 2 is the𝐫𝐫 angle between0 � 𝑟𝑟 and 𝛼𝛼 . 𝜌𝜌 𝐫𝐫 𝑑𝑑𝜏𝜏 moment′ of a′ charge′ distribution 4𝜋𝜋𝜖𝜖 𝑟𝑟 𝐩𝐩 ≡ ∫ 𝐫𝐫 𝜌𝜌 𝐫𝐫 𝑑𝑑𝜏𝜏 So, cos = 𝛼𝛼 𝐫𝐫 𝐫𝐫′ • If = 0, dipole term is zero. ′ 1 1 V 𝑟𝑟 = 𝛼𝛼 𝐫𝐫� ⋅ 𝐫𝐫� • For𝐩𝐩 a collection of point charges. ′ ′ ′ dip 2 𝐫𝐫 10 𝐫𝐫� ⋅ � 𝐫𝐫 𝜌𝜌 𝐫𝐫 𝑑𝑑𝜏𝜏 = 𝑛𝑛 V =4𝜋𝜋𝜖𝜖 𝑟𝑟 𝐢𝐢 𝑖𝑖 12 𝐩𝐩 ⋅ 𝐫𝐫� 𝐩𝐩 � 𝐫𝐫 ′𝑞𝑞 dip 𝐫𝐫 2 𝑖𝑖=1 4𝜋𝜋𝜖𝜖0 𝑟𝑟 Multipole Expansion (Monopole and Dipole terms)
Monopole term: 1 1 1 (for point V = = 𝑛𝑛 charges) ′ ′ 𝑄𝑄 Dipolemono term:𝐫𝐫 � 𝜌𝜌 𝐫𝐫 𝑑𝑑𝜏𝜏 → 𝑄𝑄 � 𝑞𝑞𝑖𝑖 4𝜋𝜋𝜖𝜖0 𝑟𝑟 4𝜋𝜋𝜖𝜖0 𝑟𝑟 𝑖𝑖=1 1 1 1 (for point V = (cos ) = 𝑛𝑛 charges) ′ ′ ′ 𝐩𝐩 ⋅ 𝐫𝐫� dip 2 2 𝐢𝐢 𝑖𝑖 𝐫𝐫 � 𝑟𝑟 𝛼𝛼 𝜌𝜌 𝐫𝐫 𝑑𝑑𝜏𝜏 → 0 𝐩𝐩 � 𝐫𝐫 ′𝑞𝑞 4𝜋𝜋𝜖𝜖0 𝑟𝑟 4𝜋𝜋𝜖𝜖 𝑟𝑟 𝑖𝑖=1 Example: A three-charge system
= 𝑛𝑛 =
𝑄𝑄 � 𝑞𝑞𝑖𝑖 −𝑞𝑞 𝑖𝑖=1 = 𝑛𝑛 = + =
𝐩𝐩 � 𝐫𝐫𝐢𝐢′𝑞𝑞𝑖𝑖 𝑞𝑞𝑞𝑞 𝒛𝒛� −𝑞𝑞𝑞𝑞 − 𝑞𝑞 −𝑎𝑎 𝒚𝒚� 𝑞𝑞𝑞𝑞 𝒛𝒛� 𝑖𝑖=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 = = 𝑛𝑛 charges) ′ ′ 𝑄𝑄 Dipolemono term:𝐫𝐫 � 𝜌𝜌 𝐫𝐫 𝑑𝑑𝜏𝜏 → 𝑄𝑄 � 𝑞𝑞𝑖𝑖 4𝜋𝜋𝜖𝜖0 𝑟𝑟 4𝜋𝜋𝜖𝜖0 𝑟𝑟 𝑖𝑖=1 1 1 1 (for point V = (cos ) = 𝑛𝑛 charges) ′ ′ ′ 𝐩𝐩 ⋅ 𝐫𝐫� dip 2 2 𝐢𝐢 𝑖𝑖 𝐫𝐫 � 𝑟𝑟 𝛼𝛼 𝜌𝜌 𝐫𝐫 𝑑𝑑𝜏𝜏 → 0 𝐩𝐩 � 𝐫𝐫 ′𝑞𝑞 Example: Origin4𝜋𝜋𝜖𝜖0 𝑟𝑟 of Coordinates 4𝜋𝜋𝜖𝜖 𝑟𝑟 𝑖𝑖=1
1 1 1 V = = V = = r 𝑞𝑞 𝑞𝑞 mono 𝑞𝑞 mono = 0 V 𝐫𝐫= 0 𝑄𝑄 𝑞𝑞 𝐫𝐫 1 0 ≠ 0 𝑄𝑄 𝑞𝑞 4𝜋𝜋𝜖𝜖0 𝑟𝑟 = V = 4𝜋𝜋𝜖𝜖 𝑟𝑟 4𝜋𝜋𝜖𝜖 dip � 𝐩𝐩 𝐫𝐫 𝐩𝐩 𝑞𝑞𝑑𝑑𝒚𝒚� dip 𝐩𝐩 ⋅ 𝐫𝐫 𝐫𝐫 2 14 4𝜋𝜋𝜖𝜖0 𝑟𝑟 Questions 2: 1 1 = V = r 𝑞𝑞 𝑞𝑞 𝑄𝑄 𝑞𝑞 mono 𝐫𝐫 1 ≠ = V = 4𝜋𝜋𝜖𝜖0 𝑟𝑟 4𝜋𝜋𝜖𝜖0 𝐩𝐩 ⋅ 𝐫𝐫� 𝐩𝐩 𝑞𝑞𝑑𝑑𝒚𝒚� dip 𝐫𝐫 2 Q: Why not calculate the potential4𝜋𝜋𝜖𝜖0 𝑟𝑟 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 = 𝑸𝑸 𝟎𝟎
𝑄𝑄 1 𝐩𝐩 ≠ 1 𝐩𝐩 𝑝𝑝𝐳𝐳�1 cos V = = = 𝐩𝐩 ⋅ 𝐫𝐫� 𝑝𝑝𝐳𝐳� ⋅ 𝐫𝐫� 𝑝𝑝 𝜃𝜃 dip 2 2 2 𝐫𝐫 0 0 0 = 4𝜋𝜋V𝜖𝜖 𝑟𝑟 4𝜋𝜋𝜖𝜖 𝑟𝑟 4𝜋𝜋𝜖𝜖 𝑟𝑟
𝐄𝐄 𝐫𝐫 −𝛁𝛁 cos = = = (2cos + sin ) 𝜕𝜕𝜕𝜕 2𝑝𝑝 𝜃𝜃 𝑟𝑟 3 𝑝𝑝 𝐸𝐸 − 0 𝐄𝐄dip 𝐫𝐫 3 𝜃𝜃 𝐫𝐫� 𝜃𝜃 𝜃𝜃̂ 𝜕𝜕1𝜕𝜕 4𝜋𝜋𝜖𝜖sin𝑟𝑟 4𝜋𝜋𝜖𝜖0𝑟𝑟 = = 𝜕𝜕𝜕𝜕 𝑝𝑝 𝜃𝜃 𝜃𝜃 3 𝐸𝐸 − 0 𝑟𝑟 𝜕𝜕1𝜕𝜕 4𝜋𝜋𝜖𝜖 𝑟𝑟 = = 0 sin 𝜕𝜕𝜕𝜕 𝐸𝐸𝜙𝜙 − 𝑟𝑟 𝜃𝜃 𝜕𝜕𝜕𝜕 16