Chapter 4: Solutions to Problems
Problem 4.1. Charge density required to produce a field.
From the given potential we calculate the electric field intensity using the gradient of potential. Then, the divergence of
E gives the charge density between plates.
∇.E = ρ(x) = ∂ Kx 3 = 3Kx 2
ε
∂x
a. E(x) = x Kx 3
ρ(x) = 3Kεx 2
Thus:
C/m3
b. The charge density is positive for x < 0, and for x > 0.
c. The charge density is constant in the y or z direction.
Problem 4.2. Charge density required to produce a given field.
a. The electric field intensity in space is given as:
E = x 10−1 4x , for x > 0
E = x 10−1 4x , for x < 0
E = 0 at x = 0
N/C
The charge density is found from the divergence of E as follows:
For x > 0:
∇.E = ρ
→
ρ = ε 0 (∇.E) = ε 0 (∇.E) = ε 0 ∂ 10−1 4x = 8.854× 10−1 2×10−1 4 = 8.854× 10−2 6
ε0
∂x
For x < 0:
∇.E = ρ
→
ρ = ε 0 (∇.E) = ε 0 (∇.E) = ε 0 ∂ 10−1 4x = 8.854× 10−1 210−1 4 = 8.854× 10−2 6
ε0
∂x
C
m3
C
m3
b. The total charge that produces this situation is infinite. Even the reference point at x = 0 is artificial. This is not very
practical although, the situation may exist for finite distances.
Problem 4.3. Electrostatic fields.
To check that the fields are electrostatic, their curl must be zero and their divergence must be either nonzero or zero.
Thus we calculate the curl and divergence of each field.
a. A = x yz + y2x
∇×A =
x
∂
∂x
y
∂
∂y
z
∂
∂z
yz
2x
0
=x 0−0 + y y−0 + z 2−z =y y + z 2−z ≠0
This is not an electrostatic field.
b. A = z5 − y2
∇×A =
x
∂
∂x
y
∂
∂y
z
∂
∂z
0
2
5
=0
∇.A =
This field is an electrostatic field for which ρ = 0.
c. A = φr . The curl in cylindrical coordinates:
73
∂(Ax) ∂(Ay) ∂(Az)
+
+
=0
∂x
∂y
∂z
r
∂
∂r
∇×A = 1
r
z
∂
∂z
φr
∂
∂φ
= r 1 0 − 0 + φ 1 0 − 0 + z 1 2r − 0 = z2
r
r
r
0
r2
0
This is not an electrostatic field because its curl is nonzero.
−8
−8
d. A = R 10 cosθ + θ 10 sinθ. The curl in spherical coordinates:
3
3
R
2R
∇×A =
R
∂
∂R
θR
∂
∂θ
φRsin θ
∂
∂φ
10−8 cosθ
R3
10−8 sinθ
2R 3
0
1
R 2 sinθ
-8
-8
= φ − 104 sin θ + 104 sin θ = 0
R
R
(Also, ∇.A = 0). This can be an electrostatic field because its curl is zero. The fact that the divergence is also zero
simply means that there is no charge anywhere in the field.
Problem 4.4. Electric field due to planar charge density.
Ez
The charge layer is positive therefore the electric
field intensity points away from the charge both
above and below the layer. A Gaussian surface in the
form of a box with top and bottom parallel to the
layer is built as shown. The area of the bottom and
top surface is S. Thus, since the electric field intensity
points in the direction of ds on each of the surfaces:
bottom
stop
s
a. Gauss' law gives:
Ez S top + −Ez −S
ds top
ε
z
−z
Figure A
= 2 Ez S = ρ0 S
ε
Ez
ρ0
sbottom
ds bottom
The electric field intensity is:
Ez = ρ0
2ε
→
E =z ρ0
2ε
N
C
for z > 0,
and
E =− z ρ0
2ε
N
C
for z < 0
b. If the dielectric is removed, the permittivity becomes ε 0 :
E =z
ρ0
2ε 0
for z > 0,
and
E =− z
ρ0
2ε 0
for z < 0
N
C
Problem 4.5. Superposition of fields of planar charges.
The electric field intensity of each layer is calculated separately. Then, the overlapping fields are summed up with due
regard to directions to produce the total fields. To simplify notation, the fields below an interface are designated with an
L and those above with an U. The positive y axis is taken upwards. From Figure A:
1. Charge layer at y = −b:
E 1L = y ρs ,
2ε 0
2. Charge layer at y = −a:
E 2L = − y ρs ,
2ε 0
3. Charge layer at y = a:
E 3L = y ρs ,
2ε 0
4. Charge layer at y = b:
E 4L = − y ρs ,
2ε 0
y < −b,
y < −a,
y < a,
y < b,
E1U = − y ρs ,
2ε 0
E 2U = y ρs ,
2ε 0
y > −b
y > −a
N
C
N
C
E 3U = − y ρs ,
2ε 0
y>a
N
C
E 4U = y ρs ,
2ε 0
y>b
N
C
74
5. Due to the volume charge density between y = − a and y = a. Here there are two domains. One inside the charge
density, one outside. Inside the charge density, we build a Gaussian surface (see Figure B) and write:
2EvS = 2xSρv
ε0
Ev = xρv
ε0
→
Thus, between y = − a and y = a:
yρ
yρ
N
E vL = − y v
−
a < y < 0,
E vU = y v
0
<y<a
ε0
ε0
C
The electric field intensity due to ρv outside the layer is found using again a Gaussian surface (Figure B) but now the
surfaces are outside the charge layer. These give:
2EvS = 2aSρv
ε0
Ev = aρv
ε0
→
The electric field intensities due to the volume charge density outside the charge layer are:
y ρv
ε0
E vL = − y
a < y < 0,
−
E vU = y
y ρv
ε0
0
N
C
<y<a
Now we sum the fields in each domain, with the aid of Figure C:
1. y < − b:
E = E 1L + E 2L + E vL + E 3L + E 4L = − y ρva
ε0
2. − b < y < − a
E = E 1U + E 2L + E vL + E 3L + E 4L = y − ρs − ρva
ε0
ε0
3. − a < y < 0:
E = E 1U + E 2U + E vL + E 3L + E 4L = − y
4. 0 < y < a:
N
C
N
C
E = E 1U + E 2U + E vU + E 3U + E 4L = y − ρs + ρva
ε0
ε0
6. b < y:
E = E 1U + E 2U + E vU + E 3U + E 4U = y ρva
ε0
E3A
− ρs
E2A
E1A
s s
s
s s
s
E2B
s s
s
E3B
− ρs
N
C
N
C
E4A
s s
s
ρs
E4B
EoA
EvA
a
ρs
s
s
EvB
E1B
Figure A
N
C
ρv y
ε0
E = E 1U + E 2U + E vU + E 3L + E 4L = y ρvy
ε0
5. a < y < b:
b
N
C
Figure B
75
s
ρv
a
s
EoB
y ρs
2ε 0
aρv
y ε0
y
b
− y ρs
2ε 0
a yρ
v
y ε0
ρv
− y |y|
ε0
−a
.
−b
.
ρs
aρ y
− y ε 0v 2ε 0
y
.
.
− y ρs
2ε 0
ρs
2ε 0
.
− y ρs
2ε 0
y
ρs
2ε 0
− y ρs
2ε 0
Figure C
Problem 4.6. Superposition of fields of planar charges.
The electric field intensity of an infinite plane is ρ0 /2ε, directed out of the plane (for positive charge density) as shown in
Figure A. Calculating this for each plane, we get the components shown in Figure B. The electric field intensity is at
45° in each quadrant. The charge will move in the direction of largest force which is in the second quadrant.
a. The electric field intensities are:
ρ
ρ
In the first quadrant:
E1 = x s + y s
2ε 1
2ε 1
ρ
ρ
In the second quadrant:
E2 = − x s + y s
2ε 0
2ε 0
ρ
ρ
In the third quadrant:
E3 = − x s − y s
2ε 2
2ε 2
ρ
ρ
In the fourth quadrant:
E4 = x s − y s
2ε 3
2ε 3
N
C
N
C
N
C
N
C
b. The electric field intensity is largest in the second quadrant. Therefore the force will also be largest in the second
quadrant.
c. The electric field intensity in the second quadrant is:
E2 = − x
The force is therefore:
The force has a magnitude:
ρs + y ρs
2ε 0
2ε 0
N
C
F = qE2 = − x qρs + y qρs
2ε 0
2ε 0
F =
2 qρs
2ε 0
2
= qρs
2 ε0
and is directed at 135° to the positive x axis.
76
N
N
E1
ρs /2ε 0 ρs /2ε 1
E= ρs /2ε
ρs
ρs /2ε 0
E= ρs /2ε
ρs /2ε 1
ε0
ε1
ε2
ε3
ρs /2ε 2
ρs /2ε 3
ρs /2ε 2
ρs /2ε 3
Figure A
Figure B
Problem 4.7. Electric field due to spherical distribution of charge.
a. R ≤ 2mm Using Gauss' law (see dotted lines in
Figure A for the Gaussian surface):
D.ds = Qenclosed
R
s
or:
D4π R = ρvdv = 4π R ρv
3
v
2
R
This gives:
D = RR
6
E=R R
6ε 0
→
R
2
.R
(
4 π 2 × 10
6
ρv+
ρv0
N
C
)
−3
Figure A
3
× 0.5 =
3
ρv+ = 0.5 C/m3
ρv0 = 0.
ρv− = −1 C/m3
ρv−
b. 2mm ≤ R ≤ 4mm. Here the total enclosed charge is
that in a sphere of radius 2 mm. Thus:
D4 π R = Qenclosed =
4
2
3
16 π × 10
−9
D=
→
3
4 × 10
3R
−9
%C(
'& m 2 *)
2
or:
D = R̂
4 × 10
3R
−9
2
4 × 10
E = R̂
→
3ε 0 R
−9
%N(
'& m *)
2
c. 4 mm ≤ R ≤ 6 mm. Now there is both the positive charge in the sphere of radius 2 mm and a negative charge which
exists beyond R = 4 mm. Thus
Qenclosed =
(
4 π 2 × 10
−3
)
3
3
( (
3
4 π R − 4 × 10
+
ρv
+
−3
) )ρ
3
−
v
3
4π
3
( 68 × 10
−9
−R
3
=
4π
3
( 8 × 10
) = D4 π R
−9
+
−9
−
3
−
)
ρ v − 64 × 10 ρ v + R ρ v =
2
or:
−9
D = − R̂
−68 × 10 + R
3R
3
−9
E = − R̂
→
2
−68 × 10 + R
3ε 0 R
3
2
%N(
&' m )*
d. R ≥ 6 mm. The total charge enclosed is the positive charge in the sphere of radius 2 mm, and the negative charge in
the shell of inner radius 4 mm and outer radius 6 mm. Thus:
D4 π R
2
=Q
enclosed
=
(
4 π 2 × 10
3
−3
)
3
+
ρv
+
(
3
3
)
−9
−
4 π 6 − 4 10 ρ v
3
That is:
77
=
4π
3
( 4 × 10
−9
− 152 × 10
−9
)=−
4π
3
× 148
D = −R̂
148 × 10
3R
−9
→
2
E = − R̂
148 × 10
3ε 0 R
−9
%N(
'& m *)
2
Note that the electric field intensity becomes negative because the total negative charge enclosed is much larger than the
total positive charge.
Problem 4.8. Application of superposition.
The field at P is due to three sources: A charge density ρv uniformly distributed on the sphere of radius 3a and equal and
negative charge densities in the hollows. The charge density is as follows:
$C'
Q
Q
Q
ρv =
=
=
& 3)
3
4
4
4
4
24π a
%m (
π (3a)3 − π (2a)3 − π (a)3
π 27a 3 − 8a 3 − a 3
3
3
3
3
Using Gauss’s law we calculate the field at P due to the uniform distribution and that of the two hollows when these are
replaced with a negative charge density. Taking first a Gaussian surface in the form of a sphere centered at the center of the
outer circle, of radius a passing through P, we get due to the uniform charge distribution (i.e., before the hollows are taken
into account):
4 3
π a ρ v aρ
$V'
Q
a Q
Q
E1 =
=3
= v=
=
& )
2
2
3
2
3ε 3ε 24π a 72π a ε
4πε a
4πε a
%m(
This field points to the left, away from the center of the geometry.
The small hollow when replaced with an identical but negative charge density has an identical field also pointing away from
the center, to the left.
4 3
π a ρ v aρ
$V'
Q
a Q
Q
E2 =
=3
= v=
=
& )
2
2
3
2
3ε 3ε 24π a 72π a ε
4πε a
4πε a
%m(
The field due to the large hollow points to the right, away from the center of the geometry and is equal to:
4
π (2a)3 ρv (2a) ρ (2a) ρ
$V'
Q
Q
2Q
3
v
v
E3 =
=
=
=
=
& )
2
2
3
2
3ε
3ε 24π a 72π a ε
4πε (2a)
4πε (2a)
%m(
Thus, the total field is:
$V'
2Q
Q
Q
Et = E3 − E1 − E2 =
−
−
=0
& )
2
2
2
72π a ε 72π a ε 72π a ε
%m(
The field at P is zero.
(
)
Gaussian surface
ρv
ε0
E s1
P
ε0
ε0
Figure A – Calculation of the field at P due to the uniform distribution, neglecting the hollows.
78
Gaussian surface
ε0
P
Q
−ρv
E s2
b
O
ε0
ε0
(2)
Figure B – Calculation of the field at P due to the negative charge density in the large hollow. The small hollw produces
zero field at P.
Problem 4.9. Electric field due to a layer of charge.
y
To find the electric field intensity we use Gauss’s
law. First, we apply Gauss’s law for the domain
|x| < d/2 (internal to the charge distribution). Then we
apply it external to the charge distribution. Inside the
charge distribution (− d/2 < x < d/2) we build a
Gaussian surface as shown. This is a box with base
surfaces S and height x, and the electric field intensity
is in the directions shown (from symmetry
considerations).
From Gauss’ law:
E
E
S
S
x
0
E
x
x
E
−∞
d/2
ρ = ρ0 |x|
Figure A
x
ρvdv
E 2S =
v
At 0 < x < d/2, this gives:
ε0
2S
=
ρ0 x'dx'
0
ε0
At x = − d/2:
→
2
E = x ρ0 x
2ε 0
At − d/2 < x < 0, this gives:
At x = d/2:
2
= Sρ0 x
2ε 0
2
E = ρ0 x
2ε 0
N
C
N
C
2
E = − x ρ0 x
2ε 0
2
E = x ρ0 d
8ε 0
N
C
N
C
2
E = − x ρ0 d
8ε 0
N
C
For |x| > d/2: Using Gauss’ law, we can see that the field remains constant and equal to that at |x’| = d/2. Thus:
For x > d/2:
2
N
E = x ρ0 d
8ε 0
C
For x < −d/2:
2
N
E = − x ρ0 d
8ε 0
C
79
Problem 4.10. Superposition of fields due to line and surface charge densities.
As superposition of the fields of the electric field intensity of the charged sheet and that of the charged line. Each of the
fields is calculated separately using Gauss’ law and then superimposed to obtain the total solution.
a. (1) Field due to the wire: The electric field intensity due to the wire may be found using Gauss’ law as shown in
Figure A. Using cylindrical coordinates, the electric field intensity is in the r direction as shown. Taking the Gaussian
surface to be a cylinder of radius r and length L, the enclosed charge density is ρlL and we have:
E12π rL = Q = ρlL
ε0 ε0
E1 2π r =
→
ρl
2π rε 0
However, it is best we write the electric field intensity in Cartesian coordinates so that we may sum up the two fields.
From Figure B, we can write (because the line is at y = 0, z = 0):
r = y 2 + z2 ,
E1y = E1 cosα = E1
Denoting this as field (1), we have:
E1 = y
y
= E1
r
y
,
y + z2
z
y + z2
E1z = E1 sinα = E1
2
ρly
ρlz
+z
2πε 0 y 2 + z 2
2πε 0 y 2 + z 2
2
N
C
(2) Field due to the infinite plane: Again, using Gauss’ law, we build a pillbox as shown in Figure C. The total charge
enclosed is ρs S and the electric field intensity is out of the box on both surfaces. Thus:
2E2 S = ρsS
ε0
Or, in vector form:
E2 = z
ρs ,
2ε 0
E2 = ρs
2ε 0
→
z > z0 ,
E2 = − z
N
C
ρs ,
2ε 0
N
C
z < z0
The total electric field intensity is the sum of the two fields as follows:
Above the charged surface (z > z0 ):
ρly
ρlz
+z
+ ρs
2πε 0 y 2 + z 2
2πε 0 y 2 + z 2
2ε 0
Below the charged surface (z < z0 ):
N
C
ρly
ρlz
+z
− ρs
2
2
2
2
2πε 0 y + z
2πε 0 y + z
2ε 0
N
C
E1 = y
E1 = y
b. At (0,0,1):
ρl ± ρs
2π 2
D 1 = ε 0 E(0,0,1) = z
C
m2
This may be negative or positive depending on the location z0 and the relative values of ρl and ρs .
c. If the whole geometry is inserted in oil, the permittivity changes from ε 0 to 4ε 0 . Thus, the electric field intensity in (a)
or (b) is lowered by a factor of 4. Above the charged surface (z > z0 ):
ρly
+z
8πε 0 y 2 + z 2
8πε 0
Below the charged surface (z < z0 ):
ρly
E1 = y
+z
8πε 0 y 2 + z 2
8πε 0
E1 = y
ρlz
+ ρs
2
2
y +z
8ε 0
N
C
ρlz
− ρs
2
2
y +z
8ε 0
N
C
y
ds a
E
Gaussian surface
E
ds c
.
ρl
r
Figure A
α
y
.
ds b
L
Ey E
y
r
E
Ez
α
z
z
z
.
.
E
E
Figure B
Figure C
80
S
S
S
x
Problem 4.11. Application: Electric field intensity in a coaxial line.
Use Gauss’ law in cylindrical coordinates. Three domains are distinguished: one inside the inner shell, one between the
two shells and one outside the outer shell. In each case, the total charge enclosed by the Gaussian surface is the source of
the electric field at the location of the Gaussian surface.
a. The three required Gaussian surfaces are shown in dotted lines (in cross-section) in Figure A. The Gaussian surface is
L [m] long.
For r < a: There is no charge enclosed in the Gaussian surface. Therefore the field is zero.
For a < r < b. A Gaussian surface in the form of a cylinder of radius r and length L, concentric with the inner conductor
is created (surface s2). The electric field intensity is uniform over this surface and the surface encloses a total charge
equal to 2πaLρs . From Gauss’ law:
2π aLρs
aρ
N
2π rLE =
→
E =r s
ε
rε
C
For r > b. Now the Gaussian surface (s3) encloses both shells. The total charge is the positive charge on the inner shell
which equals 2πaLρs minus the negative charge on the outer surface which equals 2πbLρs. Thus, again from Gauss’ law:
2πrLE = 2π aLρs − 2π bLρs
ε0
ε0
E = r ρs a − b
rε 0
→
N
C
b. See Figures B.
c. To cancel the electric field intensity in (b), a line of charge with line charge density ρl is placed at the center. The
electric field intensity due to a line of charge is:
2πrLEl = Lρl
→
ε0
Adding this to the field in (b) should produce zero:
r
ρs a − b + ρl
rε 0
2π rε 0
El = r
ρl
2π rε 0
N
C
ρ l = − 2π ρs a − b
→
The charge density required on the line of charge is positive (b > a) and equals:
ρl = − 2πρs a − b = 2πρs b − a
ρ
E = εs
r
b
r
ρs
r
E
C
m
ρ
E = εs ar
ρ
E = εs a
b
a
s1
0
r
a
b
s2
s3
− ρs
ρ
E = εs r(a−b)
ρ
E = εs (a−b)
b
Figure A
Figure B
Problem 4.12. Application: Electric field intensity in a two-conductor cable.
Use a cylindrical Gaussian surface, concentric with each cylinder and at a distance r from the center of each cylinder.
a. Starting with the left cylinder, we get from Gauss' law using a Gaussian surface of radius r and arbitrary length L:
2π rELL = ρe
ε0
→
EL = ρe
2πrε0
At r = d ,
2
EL =
ρe
= ρe
2π (d/2)ε0 πdε0
81
→
EL = x ρe
πdε0
Due to the right cylinder we get an identical solution except that the direction of the electric field is in the negative x
direction since both cylinders are positively charged.
N
E R = − x ρe
π dε 0
C
Thus, the electric field intensity at the center midway between the two cylinders is zero.
E P1 = 0
b. Using Figure B, the electric field intensity at P2 has only a vertical component (the horizontal components oppose
each other and therefore cancel). The vertical component's magnitude is:
Ey = 2E sin α = 2E h
r
N
C
Since:
r=
The electric field intensity is:
d 2 + h2
2
m
2ρlh
Ey = 2 ρl sin α = 2ρlh 2 =
2πε 0 r
2πε 0 r
2πε 0 d2 /4 + h 2
For d = 4 m, h = 2 m, ρ l = 0.5 × 10
E = −ŷ
−6
N
C
C/m, and, since the field is in the negative y direction (see Figure B):
2 ρl h
2 × 0.5 ×10 −6 × 2
=
−
ŷ
= −ŷ4493.87
2 πε 0 ( d 2 / 4 + h 2 )
2 π × 8.854 ×10 −12 × ( 4 2 / 4 + 2 2 )
&N)
(' C +*
Gaussian surfaces
y
0.5 μC/m
0.5 μC/m
r
r
ER
a
2m
(d/2)
.P
1
Gaussian surfaces
EL
a
α
2m
(d/2)
0.5 μC/m
a
0.5 μC/m
a
r
x
2m
(d/2)
E
h=2m
x
r
.P
1
90−α 90−α E
Ey
Figure A
Figure B
Problem 4.13. Fields of point and distributed charges.
The field is found as the superposition of the field of a point charge and that due to a spherical surface of charge. Each is
calculated separately and superimposed to obtain the total field.
a. For any radial position smaller than a, the electric field intensity of the surface charge density is zero. The only field is
due to the point charge Q:
N
E=R Q 2
R<a
C
4πε 0 R
For R > a, the electric field intensity is the superposition of the field above and that due to the surface of charge. Using a
spherical Gaussian surface of radius R > a, the total charge enclosed by the surface is ρs4πa2 and we get due to the
surface charge:
2
2
E4π R 2 = 4π a ρs
→
E s = R a ρs2
ε0
ε0R
Thus, the total field is the sum of this and the point charge field:
82
2
Q
E = R a ρs2 +
ε0R
4πε 0 R 2
N
C
R>a
b. For the electric field intensity to be zero, the surface charge density is:
a2 ρs +
Q
=0
2
ε0R
4πε 0 R 2
ρs = − Q 2
4π a
→
C
m2
Problem 4.14. Charge distribution required to produce a field.
a. Calculate the charge density inside the dielectric using the divergence of the electric field intensity. Note: you cannot
use Gauss' law unless the charge density is constant or is known and can be integrated. Then calculate the electric field
intensity from the charge density.
∇.E =
ρv
ε
ρ v = ε ∇.E = ε dE = ε y
dy
⇒
C
m3
Since E is negative for negative values of y, the answer throughout the dielectric is ρv = ε y
b. The electric field intensity outside the dielectric can now be calculated since the charge density is known: Using a
Gaussian surface as shown in Figure A, the electric field intensity outside the dielectric is calculated as
d/2
2
2ES =
Or, after dividing both sides by 2S:
0
2
ε0
ε ySdy
0
=
ε0
2
E = εd
8ε 0
y
dielectric
d/2
ρvSdy
2
= 2ε d S
8ε 0
N
C
E
S
x
d
ε
S
E
Figure A.
Problem 4.15. Volume charge density needed to produce a field.
The electric charge density in the volume, required to produce the given field is calculated from the divergence of the
electric field intensity.
a. The volume charge density may be found anywhere in the sphere as:
∇.E = ρv
ε0εr
At R = b/4 we get:
→
ρ v = ε 0 ε r(∇.E) = ε 0 ε r
ρv = 16ε 0 ε rb = 4ε 0 ε rb
4
b. The electric field intensity at R = b/4 is:
2
1 ∂ (4 R 4 ) = 16ε 0 ε rR
R 2 ∂R
C
m3
2
E = R 4R 2 = R 4b = R b
16
4
N
C
c. To calculate the total flux leaving the sphere of radius b/2, we integrate the flux density over the sphere. The flux
density at R = b/2 is:
2
C
D = ε 0 ε rE = R 4ε 0 ε rb = R ε 0 ε rb 2
4
m2
Since this is constant on the surface of the sphere, we get:
83
D.d s = D×4π(b/2)2 = ε0εrb 24π(b/2)2 = ε0εrb 4π
C
S
Problem 4.16. Maximum allowable charge density on power lines.
We calculate the electric field intensity of each conductor separately and add them up to find the maximum electric field
intensity. Because the conductors are small and widely separated, the maximum electric field intensity will occur at the
surface of one of the conductors. Also, it is sufficient to calculate the electric field intensity on the line connecting the
centers of the two lines. Using Gauss’ law around the positively charged conductor (Figure A), and using a cylindrical
surface of arbitrary length L and radius x’ we get:
E 2π x'L = 2π aLρs
ε0
E = aρs
ε 0 x'
→
E + = x aρs
ε 0 x'
→
N
C
Repeating the process and calculating the electric field intensity at the same location due to the negative line, we get
(now the distance from the center of the line is x’−d):
E 2π (d−x')L = − 2π aLρs
ε0
aρs
ε 0 (d−x')
E=−
→
E− = x
→
aρs
ε 0 (d−x')
N
C
This is clearly in the negative x direction to the right of the negative line but it is in the positive x direction to the left (the
electric field intensity points towards the negative charge). Now we add the two fields up:
1
E = E + + E − = x aρs + x aρs = xaρs 1 +
ε 0 x'
ε 0 x'
ε 0 (d−x')
(d−x')
N
C
This function has a minimum midway between the conductor (at x'=d/2) and is maximum at x’ = a (or x’ = d−a). (A
quick plot should convince you of these points). Taking the first:
E max = xaρs 1 + 1
ε 0 a d−a
N
C
6
Setting the maximum electric field intensity to 3×10 V/m, we find the surface charge density:
3×106 = aρs 1 + 1
ε 0 a d−a
This gives:
ρs =
20mm
3×106 ε 0 d−a
d
3×106 ×8.854× 10−1 2× 6−0.01
= 2.65× 10−5
6
ρs
x'
ρs =
→
.
d=6m
E
− ρs
−
E
C
m2
x
+
Figure A
Problem 4.17. Potential between two points in a field.
Use the general definition of potential difference between two points:
b
Vba = −
E.dl
V
a
where a refers to point P1 (2,2,2) and b to point P2 (0,2,2). Thus we need to write dl in Cartesian coordinates and find the
product E.dl:
dl = x dx + ydy + zdz
E.dl = x 4x +y3y . x dx + ydy + zdz = 4xdx + 3ydy
→
The potential difference is:
x=0
V P2P1 = −
y=2
4xdx −
x=2
3ydy = 8
y=2
84
V
Note: the integration on x and y are independent and we integrate between x = 2 to x = 0 while y = 2.
Problem 4.18. Potential due to a system of point charges.
Eight equal charges q=3 nC each are placed at the vertices of a cube in a vacuum. Place the cube in a Cartesian system of
coordinates so that the faces are parallel to the x-y, x-z and y-z planes, with one vertex at the origin. The cube is in the positive
octant of the system of coordinates. The side of the cube is a = 0.5 m. Find
(a) The potential and the electric field intensity at the center of the cube;
(b) The potential and electric field intensity at the center of the face parallel to the x-y plane (at x=a/2, y=a/2, z=a).
Solution. Calculate the electric field intensity and electric potential directly but judiciously.
a. The distance between any of the vertices of the cube to its center is the same and equal:
a
2
R=
2
Thus, the total potential at the center of the cube is:
V=
+ a
2
2
+ a
2
2
=a 3
2
4×3×10−9
8q
16q
4q
=
=
=
= 498.15
4πε0 R 4πε0 a 3 πε0 a 3 π×8.854×10−12×0.5× 3
V
The electric field intensity must be zero at the center of the cube from simple symmetry considerations.
b. The potential at the center of the surface of the cube is calculated as follows:
The four charges on the vertices of that face are a distance R1 from the center of the face. The four charges on the opposite
face are a distance R2 from the center of the given face (See Figure A).
R1 =
Thus, we have:
V=
a
2
2
+ a
2
2
= a 2,
2
R2 =
a
2
2
+ a
2
2
+ a2 = a 3
2
4q
4q
q
q
2 + 2 = q 2 1+ 1
+
= πε 1 + 1 = πε
πε0 a
0
0
4πε 0R1 4πε0 R2
R1 R1
a 2 a 3
3
−9
2×3×10
=
1 + 1 = 481.18
V
−12
3
π×8.854×10 ×0.5
c. To calculate the electric field intensity we first note that the four charges on the face on which we wish to calculate the
field produce zero field at the center of the face. Thus, we only need to calculate the field due to the remaining four charges.
Using Figure B, we write:
R=
a
2
2
+ a
2
2
+ a2 =a 3
2
The electric field intensity has a component vertical to the face (in the x direction) and a component on the face (tangential to
the face). This component cancels out since the two charges in opposite corners produce opposite tangential components. The
magnitude of the field produced by each of the four charges is:
E=
q
2q
q
=
=
4πε 0R 2 4πε0 3a 2 6πε0 a 2
V
m
The normal component of the field is:
q 2
Ex = Ecosα = E a = E 2 =
R
3 6πε0 a 2 3
V
m
The total field is four times larger:
Et =
2× 2×3×10−9
4q 2
2 2q
=
=
= 234.8
6πε 0a 2 3 3 3 πε 0a 2 3× 3×π×8.854×10−1 2× 0.5 2
85
V
m
y
a
q
a
q
R2
a/2
a
R1
q
z
a/2
q
q
q
x
q
q
a
Figure A.
y
a
q
a
q
a/2
q
q
R
a/2
Ex
α
a
q
q
z
q
x
q
a
E
Figure B.
Problem 4.19. Potential due to point charge and spherical charge distribution.
a. The electric field is calculated as a superposition of the electric field intensity of a point charge and a uniform
spherical charge distribution. Then the potential is found by integrating from infinity to the point at which the potential is
required. The electric field of a point charge anywhere in space is:
E1 = R
Q
4πε 0 R 2
V
m
The electric field intensity of a spherical charge distribution is:
3
3
E 2 = R (4/3)π a 2ρv = R a ρv2
4πε 0 R
3ε 0 R
R > a,
3
E 2 = R (4/3)π R 2ρv = R R ρv
3ε 0
4πε 0 R
R > a,
Q
E = R R ρv +
3ε 0 4πε 0 R 2
V
m
R<a
The total electric field intensity is:
E=R
a3 ρv +
Q
2
3ε 0 R
4πε 0 R 2
R<a
V
m
The potential is now found as:
R
V(R ) = −
∞
3
a3 ρv +
Q
dR' = a ρv + Q
2
2
3ε 0 R 4πε 0 R
3ε 0 R'
4πε 0 R'
R>a
V
Inside the charge density, we must first integrate from infinity to the surface of the sphere using the external field and
then from the surface of the sphere to a point R 0 using the internal field:
86
a
R0
a 3 ρv +
Q
Rρv +
Q
dR' −
dR'
2
2
3ε 0 4πε 0 R 2
∞
3ε 0 R'
4πε 0 R'
a
2
2
2
2
2
= a ρv + Q − R 0 ρv + Q + a ρv − Q = − R 0 ρv + Q + a ρv
3ε0 4πε 0a
6ε0
4πε 0R0 6ε0 4πε 0a
6ε0
4πε 0R0 2ε0
V(R0) = −
Thus, setting R 0 = R:
R < a,
V
2
2
V(R ) = − R ρv + Q + a ρv
6ε 0
4πε 0 R
2ε 0
Check:
R < a,
2
V(R = a) = a ρv + Q
3ε 0 4πε 0 a
V
V
V
a2 ρv
Q
3 ε 0 + 4π ε 0 a
b. See Figure A.
0
.5a
a
2a
1.5a
R
Figure A.
Problem 4.20. Application: Potential and field between charged plates.
Use a Gaussian surface on each of the plates in the form of a box with sides parallel to the surfaces as shown in dotted
lines in Figure A.
a. Assuming a surface S for the sides of the Gaussian surface, For the right plate:
2ES = ρsS
ε0
Thus:
E = ρs
2ε 0
V
m
ρs
2ε 0
+
ρ
E =− x s
2ε 0
V
m
V
m
to the right of the right plate
ρs
2ε 0
−
ρ
E =x s
2ε 0
V
m
V
m
to the right of the left plate
+
E =x
Similarly for the left plate:
→
−
E =− x
to the left of the right plate
to the left of the left plate
Between the two plates the field points from the positive to the negative plate:
E = E + + E −= − x ρs
ε0
Outside the plates, the electric field intensity is zero.
V
m
b. If the potential is zero on the left plate, then by integrating against the field we can write for a point x between the
plates:
x
x
V =−
E .dl = −
− x ρs .x dx' = ρsx
V
ε0
ε0
0
0
The potential increases linearly and at x = d equals:
V = ρsd
ε0
V
To the left of the left plate the potential is zero, to the right it remains constant at ρsd/ε 0 .
the right plate increases by V 0 (as does the potential
to the right of the right plate). Between the plates, the
c. If a potential difference is added across the plates,
say V 0 , then the left plate remains at zero potential,
87
additional potential is V 0 x/d. This adds to the solution
above
V = ρsx + V 0 x
ε0
d
d
E−
V
E−
S
− ρs
+
E
E
S
ε0
+
+ ρs
E=0
E=0
x
Figure A
Problem 4.21. Application: Charge density in coaxial cable connected to a battery.
Use Gauss' law to calculate the electric field intensity
at a general point a < r < b. The electric field
intensity is calculated in terms of the unknown
charge density ρs . Integrating the electric field
intensity from b to a along r gives the potential
difference V which is known.
a. From Gauss' law: We define a concentric
cylindrical surface of radius r and length L. From
symmetry considerations, the electric field intensity
is radial, pointing from outer to inner conductor and
uniform around the circumference (see Figure A).
From this:
E.ds = Qenclosed
⇒
E2 π rL = 2π bLρs
ε
ε0
S
E
a
− ρs
ρs
E
b
r
ε0
E
Figure A
E = bρs
ε0r
⇒
V
m
where Q is the total charge enclosed by this the Gaussian surface. Since the potential between the plates is known, we
can write:
a
Vab = −
b
bρs dr = bρs (lnb − lna) = bρs lnb
ε0r
ε0
ε0
a
V
Note that V a b is negative since b < a. The charge density on the inner conductor is
ρs b = V a bε 0 = V ε 0
b ln(b/a) b ln(b/a)
C
m2
and, as required, is positive (b < a and therefore V a b is negative).
b. The total charge on the inner conductor is equal to that on the outer surface in magnitude but, unlike the inner surface,
the charge density is positive on the outer surface.
b ρs b = a ρs a
Therefore
ρ sa = V ε 0 / a ln(b / a)
#$C / m 2 %&
→
2
ρ sa = V ε 0 / a ln(a / b) #$C / m %&
and is clearly negative since a < b.
Problem 4.22. Application: Potential due to a charged disk.
z
Define an element of area on the disk and calculate the
elemental potential due to the charge on the elemental
area. Then integrate over the area of the disk to obtain
the total field at the required point (see Figure A).
Using cylindrical coordinates, we define an element of
area and an element of charge as:
ds' = r'dr'dφ '
dq = ρ0 ds' = ρ0 r'dr'dφ'
m2
C
The distance between this area and point d is denoted as R and equals:
d
a
R
r'
x
dr'
φ'
r'dφ'
ρs =ρ0
dφ'
y Figure A
2
R=
r! + d
2
m
The electric potential due to this elemental charge is:
dV =
dq =
ρ0 r'dr'dφ'
4πε 0 R 4πε 0 r % 2 + d 2
Integrating this over the surface of the disk:
V = ρ0
4πε 0
2π
a
φ'=0
r'=0
r %dφ %dr % = ρ0
r % 2 + d 2 2ε 0
a
r %dr %
2
r% + d
0
2
= ρ0
2ε 0
r %2 + d 2
a
0
= ρ0
2ε 0
a2 + d 2 − d
V
Problem 4.23. Electric field and potential of a short, charged segment.
Set up an element of length dx' and therefore a point
charge dq = ρldx'. Calculate the distance between the
element of length dx’ and the point P1 or P2 in terms
of the coordinate x' and coordinates of point P1 or P2 .
Integrate along the line from x' = −1 to x'=+1 either
for electric field or potential as required. Refer to
Figure A.
2
.P
1
R
.P
2
1
R
−1
a. The distance between the element of length dx’
and a general point P(x,y,z) is:
R = (x−x')2 + y 2 + z 2
2
dx' 1
Figure A
m
(1)
where we took advantage of the fact that the segment is placed at x’, y = 0, z’ = 0. The magnitude of the electric field
intensity due to an element of length dx’ (and therefore an element of charge ρldx’) is:
dE =
dq
ρldx'
=
2
4πε 0 R
4πε 0 (x−x')2 + y 2 + z 2
(2)
The electric field intensity at a general point in space will have x, y, and z components. However, since we need to
calculate the field in the x-y plane, the z component of the electric field intensity is zero. The x and y components of the
electric field intensity are:
ρl(x − x')dx'
dEx = dEcosα =
(3)
3/2
4πε 0 (x−x')2 + y 2 + z 2
dEy = dEsinα =
ρlydx'
4πε 0 (x−x')2 + y 2 + z 2
(4)
3/2
Now we set x = 0, y = 2 before we integrate from x’ = −1 to x’ = +1 to find the electric field intensity at P1 : Setting
x = 0, y = 2, z = 0 in Eqs. (3) and (4):
dEx =
− ρlx'dx'
2
4πε 0 x' + 4
dEy =
3/2
2ρldx'
4πε 0 x'2 + 4
(5)
3/2
Integrating:
Because of symmetry, the only nonzero field component at P1 is in the y direction. Thus we can write:
Ey =
2ρl
4πε 0
x'=1
x'=−1
dx'
x' + 4
2
3/2
=
2ρl
x'
4πε 0 4 x'2 + 4
1
−1
=
ρl
ρl
1
1
+
=
2πε 0 4 1 + 4 4 1 + 4
4 5 πε 0
V
m
b. Solution is essentially the same but for potential and different dimensions. We start with the potential due to an
element of charge dq = ρldx’ at a point P(x,y,z):
ρldx'
dV(x,y,z) = dq =
4πε 0 R 4πε 0 (x−x')2 + y 2 + z 2 1/2
where R was calculated in Eq. (1). This is now integrated along the line of charge. Before we do so, we set x = 2, y = 1, z
= 0 to find the potential at this point:
89
x'=+1
V(2,1,0) =
ρldx'
= ρl
4πε 0
x'=+1
dx'
= ρl ln 2 (2−x')2 + 1 +2x' − 4
2
2 1/2
2
x'=−1 4 πε 0 (2−x') + 1
x'=−1
(2 −x') + 1 4πε 0
ρ
2
l
=
ln 2 1 + 1 − 2 − ln 2 3 + 1 − 6 = ρl ln 2 − 1
V
4πε 0
4πε 0
10 − 3
1
−1
Problem 4.24. Application: Electric field of parallel wires.
Use Gauss’ law to calculate the electric field intensity at a distance d/2 from each wire. Assuming a Gaussian surface in
the form of a cylinder of radius d and length L, we can write for (say) the lower conductor (see Figure A):
s
E.ds = Q
ε0
E2 π rL = ρlL
ε0
→
E lower(y = 0) = x ρl
2πε 0 d
or, as a vector at y = 0:
E=
→
ρl
2πε 0 r
V
m
V
m
The electric field due to the upper conductor at y = 0 is (using the same considerations as for the lower conductor):
E upper(y = 0) = x
ρl
2πε 0 d
V
m
Thus, the total electric field intensity midway between the two conductors is
E(y=0) = x
ρl
πε 0 d
V
m
The potential midway is zero. This must be so midway between equal and opposite charges.
Eupper
ρl
d
d
x
− ρl
L
y
d
Eupper
x
Figure A
ρl
y
Elower
L
Elower
d
− ρl
Figure B
Problem 4.25. Field and potential due to charged dielectric. .
x
Using Gauss' law inside and outside the plates, we
get the electric field intensity. Integration over the
electric field intensity gives the potential.
a
a. Between the plates: a Gaussian surface in the form
of a box is defined at the center of the charge
distribution as shown in Figure A. Because the
charge density is positive, the electric field intensity
points out of the box on its top and bottom surfaces.
These surfaces equal S and we get
ρ S2x
2ES = v
→
ε
0
−a
E
V=0
ρ=ρv
E
S x'
x'
S
E
V=0
Figure A
ρx
E=x v
ε
S
y
S
E
V
m
where x is the position of the upper surface. The lower surface is at −x and therefore the electric field is negative for
x < 0 . At x = a (just below the plates), the electric field intensity is:
2ES =
ρvS2a
ε
→
E=x
ρva
ε
V
m
Outside the charge distribution, at x = ±a, we draw a new Gaussian surface as shown in Figure A:
90
2ES = ρvS2a
ε0
E = x ρva for
ε0
→
E = − x ρva for
ε0
x > 0,
x<0
V
m
Note: at x = a, x = − a, there is a jump in E because ε changes to ε 0 (see Figure B).
b. To calculate the potential between the two plates, start at x' = − a since it is zero at that point and integrate to a general
point x.
x'=x
V=−
x'=−a
ρvx' dx' = − ρvx 2 + ρva2
ε
2ε
2ε
V
at x = −a, V = 0, at x = +a, V = 0. At x = 0, V = V max
V max =
The solution can be summarized as:
(
)
V = ρv a 2 − x 2 / 2ε ,
ρva2
2ε
V
[ V]
−a≤x≤a
For x > a or x < − a, the electric field intensity is constant and therefore the potential decreases linearly:
For x > a:
x'=x
V=−
x'=a
2
x ρva.xdx' = − ρvx + ρva ,
ε
2ε
2ε
x>a
V
For x < −a:
x' = x
V=−
x' = − a
2
− x ρva . xdx' = ρvax + ρva ,
ε
2ε
2ε
x < −a
V
See Figure C for a plot of the potential.
These can be written as follows:
V = ρv a ( a − x ) / ε 0 ,
x > a,
x<a
E
ρv a/ε 0
ρv a/ε
−a
V
a
v
a 2/ 2
x
a
−ρv a/ε
−ρv a/ε 0
Figure B
a
0
x
Figure C
Problem 4.26. Work in the electric field.
We write the general expression for work in the field and equate to the expression given. From this, the electric field
intensity is found. Suppose the electric field intensity is known. Then the energy required to move the charge from x’ = 0
to x’ = x gives:
x
W =−
x
qE.dl = −
x'=0
qEdx = − qEx
J
(1)
x'=0
where we assumed that E and dl are in the same (x) direction. Now, this energy is proportional to the distance x squared:
W = − qEx = Kx 2
where K is the proportionality factor. From this, and evaluating the expression at x = 2: (E = 100 V/m at x = 2)
91
−6
K = − qE = − 4×10 ×100 = − 2× 10−4
x
2
or:
−4
E = − Kx = 2×10 −6x = 50x
q
4×10
→
E = x50x
V/m
Substitution in (1) will verify that this is correct.
Problem 4.27. Charge density and electric field intensity due to potential.
To find the charge density in the volume, we must first calculate the electric field intensity. Since the electric potential is
given, the electric field intensity is found by calculating the gradient of the potential.
a. The gradient in spherical coordinates is:
∇V = R ∂V + θ 1 ∂V + φ 1 ∂V
∂R
R ∂θ
Rsinθ ∂φ
Here we only have an R component., Thus, inside the sphere:
E = −∇ V = − R
∂ R2
∂R
V
m
E = − R2R
→
Now the volume charge density is found from the divergence as follows: Assume that at any radius R < b, we apply
Gauss' law:
3
C
4π R 2 E = 4π R 2 2R = (4/3)π R ρv
→
ρ v = − 6ε
ε
m3
Note that the charge density must be negative to produce an electric field intensity in the negative R direction. Outside
the sphere, the electric field intensity is that of a point charge equal to the total charge in the sphere. That is:
Q = 4 π b 3 ρv = − 4 π b 3 6ε = − 8πε b 3
3
3
C
b. The electric field intensity outside the sphere is:
E=
3
3
Q
= − R 8πε b 2 = − R 2ε b 2
2
4πε 0 R
4πε 0 R
ε0R
V
m
The electric field intensity inside the sphere was calculated in (a) as:
V
m
E = − R2R
Problem 4.28. Charge density and electric field intensity due to potential.
The electric field intensity may be found from the expression for potential by calculating the negative of the gradient of
the potential. Then, by calculating the divergence of the electric field intensity we obtain the charge density.
Alternatively, we may use Poisson’s equation to find the charge density directly from potential without first calculating
the electric field intensity. The normal derivative of potential is found from its gradient and the work done in moving a
point charge in the electric field is found by direct line integration on the path we choose.
a. The electric field intensity is calculated from the gradient of potential:
E (x,y,z ) = − x ∂V − y ∂V − z ∂V = − x y + 2 z − 1 − y x − 2 z − 1 − z x − 2 y + 2
∂x
∂y
∂z
V
m
(1)
At (0,0,0) this gives:
E (0,0,0) = x2 − y2 + z4
b. To calculate the charge density we use:
Calculating the divergence of E:
∇.E = ρv
ε0
92
V/m
(2)
∂E
∇.E = ∂Ex + y + ∂Ez = ∂(2) + ∂(−2) + ∂(4) = 0
∂x
∂y
∂z
∂x
∂y
∂z
Note: we could also use Poisson’s equation as follows:
→
ρv = 0
2
2
2
∇2 V = ∂ V2 + ∂ V2 + ∂ V2 = − ρv
ε0
∂x
∂y
∂z
Since the potential is first order in each of the coordinates, this also produces zero as expected.
c. The derivative in the normal direction (that is, the normal component of the electric field intensity) is found from the
gradient as follows:
∂V = n.∇V = n. n ∂V
∂n
∂n
This is because the gradient is the normal derivative to a surface of constant potential. Thus:
n ∂V = ∇V = x ∂V + y ∂V + z ∂V = x y + 2 z − 1 + y x − 2 z − 1 + z x − 2 y + 2
∂n
∂x
∂y
∂z
At (0,0,0):
n ∂V
= ∇V (0,0,0) = − x2 + y2 − z4
∂n (0,0,0)
Thus:
∂V
= n. n ∂V
= − x2 + y2 − z4 . − x2 + y2 − z4 = 24 = 4.9
∂n (0,0,0)
∂n (0,0,0)
4 + 4 + 16
[V/m]
d. First we must check if the energy is path dependent. If it is not path dependent, we are free to choose the most
convenient path. If it is, then our answer will depend on the path we choose. To see if the integration is path independent
we calculate the curl of E:
∇× E=
x
∂
∂x
y
∂
∂y
z
∂
∂z
x
∂
∂x
=
y
∂
∂y
z
∂
∂z
=
Ex
Ey
Ey
− y+2 z− 1 − x− 2 z− 1 − x− 2 y+2
x (−x + 2) + y(− y − 2) + z(1 − z) + z(z − 1) + x(x − 2) + y(y + 2) =0
Thus, the energy in moving the charge is path independent and we can choose any path we wish. For simplicity in
calculation, we choose a path so as to move from (0,0,0) to (1,0,0) then from (1,0,0) to (1,1,0) and then from (1,1,0) to
(1,1,1). The differential of path and the electric field intensity everywhere in space are (see Eq, (1) for the electric field
intensity):
dl = xdx + ydy + zdz
E = −x y + 2 z − 1 − y x − 2 z − 1 − z x − 2 y + 2
The work necessary is:
W = − q E.dl = q
l
y + 2 z − 1 dx + q
l
x − 2 z − 1 dy + q
l
x − 2 y + 2 dz
l
Note that the integrands in each of the integrals is independent of the integration variable but we will write the integrals
explicitly for the sake of generality:
x=1,y=0,z=0
W = − q E.dl = q
l
x=1,y=0,z=0
q
x=0,y=0,z=0
x=1,y=1,z=0
− 2 dx + q
x=0,y=0,z=0
x=1,y=1,z=0
0 + 2 0 − 1 dx + q
x=1,y=0,z=0
1 − 2 1 + 2 dz =
x=1,y=1,z=0
x=1,y=1,z=1
− 3 dz = q(−2+1 − 3) = − 4q = − 4×5×10−6 = − 20×10−6
1dy + q
x=1,y=0,z=0
x=1,y=1,z=1
1 − 2 0 − 1 dy + q
x=1,y=1,z=0
This requires −20 microJoules. That is, since the energy is negative, it is done by the system, rather than against the
system. Moving the charge from the origin to (1,1,1) reduces the potential energy of the system by 20 microJoules.
93
J
Problem 4.29. Conductor in a uniform electric field.
The electric field intensity in (a) is found by dividing the potential difference by the distance between the plates. When
the conductor is inserted, the distance is reduced by the thickness of the conductor (the potential on the conductor
remains constant).
a. Before the conductor is inserted, the electric field intensity is:
V
E= V
E=− yV
→
m
2b
2b
This electric field intensity is directed downwards.
The potential at a distance y from the lower plate (see Figure A) is:
V(y) = E y = V y
2b
V
m
(1)
V
(2)
b. When the conductor is inserted, the potential difference between the plates remains the same but the electric field
changes because of reduction in the size of the dielectric. The electric field now is:
E=
V
2b − 2c
V
m
E=− y
→
V
2b − 2c
V
m
(3)
and the potential is as follows (Figure B):
V(y) = E y =
V(y) = E y =
V y
2b − 2c
0<y≤d
V d+ V
y − (d + 2c)
2b − 2c
2b − 2c
V
(4)
d + 2c < y ≤ 2b
V
(5)
c. The position of the conductor is not important. Moving the conductor up or down does not change the electric field
intensity between the plates. It does change the value of d in Eqs. (4) and (5) and therefore the potential at y.
y
y
V
2b
V
E
V
y
2b
E
E=0
d
V
2c
E
V=0
V=0
Figure A
Figure B
Problem 4.30. Potential in a conductor.
Since there is a single charge it will stay at the center. E outside the sphere (using Gauss' Law in the form of a concentric
spherical surface of radius R) is:
V
E=R Q 2
m
4πε 0 R
The potential just outside the surface of the conducting sphere is:
V=
Q
4πε 0 R
V
Since the potential in a conductor is constant, we have
V r=R/2 =
Q
4πε 0 R
V
That is, the potential anywhere within the conductor remains constant and equal to its surface potential.
Problem 4.31. Layered structures.
Assume a charge on the outer surface of the inner conductor. This charge is −Q. We calculate the electric field intensity
this charge produces in both materials, then integrate the electric field to find the potential. Since the potential is given,
94
this allows the solution for Q. The charge on the inner surface of the outer conductor is then Q and the charge density is
found by dividing the charge over the area of the corresponding conductor.
In material 1 (a < R < b), ε 1 ), the electric field intensity is (using Gauss's law on a spherical surface of radius R and
enclosing a total charge −Q which resides on the surface of the inner sphere):
E1 = − R
Q
4πε 1 R 2
V
m
In material 2 (b < R < c), ε 2 ), the electric field intensity is:
E2 = − R
Q
2
4πε 2 R
V
m
The potential difference between the outer and inner conductors is given by:
c
V ca = Vc − Va = −
r=b
E.dl = −
a
r=c
E 1 .dR −
r=a
E 2 .dr = V 0
V
r=b
Substituting the two fields found above
r=b
V0 = −
r=a
− R Q 2 . RdR −
4πε 1 R
r=c
r=b
− R Q 2 . RdR = − Q
4πε 1 R
4πε 2 R
b
−
a
Q
4πε 2 R
c
V
b
After evaluation and simplification we have
V0 = Q 1 − 1 + Q 1 − 1 = Q 1 − 1 + 1 − 1
4πε 1 a b
4πε 2 b c 4π aε 1 bε 1 bε 2 cε 2
V
Now, the charge on the outer sphere is (we can calculate this charge because the potential is given as positive on the
outer conductor):
V 0 4π
Q=
C
1 − 1 + 1 − 1
aε 1 bε 1 bε 2 cε 2
The charge density on the outer sphere is equal to this charge divided by the area of the inner surface of the outer sphere.
The charge density on the inner sphere is equal to the negative of this charge divided by the area of the inner sphere of
the outer conductor:
V0
ρsc = Q 2 =
2 1
4π c
1
c
−
+ 1 − 1
aε 1 bε 1 bε 2 cε 2
V0
ρsa = − Q 2 = −
2 1
4π a
1
a
−
+ 1 − 1
aε 1 bε 1 bε 2 cε 2
C
m2
As mentioned, the charge is distributed on the outer surface of the inner conductor and on the inner surface of the outer
conductor. The magnitude of the charge on each surface is the same but the charge density is not, with the higher charge
density on the inner conductor.
Problem 4.32. Electric field and potential in spherical shells and conductors.
Using the electric field intensity of a point charge, or Gauss’ law on a concentric sphere of radius R and calculating the
total charge enclosed in the Gaussian surface.
a. For the configuration in Figure 4.68a: Inside the smaller shell (from Gauss' law): 0 < R ≤ a:
Q
V,
E=− R
0 ≤ R<a
m
4πε 0 R 2
Note: at R = 0 both E tends to infinity (negative). Between the two shells: a < R < b:
E=− R
Outside the outer shell R ≥ b:
ρsa2
+
R
4πε 0 R 2
ε0R 2
Q
V,
m
95
a<R<b
2
E=− R
Q
ρsa2 − R ρsb
+
R
4πε 0 R 2
ε0R 2
ε0R 2
V,
m
R>b
To calculate the potential, we start from infinity and integrate up to the required point:
Outside the outer shell R ≥ b:
R
V(R) = −
R
E.dR' = −
∞
−R
∞
Q
ρsa2 − R ρsb 2 .RdR' = − Q + ρsa2 − ρsb 2
+
R
4πε 0 R ε 0 R
ε0R
4πε 0 R'2
ε 0 R'2
ε 0 R'2
V, R>b
Between the two shells:
b
R
R
2
Q + ρsa2 − ρsb 2 −
Q
−R
+ R ρsa 2 .dR'
2
4πε 0 b ε 0 b
ε0b
∞
b
b
4πε 0 R
ε0R
2
2
2
2
2
Q
ρ
ρ
Q
ρ
Q
ρ
Q
ρ
ρ
sa
sb
sa
sa
sa
sb
=−
+
−
−
+
+
−
=−
+
−
V,
a<R<b
ε0
4πε 0 b ε 0 b
ε 0 b 4πε 0 R ε 0 R 4πε 0 b ε 0 b
4πε 0 R ε 0 R
V(R) = −
E(R>b).dR' = −
E(a<R<b).dR' = −
Inside the inner shell:
b
V(R) = −
a
E(R>b).dR' −
R
E(a<R<b).dR' −
b
∞
R
E(R<a).dR' = V a −
a
E(R<a).dR'
a
a
Q + ρsa2 − ρsb −
Q
−R
.dR'
ε0
4πε 0 a ε 0 a
b
4πε 0 R 2
2
= − Q + ρsa − ρsb − Q + Q = − Q + ρsa − ρsb
ε0
ε0
ε0
4πε 0 a ε 0 a
4πε 0 R 4πε 0 a
4πε 0 R
=−
V,
R<a
Note that these potentials are everywhere negative since the total net charge is negative. However, between the two
shells, the electric field intensity may be positive, depending on the magnitudes of Q and ρs. For the configuration in
Figure 4.68b we have:
Q , R < a,
Q , R> b
V
E=− R
E = 0, a < R < b,
E=− R
2
m
4πε 0 R
4πε 0 R 2
The potential is calculated as above:
Q
V
R≥ b
4πε 0 R
V
(V is constant)
V(R) = −
Q
4πε 0 b
V(R) = − Q + Q − Q
4πε 0 R 4πε 0 a 4πε 0 b
V(R) = −
V
a≤ R≤ b
R≤a
b. Plots: Figures A and B show the electric field intensity and potential for the configuration in Fig. 4.68a. Figures C
and D show the electric field intensity and potential for the configuration in Fig. 4.68b.
E
−
−
a
Q
4πε 0 a 2
−
V
ρ
Q
+ s
4πε 0 a 2 ε 0
ρ 2
Q
−
+ sa
4πε 0 b 2 ε0 b 2
b
a
R
b
−
−
ρ 2 ρ
Q
+ sa − s
4πε 0 b 2 ε0 b 2 ε0
Figure A
ρs a ρs b
Q
−
ε0
4πε 0 a + ε 0
Figure B
96
ρs a 2 ρs b
Q
−
4πε 0 b + ε 0 b ε 0
R
E
a
Q
−
4πε 0 b 2
Q
−
4πε 0 a 2
V
R
b
−
a
R
b
Q
4πε 0 b
Figure C
Figure D
Problem 4.33. Electric field and potential of point charge in the presence of conductors and
dielectrics.
For purposes of calculating the electric field intensity at point P, Gauss’s law disregards any material inside the Gaussian
surface and only takes into account the total charge enclosed by the surface. Taking a Gaussian surface in the form of a
sphere of radius R = r4 and since the total charge enclosed by this surface is −q we get:
E4π r42 = − q
ε0
E =− R
→
q
4πε 0 R 42
V
m
The potential at P (with reference to infinity) is:
r4
r4
VP = −
E.dl = −
∞
−R
∞
q
q
. RdR = −
2
4πε 0 R 4
4πε 0 R 4
V
Problem 4.34. Polarization in dielectrics.
The polarization in the dielectric is due to the electric field intensity generated by the central charged wire. The
concentric cylinders have no effect other than to contain the dielectric. That is, the electric field intensity in the dielectric
is the same with or without the conducting cylinders. The electric field anywhere in space, due to a charged line is found
using Gauss' law in cylindrical coordinates. A Gaussian surface in the form of a cylinder of radius a < r < b and arbitrary
length L is stipulated and the electric field on its surface integrated and equated to the total charge enclosed by the
surface:
In dielectric No. (1) a < r < b:
E.ds = 2π rLE = ρlL
ε1
S
E=
→
The electric flux density is:
D = ε1E = r
Now the polarization vector may be written as:
ρl
2πε 1 r
→
ρl
2π r
C
m2
E= r
ρl
2πε 1 r
V
m
(2)
P = D − ε 0 Ε = r ρl − r ε 0 ρl = r ρl 1 − ε 0 = r ρl
ε1 − ε 0
ε1
2π r
2πε 1 r
2π r
2πε 1 r
Thus, in dielectric (1):
C
P = r ρl
ε1 − ε 0
a<R<b
2πε 1 r
m2
C
m2
In dielectric No. (2) b < r < c, using a Gaussian surface at b < r < c:
E.ds = 2π rLE = ρlL
ε2
S
→
E=
The electric flux density is:
D = ε2E = r
and the polarization vector is:
ρl
2πε 2 r
→
ρl
2π r
C
m2
97
(1)
E= r
ρl
2πε 2 r
V
m
P=r
ρl ε − ε
2
0
2πε 2 r
C
m2
Problem 4.35. Polarization in dielectrics.
The electric field intensity between two parallel plates equals the potential difference divided by distance. The
polarization in the dielectric is then found from the general relation D = ε 0 E + P.
12
E=V =
= 120
d 0.1× 10−3
a.
kV
m
b.
D = ε 0E + P = ε E
→
With ε r = 4, V = 12 V, d = 0.1 mm, we get:
P = ε0 3
P = ε E − ε 0E = ε 0 εr − 1 E = ε 0 εr − 1 V
d
C
m2
12
36
= ε0
= 3.6× 105 ε 0 = 3.6× 105 ×8.854× 10−1 2 = 3.187× 10−6
0.1× 10−3
0.1× 10−3
C
m2
Problem 4.36. Electric flux density in polarized medium.
a. The electric flux density due to the surface charge:
D s = x ρs
2
C ,
m2
D s = −xρs
2
x > 1,
C ,
m2
x < 1,
C ,
m2
Ds=0
at x = 1 (1)
The electric field intensity and electric flux density due to the line of charge is:
E=r
ρl
2πε 0 r
⇒
Dl = r
ρl
2π r
C
m2
(2)
where r is the distance between the line and a general point P(x,y,z). Due to line: Since we have chosen to use cylindrical
coordinates for the line, and placing the line on the z axis as in Figure A, the electric flux density may be separated into
its components:
D lx =
x Dlx ,
x2 +y 2
Dly =
y Dly
x2 +y 2
C
m2
(3)
Ds
where Dl is the magnitude of Dl above. As a vector:
x ρlx
y ρly
Dl =
+
2
2
2π x + y
2π x 2 + y 2
C
m2
Dly
x
ρs
(4)
.
x=1
r
.P(x,y,z)D
lx
y
ρl
Ds
Dl
Figure A
Thus, at a general point (x,y,z), the total electric flux density is:
D=x
ρlx
ρly
+ ρs + y
,
2
2
2π x + y
2
2π x 2 + y 2
x >1
C
m2
(5)
D=x
ρlx
ρly
− ρs + y
,
2
2
2π x + y
2
2π x 2 + y 2
x <1
C
m2
(6)
The flux density is independent of z.
b. At x = 1, y = 0, the electric field due to the surface charge density is zero. Substituting this into either Eq. (5) or (6):
D=x
ρl1
2π 1 + 0
+y
−6
ρl0
= x ρl = x 4×10 = x6.36710−7
2π 1 + 0
2π
2π
C
m2
c. If space is replaced with a dielectric of relative permittivity ε r, the polarization depends on the value of ε r as follows:
P = D − ε0E = ε0εr − ε 0 = ε 0 εr − 1 E
98
C
m2
From Eqs. (5) and (6), the electric field intensity is:
ρlx
ρly
E = D =x
+ ρs + y
2
2
ε rε 0
2πε rε 0 x + y
2ε rε 0
2πε rε 0 x 2 + y 2
ρ
x
ρ
ρly
l
s
E = D =x
−
+y
2
2
ε rε 0
2πε rε 0 x + y
2ε rε 0
2πε rε 0 x 2 + y 2
x >1
x <1
Thus, the polarization vector is:
P = ε0 εr − 1 E = x
ρlx(ε r − 1) + ρs(ε r − 1) + y ρly(ε r − 1)
2πε r x 2 + y 2
2ε r
2πε r x 2 + y 2
x >1
C
m2
P = ε0 εr − 1 E = x
ρlx(ε r − 1) − ρs(ε r − 1) + y ρly(ε r − 1)
2πε r x 2 + y 2
2ε r
2πε r x 2 + y 2
x <1
C
m2
For ε r = 4:
P=x
3ρlx
3ρly
+ 3ρs + y
8π x 2 + y 2
8
8π x 2 + y 2
x >1
C
m2
P=x
3ρlx
3ρly
− 3ρs + y
2
2
8π x + y
8
8π x 2 + y 2
x <1
C
m2
Problem 4.37. Application: Multilayer devices.
Use: Dair = Ddielectric:
a.
ε 0 Eair = 80ε 0 Ediel.
Eair = 80Ediel.
→
V/m
From the circulation of E:
V=
E.dl
⇒
V = Eair×0.005 + Edie×0.0025=80Edie×0.005 + Edie×0.0025=1
Thus:
Edie = 2.4845
V/m
Eair = 198.76
V/m
Emax = 198.76 V/m in air.
b. Approximately 15,000 V. That is, if the potential in air is allowed to be at breakdown, a total of 5×3000 = 15,000 V is
allowed due to the air region. In addition, the electric field intensity in the dielectric region is:
Ediel. = Eair = 3000 = 37.5
80
80
V
mm
The voltage across the dielectric region is therefore: V diel. = E diel.×2.5 = 37.5× 2.5 = 93.75
The total potential difference allowed across the plates is 15,093.75 [V].
V
Problem 4.38. Application: Maximum charge and maximum potential on a sphere.
The maximum allowable potential on a sphere is that which will cause breakdown at the surface. The maximum charge
density is calculated from Gauss’ law so that the electric field intensity at the surface equals the breakdown electric field
intensity of 3000 V/mm.
a. From Gauss law, the electric field intensity at a distance R from the center of the sphere due to a uniform charge
density ρs on the surface of the sphere of radius a is:
E.ds = Q
ε0
s
or:
→
2
E = ρsa 2
ε0R
E4 π R 2 = ρs4π a
ε0
2
V
m
6
This field is in the radial direction. The maximum electric field intensity occurs at R = a, and must equal 3×10 V/m:
99
2
Emax = ρsa2 = ρs = 3×106
ε0
ε0a
V
m
Thus, the maximum charge density on the surface of the sphere is:
C
m2
ρs max = ε 0 ×3×106 = 8.854× 10−1 2×3×106 = 26.55× 10−6
Beyond this, there will be breakdown and the charge will be reduced.
b. The maximum potential on the sphere is that due to the maximum charge density. Assuming maximum charge density
on the surface, the total charge is 4π a2 ρsmax. The potential at R = a equals:
−6
2
V max = Qmax = 4π a ρsmax = 26.55× 10 −1a2 = 3×106 a
4πε 0 a
4πε 0 a
8.854× 10
V
That is, the maximum potential equals the breakdown electric field in the surrounding medium (air) multiplied by the
radius of the sphere. This is the reason for the large, spherical terminals on electrostatic machines such as Van de Graaf
generators.
Problem 4.39. Maximum charge on the globe.
Given the maximum electric field that the surface charge layer on the surface of the globe can cause, we can calculate the
charge density that will cause this field and then integrate the charge density to obtain the total charge. From Gauss' law,
the electric field intensity at a distance R from the center of the sphere due to a uniform charge density ρs on the surface
of the sphere of radius a is:
2
2
V
E.ds = Q
→
E4 π R 2 = ρs4π a
→
E = ρsa 2
ε0
ε0
m
s
ε0R
6
This field is in the radial direction. The maximum electric field intensity occurs at R = a, and must equal 3×10 V/m:
2
Emax = ρsa2 = ρs = 3×106
ε0
ε0a
V
m
Thus, the maximum charge density on the surface of the globe is:
C
m2
ρs max = ε 0 ×3×106 = 8.854× 10−1 2×3×106 = 26.562× 10−6
Beyond this, there will be breakdown and the charge will be reduced. note that this is a constant: it does not depend on
the size of the globe. Because the charge density on the surface is constant, the total charge is simply this charge density
multiplied by the area of the globe:
2
Q = ρs maxS = ρs max4π R 2 = 26.55× 10−6 ×4π ×(6.4×106 ) = 1.367× 101 0
C
Problem 4.40. Breakdown in power lines.
To calculate the potential difference between the two cylinders we assume a positive charge density on one cylinder and
an equal but negative charge density on the other. Then, we calculate the electric field intensity anywhere on the line
connecting their centers, in terms of the unknown charge density. By setting the electric field intensity to the maximum
breakdown field at the surface of one cylinder we can evaluate the maximum charge density and the electric field
intensity anywhere between the cylinders. Integrating the electric field along the path between the two cylinders gives
the maximum potential difference between the cylinders.
a. Assuming charge densities as shown in Figure A, the electric field of a cylinder is:
E = 2π aρs = aρs ,
2πε 0 r ε 0 r
r >a
V
m
Maximum electric field is at the surface of one of the conductors: Suppose at r = x = a
E = x ρs a
ε0 a
V
m
due to left cylinder,
E=x
ρsa
ε0 d − a
100
V
m
due to the right cylinder
The total field at the surface of the left cylinder (note that the fields of both cylinders are in the positive x direction):
a
E total = x ρs 1 +
= x ρs d
ε0
ε0 d − a
d −a
V
m
6
This cannot exceed 3⋅10 V/m. Therefore, the surface charge density is:
6
E max = 3×10 =
6
ρs
d
ε0 d − a
ρs =
⇒
3×10 × ε 0 d − a
d
C
m2
The electric field intensity at any general point between the cylinders is:
3×106 a d − a 1
E = x ρsa + x ρsa
=x
+ 1
ε0x
x
ε0 d − x
d
d− x
V
m
Now, to find the potential difference we integrate from the negative cylinder to positive cylinder (so as to obtain a
positive potential difference, which in turn complies with the notation we chose):
a
3×106 a d − a
3×106 a d − a
a
1 + 1
=−
ln x − ln d − x
x
d
d−x
d
d−a
d−a
d− a
6
3×106 a d − a
2×3×10
a
d
−
a
=−
ln a − ln d − a + ln a = −
ln a
d
d−a
d
d−a
= − 2× 3×106 ×0.02× 0.98 ln 0.02 = 4.577× 105
V
0.98
b) The solution remains the same but now a ! = 0.022 m because the dielectric extends the conductor diameter by 2 mm
(see Figure B). That is, because the insulation has a much higher breakdown electric field intensity, breakdown will
occur again in air and for this purpose, the radius of the cylinders has increased by 2 mm. Although the electric field
intensity in the insulation is different than that in air, it is close to that of air because their permittivity is close. Thus, to a
first approximation:
V =−
6
V = − 6×10 a #(d - a #) ln a = − 6×106 ×(0.022)×(0.978) ln 0.02 = 5.024× 105
d
d−a
0.98
V
The maximum voltage increases because the location of maximum E has been extended outwards. An exact calculation
is only possible if the permittivity and breakdown voltage of the insulation are known.
a
(2)
E1
ρs
E2
x
− ρs
d
(1) a'
ρs
a
Figure A
(2) a'
E1
E2
− ρs
d
Figure B
Problem 4.41. Interface conditions for D. Using Figure A as a reference, the interface conditions between the
two materials for E and D are:
E1t = E2t,
D1n = D2n
The interface is on the x = const. plane. Thus, the x component of E is normal and the y component is tangential:
E1t = 3,
E1n = 5
V/m
D1t = ε 0 E1t = 3ε 0 ,
In material (2):
Thus:
101
D1n = ε 0 E1n = 5ε 0
C/m2
D2n = D1n = 5ε 0
C/m2 but:
(1)
ε=ε 0
ε=2ε 0
C/m2
D2t = 2ε 0 E2t = 2ε 0 E1t = 6ε 0
y
(2)
E=x 5 + y 3
Thus, the electric flux density in material (2) is:
D 2 = x5 ε 0 +y6ε 0
x
C/m2
n
Figure A
Problem 4.42. Interface conditions in layered structures.
d
Refer to Figure A. From the continuity of the normal
components of the electric flux density we have:
D2n = D4n = D5n = D1n = ε 0 E ,
D3n = 0
In material (1), E is given and equals −x E.
In material (2):
D2n = ε 0 E → E2n = ε 0 E → E 2 = − x ε 0 E
ε1
ε1
D2n = ε 0 E
D4n = ε 0 E
E 2 = − x ε0E
ε1
→
E 4 = − x ε0E
ε1
→
D5n = ε0E
In material (5):
d
−
−
−
−
−
−
−
−
E4
E5
ρs=− ε0E
Figure A
E2n = ε 0 E
ε1
→
In material (3), E=0.
In material (4):
ρs=ε0E
V
m
t
E3=0
E2
E
a. The flux density in the conductor must be zero.
+
+
+
+
+
+
+
+
V
m
V
m
E5 = − x ε0E = − xE
ε0
→
V
m
Note: on the interface between conductor and dielectric there must be a charge density equal in magnitude to:
ρs = D2 − D3 = D2 − 0 = ε 1 E
C/m2
ρ s =ε 1 E
→
The charge density is positive at the interface between material (2) and (3) and negative between material (3) and (4).
b. To calculate the potential difference assume you start from left to right (going against the field) and calculate:
d
V = − E.dl = −
2d+t
E 2 .dl −
0
E 4 .dl
V
d+t
Because dl = x dx, and the electric field intensity is constant in each of the materials:
V = ε 0 Ed + ε 0 Ed = 2ε 0 Ed
ε1
ε1
ε1
V
This is the potential difference between the right and left side of the structure.
PROBLEM 4.43. Interface conditions with charge density on the interface
Since a charge density at the surface exists, the general relation in Eq. (4.76) must be used. Using Figure 4.34 for reference,
with the ground denoted as material (1) (see Figure 4.32), we get
ε1 E1n − ε 2 E2n = ρs → E1n →
ρs + ε 2 E2n
ε1
(a). With ε2 = ε0 = 8.854 × 10–12 F/m and ε1 = 2ε0, we get
102
E1n =
#V&
% (
$m'
−5×10−10 + 8.854 ×10−12 ×100 3.854 ×10−10
=
= 21.76
2 × 8.854 ×10−12
17.708×10−12
(b). If ρs = 0, we get
E1n =
8.854 ×10−12 ×100
= 50
2 × 8.854 ×10−12
#V&
% (
$m'
Note. The Earth is not a perfect dielectric because it has a small conductivity. However, some soils, like dry sand, are
close to perfect dielectrics.
Problem 4.44. Interface conditions in layered dielectrics.
The electric field intensity is calculated using the continuity of the normal and tangential components of the electric field
intensity according to the interface conditions.
a. The normal and tangential components of the electric field intensities are (magnitudes):
E2t = E3t = E4t = E1 sinθ
E2n = ε 0 E1 cos θ ,
ε1
V/m
E3n = ε 0 E1 cos θ ,
ε2
E4n = E1n
The electric field intensities in the various materials are:
E 2 = − yE 1 sinθ − x ε 0 E1 cos θ ,
E 3 = − yE 1 sinθ − x ε 0 E1 c o s θ ,
ε1
ε2
E 4 = E1 = − yE 1 sinθ − xE 1 cosθ
V
m
b. The potential is due to the normal components and equals:
0
V (a, −a) = −
a
E2n dx −
x=−a
E3n dx = − a(E2n + E3n )
x=0
Since E is negative, we have
V = E1 aε 0 cosθ 1 + 1
ε1 ε2
(2)
(1)
V
ε0
En
The potential is higher at x = a as expected.
E
c. From the calculation of the electric field, it is clear
that the electric field in materials (4) and (1) are the
same. Thus the angle is also the same and equals θ.
θ
y
(3)
d
d
ε1
ε2
(4)
ε0
Et
x =− a x =
0
x =+a
x
Problem 4.45. Interface conditions for the polarization vector.
Write the interface conditions for the electric flux density and then the electric flux density in terms of the polarization
vectors on both sides of the interface.
a. Since ρs = 0, we get:
D1n = D2n
E1t = E2t
The electric flux density in a polarized material is:
D = ε 0E + P
and
ε 0 ε rE = ε 0 E + P
→
P = Eε 0 ε r − 1
Since E1t = E2t:
Similarly for the normal components:
P1n = D1n − ε 0 E1n
Since D1n = D2n :
P1t = P2t
ε1 − ε 0 ε2 − ε 0
P2n = D2n − ε 0 E2n
C/m2
C/m2
ε 1 P1n = ε 2 P2n
ε1 − ε 0 ε2 − ε 0
If there is a charge density at the interface, the condition on the tangential component remains the same since (E1t = E2t
in either case):
P1t = P2t
ε1 − ε 0 ε2 − ε 0
but:
D1n − D1n = ρs
and, accordingly:
ε 1 P1n − ε 2 P2n = ρs
ε1 − ε 0 ε2 − ε 0
b. From (a) we can write:
P1t = ε 1 − ε 0
P1n = ε 2 ε 1 − ε 0
and
P2t ε 2 − ε 0
P2n ε 1 ε 2 − ε 0
or:
P1n = ε 2 P2n
P1t ε 1 P2t
Problem 4.46. Application: Parallel plate capacitor.
The easiest way is to view the two capacitors in series as shown:
C 1 = ε 1 ×area = 20ε 0
d1
d1
For two capacitors in series:
C2 = ε 2 ×area = 3ε 0 = 3ε 0
d2
d−d1
d2
20ε 0 3ε 0
C
C
d1
d2
C total = 1 2 =
= 60ε 0
C 1 +C 2
20ε 0 + 3ε 0
3d1 + 20d2
d1
d2
In more general terms:
F
C1 = ε 1 A 1
d1
Substituting for ε 1 and ε 2 :
C2 = ε 2 A 2
ε 1 = 20ε 0
d2
ε1 ε2
ε1 ε2
C total = C 1 C 2 = d1 d2 =
C 1 +C 2 ε 1 + ε 2 ε 1 d2 + ε 2 d1
d1 d2
C total =
60ε 02
= 60ε 0
20ε 0 d2 + 3ε 0 d1 20d2 + 3d1
F
ε 2 = 3ε 0
F
F
Problem 4.47. Capacitance per unit length.
As with any capacitance, we must assume a potential difference between the two conductors (or a charge density on the
surface of each of the conductors and find the electric field, followed by potential). Either method is suitable.
Method A. Assume a charge density ρs on the surface of the inner conductor. We use Gauss’ law in cylindrical
coordinates assuming a Gaussian surface of length L = 1 m at a distance r from the center of the cable. The electric field
intensity is uniform over this surface and the surface encloses a total charge equal to 2πbLρs. Thus, from Gauss’ law.
Figure A shows the Gaussian surface in cross-section.
2π rLE = 2π bLρs
ε0
→
E = r bρs
rε 0
V
m
Since the field points in the direction of r, we integrate from r = a to r = b (against the field) to find the potential
difference between the shells:
b
Vba = −
b
E.dr = −
a
a
r bρs .r dr = − bρs lnb = bρs ln a
rε 0
ε0 a ε0 b
104
V
The total charge on the inner (or outer shell) is Q = 2πbρs . The capacitance is therefore
C = Q = 2π bρs = 2πε 0
V b a bρs ln a
ln a
b
ε0 b
F
m
Method B. Assume a potential difference V b a exists between the conductors. Now we need to calculate the charge on
the inner plate. To do so, we assume an unknown charge density ρs exists on the surface of the inner conductor.
Repeating the result of the previous method, we get:
C
V b a = bρs ln a
→
ρ s = ε0V b a
ε0 b
bln(a/b)
m2
The total charge on the surface of the inner conductor
E
is:
a
Q = 2π bρs = 2πε 0 V b a
ln(a/b)
− ρs
C
E
ρs
b
r
ε0
Again dividing Q by V b a we get:
C = Q = 2πε 0
V b a ln(a/b)
E
Figure A
F
m
as in method A.
Problem 4.48. Application: Spherical capacitor.
We first assume a charge Q on the inner surface and calculate the electric field intensity between the plates using Gauss’
law. Then we obtain the potential difference between the plates and obtain C = Q/V.
Assuming a total charge Q on the inner shell, the electric field intensity anywhere between the shells is:
E=R
Q
4πε R 2
a< R< b
The potential difference between the two shells is:
a
Vab = −
E.dl
V
b
E is in the positive R direction and since we integrate between R = b and R = a, dl is in the direction of increasing field.
Thus:
a
, Q /
a$
a
Q '
Q
Q , 1 1/
Q , 1 1/
Vab = − ∫ & R̂
⋅ R̂dR = − ∫
dR = .
=
− 1=
)
1
.
. − 1
2
2
b
b
4πε R
- 4πε R 0b 4πε - a b 0 13.2πε 0 - a b 0
% 4πε R (
(
C=
C=
)
Q
=
Vab
13.2πε 0 13.2πε 0 ab
Q
=
=
b− a
Q $ 1 1' $ 1 1'
& − ) & − )
13.2πε 0 % a b ( % a b (
13.2 × π × 8.854 ×10−12 × 0.04 × 0.05
= 73.43×10−12
0.05− 0.04
,-V/0
$%F'(
$%F&'
Problem 4.49. Layered spherical capacitor.
We first assume a charge Q on the inner surface and calculate the electric field intensity between the plates using Gauss’
law. Then we obtain the potential difference between the plates and obtain C = Q/V. This gives the capacitance in
Figure 4.74 but without the dielectrics. Now we move to Figure 4.74a and use two layers of dielectrics. One is x [m]
thick, of relative permittivity 5 and the other is b − a − x thick and of relative permittivity 2. We calculate the electric
field intensity in each dielectric using the same charge Q on the inner plate, and then integrate to find the potential
difference between the conducting shells. Assuming a total charge Q on the inner plate, the electric field intensity
anywhere between the plates is:
105
E=R
Q
4πε R 2
a< R< b
(1)
V
(2)
The potential difference between the two plates is:
a
Vab = −
E.dl
b
E is in the positive R direction and since we integrate between R = b and R = a, dl is in the direction of increasing field.
Thus:
a
Vab = −
R
b
Q
. RdR = −
4πε R 2
a
b
Q
dR = Q
2
4πε R
4πε R
a
= Q 1−1
4πε a b
b
V
(3)
The capacitance is:
Q
C= Q =
= 4πε
Vab
Q 1−1
1−1
a b
a
4πε
b
F
(4)
This is the capacitance between two concentric spheres of radius a and b (b > a). For the conditions given here, ε = 2ε 0 :
C = 4πε = 8πε 0 = 8πε 0 ab
1−1
1−1
b−a
a b
a b
F
(5)
Numerically this gives:
−1 2
C = 8πε 0 ab = 8× π ×8.854× 10 ×0.01× 0.05 = 2.78× 10−1 2
b−a
0.05 − 0.01
F
(6)
a. The configuration we treat here is shown in Figure A. The two layers may be viewed as two capacitors in series.
Calculating each capacitance separately based on the calculations above we get: The thickness of the high permittivity
dielectric is x = d − a. The two capacitances are:
C 1 = 4πε 2
1 − 1
a d
C 2 = 4πε 1
1 − 1
d b
F
The total capacitance again must equal twice the capacitance in Eq. (4):
1 − 1
1 − 1 1 − 1
1 + 1 =a d +d b =a b
C1 C2
4πε 2
4πε 1
8πε 1
From this:
adb 2ε 1 − 2ε 1 + 2ε 2 − 2ε 2 = ε 2 − ε 2
a
a b
d
d
b
or:
2ε 1 bd − 2ε 1 ab + 2ε 2 ab − 2ε 2 ad = ε 2 bd − ε 2 ad
Solving for d after substituting ε 1 = 5ε 0 , ε 2 = 2ε 0 , we get:
d=
2ab ε 2 − ε 1
ε 2 b − ε 2 a − 2bε 1 + 2ε 2 a
=
2ab ε 2 − ε 1
ε 2 a + ε 2 b − 2bε 1
=
2ab 3ε 0
5aε 0 + 5bε 0 − 4bε 0
= 6ab
5a + b
m
and the high dielectric layer thickness is:
Numerically this gives:
2
5a b − a
x = d − a = 6ab − 5a − ab =
5a + b
5a + b
x=
5a b − a
5×0.01 0.05 − 0.01
=
= 0.02
5a + b
5×0.01 + 0.05
m
m
b. The configuration we treat here is shown in Figure B. Now we repeat the process but interchange between the
dielectric so that the high dielectric material is on the inner conductor. Its thickness now is x = b − d. The two
capacitances are: For the inner layer (1) and outer layer (2) we get:
106
C 1 = 4πε 1
1 − 1
a d
C 2 = 4πε 2
1 − 1
d b
V
Note that the thickness of the high dielectric material is x = b − d.
The total capacitance of these two capacitors in series must be twice the capacitance calculated above in Eq. (4).
1 − 1
1 − 1
1 − 1
1 = 1 + 1 =a d +d b =a b
Ct C1 C2
4πε 1
4πε 2
8πε 1
From this we get:
adb 2ε 2 − 2ε 2 + 2ε 1 − 2ε 1 = ε 2 − ε 2
a
a
d
d
b
b
or:
2ε 2 bd − 2ε 2 ab + 2ε 1 ab − 2ε 1 ad = ε 2 bd − ε 2 ad
Solving for d after substituting ε 1 = 2ε 0 , ε 2 = 5ε 0 , we get:
d=
2ab ε 2 − ε 1
2ab ε 2 − ε 1
2ab 3ε 0
=
=
= 6ab
2bε 2 − 2aε 1 − bε 2 + aε 2 bε 2 − 2aε 1 + aε 2 5bε 0 − 4aε 0 + 5aε 0 5b + a
The thickness of the high dielectric layer is therefore:
2
5b b − a
x = b − d = 5b + ab − 6ab =
5b + a
5b + a
Numerically this gives
5b b − a
5×0.05 0.05 − 0.01
x=
=
= 0.0385
5b + a
5×0.05 + 0.01
b
a
a
d
ε r=5
ε r=2
m
m
b
x
x
m
a=10 mm
b =50 mm
Figure A
ε r=2
ε r=5
d
a=10 mm
b =50 mm
Figure B
Problem 4.50. Layered spherical capacitor.
The simplest way is to calculate the electric field intensity everywhere between the two spherical shells due to an
assumed total charge Q on one of the plates. After the electric field intensity is found, the potential difference between
the plates is found by integrating along the field. The ratio between charge and potential gives the capacitance.
We start by assuming a total charge Q on the inner shell. Because of this, the electric flux density everywhere between
the plates is:
C
D=R Q 2
a< R<c
m2
4π R
The electric field intensity in the two dielectrics is:
Q
E1 = D = R
ε1
4πε 1 R 2
a < R < b,
Q ,
E2 = D = R
ε2
4π ε 2 R 2
Now we calculate the potential between R = a and R = c as follows:
107
b< R<c
V
m
b
V ac = −
b
a
E 2 .dl −
E 1 .dl = −
c
b
c
Q
dR −
4πε 2 R 2
a
b
Q dR =
Q
4πε 2 R
4πε 1 R 2
Q 1− 1 + Q 1− 1
4πε 2 b c
4πε 1 a b
The capacitance is:
b
+
c
Q
4πε 1 R 2
a
=
b
V
Q
4πε 1 ε 2 bca
C= Q =
=
V ac
Q 1−1 + Q 1−1
ε 1 a(c − b) + ε 2 c(b − a)
4πε 2 b c
4πε 1 a b
F
Problem 4.51. Potential energy of a system of point charges.
To calculate the potential energy in the system we must calculate first the potential at the location of each charge in its
absence. Then the general formula for energy in a system of charges may be used.
Q
−Q
(2) 2Q
To simplify calculation we denote the charges as (1)
(3)
(1)
a
b
through (3) as shown in Figure A.
Q1
Q3
Q2
1. The potential at the location of charge (1) due to
V1
V3
V2
charge (2) and (3) is:
Figure A
Q
Q
2Q
Q
2
3
V1 =
+
=
−
V
4πε 0 R 1 2 4πε 0 R 1 3 4πε 0 a 4πε 0 (a +b)
2. The potential at the location of charge (2) due to charge (1) and (3) is:
V2 =
Q1
Q3
+
= Q − Q
4πε 0 R 1 2 4πε 0 R 3 1 4πε 0 a 4πε 0 b
V
3. The potential at the location of charge (3) due to charge (1) and (2) is:
V3 =
Q1
Q2
Q
+
=
+ 2Q
4πε 0 R 1 3 4πε 0 R 2 3 4πε 0 (a +b) 4πε 0 b
V
The energy in the system now is:
W = 1 Q1 V 1 + Q2 V 2 + Q3 V 3 =
2
2
2
2
2
2
Q2
2Q
Q2
1 2Q −
2 − 1 − 2
+
− 2Q −
− 2Q
= Q
2 4πε 0 a 4πε 0 (a +b) 4πε 0 a 4πε 0 b 4πε 0 (a +b) 4πε 0 b
4πε 0 a a + b b
J
Problem 4.52. Potential energy of distributed charges.
The energy stored may be viewed as being stored in the electric field associated with the spherical charge distribution.
Thus, we calculate the electric field intensity inside and outside the sphere and then the energy density inside and outside
the sphere. These are then integrated to find the stored energy inside and outside the sphere. The sum of the two energies
is the total energy and equals that found in Example 4.28.
The electric field intensity at a point R inside the sphere is calculated using a Gaussian surface in the form of a sphere of
radius R < a. Because the field is radial, we get:
E.ds =
SR
ρvdv
vR ε 0
→
2
E 4π R =
ρv 4π R 3
ε0 3
or:
E =R
ρvR
3ε 0
The electric field intensity outside the sphere at a point R > a is calculated as:
3
ρvdv
2 ρ
ρ a3
E.ds =
→
E 4π R = v 4π a
or:
E =R v 2
ε0 3
SR
va ε 0
3ε 0 R
R<a
R>a
Due to these, the energy density inside and outside the sphere are:
2 2
w i = ε 0 E.E = ρv R
2
18ε 0
R < a,
2 6
wo = ε 0 E.E = ρv a 4
2
18ε 0 R
108
R>a
J
m3
2
a. Now we define an element of volume inside the sphere as dv = R sinθdRdθdφ . Substituting and integrating over the
volume of the sphere, we have:
Wi =
w idv =
vi
2πρv2
18ε 0
a
π
vi
4
R sinθdRdθ =
r=0
a
ρv2 R 2 R 2 sinθdRdθdφ = ρv2
18ε 0
18ε 0
θ=0
18ε 0
r=0
a
2πρv2
R
4
−cosθ π0 dR
=
r=0
2π
π
θ=0
φ=0
a
4πρv2
18ε 0
R 4 sinθdRdθdφ =
2 5
2 5
R dR = 4πρv a = 2πρv a
90ε 0
45ε 0
J
4
r=0
The energy stored within the volume of the sphere is:
2 5
W in = 2πρv a
45ε 0
J
b. The energy stored outside the sphere is found in identical manner except that the integration is on the whole space
outside the sphere and we use the energy density outside the sphere. Thus:
Wo =
w o dv =
vo
2πρv2 a6
18ε 0
∞
π
r=a
θ=0
vo
ρv2 a6 R 2 sinθdRdθdφ = ρv2 a6
18ε 0
18ε 0 R 4
2πρv2 a6
sinθdRdθ =
18ε 0
R2
∞
∞
π
2π
r=a
θ=0
∞
φ=0
4πρv2 a6
1 −cosθ π0 dR =
18ε 0
R2
r=a
r=a
sinθdRdθdφ =
R2
2 6∞
2 5
1 dR = − 4πρv a = 4πρv a
2
18ε 0 R a
18ε 0
R
J
The energy stored outside the volume of the sphere is:
2 5
W out = 4π ρv a
18ε 0
J
c. The total energy associated with the charge distribution is:
2 5
2 5
2 5
W total = Win + Wout = 2πρv a + 4πρv a = 4πρv a
45ε 0
18ε 0
15ε 0
J
Note: This is the same energy we found in example 4.28.
Problem 4.53. Energy stored in a capacitor.
Assume the sphere has a charge Q distributed on its surface. Calculate the electric potential due to this charge at a point
in space, at a distance R from the center of the sphere, and outside the sphere. Now set R = a to obtain the potential on
the surface. The energy then is the product of charge and voltage. Using Gauss's law (create a spherical surface at a
radius R, enclosing the charge Q) the electric field intensity at a distance R is
4π R 2 E = Q
ε0
⇒
E=
Q
4πε 0 R 2
V
m
This electric field is the same as that of a point charge of magnitude Q. The electric potential is therefore,
V(R) =
At R = a:
V(a) =
and the energy is
Q
4πε 0 a
Q
4πε 0 R
⇒
V
Q = 4πε 0 aV
W = QV = 2πε 0 aV 2
2
C
J
Problem 4.54. Application: Energy stored in parallel plate capacitor.
Calculate the electric field intensity in each capacitor as a parallel plate capacitor. From the electric field, calculate the
energy density and total energy and subtract the energy of the capacitor in Figure 4.76b from that in Figure 4.76a.
a. The electric field intensities in the two capacitors are
109
E = V in (a)
d
E = V in (b)
d− a
and
V
m
Note that the conductor has the simple effect of reducing the distance between the plates or, in other words, of increasing
the capacitance. The energy density due to an electric field intensity is
2
w = ε0E
2
Energy densities in the two capacitors are
2
J
m3
2
wa = ε 0 V
2d 2
wb = ε 0 V
2(d − a)2
J
m3
To calculate the total energy stored in the capacitors, we multiply the above by a volume which is equal to v = 1×d in
case (a) and v = 1×(d − a) in case (b) (that is, a volume of unit area and height equal to d or (d − a)). This gives
2
2
Wb = ε 0 V (d −2a)
2(d − a)
W a = ε0V 2 d
2d
J
The change in potential energy is the difference between the two
2
2
2
1 − 1 = εV 2
a
∇W = Wb − Wa = ε V
− εV = εV
2(d − a)
2d
2 (d − a) d
2 d(d − a)
J
b. With the result in (a), we get by multiplying by the actual area of 0.1×0.1 m2 :
2
2
a
a
∇W = 10−2ε V
=ε V
2 d(d − a) 200 d(d − a)
J
c. The change in energy represents a change in capacitance ∇W = ΔCV 2 /2 J . This can be used to sense the conductor
either in a position sensor (position relative to the plates) or as a proximity sensor.
Problem 4.55. Energy stored in a spherical capacitor.
Assume a charge +Q on the inner shell and a charge −Q on the outer shell. Use Gauss' law to calculate the electric field
intensity E. From that calculate the potential V everywhere between the shells. Calculate the capacitance from Q and V.
2
Now use W = CV /2 to calculate the energy. Subtract the energy without the dielectric (final energy) from the energy
with the dielectric (initial energy) to obtain the change in energy.
Using Gauss' law on a sphere of radius a < r < b:
E4π R 2 = Q
ε
To calculate the electric potential:
⇒
R=a
Vab = −
R=b
E=
Q
4πε R 2
V
m
Qdr = Q 1 − 1
4π R 2 ε 4πε a b
V
Note that with the assumption of positive charge on the inner shell, the potential on the inner shell is positive.
Capacitance is calculated from C = Q/V or and since the potential difference is negative we write
C = Q = 4πε = 4πε ab
V
1−1
b−a
a b
and the energy stored in the capacitor is
2
W ε =C V = 4πε
2
2 1−1
a b
Q 1−1
4πε a b
2
F
2
= Q 1−1
8πε a b
J
If the dielectric is removed, the only change is that the permittivity is changed from ε to ε 0 . The energy is therefore
2
1− 1
W ε0 = Q
8πε 0 a b
J
110
The change in potential energy is
2
2
1 − 1 εr − 1
Δ W = Wε0 − Wε = Q 1 − 1 1 − 1 = Q
a
ε
ε
εr
0
8π
b
8πε 0 a b
J
where ε r is the relative permittivity of the dielectric. Note that this energy is positive. Removing the dielectric while
keeping the charge constant increases the potential energy of the system.
Problem 4.56. Energy density.
Calculate the electric field intensity everywhere using Gauss' law. From the electric field intensity calculate the energy
density and, integrating the energy density over the volume between the two shells, gives the energy stored.
a. For 0 < R < a: the electric field intensity is due to a point charge
E=R
q
4πε 0 R 2
V
m
For a < R < b, use a Gaussian surface that includes the inner shell. Gauss' law now reads:
q + 4π a2 ρ0
ε0
The electric field intensity is again in the R direction and is given as
2
E4π R =
2
E = R q + 4π2a ρ0
4π R ε 0
V
m
For R > b, we get using identical considerations (a Gaussian surface that includes both shells):
Energy density is given as
2
2
E = R q + 4π a ρ0 2− 4π b ρ0
4π R ε 0
w = εE
2
0 < R < a:
w=
a<R<b
w=
2
J
m3
q2
32π 2 ε 0 R 4
q + 4π a2 ρ0
2
32π ε 0 R
R > b:
J
m3
2
J
m3
4
2
w=
V
m
q + 4π a2 ρ0 − 4π b ρ0
2
J
m3
32π 2 ε 0 R 4
w [W/m3]
A plot of these relation is shown in Figure A.
q2
b. To integrate we note that the volume of a shell of
2
radius R and thickness dR has volume dv = 4πR dR.
The total energy stored between the two shells is:
R=a
W=
R=b
q + 4π a2 ρ0
8π ε 0 R 2
2
dR =
q + 4π a2 ρ0
8π ε 0
2
1− 1
a b
32π 2ε
0R
J
0
a
b
4
(q +4πa2 ρ0) 2
32π 2ε0 a 4
(q +4πa2 ρ0 −4πb2 ρ0) 2
32π 2ε0 a 4
R
Figure A
Problem 4.57. Work in the electrostatic field.
The total work needed to pull the dielectric out equals the energy of the capacitor after the dielectric has been pulled out
minus the energy of the capacitor before the dielectric has been pulled out. Thus we calculate the energy stored with and
without the dielectric. Assuming infinite plates (these are finite but we neglect fringing of the field at edges. This is
allowed if d << b):
111
E=V
d
ε E 2 dv
2
W=
⇒
v
With the dielectric, the total stored electric energy is
W
1
ε E 2 dv = ε V 2 b 2 d = ε V 2 b 2
2
2d
2d 2
=
v
J
since the energy density is uniform throughout the capacitor. Without the dielectric, ε changes to ε 0 and we get:
W
2
=
ε 0 E 2 dv = ε 0 V 2 b 2 d = ε 0 V 2 b 2
2d
v 2d
2d 2
J
The change in energy is the work required:
2 2
Δ W = W 2 − W 1 = ε 0 V b (1 − ε r)
2d
J
and since ε r > 1, the change in energy is negative. This is work performed by the system (reduces the potential energy of
the system).
Problem 4.58. Energy per unit length.
We calculate first the electric field intensity and then the energy per unit length from the electric field intensity.
From Gauss' law:
E.ds = ρlL
ε
s
E.ds = 2π rLE = ρlL
ε
S
→
⇒
E = ρl
2πε r
V
m
where L is an arbitrary length of the line of charge. The electric field intensity at a distance r from the line, inside the
dielectric layer is therefore:
To calculate energy we use:
W = 1 ε E 2 dv
where:
dv = 2 π rdr
2 v
(length of cylinder = 1 m). Thus, the energy per unit length is:
b
W = 1 ε E 2 dv = 1
2 v
2
ε
a
ρl 2 2π rdr = ρl2
2π rε
4πε
b
a
2
1 dr = ρl lnb
r
4πε a
J
Problem 4.59. Pressure in the electrostatic field.
Pressure is force per unit area. Calculate the force per unit area of the plates by calculating the energy stored between the
plates of area 1 m2 and then allowing a virtual displacement of one of the plates. The gradient of the change in work then
gives the force per unit area which equals the pressure.
The energy density in the space between the plates is
w = 1ε E2 = 1ε V
2
2 d
2
J
m3
This is an energy per unit volume. Now suppose, the plates are moved a distance dy apart. The change in energy due to
this movement is:
2
2
2
dW = wdv = 1 ε V (1×1)dy = 1 2ε 0 V dy = ε 0 V dy
2 d
2
d
d
J
where 1×1 signifies the unit area of the plates. The force per unit area (that is, pressure) is:
p =dW = ε 0 V
dy
d
The unit [N/m2] is designated as the pascal [Pa]
2
N
m2
112
Problem 4.60. Force in the electrostatic field.
Calculate the force per unit area of the plates from the potential difference by first calculating the electric field intensity
between the plates. Then find a relation between potential and P so that the plate balances.
The energy density in the space between the plates is
2
w = 1 ε0E 2 = 1 ε V
2
2 d
J
m3
This is an energy per unit volume. Now suppose, the plates are moved a distance dy apart. The change in energy
contained in the volume due to this movement is:
2
dW = wdv = 1 ε V (1×1)dy
2 d
J
where 1×1 is the total volume under a 1 m2 area of the upper plate. This is the change in energy per unit area of the
plates. Then, the force per unit area of the upper plate is:
2
f = dW = 1 ε V 2
dy 2 d
N
m2
The total force on the upper plate is equal to this force multiplied by the area of the plate and this must be equal to P.
2 2
P = f b 2 = 1 ε b V2
2 d
2 2
P = ε b V2
2d
→
V=d
b
→
2P
ε
V
Problem 4.61. Force in the electrostatic field.
Calculate the electric field intensity as for parallel plate capacitors. Calculate the energy density. Multiply by the volume
of the overlapping plates to get the total energy between the plates. Allow a differential movement of the plate to the left.
Calculate the change in energy due to the virtual displacement. This gives the force
The energy density between the two plates is:
E=V
d
V
m
2
w = ε 0 ε rE = ε 0 ε rV
2
2d 2
2
J
m3
Due to a displacement dx, say to the left, the change in energy is wbddx:
2
and the force is:
dw = ε 0 ε rV
bddx
2d 2
J
2
F = − dw = − ε 0 ε rV b
dx
2d
N
Problem 4.62. Force on dielectric in a capacitor.
We use the energy approach to calculate force (virtual displacement). First we calculate the change in energy due to the
displacement of the dielectric a small distance dx. The energies per unit volume in air and in the dielectric are:
ε E2
εE2
wa = 0 a ,
wd = d
[J]
2
2
where the index d indicates the dielectric and a indicates air. If the dielectric moves a distance dx, say to the left: the air space
decreases by a small volume dv=bddx whereas the dielectric volume increases by the same volume. The change in energy is:
⎛ εE2 ε E2 ⎞
dW = wd dv − wa dv = ⎜ d − 0 a ⎟ bddx
[J]
2 ⎠
⎝ 2
The force therefore is:
dW ⎛ ε Ed2 ε 0 Ea2 ⎞
=
−
bd
[N]
dx ⎜⎝ 2
2 ⎟⎠
Now we calculate the electric field intensity in air and in the dielectric:
113
Ea =
V0
,
d
Ed =
⎡V⎤
⎢⎣ m ⎥⎦
V0
d
The force therefore is:
⎛ εV 2 ε V 2 ⎞
V 2b
F = ⎜ 0 − 0 0 ⎟ b = 0 (ε − ε 0 )
[N]
2d ⎠
2d
⎝ 2d
This force is to the left. Note that when the dielectric is all the way in, there is only one dielectric and there can be no force.
On the other hand, this force is constant (independent of x).
Problem 4.63. Inflation of a balloon by electrostatic forces.
The charge is uniformly distributed on the surface of the balloon. Opposite charges generate a net force pointing out of the
balloon keeping the balloon inflated. To calculate the force and hence the pressure we use the idea of virtual displacement.
a. The electric field intensity at the surface of the balloon is (using a Gaussian surface of radius a):
$N'
4π a 2 ρ0
ρ
Q
E = R̂
=
R̂
= R̂ 0
& )
2
2
ε0
4πε 0 a
4πε 0 a
%C(
The energy density at the surface of the balloon is therefore
# J &
ε E2 ρ2
w= 0 = 0
% 3(
2
2ε 0
$m '
Now suppose the radius of the balloon increases by dR. The change in volume is:
"m 3 $
dv = 4π a 2 dR
# %
The change in energy is therefore:
dW = Wdv =
ρ02
2πρ02 a 2
4π a 2 dR =
dR
2ε 0
ε0
$%J &'
Hence, the force on the balloon is
2
F=
dW 2πρ0 a
=
dR
ε0
2
$%N &'
Dividing the force by the surface of the balloon we get the pressure:
P=
2 2
ρ02
F 2πρ0 a
=
=
S 4π a 2ε 0 2ε 0
$N'
& 2)
%m (
b. The charge needed is:
%C(
' 2*
&m )
This is a relatively large charge density but may be easier to generate than the approximately 4,200 cubic meters of gas
needed to inflate the balloon by conventional means. Air at normal pressure weighs 1.2 kg/m3 so a balloon of that size,
inflated with air would weigh about 5 ton. Of course, the weight is not an issue but the amount of air needed is.
Note as well that this configuration produces an electric field intensity of 4.75x106 V/m that is above breakdown in air but in
vacuum the breakdown voltage is much higher (see Table 4.2 and part (c) of this problem) so this is not an issue.
c. In vacuum, the maximum electric field intensity is 109 V/m. From (a), the electric field intensity and charge density are:
$C'
ρ
E = R̂ 0
→
ρ0 = ε 0 E
& 2)
ε0
%m (
This represents the maximum charge density for a maximum electric field intensity. Therefore, from (b):
ρ0 = 2ε 0 P = 2 × 8.854 ×10−12 ×100 = 4.2 ×10−5
2
ρ02 ε 0 $ ρ0 ' ε 0 E 2 8.854 ×10−12 ×1018
,-Pa ./
= & ) =
=
= 4.427 ×106
2ε 0 2 &% ε 0 )(
2
2
This is a significant pressure (over 44 atm.) and, of course, there is no theoretical reason the pressure cannot be increased (see
Table 4.2 for the electric field intensity in vacuum).
Of course, the charge needed to do so is large but this problem shows that the idea can be implemented.
ρ0 = 2ε 0 P
→
P=
114