Phys 685 Electrodynamics Homework #6: Solutions to Q1, Q4, Q5.
Q 1. A conducting spherical shell is placed in a uniform electrical field E⃗0 . The outer and inner radii of the shell are R1 and R2 , respectively. An electric dipole (with dipole moment ⃗p ) is placed at the center of the sphere. The angle between ⃗p and E⃗0 is α . Suppose the electrical potential of the sphere is V0 . (1) Find the potential inside and outside the sphere; and (2) Calculate the energy and force of the dipole in the uniform field.
Solution:
The potential outside the conducting sphere Φout ( r>R1 ) and the potential inside the sphere
Φin ( r<R2 ) satisfies the Laplace equation. Based on the symmetry and the general solution to the
Laplace equation we can write:
Φout=−E0rcosθ+
∑
n=0
∞ b
n
rn+1Pn(cosθ) ( r>R1 )
Inside the sphere, the potential due to the dipole is 1
4π ϵ0 ⃗ p⋅⃗r
r3 , or 1 4π ϵ0
pcos(θ−α)
r2 (depending
on the orientation of the dipole, the potential can also be written as 1
4π ϵ0
pcos(θ+α) r2 )
Therefore,
Φin=
∑
n=0
∞
an' rnPn(cos(θ−α))+
b0' r +
1 4π ϵ0
pcos(θ−α)
r2 ( r<R2 )
For r=R1 , Φout(R1,θ)=V0 :
Φout=−E0rcosθ+
∑
n=0
∞ b
n
rn+1Pn(cosθ)=−E0rcosθ+ b0
r + b1
r2cosθ+(vanishing terms)=V0 for (r=R1) α
⃗ E0
So: b0
R1=V0 and b1 R12
cosθ=E0R1cosθ , we derive b0=V0R1 and b1=E0R1 3
Φout(r ,θ)=−E0rcosθ+V0 R1
r +E0 R13 r2 cosθ
For r=R2 , Φin(R2,θ)=V0
Φin(R1,θ)=
∑
n=0
∞
an' R2nP
n(cos(θ−α))+
b0' R2
+ 1
4π ϵ0
pcos(θ−α) R22 =V0
a0'+b0' R2
+a1' R2cos(θ−α)+(vanishing terms)+ 1 4πϵ0
pcos(θ−α) R22
=V0
Therefore, we must have: a0'+b0'
R2
=V0 and a1' R2=− p
4πϵR22 , that is,
Φin(r ,θ)=a0'+(V0−a0') R2
r + p 4πϵ0
(
1 r2−
r
R32
)
cos(θ−α)a0' can be solved by considering the total charge on the interior surface of the conductor, which is 0.
∮
S(
∂Φin ∂r
)
R2d S=0 , or
∮
Ω[
a0'−(V0−a0') R2 −
1 4πϵ0
(
1 R23−
1
R23
)
cos(θ−α)]
R2 2dΩ=0
We derive:
a0'=V0
Φin(r ,θ)=V0+ p 4πϵ0
(
1 r2−
r R2
3
)
cos(θ−α) or,Φin(r ,θ)=V0+ 1 4πϵ0
pcos(θ−α) r2 −
1 4π ϵ0
pcos(θ−α)r R23
Now, it is easy to see that Φin(r ,θ) has two contributions from the dipole 1
4π ϵ0
pcos(θ−α)
r2 and
an effective external potential Φ=V0−
1 4π ϵ0
pcos(θ−α)r R2
The energy of a dipole in an external field can be calculated from:
W=−⃗p⋅⃗E (where E⃗ is the external electrical field ( E⃗=−∇ Φ )
For this, please refer to Jackson eqn. 4.24 and eqn 4.94 as well as the discussions below eqn. 4.94 on page 167.
Here, E⃗=−∂ Φ
∂r r̂− 1 r ∂ Φ∂ θ ̂θ
⃗ E= 1
4π ϵ0
pcos(θ−α) R2
3 ̂r+ 1 4π ϵ0
psin(θ−α) R2
3 ̂θ=
⃗p 4π ϵ0R2
3
The energy of the dipole in the external field is:
W=−⃗p⋅⃗E=− p2 4πϵ0R2
3
The force on the dipole is
⃗
F=−∇W=0
Q2. A solid infinitely long cylindrical dielectric with radius R is placed in a uniform electrical field ⃗
E0 . The electrical permeability of the cylinder is ϵ . The axis of the cylinder is perpendicular to ⃗
E0 . Calculate the electrical potential and electric field within the cylinder.
See handouts for solution
Q3. A total amount of charge q is uniformly distributed on a ring with radius a. Suppose the ring is perpendicular to the z axis, the charge distribution has the following form:
λ= q
2πa(1+sinϕ)
Calculate the electric dipole moment and quadruple moment of the total charge with respect to the center of the ring.
Q4.
(a) Prove the following theorem: For an arbitrary charge distribution ρ(⃗x) the values of the
(2l+1) moments of the first non-vanishing multipole are independent of the origin of the coordinate axes, but the values of all higher multipole moments do in general depend on the choice of origin. (b) A charge distribution has multipole moments q ,⃗p ,Qij … with respect to one set of coordinate
axes, and moments q ' ,⃗p ' ,Qij' … with respect to another set whose axes are parallel to the first but
whose origin is located at the point R=(X ,Y , Z) relative to the first. Determine explicitly the connections between the monopole, dipole, and quadrupole moments in the two coordinate frames. (c) If q≠0 , can R be found so that ⃗p '=0 ? if q≠0 , ⃗p≠0 , or at least ⃗p≠0 , can R be found so that Qij=0 ?
Solution
a)
Multipole moments are defined as: qlm=
∫
Ylm*
(θ' ,ϕ')r 'lρ(x')d3x ' . For the l -th multipole
moment , there are 2l+1 coefficients ( m=−l ,−1+1,...l ).
Suppose the (l−1) -th multipole moments are the terms that vanish, that is
ql−1,m=
∫
Yl−1,m*
(θ' ,ϕ')r 'l−1ρ(x')d3x '=0 .
for arbitrary (θ,ϕ) ,
ql−1,mYlm(θ,ϕ)=
∫
r ' l−1ρ(r ' ,θ,ϕ)d r'=0 . Hence, the following expression is satisfied:
∫
r'n−1ρ(r ' ,θ,ϕ)d r '=0 ( n=1,...l ).Now, with an arbitrary constant d , we have
∫
(r '+d)l−1ρ(x')d3x '=0 (this can also be derived from the bimodal theorem(a+b)n
=
∑
k
n ! k !(n−k)!a
k
bn−k ).
Therefore,
∫
(r '+d)lρ(r ' ,θ,ϕ)d r'=∫
(r '+d)l−1(r '+d)ρ(r' ,θ,ϕ)d r ' =∫
(r '+d)l−1(r '+d)ρ(r ' ,θ,ϕ)dr ' =
∫
(r '+d)l−1(r ')ρ(r ' ,ϕ,θ)d3x '=... =
∫
(r ')lρ(r ' ,θ,ϕ)d r '
Multiply both sides by Ylm
*
∫
Ylm*
(θ' ,ϕ')(r '+d)lρ(x')d3x '=
∫
Ylm*
(θ' ,ϕ')r 'lρ(x')d3x '
The multipole moment qlm is thus independent of the choice of origin ( arbitrary constant d).
q.e.d.
For higher multipole moments (e.g., l+1 th multipole moment),
∫
(r '+d)l+1ρ(r ' ,θ,ϕ)d r '=
∑
k=0
l+1
(l+1)!
(l+1−k)! k !
∫
r 'k
dl+1−kρ(r ' ,θ,ϕ)d r '
for k=1,...l
∫
r'kdl+1−kρ(r ' ,θ,ϕ)d r'=0Hence,
∫
(r '+d)l+1ρ(r ' ,θ,ϕ)d r '=∫
(d)l+1ρ(r ' ,θ,ϕ)d r 'Obviously, high-order multipole moments are dependent on the choice of origin.
q.e.d.
(b) In the first coordinate system, multipole moments are written as
qlm=
∫
Ylm* (θ' ,ϕ')r 'lρ(x')d3x 'and the shifted coordinate system, ρ(x')→ρ(x'+R) .
Denote the distance from R(X ,Y , Z) to the origin of the first coordinate system to be d ,
multipole moments in the shifted coordinate system become:
qlm R
=
∫
Ylm*
(θ' ,ϕ')(r '+d)lρ(x')d3x '
For monopole moment ( l=0 )
q=q ' (the value of the monopole moment is unchanged).
For dipole moment
p=
∫
x'ρ(x')d3x 'p'=
∫
(x'+R)ρ(x')d3x=p+RqFor quadruple moment
Qij=
∫
(3xi' xj'−r2
δi j)ρ(x')d
3 x '
Qij'=
∫
[3(xi'−Xi)(xj−Xj)−(r+d)2
δi j]ρ(x')d
3 x ' =Qij+
∫
[3(xi' X j+xj' Xi+XiXj)−2r dδi j−d2
δi j]ρ(x)d
3 x ' =Qij+3piXj+3pjXi+(3XiX j−d2
δij)q−2R pδi j
(c) if q≠0 , R can be found to make p'=0 . R=p/q .
Likewise, if q=0 and p≠0 , we find
Qij'=Qij+3piXj+3pjXi−2R pδi , j
X1=−Q12p1+Q13p2+Q12p3 6p2p3
X2=−Q23p2+Q12p3+Q23p1 6 p1p3
X3=−Q13p3+Q23p1+Q31p2 6p2p1
Q5. Two concentric conducting spheres of inner and outer radii a and b , respectively, carry charges ±Q . The empty space between the spheres is half-filled by a hemispherical shell of dielectic (of dielectric constant ϵ/ϵ0 ), as shown in the figure.
(a) Find the electric field everywhere between the spheres. (b) Calculate the surface charge distribution on the inner sphere.
(c) Calculate the polarization-charge density induced on the inner surface on the dielectric r=a .
Solution
Due to azimuthal symmetry, the general solution of the potential can be written as follows:
Φϵ0(r ,0<θ<π /2)= 1 4π ϵ0
∑
l=0∞
[Alrl
+Blr−l−1
]Pl(cosθ) and
+Q
−Q
ϵ
θ
Φϵ(r ,π/2<θ<π)=41
πϵ
∑
l=0∞
[Clrl
+Dlr−l−1
]Pl(cosθ)
Since the two spheres are conductors, the potentials on both spheres are constant values. Based on this,
we can simplify the potentials
Φϵ0(r ,0<θ<π /2)= 1
4π ϵ0[A0+B0/r] and
Φϵ(r ,π/2<θ<π)=41
πϵ[C0+D0/r]
For the left half ( θ<π/2 )
4π ϵ0Φϵ0(r)=A0+B0/r
for the right half ( π/2<θ<π )
4π ϵΦϵ(r)=C0+D0/r
Consequently, for the left hemisphere ( r= b)
4π ϵ0Φϵ0(b)=A0+B0/b (1) 4π ϵ0Φϵ0(a)=A0+B0/a (2)
(1) – (2):
4π ϵ0[Φϵ0(b)−Φϵ(b)]=B0( 1 b−
1 a)
and for the right hemispheres
4π ϵΦϵ(b)=C0+D0/b (3)
4π ϵΦϵ(a)=C0+D0/a (4)
(3) - ( 4):
4π ϵ[Φϵ0(b)−Φϵ(b)]=D0( 1 b−
1 a)
Also because of the condition that the spheres are conducting, we derive D0/B0=ϵ/ϵ0
We now apply boundary conditions:
(1) ∂Φϵ0
∂r ∣r=a=
Qleft 2πa2=
1 4π ϵ0
(2) ∂Φϵ
∂r ∣r=a=
Qright 2πa2=
1 4πϵ
D0 a2
There, we derive the charge on the left half of the inner sphere
Qleft= Q
ϵ/ϵ0+1 and Qright= Q ϵ0/ϵ+1
The electric field everywhere is:
E(r)= Q
2π(ϵ+ϵ0)r2r
(3) The polarization charge density induced on the surface of the dielectric at r=a:
σpol=−( ⃗P2− ⃗Pa)⋅̂n=− ⃗P2⋅̂r
σpol=
Q(ϵ0−ϵ) 2π(ϵ0+ϵ)a
