Problem 4.48 With reference to Fig. 4-19, find E1 if E2 = x̂3 − ŷ2 + ẑ2 (V/m),
ε1 = 2ε0 ,
ε2 = 18ε0 , and the boundary has a surface charge density
ρs = 3.54 × 10−11 (C/m2 ). What angle does E2 make with the z-axis?
Solution: We know that E1t = E2t for any 2 media. Hence, E1t = E2t = x̂3 − ŷ2.
Also, (D1 − D2 ) · n̂ = ρs (from Table 4.3). Hence, ε1 (E1 · n̂) − ε2 (E2 · n̂) = ρs , which
gives
E1z =
ρs + ε2 E2z 3.54 × 10−11 18(2)
3.54 × 10−11
=
+
=
+ 18 = 20 (V/m).
ε1
2ε0
2
2 × 8.85 × 10−12
Hence, E1 = x̂3 − ŷ2 + ẑ20 (V/m). Finding the angle E2 makes with the z-axis:
¶
µ
√
2
−1
√
= 61◦ .
2 = 9 + 4 + 4 cos θ ,
θ = cos
E2 · ẑ = |E2 | cos θ ,
17
Problem 4.56 Figure P4.56(a) depicts a capacitor consisting of two parallel,
conducting plates separated by a distance d. The space between the plates contains
two adjacent dielectrics, one with permittivity ε1 and surface area A1 and another
with ε2 and A2 . The objective of this problem is to show that the capacitance C of the
configuration shown in Fig. P4.56(a) is equivalent to two capacitances in parallel, as
illustrated in Fig. P4.56(b), with
C = C1 +C2
(19)
where
ε1 A1
d
ε2 A2
C2 =
d
C1 =
(20)
(21)
To this end, proceed as follows:
(a) Find the electric fields E1 and E2 in the two dielectric layers.
(b) Calculate the energy stored in each section and use the result to calculate C1
and C2 .
(c) Use the total energy stored in the capacitor to obtain an expression for C. Show
that (19) is indeed a valid result.
A1
d
ε1
A2
+
V
−
ε2
(a)
+
C1
C2
−
V
(b)
Figure P4.56: (a) Capacitor with parallel dielectric
section, and (b) equivalent circuit.
Solution:
ε2
ε1
E1
E2
d
+
V
-
(c)
Figure P4.56: (c) Electric field inside of capacitor.
(a) Within each dielectric section, E will point from the plate with positive voltage
to the plate with negative voltage, as shown in Fig. P4-56(c). From V = Ed,
E1 = E2 =
V
.
d
(b)
We1 =
1
1 V2
1
A1
ε1 E12 · V = ε1 2 · A1 d = ε1V 2 .
2
2 d
2
d
But, from Eq. (4.121),
We1 =
1
C1V 2 .
2
A1
A2
Hence C1 = ε1
. Similarly, C2 = ε2
.
d
d
(c) Total energy is
We = We1 +We2 =
Hence,
C=
1 V2
1
(ε1 A1 + ε2 A2 ) = CV 2 .
2 d
2
ε1 A1 ε2 A2
+
= C1 +C2 .
d
d
Problem 5.12 Two infinitely long, parallel wires are carrying 6-A currents in
opposite directions. Determine the magnetic flux density at point P in Fig. P5.12.
I1 = 6 A
I2 = 6 A
P
0.5 m
2m
Figure P5.12: Arrangement for Problem 5.12.
Solution:
B = φ̂φ
µ0 I1
µ0 I2
µ0
8µ0
(6 + 2) = φ̂φ
+ φ̂φ
= φ̂φ
2π (0.5)
2π (1.5)
π
π
(T).
Problem 5.14 Two parallel, circular loops carrying a current of 40 A each are
arranged as shown in Fig. P5.14. The first loop is situated in the x–y plane with its
center at the origin, and the second loop’s center is at z = 2 m. If the two loops have
the same radius a = 3 m, determine the magnetic field at:
(a) z = 0
(b) z = 1 m
(c) z = 2 m
z
a
z=2m
I
a
0
y
I
x
Figure P5.14: Parallel circular loops of Problem 5.14.
Solution: The magnetic field due to a circular loop is given by (5.34) for a loop in
the x–y plane carrying a current I in the +φ̂φ-direction. Considering that the bottom
loop in Fig. is in the x–y plane, but the current direction is along −φ̂φ,
H1 = −ẑ
Ia2
,
2(a2 + z2 )3/2
where z is the observation point along the z-axis. For the second loop, which is at a
height of 2 m, we can use the same expression but z should be replaced with (z − 2).
Hence,
Ia2
.
H2 = −ẑ 2
2[a + (z − 2)2 ]3/2
The total field is
H = H1 + H2 = −ẑ
Ia2
2
·
1
1
+
(a2 + z2 )3/2 [a2 + (z − 2)2 ]3/2
¸
A/m.
(a) At z = 0, and with a = 3 m and I = 40 A,
·
¸
1
40 × 9 1
+
H = −ẑ
= −ẑ 10.5 A/m.
2
33 (9 + 4)3/2
(b) At z = 1 m (midway between the loops):
¸
·
40 × 9
1
1
H = −ẑ
= −ẑ 11.38 A/m.
+
2
(9 + 1)3/2 (9 + 1)3/2
(c) At z = 2 m, H should be the same as at z = 0. Thus,
H = −ẑ 10.5 A/m.
Problem 4.49 An infinitely long conducting cylinder of radius a has a surface
charge density ρs . The cylinder is surrounded by a dielectric medium with εr = 4
and contains no free charges. The tangential component of the electric field in the
region r ≥ a is given by Et = −φ̂φ cos φ /r2 . Since a static conductor cannot have any
tangential field, this must be cancelled by an externally applied electric field. Find
the surface charge density on the conductor.
Solution: Let the conducting cylinder be medium 1 and the surrounding dielectric
medium be medium 2. In medium 2,
E2 = r̂Er − φ̂φ
1
cos φ ,
r2
with Er , the normal component of E2 , unknown. The surface charge density is related
to Er . To find Er , we invoke Gauss’s law in medium 2:
∇ · D2 = 0,
or
which leads to
1 ∂
1 ∂
(rEr ) +
r ∂r
r ∂φ
∂
∂
(rEr ) =
∂r
∂φ
µ
µ
¶
1
− 2 cos φ = 0,
r
1
cos φ
r2
¶
=−
1
sin φ .
r2
Integrating both sides with respect to r,
Z
∂
1
(rEr ) dr = − sin φ
dr
∂r
r2
1
rEr = sin φ ,
r
Z
or
Er =
1
sin φ .
r2
Hence,
E2 = r̂
1
sin φ .
r2
According to Eq. (4.93),
n̂2 · (D1 − D2 ) = ρs ,
where n̂2 is the normal to the boundary and points away from medium 1. Hence,
n̂2 = r̂. Also, D1 = 0 because the cylinder is a conductor. Consequently,
ρs = −r̂ · D2 |r=a
= −r̂ · ε2 E2 |r=a
·
¸¯
¯
1
= −r̂ · εr ε0 r̂ 2 sin φ ¯¯
r
r=a
4ε0
= − 2 sin φ (C/m2 ).
a
Problem 4.50 If E = R̂150 (V/m) at the surface of a 5-cm conducting sphere
centered at the origin, what is the total charge Q on the sphere’s surface?
Solution: From Table 4-3, n̂ · (D1 − D2 ) = ρs . E2 inside the sphere is zero, since we
assume it is a perfect conductor. Hence, for a sphere with surface area S = 4π a2 ,
Q
ρs
=
,
ε0 Sε0
3πε0
Q = ER Sε0 = (150)4π (0.05)2 ε0 =
2
D1R = ρs ,
E1R =
(C).
Problem 4.58 The capacitor shown in Fig. P4.58 consists of two parallel dielectric
layers. Use energy considerations to show that the equivalent capacitance of the
overall capacitor, C, is equal to the series combination of the capacitances of the
individual layers, C1 and C2 , namely
C=
where
C1 = ε1
C1C2
C1 +C2
A
,
d1
C2 = ε2
(22)
A
d2
(a) Let V1 and V2 be the electric potentials across the upper and lower dielectrics,
respectively. What are the corresponding electric fields E1 and E2 ? By
applying the appropriate boundary condition at the interface between the two
dielectrics, obtain explicit expressions for E1 and E2 in terms of ε1 , ε2 , V , and
the indicated dimensions of the capacitor.
(b) Calculate the energy stored in each of the dielectric layers and then use the sum
to obtain an expression for C.
(c) Show that C is given by Eq. (22).
d1
A
d2
+
−
ε1
V
ε2
(a)
C1
+
V
−
C2
(b)
Figure P4.58: (a) Capacitor with parallel dielectric
layers, and (b) equivalent circuit (Problem 4.58).
Solution:
+
d1
ε1
d2
ε1
E1
+
V1
-
+
E2 V1
-
V
Figure P4.58: (c) Electric fields inside of capacitor.
(a) If V1 is the voltage across the top layer and V2 across the bottom layer, then
V = V1 +V2 ,
and
E1 =
V1
,
d1
E2 =
V2
.
d2
According to boundary conditions, the normal component of D is continuous across
the boundary (in the absence of surface charge). This means that at the interface
between the two dielectric layers,
D1n = D2n
or
ε1 E1 = ε2 E2 .
Hence,
V = E1 d1 + E2 d2 = E1 d1 +
which can be solved for E1 :
E1 =
V
ε1 .
d1 + d2
ε2
Similarly,
E2 =
V
ε2 .
d2 + d1
ε1
ε1 E1
d2 ,
ε2
(b)
2

·
¸
1 2
1
ε1 ε22 Ad1
1 
V

2
We1 = ε1 E1 · V 1 = ε1 
ε1  · Ad1 = 2 V (ε2 d1 + ε1 d2 )2 ,
2
2
d1 + d2
ε2
2

¸
·
1 2
1
ε12 ε2 Ad2
1 
V

2
We2 = ε2 E2 · V 2 = ε2 
ε2  · Ad2 = 2 V (ε1 d2 + ε2 d1 )2 ,
2
2
d2 + d1
ε
· 2 1 2
¸
1 2 ε1 ε2 Ad1 + ε1 ε2 Ad2
We = We1 +We2 = V
.
2
(ε1 d2 + ε2 d1 )2
But We = 12 CV 2 , hence,
C=
ε1 ε22 Ad1 + ε12 ε2 Ad2
ε1 ε2 A
(ε2 d1 + ε1 d2 )
= ε1 ε2 A
=
.
2
2
(ε2 d1 + ε1 d2 )
(ε2 d1 + ε1 d2 )
ε2 d1 + ε1 d2
(c) Multiplying numerator and denominator of the expression for C by A/d1 d2 , we
have
ε1 A ε2 A
·
C1C2
d
d2
=
,
C= 1
ε1 A ε2 A C1 +C2
+
d1
d2
where
C1 =
ε1 A
,
d1
C2 =
ε2 A
.
d2
Problem 5.9 The loop shown in Fig. P5.9 consists of radial lines and segments of
circles whose centers are at point P. Determine the magnetic field H at P in terms of
a, b, θ , and I.
I
b
θ
a
P
Figure P5.9: Configuration of Problem 5.9.
Solution: From the solution to Example 5-3, if we denote the z-axis as passing out
of the page through point P, the magnetic field pointing out of the page at P due to
the current flowing in the outer arc is Houter = −ẑI θ /4π b and the field pointing out
of the page at P due to the current flowing in the inner arc is Hinner = ẑI θ /4π a. The
other wire segments do not contribute to the magnetic field at P. Therefore, the total
field flowing directly out of the page at P is
µ
¶
Iθ 1 1
I θ (b − a)
H = Houter + Hinner = ẑ
−
.
= ẑ
4π a b
4π ab
Problem 5.10 An infinitely long, thin conducting sheet defined over the space
0 ≤ x ≤ w and −∞ ≤ y ≤ ∞ is carrying a current with a uniform surface current density
Js = ŷ5 (A/m). Obtain an expression for the magnetic field at point P = (0, 0, z) in
Cartesian coordinates.
Solution:
z
|R| =
x2 + z2
P = (0, 0, z)
R
x
x
0
w
Figure P5.10: Conducting sheet of width w in x–y plane.
The sheet can be considered to be a large number of infinitely long but narrow wires
each dx wide lying next to each other, with each carrying a current Ix = Js dx. The
wire at a distance x from the origin is at a distance vector R from point P, with
R = −x̂x + ẑz.
Equation (5.30) provides an expression for the magnetic field due to an infinitely long
wire carrying a current I as
B
φ̂φI
H=
=
.
µ0 2π r
We now need to adapt this expression to the present situation by replacing I with
Ix = Js dx, replacing r with R = (x2 + z2 )1/2 , as shown in Fig. P5.10, and by assigning
the proper direction for the magnetic field. From the Biot–Savart law, the direction
× R, where l is the direction of current flow. In the present case,
of H is governed by l×
l is in the ŷ direction. Hence, the direction of the field is
× (−x̂x + ẑz)
×R
ŷ×
x̂z + ẑx
l×
=
= 2 2 1/2 .
× R| |ŷ×
× (−x̂x + ẑz)| (x + z )
|l×
Therefore, the field dH due to the current Ix is
dH =
x̂z + ẑx
(x2 + z2 )1/2
Ix
(x̂z + ẑx)Js dx
=
,
2π R
2π (x2 + z2 )
and the total field is
H(0, 0, z) =
=
=
=
=
=
Z w
Js dx
2π (x2 + z2 )
x=0
Z w
dx
Js
(x̂z + ẑx) 2 2
2π x=0
x +z
µ Zw
¶
Z w
Js
dx
x dx
x̂z
+ ẑ
2
2
2
2
2π
x=0 x + z
x=0 x + z
µ ¶¶¯w
µ µ
¶
¯
¢¯w
¡1
Js
1 −1 x
2
2
¯ + ẑ ln(x + z ) ¯
tan
x̂z
2
¯
x=0
2π
z
z
x=0
·
¸
µ ¶
5
2
2
2
−1 w
1
x̂2π tan
for z 6= 0,
+ ẑ 2 (ln(w + z ) − ln(0 + z ))
2π
z
·
µ 2 2 ¶¸
µ ¶
w +z
5
−1 w
1
x̂2π tan
(A/m) for z 6= 0.
+ ẑ 2 ln
2π
z
z2
(x̂z + ẑx)
An alternative approach is to employ Eq. (5.24a) directly.
Problem 5.11 An infinitely long wire carrying a 25-A current in the positive
x-direction is placed along the x-axis in the vicinity of a 20-turn circular loop located
in the x–y plane (Fig. P5.11). If the magnetic field at the center of the loop is zero,
what is the direction and magnitude of the current flowing in the loop?
1m
d=2m
x
I1
Figure P5.11: Circular loop next to a linear current
(Problem 5.11).
Solution: From Eq. (5.30), the magnetic flux density at the center of the loop due to
I2
Figure P5.11: (b) Direction of I2 .
the wire is
µ0
I1
2π d
where ẑ is out of the page. Since the net field is zero at the center of the loop, I2 must
be clockwise, as seen from above, in order to oppose I1 . The field due to I2 is, from
Eq. (5.35),
µ0 NI2
B = µ0 H = −ẑ
.
2a
Equating the magnitudes of the two fields, we obtain the result
B1 = ẑ
I1
NI2
=
,
2a
2π d
or
I2 =
2aI1
1 × 25
=
= 0.2 A.
2π Nd π × 20 × 2
Problem 5.13 A long, East-West–oriented power cable carrying an unknown
current I is at a height of 8 m above the Earth’s surface. If the magnetic flux density
recorded by a magnetic-field meter placed at the surface is 15 µ T when the current is
flowing through the cable and 20 µ T when the current is zero, what is the magnitude
of I?
Solution: The power cable is producing a magnetic flux density that opposes Earth’s,
own magnetic field. An East–West cable would produce a field whose direction at
the surface is along North–South. The flux density due to the cable is
B = (20 − 15) µ T = 5µ T.
As a magnet, the Earth’s field lines are directed from the South Pole to the North
Pole inside the Earth and the opposite on the surface. Thus the lines at the surface are
from North to South, which means that the field created by the cable is from South
to North. Hence, by the right-hand rule, the current direction is toward the East. Its
magnitude is obtained from
5 µ T = 5 × 10−6 =
which gives I = 200 A.
4π × 10−7 I
µ0 I
=
,
2π d
2π × 8