Jose G. Chiriboga Alcivar
TCET 2220
Prof. Viviana
Homework 1
09-15-13
(6.1) The voltage between two parallel plates separated by a distance of 5 mm
is 200 V. Determine the electric field intensity
𝐸=
𝑉
200𝑉
=
= 40000 𝑉 ⁄𝑚
𝑑 5 𝑥 10−3 𝑚
(6.2) The Voltage between two parallel plates separated by a distance of 0.4
inch is 60 V. Determine the electric field intensity.
x=
0.4 inch x 2.5cm
= 1 cm = 0.01 m
𝑖𝑛𝑐ℎ
𝐸=
𝑉
60 𝑉
=
= 6000 𝑉 ⁄𝑚
𝑑
1 𝑥 10−2 𝑚
(6.3) The electric field intensity in the region between two parallel plates
separated by a distance of 4 cm is 2 kV/m. Determine the voltage between the
plates.
4 cm = 0.04m
V = E x d = 2000
𝑉
∗ 0.04 m = 80 V
𝑚
(6.4) The electric field intensity in the region between two parallel plates
separated by a distance of 8 mm is 200 V/mm. Determine the voltage between
the plates.
V
V = E x d = 200
∗ 8 mm = 1600V
mm
(6.5) A direct current of 5 A is flowing in a conductor. Determine the magnetic
field intensity at a distance of 3 m from the conductor.
H=
I
5A
=
= 265.26 mA/m
2πd 2 ∗ π ∗ 3 m
(6.6) A direct current of 40 mA is flowing in a conductor. Determine the
magnetic field intensity at a distance of 5 ft. from the conductor.
5ft = 60 inches
Chiriboga 1
x=
60 inch x 2.5cm
= 150 cm = 1.5 m
inch
H=
I
40 mA
=
= 4.244 mA/m
2πd 2 ∗ π ∗ 1.5 m
(6.7) For the parallel plates of Problem (6.3), determine the electric flux density
if the dielectric is polyethylene (∈𝒓 = 𝟐. 𝟐𝟓).
∈=∈𝑟 ∗∈𝟎 = 2.25 ∗ 8.842 ∗ 10−12 = 19.89 ∗ 10−12 F/m
D =∈∗ E = 19.89 ∗ 10−12
F
𝑉
∗ 2000 = 39.78 nC/𝑚2
m
𝑚
(6.8) For the parallel plates of Problem (6.4), determine the electric flux density
if the dielectric is air.
∈=∈𝟎 = 8.842 ∗ 10−12 F/m
D =∈∗ E = 8.842 ∗ 10−12
F
V
∗ 200
= 1.768 nC/mm2
m
mm
(6.9) For the current-carrying conductor of Problem (6.5), determine the
magnetic flux density at a distance of 3 m from the conductor if the medium is
air.
μ = 𝜇0 = 4π ∗ 10−7 = 1.2566 ∗ 10−6 H/m
B = μ ∗ H = 1.2566 ∗ 10−6
H
mA
∗ 265.26
= 333.32 nWb/m2
m
m
(6.10) For the current-carrying conductor of Problem (6.6), determine the
magnetic flux density at a distance of 5 ft from the conductor if the medium is
air.
μ = 𝜇0 = 4π ∗ 10−7 = 1.2566 ∗ 10−6 H/m
B = μ ∗ H = 1.2566 ∗ 10−6
H
mA
∗ 4.244
= 5.333 nWb/m2
m
m
(6.11) The electric flux density normal to a rectangular surface with dimensions
8 m x 75 cm is 4 𝝁𝑪/𝒎𝟐 . Determine the value of the electric flux across the area.
A = 8m ∗ 0.75m = 6 m2
ψ = A ∗ D = 6 m2 ∗ 4
𝜇𝐶
= 24 μC
m2
Chiriboga 2
(6.12) The electric flux density normal to a circular surface with a diameter of
3m is 8 𝝁𝑪/𝒎𝟐 . Determine the value of the electric flux across the area.
A = π𝑟 2 = (π) ∗ 1.52 = 7.06 m2
ψ = A ∗ D = 7.06 m2 ∗ 8
𝜇𝐶
= 56.55 μC
𝑚2
(6.13) The magnetic flux density normal to a circular surface with a radius of
5m is 4 𝒏𝑾𝒃/𝒎𝟐 . Determine the value of the magnetic flux across the area.
A = π𝑟 2 = (π) ∗ 52 = 78.539 m2
ϕ = A ∗ B = 78.539 m2 ∗ 4
𝑛𝑊𝑏
= 314.159 nWb
𝑚2
(6.14) The magnetic flux density normal to a rectangular surface with
dimensions 30 cm x 60 cm is 12 𝒏𝑾𝒃/𝒎𝟐 . Determine the value of the magnetic
flux across the area.
A = 0.30m ∗ 0.60m = 0.18 m2
ϕ = A ∗ B = 0.18 m2 ∗ 12
𝑛𝑊𝑏
= 2.16 nWb
𝑚2
(6.15) A current of 8 A is uniformly distributed over a rectangular conductor
with dimensions 5 mm x 4 mm. Determine the current density.
A = 0.005m ∗ 0.004m = 0.00002 m2
J=
I
8A
A
=
= 400000 2
2
A 0.00002 m
m
(6.16) A current of 4 A is uniformly distributed over a circular conductor with a
dimensions 3 cm. Determine the current density.
A = π𝑟 2 = (π) ∗ (0.0152 ) = 706.85 ∗ 10−6 m2
J=
I
4A
kA
=
=
5.658
A 706.86 ∗ 10−6 m2
m𝟐
(6.17) Assume that the conductivity for the conductor of Problem (6.15) is 5
MS/m. Determine the electric field intensity.
E=
J 400000 A⁄m2
=
= 80 mV⁄m
σ
5 MS/m
Chiriboga 3
(6.18) Assume that the conductivity for the conductor of Problem (6.16) is
𝟔 𝒙 𝟏𝟎𝟕 S/m. Determine the electric field intensity.
J 5.658 kA⁄m2
E= =
= 94.3 μ V⁄m
σ
6 ∗ 107 S/m
(6.19) The rms magnitude of the magnetic field of a plane wave in air is 𝑯𝒚 =
𝟐𝟎𝟎 𝝁𝑨⁄𝒎. Assuming that E is in the positive x-direction, determine the
following for a circular surface of diameter 50 m in the x-y plane over which
the fields are constant:
(a)𝑬𝒙
𝜂0 = 120𝜋 = 377Ω
𝐸𝑥 = 𝜂𝐻𝑦 = 377 Ω ∗ 200
(b)𝑷𝒛
Pz =
𝜇𝐴
= 75.4 𝑚𝑉/𝑚
𝑚
Ex 2 75.4 2 mV/m
=
= 15.08 μW/m2
n0
377
(c) Total power transmitted through area
A = π ∗ r 2 = π ∗ 252 = 1.963 ∗ 103 m2
P = Pz ∗ 𝐴 = 15.08
μW
∗ 1.963 ∗ 103 m2 = 29.61 𝑚𝑊
m2
(6.20) The rms magnitude of the magnetic field of a plane wave in sea water
(∈𝒓 = 𝟖𝟎) is 𝑬𝒙 = 𝟑 𝑽/𝒎. Assuming that H is in the positive y-direction,
determine the following for a square surface with sides of 15 m each in the x-y
plane over which the fields are constant:
(a)𝑯𝒚
η=
𝐻𝑦 =
(b)𝑷𝒛
𝜂0
377
=
= 42.15Ω
√∈𝑟 √80
Ex
3 V/m
=
= 71.17 mA/m
η0 42.15Ω
Ex 2
32 V/m
Pz =
=
= 126.46 W/m2
n0
71.17 mA/m
(c) Total power transmitted through area
Chiriboga 4
𝐴 = 15m ∗ 15m = 225 m
P = Pz ∗ 𝐴 = 126.46
W
∗ 225 m = 28.45 kW
m2
(6.21) In a lossless dielectric medium, the rms electric and magnetic field
intensities are 𝑬𝒙 = 𝟓𝟎 𝒎𝑽/𝒎 and 𝑯𝒚 = 𝟏𝟎𝟎 𝝁𝑨/𝒎. Determine the following:
(a) Intrinsic impedance
η=
𝐸𝑥
50 mV/m
=
= 500Ω
𝐻𝑦 100 μA/m
(b) Power density
Pz = Ex ∗ Hy = 50
mV
μA
∗ 100
= 5 μW/m2
m
m
(c) Dielectric constant
η=
∈r =
𝜂0
377
=
√∈𝑟 √∈𝑟
3772
3772
=
= 568.51 ∗ 10−3
η2
500Ω2
Chiriboga 5