MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
Physics 8.07: Electromagnetism II
October 18, 2012
Prof. Alan Guth
QUIZ 1
Reformatted to Remove Blank Pages∗
THE FORMULA SHEETS ARE AT THE END OF THE EXAM.
Your Name
∗
Problem
Maximum
1
30
2
20
3
20
4
30
TOTAL
100
Score
The errors that were corrected on the blackboard during the quiz are incorporated
here into the text.
8.07 QUIZ 1, FALL 2012
p. 2
PROBLEM 1: SOME SHORT EXERCISES (30 points)
× (sA),
where s is a scalar
(a) (10 points) Use index notation to derive a formula for ∇
is a vector field A(
r ).
field s(r ) and A
(b) (10 points) Which of the following vector fields could describe an electric field in
electrostatics? Say yes or no for each, and give a very brief reason.
r ) = x êx − y êy .
(i) E(
r ) = y êx + x êy .
(ii) E(
r ) = y êx − x êy .
(iii) E(
(c) (10 points) Suppose that the entire x-z and y-z planes are conducting. Calculate the
force F on a particle of charge q located at x = x0 , y = y0 , z = 0.
PROBLEM 2: ELECTRIC FIELDS IN A CYLINDRICAL GEOMETRY (20
points)
A very long cylindrical object consists of a solid inner cylinder of radius a, which has
a uniform charge density ρ, and a concentric thin cylinder, of radius b > a, which has an
equal but opposite total charge, uniformly distributed on the surface.
(a) (7 points) Calculate the electric field everywhere.
(b) (6 points) Calculate the electric potential everywhere, taking V = 0 on the outer
cylinder.
(c) (7 points) Calculate the electrostatic energy per unit length of the object.
PROBLEM 3: MULTIPOLE EXPANSION FOR A CHARGED WIRE (20
points)
A short piece of wire is placed along the z-axis, centered at the origin. The wire
carries a total charge Q, and the linear charge density λ is an even function of z: λ(z) =
λ(−z). The rms length of the charge distribution in the wire is l0 ; i.e.,
1
2
z 2 λ(z) dz .
l0 =
Q wire
(a) (10 points) Find the dipole and quadrupole moments for this charge distribution.
Note that the dipole and quadrupole moments are defined on the formula sheets as
pi = d3 x ρ(r ) xi ,
Qij =
d3 x ρ(r )(3xi xj − δij |r |2 ) .
(b) (10 points) Give an expression for the potential V (r, θ) for large r, including all terms
through the quadrupole contribution.
8.07 QUIZ 1, FALL 2012
p. 3
PROBLEM 4: A SPHERICAL SHELL OF CHARGE (30 points)
(a) (10 points) A spherical shell of radius R, with an unspecified surface charge density,
is centered at the origin of our coordinate system. The electric potential on the shell
is known to be
V (θ, φ) = V0 sin θ cos φ ,
where V0 is a constant, and we use the usual polar coordinates, related to the Cartesian coordinates by
x = r sin θ cos φ ,
y = r sin θ sin φ ,
z = r cos θ .
Find V (r, θ, φ) everywhere, both inside and outside the sphere. Assume that the
zero of V is fixed by requiring V to approach zero at spatial infinity. (Hint: this
problem can be solved using traceless symmetric tensors, or if you prefer you can
use standard spherical harmonics. A table of the low- Legendre polynomials and
spherical harmonics is included with the formula sheets.)
(b) (10 points) Suppose instead that the potential on the shell is given by
V (θ, φ) = V0 sin2 θ sin2 φ .
Again, find V (r, θ, φ) everywhere, both inside and outside the sphere.
(c) (10 points) Suppose instead of specifying the potential, suppose the surface charge
density is known to be
σ(θ, φ) = σ0 sin2 θ sin2 φ .
Once again, find V (r, θ, φ) everywhere.
8.07 FORMULA SHEET FOR QUIZ 1, FALL 2012
p. 4
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
Physics 8.07: Electromagnetism II
October 18, 2012
Prof. Alan Guth
FORMULA SHEET FOR QUIZ 1
Exam Date: October 18, 2012
∗∗∗
Some sections below are marked with asterisks, as this section is. The asterisks
indicate that you won’t need this material for the quiz, and need not understand it. It is
included, however, for completeness, and because some people might want to make use
of it to solve problems by methods other than the intended ones.
Index Notation:
·B
= Ai Bi ,
A
i = ijk Aj Bk ,
×B
A
ijk pqk = δip δjq − δiq δjp
det A = i1 i2 ···in A1,i1 A2,i2 · · · An,in
Rotation of a Vector:
Ai = Rij Aj ,
Orthogonality: Rij Rik = δjk
j=1
Rotation about z-axis by φ: Rz (φ)ij
Rotation about axis n
ˆ by φ:∗∗∗
j=2

i=1  cos φ − sin φ


sin φ cos φ
= i=2 


i=3
0
0
(RT T = I)
j=3
0
0
1







R(n̂, φ)ij = δij cos φ + n̂i n̂j (1 − cos φ) − ijk n̂k sin φ .
Vector Calculus:
Gradient:
ϕ)i = ∂i ϕ ,
(∇
Divergence:
·A
≡ ∂i A i
∇
Curl:
× A)
i = ijk ∂j Ak
(∇
Laplacian:
· (∇
ϕ) =
∇2 ϕ = ∇
∂i ≡
∂
∂xi
∂ 2ϕ
∂xi ∂xi
8.07 FORMULA SHEET FOR QUIZ 1, FALL 2012
p. 5
Fundamental Theorems of Vector Calculus:
b
Gradient:
ϕ · d = ϕ(b) − ϕ(a)
∇
a
·A
d x=
∇
· da
A
3
Divergence:
V
Curl:
S
S
where S is the boundary of V
· d
(∇ × A) · da =
A
P
where P is the boundary of S
Delta Functions:
ϕ(r )δ 3 (r − r ) d3 x = ϕ(r )
ϕ(x)δ(x − x ) dx = ϕ(x ) ,
d
dϕ ϕ(x) δ(x − x ) dx = −
dx
dx x=x
δ(x − xi )
, g(xi ) = 0
δ(g(x)) =
|g (xi )|
i
∇2
1
= −4πδ 3 (r − r )
|r − r |
Electrostatics:
1 (r − r ) qi
1
(r − r )
r ) d3 x
E(r ) =
=
3 ρ(
4π0
4π0 i |r − r |3
|r − r |
r
1
ρ(r ) 3 E (r ) · d =
V (r ) = V (r 0 ) −
d x
4π0
|r − r |
r0
V
·E
= ρ ,
×E
= 0,
= −∇
∇
∇
E
0
ρ
∇2 V = −
(Poisson’s Eq.) ,
ρ = 0 =⇒ ∇2 V = 0 (Laplace’s Eq.)
0
Laplacian Mean Value Theorem (no generally accepted name): If ∇2 V = 0, then
the average value of V on a spherical surface equals its value at the center.
Energy:
1 1 qi qj
ρ(r )ρ(r )
1 1
W =
=
d3 x d3 x
2 4π0
rij
2 4π0
|r − r |
1
W =
2
ij
i=j
1
d xρ(r )V (r ) = 0
2
3
2 3
E
d x
8.07 FORMULA SHEET FOR QUIZ 1, FALL 2012
p. 6
Conductors:
= σ n̂
Just outside, E
0
Pressure on surface:
1
outside
σ |E|
2
Two-conductor system with charges Q and −Q: Q = CV , W = 12 CV 2
N isolated conductors:
Vi =
Pij Qj ,
Pij = elastance matrix, or reciprocal capacitance matrix
Cij Vj ,
Cij = capacitance matrix
j
Qi =
j
Image charge in sphere of radius a: Image of Q at R is q = −
a
a2
Q, r =
R
R
Separation of Variables for Laplace’s Equation in Cartesian Coordinates:
V =
cos αx
sin αx
cos βy
sin βy
cosh γz
sinh γz
where γ 2 = α2 + β 2
Separation of Variables for Laplace’s Equation in Spherical Coordinates:
Traceless Symmetric Tensor expansion:
1
1 ∂
2 ∂ϕ
+ 2 ∇2θ ϕ = 0 ,
r
∇ ϕ(r, θ, φ) = 2
∂r
r
r ∂r
where the angular part is given by
1 ∂
∂ϕ
1 ∂ 2ϕ
2
sin θ
+
∇θ ϕ ≡
sin θ ∂θ
∂θ
sin2 θ ∂φ2
2
()
()
∇2θ Ci1 i2 ...i n̂i1 n̂i2 . . . n
ˆ i = −( + 1)Ci1 i2 ...i n
ˆ i1 n
ˆ i2 . . . n
ˆ i ,
()
where Ci1 i2 ...i is a symmetric traceless tensor and
n̂ = sin θ cos φ ê1 + sin θ sin φ ê2 + cos θ ê3 .
General solution to Laplace’s equation:
∞ B
()
A r + +1 Ci1 i2 ...i n
ˆ i1 n
ˆ i2 . . . n
ˆ i ,
V (r ) =
r
=0
where r = rn
ˆ
8.07 FORMULA SHEET FOR QUIZ 1, FALL 2012
p. 7
Azimuthal Symmetry:
∞ B
A r + +1 c { ẑi1 . . . ẑi } n̂i1 . . . n
V (r ) =
ˆ i
r
=0
where { . . . } denotes the traceless symmetric part of . . . .
Special cases:
{1} = 1
{ ẑi } = ẑi
{ ẑi ẑj } = ẑi ẑj − 13 δij
zˆi δjk + ẑj δik + ẑk δij
{ ẑi ẑj ẑk ẑm } = ẑi ẑj ẑk ẑm − 17 ẑi ẑj δkm + ẑi ẑk δmj + ẑi zˆm δjk + ẑj ẑk δim
1
δij δkm + δik δjm + δim δjk
+ ẑj ẑm δik + ẑk ẑm δij + 35
{ ẑi ẑj ẑk } = ẑi ẑj ẑk −
1
5
Legendre Polynomial / Spherical Harmonic expansion:
General solution to Laplace’s equation:
∞ Bm
Am r + +1 Ym (θ, φ)
V (r ) =
r
=0 m=−
2π
π
dφ
Orthonormality:
0
0
sin θ dθ Y∗ m (θ, φ) Ym (θ, φ) = δ δm m
Azimuthal Symmetry:
∞ B
A r + +1 P (cos θ)
V (r ) =
r
=0
Multipole Expansion:
First several terms:
1 Q p · r̂ 1 r̂i r̂j
V (r ) =
+ 2 +
Qij + · · · , where
r
2 r3
4π0 r
3
3
Q = d x ρ(r ) , pi = d x ρ(r ) xi Qij = d3 x ρ(r )(3xi xj −δij |r |2 ) ,
dip =
E
1 1
1
3(p
·
r̂)r̂
−
p
−
pi δ 3 (r )
4π0 r 3
30
8.07 FORMULA SHEET FOR QUIZ 1, FALL 2012
p. 8
Traceless Symmetric Tensor version:
∞
1 1
()
Ci1 ...i n
ˆ i1 . . . n̂i ,
V (r ) =
+1
4π0
r
=0
where
()
Ci1 ...i
(2 − 1)!!
=
!
ρ(r ) { r i1 . . . r i } d3 x
Griffiths version:
∞
1 1
V (r ) =
r ρ(r )P (cos θ ) d3 x
+1
4π0
r
=0
where θ = angle between r and r .
∞
r
1
<
=
P (cos θ ) ,
+1
|r − r |
r>
=0
1
P (x) = 2 !
d
dx
∞
1
√
=
λ P (x)
2
1 − 2λx + λ
=0
(x2 − 1) ,
(Rodrigues’ formula)
P (1) = 1
P (−x) = (−1) P (x)
1
−1
dx P (x)P (x) =
2
δ 2 + 1
Spherical Harmonic version:∗∗∗
∞
1 4π qm
Ym (θ, φ)
4π0
2 + 1 r +1
=0 m=−
∗ r ρ(r ) d3 x
where qm = Ym
V (r ) =
Connection Between Traceless Symmetric Tensors and Legendre Polynomials
or Spherical Harmonics:
P (cos θ) =
(2)!
{ ẑi1 . . . ẑi } n̂i1 . . . n̂i
2 (!)2
For m ≥ 0,
(,m)
Ym (θ, φ) = Ci1 ...i n
ˆ i1 . . . n̂i ,
(,m)
ˆ+
where Ci1 i2 ...i = dm { û+
i1 . . . u
im ẑim+1 . . . ẑi } ,
2m (2 + 1)
(−1)m (2)!
with dm =
,
4π ( + m)! ( − m)!
2 !
8.07 FORMULA SHEET FOR QUIZ 1, FALL 2012
p. 9
1
and û+ = √ (êx + iêy )
2
∗
(θ, φ)
Form m < 0, Y,−m (θ, φ) = (−1)m Ym
More Information about Spherical Harmonics:∗∗∗
Ym (θ, φ) =
2 + 1 ( − m)! m
P (cos θ)eimφ
4π ( + m)!
where Pm (cos θ) is the associated Legendre function, which can be defined by
Pm (x) =
+m
(−1)m
2 m/2 d
)
(x2 − 1)
(1
−
x
2 !
dx+m
Legendre Polynomials:
8.07 FORMULA SHEET FOR QUIZ 1, FALL 2012
p. 10
SPHERICAL HARMONICS Ylm(θ , φ)
l=0
1
Y00 =
4π
3
sin θeiφ
8π
Y11 = l=1
3
cos θ
4π
Y10 =
Y22 =
l=2
1
4
15
sin2 θe2iφ
2π
15
sin θ cosθeiφ
8π
Y21 = -
Y20 =
5
( 32 cos2θ
4π
1
)
2
35
sin3 θe3iφ
4π
Y33 = -
1
4
Y32 =
1
4
105
sin2 θ cos θe2iφ
2π
Y31 = -
1
4
21
sinθ (5cos2θ -1)eiφ
4π
l=3
Y30 =
7
( 5 cos3θ
4π 2
3
2
cos θ)
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8.07 Electromagnetism II
Fall 2012
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