Download Physics 417G : Solutions for Problem set 3

Physics 417G : Solutions for Problem set 3
Due : February 5, 2016
Please show all the details of your computations including intermediate steps.
1
Problem 1
A thick spherical shell of inner radius a and outer radius b carries a polarization P~ (~r) = k~r, where k
is a constant and ~r is the vector from the center.
a) Calculate the surface bound charges σa at r = a and σb at r = b and ρ inside the thick spherical
shell.
b) What is the total charge?
c) Find the electric fields in the three regions r < a, a < r < b and r > b.
Sol: a) The bound charges are given by σ = P~ · n̂ with the unit vector n̂ outward normal to the surface
~ P~ (~r). Thus
and ρ = −∇
σa = P~ · (−r̂) = −ka ,
σb = P~ · (r̂) = kb ,
~ · P~ (~r) = − 1 ∂r r2 kr = −3k .
ρ = −∇
2
r
b) The total charge should be 0. This can be checked by adding all the charges
4πa3
4πb3
2
2
−
ρ=0.
Qtotal = 4πa σa + 4πb σb +
3
3
H
~ · dS
~ = Qencl /0 and the charges, we can get
c) Using the Gauss law E

~ =0,

E
r<a

~
~
E=
E = − k0 ~r
a<r<b ,


~ =0
E
b<r
3
3
3
where inside the shell, the total charge is Qencl = −4πka3 − 3k 4π
3 [r − a ] = −4πkr , which is the
combination of the total surface charge at r = a and the volume charge between the radius a and r.
For the region r > b, the total charge enclosed is 0, thus the electric field vanishes.
2
Problem 2
Let us also consider the same thick spherical shell of inner radius a and outer radius b, which now
carries a different polarization vector P~ (~r) = kr r̂, where k is a constant and ~r is the vector from the
center. We add a metal sphere with radius a inside the inner shell with the same origin of the spherical
shell. The inner sphere carries total charge Q.
~ for the three regions, r < a, a < r < b
a) Use a Gauss’ law to compute the displacement current D
and r > b.
b) Using the results of a), compute the electric field for the corresponding regions.
c) Calculate the surface bound charges σa at r = a and σb at r = b and ρ inside the thick spherical
shell.
d) Find the electric fields in the three regions r < a, a < r < b and r > b using the bound charges
along with the charge Q. Compare the results with b).
H
~ goes D
~ · dS
~ = Qf ree . The charge in the conductor is a free charge and
Sol: a) Gauss’ law for the D
is located in the surface. Thus we have

~ =0,
D
r<a

~
~ = Q r̂2
D=
a
<
r<b .
D
4π r

Q r̂
~
D = 4π r2
b<r
Note : For the rest of the problem, we need to think about the situation carefully. You can think
about the problem as the thick spherical shell (a dielectric material) responds to the electric field due
to the charge Q of the metal sphere and produce P~ .
~ and D
~ = 0 E
~ + P~ , we get
b) From the relation between the electric field E

~ =0,
E
r<a


Q 1
k
1
~
~
a<r<b .
E = 0 4π r2 − r r̂
E=


Q r̂
~
E=
b<r
2
4π0 r
~ · P~ = − 12 ∂r r2 k = − k2 and
c) The bound charges are ρb = −∇
r
r
r
k/b ,
r=b
σb = P~ · n̂ =
.
−k/a
r=a
d) The electric fields are the same as

~ =0,
E


Q 1
k
1
~ =
~ =
−
E
E
0
4π r 2
r r̂


Q
r̂
~ =
E
4π0 r 2
r<a
a<r<b .
b<r
For instance, in the middle region a < r < b, the charge enclosed is
Z r
Qencl = Q + (−k/a)πa2 + 4π
r2 dr(−k/r2 ) = Q − 4πkr .
a
This gives the electric field for a < r < b.
3
Problem 3
In this problem we are going to find the electric field produced by a uniformly polarized sphere of
radius R. For this you can start with
I
Z
1
1
σb
ρb
~ ·P ,
φ(~r) =
dS 0
+
d3 x0
,
σb = P~ · n̂ , ρb = −∇
4π0 S 0
|~x − ~x0 | 4π0 V 0
|~x − ~x0 |
where V 0 is a localized volume and S 0 is the boundary with its unit normal vector n̂.
Part I:
~ = −∇φ
~ can be rewritten as
a) Show that the electric field E
Z
~x − ~x0
1
~ r) = −P~ · ∇
~ E~ ,
d3 x0
.
E(~
E~ =
4π0 V 0
|~x − ~x0 |3
b) Evaluate E~ inside and outside of the uniform sphere.
~ r).
c) Now compute the electric field E(~
Part II:
For any charge configuration with Azimuthal symmetry (potential is independent of the angle φ). we
can express the potential as
φ(r, θ) =
∞ X
l=0
Bl
Al r + l+1
r
l
Pl (cos θ) ,
in the spherical polar coordinate, where Pl represent Legendre Polynomial.
d) For uniformly polarized sphere, compute the bound charges. Are they familiar?
e) Find the resulting potential inside and outside the sphere.
Hint: It is important to have finite potentials both inside and outside. This will determine possible
expansions inside and outside. This allows us to determine Bl in terms of Al . The surface charge
provides a discontinuity on the radial derivative of the potential at r = R. This determines Al .
f) Compute electric field inside and outside of the sphere. Are they the same as c)?
H
σb
1
Sol: a) One can easily check the ρb = 0 for uniform polarization. Thus φ(~r) = 4π
dS 0 |~x−~
x0 | .
S0
0
For this you can start with
I
I
1
~x − ~x0
1
~ 0 · P~ −∇
~
~ 0 · P~
~ = −∇φ(~
~ r) = 1
dS
d
S
=
E
4π0 S 0
|~x − ~x0 |
4π0 S 0
|~x − ~x0 |3
Z
Z
0
1
1
x − ~x0
3 0 ~
~ ·∇
~
~ 0k P~k ~x − ~x
=
=
−
P
,
d3 x0 ∇
d
x
4π0 V 0
|~x − ~x0 |3
4π0 V 0
|~x − ~x0 |3
~ 0 f (~x − ~x0 ) = −∇f
~ (~x − ~x0 ) in the
where we used the divergence theorem, the constant nature of P~k , ∇
second line.
b) E~ is the electric field due to a sphere uniformly charged ρ = 1. Using the Gauss’ law, we can get
(
~
r
r<R
30
3
.
E~ =
4πR /3 r̂
r>R
4π0 r 2
~ r) = −P~ · ∇
~ E.
~
c) Then the electric fields are E(~
~
P
(
~ =
E
4πR3 /3
4π0 r 3
− 30
3(P~ · r̂)r̂ − P~
r<R
r>R
.
d) The bound charges are ρb = 0, σb = P cos θ.
e) Inside and outside the sphere, we need to have finite potentials. Thus we have
φin (r, θ) =
φout (r, θ) =
∞
X
l=0
∞
X
l=0
Al rl Pl (cos θ)
r≤R,
Bl
Pl (cos θ)
rl+1
r≥R.
(1)
For example, φout is finite everywhere for r ≥ R, especially at r → ∞. These two functions should be
continuously connected at r = R. Thus we have
∞
X
Al Rl Pl (cos θ) =
l=0
∞
X
Bl
Pl (cos θ) ,
Rl+1
l=0
which should be valid for each l because the Legendre polynomial Pl (cos θ) are orthogonal each other.
To prove this, we multiply Pl0 (cos θ) and integrate to use the orthogonality properties, given by
Z π
2
d(cos θ)Pl (cos θ)Pl0 (cos θ) = −
δll0 .
2l + 1
0
Thus we have
Bl = Al R2L+1 .
To determine Al , we use the fact that the surface charge produces the discontinuity in the electric
field, only the normal component of the electric field with respect to the surface. This is radial direction.
We can express this as n̂ · E = r̂ · E = −∂r φ. Thus we have
σb
(∂r φout − ∂r φin )
=− ,
0
r=R
which gives
−
∞
X
∞
(l + 1)
l=0
−→
X
Bl
σb
Pl (cos θ) −
lAl Rl−1 Pl (cos θ) = − ,
l+2
R
0
l=0
∞
X
l=0
(2l + 1)Al Rl−1 Pl (cos θ) =
σb
,
0
(2)
where we use Bl = Al R2L+1 . By multiplying Pl0 (cos θ) and using the orthogonality properties, we get
Z π
1
Al = − l−1
σb (θ)Pl (cos θ)d(cos θ) .
2R 0 0
Until now we use a general formula that will hold with the Azimuthal symmetry in spherical polar
coordinate.
For σb (θ) = P cos θ = P P1 (cos θ), we know that only A1 will be non-zero due to the orthogonality
property of Legendre polynomial. Thus
Z π
P
1
P1 (cos θ)Pl (cos θ)d(cos θ) =
δl1 .
Al = −P
2Rl−1 0 0
30
Thus we determine the potential as
P
P
rP1 (cos θ) =
r cos θ
r≤R,
30
30
P R3 1
P R3 1
φout (r, θ) =
P
(cos
θ)
=
cos θ
r≥R.
1
30 r2
30 r2
φin (r, θ) =
~ r) = −∇φ.
~
f) Then the electric fields are E(~
(
~
P
− 30
~
3
E=
4πR /3
~ · r̂)r̂ − P~
3(
P
3
4π0 r
4
r<R
r>R
.
Problem 4
a) A short cylinder, of radius a and length L, carries a ‘frozen’ uniform polarization P~ , parallel to its
axis. Find the bound charges. Now sketch the electric fields (i) for L a, (ii) for L a, and (iii) for
L ≈ a.
b) A very long cylinder, of radius a, carries a uniform polarization P~ perpendicular to its axis, not
radial direction. Find the electric field inside and outside the cylinder (using one of the two methods
below).
Hint : Method (I). Think of it as two cylinders of opposite uniform charge density ±ρP (we reserve
ρ for the cyliderical coordinate), whose centers are separated by a vector d~ with d < R. see the figure,
which is a view of the cross section of the cylinder for method (I). One can use Gauss’ law to compute
the electric fields of the two cylinders and add them to get the result we want. Now it remains to
~ The polarization vector P~ is related to d~ because the total electric dipole moment is
connect P~ and d.
nothing but volume times the polarization vector.
Method (II). You can find the bound charges and solve the scalar potential as the second part of the
problem 3.
Sol: a) Simple computations give ρb = 0, σb = P~ · n̂ = ±P respectively at each end.
(i) For L a, the ends look like point charges, and the whole thing is like a physical dipole, of length
L. See Fig. (a).
(ii) For L a, it is like a circular parallel-plate capacitor. Field is nearly uniform inside; non-uniform
“fringing field” at the edges. See Fig. (b).
(iii) For L ≈ a, See Fig. (c).
b) Think of it as two cylinders of opposite uniform charge density ±ρP . Inside, the field at a
distance ρ from the axis of a uniformly charge cylinder is given by Gauss’s law
I
2
~ 0 · E(~
~ ρ) = Qencl −→ 2πρlE(ρ) = ρP πρ l −→ E(~
~ ρ) = ρP ρ
dS
~,
0
0
20
S0
where l is the length of Gaussian surface with radius ρ. For two such cylinders, one plus and one
minus, the net field (inside) is
ρP ~
~ ρ) = E
~+ − E
~ − = ρP (~
E(~
ρ+ − ρ
~− ) = −
d,
20
20
~ which is a vector from negative to positive axis. Now it remains to
where we use ρ
~+ − ρ
~− = −d,
~ The polarization vector P~ is related to d~ because the total electric dipole moment is
connect P~ and d.
nothing but volume times the polarization vector
P~ (πa2 L) = (ρP πa2 L)d~
−→
P~ = ρd~ ,
where L is the total dipole moment of a chunk of length L. Thus
~
~ ρ) = − P .
E(~
20
For outside, the Gauss’s law gives Gauss’s law
2πρlE(ρ) =
ρP πa2 l
0
−→
2
~ ρ) = ρP a ρ̂ ,
E(~
20 ρ
where l is the length of Gaussian surface with radius ρ. For two such cylinders, one plus and one
minus, the net field (outside) is
!
2
2
~
ρ
a
ρ̂
ρ̂
a
1
ρ
~
·
d
P
+
−
~ ρ) = E
~+ − E
~− =
E(~
(
−
)=
2~
ρ 2 − d~ ,
20 ρ+
ρ−
20 ρ2
ρ
where we use ρ
~± = ρ
~∓
d~
2
and
ρ̂+
ρ
~+
= 2 = (~
ρ−
ρ+
ρ+
d~ 2 d2
~ −1 ≈ 1
)(ρ +
−ρ
~ · d)
2
4
ρ2
ρ
~ · d~ d~
ρ
~+ρ
~ 2 −
ρ
2
where we keep on the terms linear in d~ and similar relation for ρ− . Thus
2
~ ρ) = a 1 2[P~ · ρ̂]ρ̂ − P~ .
E(~
20 ρ2
!
,