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–1–
Solution to HW Problems
52. Griffiths 4.15 (a) First we find the bound charges using the spherical coordinates. The volume
charge density is
~ · P~ = − 1 ∂ (r 2 Pr ) = − 1 ∂ (kr) = − k
ρb = − ∇
r 2 ∂r
r 2 ∂r
r2
The surface charge density at the inner radius is
k
k
σb (a) = P~ · n̂|r=a = r̂ · (−r̂) = −
a
a
And the surface charge density at the outer radius is
k
k
σb (b) = P~ · n̂|r=b = r̂ · r̂ =
b
b
Note that both the volume and surface charge distributions show spherical symmetry, so we can
use Gauss’s law to find the electric field which is only radial. The Gaussian surface is a concentric
spherical shell.
At r < a, since the total charge enclosed by the Gaussian surface is zero, the electric field is zero.
Inside the spherical shell a < r < b Gauss’s law gives
I
Z
Z
Z r
Qenc
4πk
1
1
k
2
′
′
~
~
E · da = Er 4πr =
− 2 r 2 dr − 4πa2 )
= ( ρb dτ + σb (a)da ) = (
ǫ0
ǫ0
ǫ0 a
r
a
=−
4πkr
4πk
(r − a + a) = −
ǫ0
ǫ0
leading to
Er = −
Or
k
ǫ0 r
~
~ = − k r̂ = − P
E
ǫ0 r
ǫ0
Outside the sphere r > b, the total charge enclosed by the Gaussian surface is zero, since the
polarized sphere has net zero charge. Therefore, the electric field outside the polarized sphere is
zero.
~ also complies
(b) Note that the polarization has a spherical symmetry, we can safely assume that D
with a spherical symmetry, so that we may choose a Gaussian surface as a concentric spherical
surface. Since there is no free charge anywhere in the space (and the polarization has a spherical
~ = ~0. Outside the spherical shell, there is no polarized material, so
symmetry!), it follows that D
~ = D/ǫ
~ 0 = ~0. Inside the shell, E
~ = (D
~ − P~ )/ǫ0 = −P~ /ǫ0 , same as from (a), but
P~ = ~0, so E
apparently this approach to obtain the answer is a lot simpler.
–2–
53. Griffiths 4.18 (a) The displacement field does not feel the material and is determined by the
~ = −σẑ.
free charge. In this setup, the displacement between two charge surface is D
~ = D/ǫ.
~
(b) The electric field in the linear dielectric material is given by E
In the upper half of the
~
~
slab, E = −σ/(2ǫ0 )ẑ, and in the lower half of the slab, E = −σ/(1.5ǫ0 )ẑ = −2σ/(3ǫ0 )ẑ.
~ = ǫ0 (ǫr − 1)E.
~ In the upper half, P~ = ǫ0 (2 − 1)E
~ = −σ/2ẑ;
(c) For the linear material, P~ = ǫ0 χE
~ = −σ/3ẑ.
in the lower half, P~ = ǫ0 (1.5 − 1)E
R
R
R
~ = − 0 −2σ/(3ǫ0 )dz − a −σ/(2ǫ0 )dz =
~ · dl
(d) The potential difference is given by V = − E
−a
0
2σa/(3ǫ0 ) + σa/(2ǫ0 ) = 7σa/(6ǫ0 ).
(e) The polarization in each half slab is constant, so the volume bound charge is zero. The surface
bound charge is given by P~ · n̂. Here we consider 3 surfaces, the top surface, the boundary between
the two dielectric materials, and the bottom surface. At the top surface, n̂ = ẑ; at the bottom
surface, n̂ = −ẑ, and at the boundary between the two materials, n̂ = −ẑ for the upper half, and
n̂ = ẑ for the lower half. Therefore, we find the surface bound charge at the top, middle boundary,
and bottom surfaces as σb = −σ/2, σ/6(= σ/2 − σ/3), σ/3, respectively. It can be easily seen that
the total bound charge is zero, as should be.
(f) From (e), we can compute all the surface charges at the top, middle, and bottom surface to
be: σtot = σ/2, σ/6, −2σ/3, respectively. All these surface charges generate the net electric field
between the two plates. Recall that a uniform surface charge generates a uniform electric field
as E = ±σ/(2ǫ0 ) above (+) and below (-) the surface, respectively. The electric field by this
configuration is a superposition of contribution by all 3 surfaces. In the upper half of the slab, the
net electric field is
σ
σ/2 σ/6 −2σ/3
+
+
=−
Ez = −
2ǫ0
2ǫ0
2ǫ0
2ǫ0
In the lower half of the slab, the net electric field is
Ez = −
2σ
σ/2 σ/6 −2σ/3
−
+
=−
2ǫ0
2ǫ0
2ǫ0
3ǫ0
The result is the same as obtained in (b).
54. Griffiths 4.21 In the configuration, if an amount of charge λ per unit length is carries by the
copper cable, and charge −λ is distributed at the outer surface, then we can find the electric field
and the potential difference V between the inner and outer radius. In this configuration, we first
find the displacement field using Gauss’s law, and the Gaussian surface is a coaxial cylidrical shell
of radius s. At s < a and s > c, the total enclosed free charge is zero and the polarization is zero
~ = ~0 and E
~ = D/ǫ
~ 0 = ~0.
(no dielectric material), so D
~ =
At a < s < b, using Gauss’s law, we find D
~ =
At b < s < c, in the same way, we get D
2πλ
s ŝ,
2πλ
s ŝ,
~ = D/ǫ
~ 0=
and E
~ = D/ǫ
~ =
but E
2πλ
ǫ0 s ŝ.
2πλ
ǫs ŝ,
where ǫ = ǫ0 ǫr .
–3–
The potential difference between the inner and outer surfaces is therefore
Z s=a
Z s=b
Z s=a
2πλ
2πλ
~
~
ds −
=
E · dl = −
V =−
ǫs
s=b ǫ0 s
s=c
s=c
2πλ 1
[ ln(c/b) + ln(b/a)]
ǫ0 ǫr
The capacitance per unit length of the cable is therefore
C=
λ
2πǫr ǫ0
=
V
ln(c/b) + ǫr ln(b/a)
55. Griffiths 4.19 Suppose the distance between the two plates is d, and the area of the plates is
A. A given potential difference V between the two plates without dielectric material in between is
given by V = σd/ǫ0 , and the capacitance is C0 = Q/V = σA/V = Aǫ0 /d.
For the first case in the problem, suppose that the top and bottom surfaces have free charge
density ±σf , the displacement field is the same in the gaps and in the material, D = σf , pointing
from the top to the bottom. The electric field is therefore E = D/ǫ0 = σf /ǫ0 in the gaps, and
E = D/ǫ = σf /ǫ in the material, where ǫ = ǫ0 ǫr , pointing from the positive charge (top) to the
R
~ =
~ · dl
negative charge (bottom). The potential difference between the two plates is V = − E
σf d/(2ǫ0 ) + σf d/(2ǫ) = σf d/(2ǫ0 )(1 + 1/ǫr ). The capacitance is therefore
C1 =
σf A
Aǫ0 2ǫr
2ǫr
Q
=
=
= C0
V
V
d 1 + ǫr
1 + ǫr
For the second case, to maintain the potential difference V , the left half of the capacitor will have
to carry more free charges than the right half of the capacitor to compensate for the bound charges
due to polarization of the dielectric material. To have the same V , in both halves, E = V /d. In the
left half, D = σl = ǫE leading to σl = ǫV /d. In the right half, D = σr = ǫ0 E leading to σr = ǫ0 V /d.
The capacitance is therefore
C2 =
Q
=
V
A
2 σl
+
V
A
2 σr
=
Aǫ0 ǫr + 1
ǫr + 1
= C0
d
2
2
It can be easily seen that C1 − C2 < 0, so C1 < C2 .
In the first case, the free charge density at the top or bottom is given as
σf =
V ǫ0 2ǫr
d ǫr + 1
We can take the positive charge at the top and the negative charge at the bottom. In the dielectric
material,
ǫr − 1
V ǫ0 2(ǫr − 1)
P =D
=
ǫr
d ǫr + 1
–4–
And
σb = P =
V ǫ0 2(ǫr − 1)
d ǫr + 1
The negative bound charge is at the top of the material.
In the second case, in the left, the free charge is given by
σl =
ǫr ǫ0 V
d
and in the right, the free charge is
σr =
ǫ0 V
d
The polarization in the left is
P =D
ǫr − 1
ǫV ǫr − 1
ǫ0 V
=
= (ǫr − 1)
ǫr
d ǫr
d
and
σb = P = (ǫr − 1)
negative at the top and positive at the bottom.
ǫ0 V
d