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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
β’ No Class on Friday; No office hour.
2
β’ HW # 4 has been posted β’ Solutions to HW # 5 have been posted
Summary of Lecture # 10:
The potential in the region above the infinite grounded conducting plane?
V π«π« =ππ
4ππππ01
x2 + y2 + (z β d)2β
1x2 + y2 + (z + d)2
The potential outside the grounded conducting sphere?
V π«π« =1
4ππππ0ππr +
ππβ²rβ²
ππ =π π 2
ππ ππβ² = βπ π ππππ
3
Question:
4
The potential in the region above the infinite grounded conducting plane?
V π«π« =ππ
4ππππ01
x2 + y2 + (z β d)2β
1x2 + y2 + (z + d)2
Q: Is kV π«π« 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.
ππ2V1 = βππππ0
In this case one has to satisfy the Poissonβs equation. But kV π«π« does not satisfy it.
Only when ππ = 0 (everywhere), kV π«π« can be a solution as well Exception: But then since in this case V π«π« = ππ, it is a trivial solution.
5
Multipole Expansion (Potentials at large distances) β’ What is the potential due to a point charge (monopole)?
V π«π« =1
4ππππ0ππππ
β’ What is the potential due to a dipole at large distance?
V π«π« =1
4ππππ0ππ
r+ β
ππrβ
rΒ±2 = ππ2 +
ππ2
2
β ππππcosππ
β ππ2 1 βππππ
cosππ
for ππ β« ππ, and using binomial expansion 1rΒ±
=
(for ππ β« ππ)
1r+β
1rβ
=
V π«π« =1
4ππππ0ππππcosππ ππ2
(goes like ππ/ππ)
(goes like ππ/ππππ at large ππ)
= ππ2 1 βππππ
cosππ +ππ2
4ππ2
β1ππ
1 Β±ππ2ππ
cosππ
So, ππππ2
cosππ 1ππ
1 βππππ
cosππβ12
6
Multipole Expansion (Potentials at large distances)
β’ What is the potential due to a quadrupole at large distance?
(goes like ππ/ππππ at large ππ)
β’ What is the potential due to a octopole at large distance?
(goes like ππ/ππππ at large ππ)
Note1: Multipole terms are defined in terms of their ππ 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 ππ-charge system can have any multipole contribution.
7
Multipole Expansion (Potentials at large distances) β’ What is the potential due to a
localized charge distribution?
V(π«π«) =1
4ππππ0οΏ½ππππr =
14ππππ0
οΏ½ππ(π«π«β²)
r ππππβ²
Using the cosine rule,
r2 = ππ2 + ππβ²2 β 2ππππβ²cosπΌπΌ
r2 = ππ2 1 +ππβ²
ππ
2
β 2ππβ²
ππcosπΌπΌ
Define: ππ β‘ππβ²
ππππβ²
ππβ 2cosπΌπΌ r =
1r
=
1r
=1ππ
1 β12ππ +
38ππ2 β
516
ππ3 + β― (using binomial expansion)
So,
Or,
Source coordinates: (ππβ²,ππβ²,ππβ²) Observation point coordinates: (ππ,ππ,ππ) Angle between π«π« and π«π«β²: πΌπΌ r = ππ 1 +
ππβ²
ππππβ²
ππβ 2cosπΌπΌ
1ππ
1 + ππ β1/2
ππ 1 + ππ
Multipole Expansion (Potentials at large distances)
1r
=1ππ
1 β12ππ +
38ππ2 β
516
ππ3 + β―
=1ππ 1 β
12
ππβ²
ππππβ²
ππ β 2cosπΌπΌ +38
ππβ²
ππ
2 ππβ²
ππ β 2cosπΌπΌ2
β5
16ππβ²
ππ
3 ππβ²
ππ β 2cosπΌπΌ3
+ β―
=1ππ 1 +
ππβ²
ππ cosπΌπΌ +ππβ²
ππ
2
3cos2πΌπΌ β 1 /2 βππβ²
ππ
3
5cos3πΌπΌ β 3cosπΌπΌ /2 + β―
=1ππ οΏ½
ππβ²
ππ
ππβ
ππ=0
ππππ(cosπΌπΌ)
V(π«π«) 8
=1
4ππππ0οΏ½
1ππ ππ+1
β
ππ=0
οΏ½ ππβ² ππ ππππ(cosπΌπΌ)ππ π«π«β² ππππβ² =1
4ππππ0οΏ½ππ(π«π«β²)
r ππππβ²
β’ What is the potential due to a localized charge distribution?
V(π«π«) =1
4ππππ0οΏ½ππππr =
14ππππ0
οΏ½ππ(π«π«β²)
r ππππβ²
ππππ(cosπΌπΌ) are Legendre polynomials
V(π«π«) =1
4ππππ0οΏ½
1ππ ππ+1
β
ππ=0
οΏ½ ππβ² ππ ππππ(cosπΌπΌ)ππ π«π«β² ππππβ²
Multipole Expansion of ππ(π«π«)
Monopole potential ( 1/ππ dependence)
Dipole potential ( 1/ππ2 dependence)
Quadrupole potential ( 1/ππ3 dependence)
9
Multipole Expansion (Potentials at large distances) β’ What is the potential due to a
localized charge distribution?
V(π«π«) =1
4ππππ0οΏ½ππππr =
14ππππ0
οΏ½ππ(π«π«β²)
r ππππβ²
=1
4ππππ0πποΏ½ππ π«π«β² ππππβ² +
14ππππ0ππ2
οΏ½ ππβ²(cosπΌπΌ)ππ π«π«β² ππππβ² +1
4ππππ0ππ3οΏ½ ππβ² 2 3
2 cos2πΌπΌ β12 ππ π«π«β² ππππβ²
+β―
Multipole Expansion (Few comments)
β’ More terms can be added if greater accuracy is required
β’ It is an exact expression, not an approximation.
β’ At large ππ, the potential can be approximated by the first non-zero term.
β’ A particular term in the expansion is defined by its ππ dependence
10
+1
4ππππ0ππ2οΏ½ ππβ²(cosπΌπΌ)ππ π«π«β² ππππβ² +
14ππππ0ππ3
οΏ½ ππβ² 2 32
cos2πΌπΌ β12
ππ π«π«β² ππππβ²
+β―
V π«π«
=1
4ππππ0πποΏ½ππ π«π«β² ππππβ²
Monopole potential ( 1/ππ dependence)
Dipole potential ( 1/ππ2 dependence)
Quadrupole potential ( 1/ππ3 dependence)
Questions 1:
Q: In this following configuration, is the βlarge π«π«β limit valid, since the source dimensions are much smaller than π«π«?
Ans: No. The βlarge π«π«β limit essentially mean π«π« β« |π«π«β²|. In majority of the situations, the charge distribution is centered at the origin and therefore the βlarge π«π«β limit is the same as source dimension being smaller than π«π«.
11
Vmono π«π« =1
4ππππ01πποΏ½ππ π«π«β² ππππβ²
Monopole term:
=1
4ππππ0ππππ
β’ ππ = β«ππ π«π«β² ππππβ² is the total charge β’ If ππ = 0, monopole term is zero. β’ For a collection of point charges
Dipole term:
Vdip π«π« =1
4ππππ01ππ2οΏ½ ππβ²(cosπΌπΌ)ππ π«π«β² ππππβ²
πΌπΌ is the angle between π«π« and π«π«β².
Vdip π«π« =1
4ππππ01ππ2π«π«οΏ½ β οΏ½ π«π«β²ππ π«π«β² ππππβ²
β’ π©π© β‘ β« π«π«β²ππ π«π«β² ππππβ² is called the dipole moment of a charge distribution
Vdip π«π« =1
4ππππ0π©π© β π«π«οΏ½ππ2
β’ If π©π© = 0, dipole term is zero.
β’ For a collection of point charges.
Multipole Expansion (Monopole and Dipole terms)
ππ = οΏ½ππππ
ππ
ππ=1
π©π© = οΏ½π«π«π’π’β²ππππ
ππ
ππ=1
So, ππβ² cosπΌπΌ = π«π«οΏ½ β π«π«β²
12
Monopole term:
Example: A three-charge system
ππ = οΏ½ππππ
ππ
ππ=1
= βππ
π©π© = οΏ½π«π«π’π’β²ππππ
ππ
ππ=1
= ππππ πποΏ½ + βππππ β ππ βππ πποΏ½ = ππππ πποΏ½
Therefore the system will have both monopole and dipole contributions 13
Multipole Expansion (Monopole and Dipole terms)
Dipole term:
β1
4ππππ0π©π© β π«π«οΏ½ππ2
ππ = οΏ½ππππ
ππ
ππ=1
π©π© = οΏ½π«π«π’π’β²ππππ
ππ
ππ=1
Vmono π«π« =1
4ππππ01πποΏ½ππ π«π«β² ππππβ² β
14ππππ0
ππππ
Vdip π«π« =1
4ππππ01ππ2οΏ½ ππβ²(cosπΌπΌ)ππ π«π«β² ππππβ²
(for point charges)
(for point charges)
Monopole term:
Dipole term:
β1
4ππππ0π©π© β π«π«οΏ½ππ2
ππ = οΏ½ππππ
ππ
ππ=1
π©π© = οΏ½π«π«π’π’β²ππππ
ππ
ππ=1
14
Multipole Expansion (Monopole and Dipole terms)
Vmono π«π« =1
4ππππ01πποΏ½ππ π«π«β² ππππβ² β
14ππππ0
ππππ
Vdip π«π« =1
4ππππ01ππ2οΏ½ ππβ²(cosπΌπΌ)ππ π«π«β² ππππβ²
Example: Origin of Coordinates
ππ = ππ
π©π© = 0
Vmono π«π« =1
4ππππ0ππππ
Vdip π«π« = 0
ππ = ππ
π©π© = πππππποΏ½
Vmono π«π« =1
4ππππ0ππππβ
14ππππ0
ππr
Vdip π«π« =1
4ππππ0π©π© β π«π«οΏ½ππ2
(for point charges)
(for point charges)
Questions 2:
ππ = ππ
π©π© = πππππποΏ½
Vmono π«π« =1
4ππππ0ππππβ
14ππππ0
ππr
Vdip π«π« =1
4ππππ0π©π© β π«π«οΏ½ππ2
Q: Why not calculate the potential directly ??
Ans: Yes, that is what should be done. For a point charge, we donβt need a multipole expansion to find the potential. This is only for illustrating the connection.
15
The electric field of pure dipole ( πΈπΈ = ππ )
Vdip π«π« =1
4ππππ0π©π© β π«π«οΏ½ππ2
=1
4ππππ0πππ³π³οΏ½ β π«π«οΏ½ππ2
=1
4ππππ0ππcosππππ2
ππ π«π« = βππV
πΈπΈππ = βππππππππ
=2ππcosππ4ππππ0ππ3
πΈπΈππ = β1ππππππππππ
=ππsinππ4ππππ0ππ3
πΈπΈππ = β1
ππsinππππππππππ
= 0
ππdip π«π« =ππ
4ππππ0ππ3(2cosππ π«π«οΏ½ + sinππ οΏ½ΜοΏ½π)
16
ππ = 0 Assume π©π© = πππ³π³οΏ½ π©π© β 0 And
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