# Jose G. Chiriboga Alcivar TCET 2220 Prof. Viviana Homework 1 09

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
πΈ=
π
200π
=
= 40000 π ⁄π
π 5 π₯ 10−3 π
(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
πππβ
πΈ=
π
60 π
=
= 6000 π ⁄π
π
1 π₯ 10−2 π
(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
π
∗ 0.04 m = 80 V
π
(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 (∈π = π. ππ).
∈=∈π ∗∈π = 2.25 ∗ 8.842 ∗ 10−12 = 19.89 ∗ 10−12 F/m
D =∈∗ E = 19.89 ∗ 10−12
F
π
∗ 2000 = 39.78 nC/π2
m
π
(6.8) For the parallel plates of Problem (6.4), determine the electric flux density
if the dielectric is air.
∈=∈π = 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 = π0 = 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 = π0 = 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 ππͺ/ππ . Determine the value of the electric flux across the area.
A = 8m ∗ 0.75m = 6 m2
ψ = A ∗ D = 6 m2 ∗ 4
ππΆ
= 24 μC
m2
Chiriboga 2
(6.12) The electric flux density normal to a circular surface with a diameter of
3m is 8 ππͺ/ππ . Determine the value of the electric flux across the area.
A = ππ 2 = (π) ∗ 1.52 = 7.06 m2
ψ = A ∗ D = 7.06 m2 ∗ 8
ππΆ
= 56.55 μC
π2
(6.13) The magnetic flux density normal to a circular surface with a radius of
5m is 4 ππΎπ/ππ . Determine the value of the magnetic flux across the area.
A = ππ 2 = (π) ∗ 52 = 78.539 m2
Ο = A ∗ B = 78.539 m2 ∗ 4
πππ
= 314.159 nWb
π2
(6.14) The magnetic flux density normal to a rectangular surface with
dimensions 30 cm x 60 cm is 12 ππΎπ/ππ . Determine the value of the magnetic
flux across the area.
A = 0.30m ∗ 0.60m = 0.18 m2
Ο = A ∗ B = 0.18 m2 ∗ 12
πππ
= 2.16 nWb
π2
(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 = ππ 2 = (π) ∗ (0.0152 ) = 706.85 ∗ 10−6 m2
J=
I
4A
kA
=
=
5.658
A 706.86 ∗ 10−6 m2
mπ
(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
π π πππ 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 π―π =
πππ ππ¨⁄π. 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)π¬π
π0 = 120π = 377Ω
πΈπ₯ = ππ»π¦ = 377 Ω ∗ 200
(b)π·π
Pz =
ππ΄
= 75.4 ππ/π
π
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 ∗ π΄ = 15.08
μW
∗ 1.963 ∗ 103 m2 = 29.61 ππ
m2
(6.20) The rms magnitude of the magnetic field of a plane wave in sea water
(∈π = ππ) is π¬π = π π½/π. 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)π―π
η=
π»π¦ =
(b)π·π
π0
377
=
= 42.15Ω
√∈π √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
π΄ = 15m ∗ 15m = 225 m
P = Pz ∗ π΄ = 126.46
W
∗ 225 m = 28.45 kW
m2
(6.21) In a lossless dielectric medium, the rms electric and magnetic field
intensities are π¬π = ππ ππ½/π and π―π = πππ ππ¨/π. Determine the following:
(a) Intrinsic impedance
η=
πΈπ₯
50 mV/m
=
= 500Ω
π»π¦ 100 μA/m
(b) Power density
Pz = Ex ∗ Hy = 50
mV
μA
∗ 100
= 5 μW/m2
m
m
(c) Dielectric constant
η=
∈r =
π0
377
=
√∈π √∈π
3772
3772
=
= 568.51 ∗ 10−3
η2
500Ω2
Chiriboga 5