Solved Problems for practice
1.
A point Q is located at (a, 0, 0), whereas another charge -Q is at (-a, 0, 0). Find E at (a)
(0, 0, 0), (b) (0, a, 0), (c) (a, 0, a).
2.
A cube is defined by 0 < x < a, 0 < y < a, and 0 < z < a. If it is charged with ρv=(ρox)/α,
where ρο is a constant, calculate the total charge in the cube.
3.
A volume charge with density ρv=5ρ2z mC/m3 exists in a region defined by 0 < ρ < 2, 0 <
z <1, 30 < φ <90. Calculate the total charge in the region.
Solved Problems for practice
4.
An annular disk of inner radius α and outer radius b is placed on the xy-plane and
centered at the origin. If the disk carries uniform charge with density ρs, find E at (0, 0,
h).
5.
Three surface charge distributions are located in free space as follows: 10 μC/m2 at x=2, 20μC/m2 at y=-3, and 30μC/m2 at z=5. Determine E at (a) P(5, -1, 4), (b) R(0, -2, 1), (c)
Q(3, -4, 10).
Solved Problems for practice
6.
If spherical surfaces a=1 m and b=2 m, respectively, carry uniform surface charge densities
8 nC/m2 and -6 nC/m2, find D at r=3 m
Gaussian Surface
a
b
r
7.
A sphere of radius αis centered at the origin. If
1/2
, 0<𝑟<𝑎
𝜌v = {5𝑟
,
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
𝑜
Determine E everywhere
Solved Problems for practice
8.
Let
10
𝜌v = {𝑟 2
𝑜
𝑚𝐶/𝑚3 , 1 < 𝑟 < 4
, 𝑟>4
(a) Find the net flux crossing surface r=2 m and r= 6 m.
(b) Determine D at r=1 m an r = 5 m.
Solved Problems for practice
9.
Three point charges 1mC, -2 mC, and 3 mC are, respectively, located at (0. 0. 4), (-2, 5,
1), and (3, -4, 6).
(a) Find the potential V at (-1, 1, 2)
(b) Calculate the potential difference VPQ if Q is (1, 2, 3)
Solved Problems for practice
10.
A circular disk of radius α carries charge ρs=1/ρ C/m2. Calculate the potential at (0, 0, h
11.
Let charge Q be uniformly distributed on a circular ring defined by a < ρ < b as shown
below, Find D at (0, 0, h).
Solved Problems for practice
Draw in 3D:
Solved Problems for practice
12.
In free space D = 20ax + 40ay -10az V/m. Calculate the work done in transferring a 2 mC
charge along the arc ρ = 2, 0 < φ < π/2 in the z =0 plane. Do multiple ways by noting that
the Electric Field is conservative.
Solved Problems for practice
13.
If V = 2x2 + 6y2 V in free space, find the energy stored in a volume defined by -1 ≤ x ≤ 1,
-1 ≤ y ≤ 1, and -1 ≤ z ≤ 1
Practice Exercises in EM waves
Problem 1
Problem 2
Non-magnetic means: μ=μ0.
From cos(2𝜋 × 109 𝑡 − 200𝑥), we can conclude that
𝜔 = 2𝜋 × 109  𝑓 = 109 Hz
EM-wave is traveling in the +x-direction (𝜶𝐤 =𝜶𝒙 ), because of the term −200𝑥 in the argument of the
cosine.
𝛽 = 200 𝑚−1
1.
2.
3.
The general expression for the E- and H-fields of an EM-wave traveling in the +x-direction are:
𝑬 = 𝐸0 𝑒 −𝑎𝑥 cos(2𝜋𝑓𝑡 − 𝛽𝑥) 𝜶𝚬
𝑯=
𝐸0 −𝑎𝑥
𝑒
cos(2𝜋𝑓𝑡 − 𝛽𝑥 − 𝜃𝜂 ) 𝜶𝚮
|𝜂|
Comparing the given expression of H with the general one we can conclude that:
α= 100 𝑚−1 (from the term 𝑒 −100𝑥 =𝑒 −𝑎𝑥 ) , we also conclude that the material is lossy
The direction of the H-field is 𝜶𝚮 = 𝜶𝐲 .
We also know that
𝑎 = 𝜔√
𝜇𝜀
𝜎 2
[√1 + ( ) − 1] ,
2
𝜔𝜀
𝛽 = 𝜔√
𝜇𝜀
𝜎 2
[√1 + ( ) + 1],
2
𝜔𝜀
and
𝜂 = 𝜔𝜀 √
𝜔𝜇
= |𝜂|𝑒 𝑗𝜃𝑛
𝜎 + 𝑗𝜔𝜀
where
|𝜂| =
√𝜇/𝜀
√1 + ( 𝜎 )
𝜔𝜀
4
2
, tan 2𝜃𝑛 =
𝜎
𝜔𝜀
Since we know α, β we can take their ratio and solve for the quantity σ/ωε
𝑎 100
=
=
𝛽 200
[√1 + (
𝜎 2
) − 1]
𝜔𝜀
𝜎 2
[√1 + ( ) + 1]
𝜔𝜀
√
(
𝜎 2 16
σ
4
) =

=
𝜔𝜀
9
ωε 3
Then:
tan 2𝜃𝑛 =
𝜎
4
= 𝜃𝑛 = 26.57
𝜔𝜀 3
and
|𝜂| =
𝜇
√
𝜀
2
√1 + ( 𝜎 )
𝜔𝜀
4
=
𝜇
√𝜀 𝜀0
0 𝑟
√1 + 16
9
=
𝜇
√𝜀 𝜀0
0 𝑟
4
√25
9
4
=
1
120𝜋√
𝜀𝑟
√5
3
We are missing the value of 𝜀𝑟 We must find 𝜀𝑟 to proceed. We can use either the expression for α or β:
𝑎 = 𝜔√
𝑎 =
𝜇𝜀
𝜎 2
𝜇0 𝜀0 𝜀𝑟
16
𝜔 𝜀𝑟 25
𝜔 𝜀𝑟 5
[√1 + ( ) − 1] 𝑎 = 𝜔√
[√ 1 +
− 1] 𝑎 = √ [√ − 1] 𝑎 = √ [ − 1]
2
𝜔𝜀
2
9
𝑐 2
9
𝑐 2 3
𝜔 𝜀𝑟 2
𝜔 𝜀𝑟 𝑎𝑐
𝜀𝑟
𝑎𝑐
100 × 3 × 108
√
𝑎 = √ 
= √ √𝜀𝑟 = √3 √𝜀𝑟 = √3
√𝜀𝑟 = 8.26
𝑐 23
𝑐 3
𝜔
3
𝜔
2𝜋109
Then
|𝜂| =
120𝜋
5
8.26 √
3
= 35.35 Ω
Hence, the complex impedance is
𝜂 = 35.35𝑒 𝑗26.57
To find direction of oscillation of the E-field. Since 𝑯 = 𝐻0 𝑒 −100𝑥 cos(2𝜋 × 109 𝑡 − 200𝑥) 𝜶𝑦 mA/m
and 𝜶𝑘 = 𝜶𝑥
𝜶𝛦 = 𝜶𝐻 × 𝜶𝑘 = 𝜶𝑦 × 𝜶𝑥 = −𝜶𝑧
We know that
𝐻0 =
𝛦0
𝛦0 = |𝜂|𝐻0
|𝜂|
Careful: We know that in the lossy material there is a phase difference between the E and H fields. We note
this by introducing a phase term of 𝜃𝑛 in the argument of the cosine of the H-field. However, our problem
gives us the H field and there is no such term in the cosine. Since the H field should follow the E field by 𝜃𝑛 ,
we should introduce a negative 𝜃𝑛 in the E-field to account for this phase difference. That is
Hence, 𝑬 = −|𝜂|𝐻0 𝑒 −100𝑥 cos(2𝜋 × 109 𝑡 − 200𝑥 + 𝜃𝑛 ) 𝜶𝑧 
𝑬 = −35.35 ∙ 50𝑒 −100𝑥 cos (2𝜋 × 109 𝑡 − 200𝑥 +
𝑬 = −1767.5 𝑒 −100𝑥 cos (2𝜋 × 109 𝑡 − 200𝑥 +
26.57
𝜋) 𝜶𝑧 
180
26.57
𝜋) 𝜶𝑧
180
mV/m
From the E-field we can conclude that the polarization of the wave is along the z-axis.
Alternative way – MUCH FASTER
We can also use
𝜂=
𝜔𝜇
𝜔𝜇
2𝜋109 4𝜋107
=
=
= 35.31𝑒 𝑗26.57°
𝑘
𝛽 − 𝑗𝛼
200 − 𝑗100
Since we have the |η| and the phase 𝜃𝑛 = 26.57, we can find the rest as above.
𝜶𝛦 = 𝜶𝐻 × 𝜶𝑘 = 𝜶𝑦 × 𝜶𝑥 = −𝜶𝑧 and 𝐸0 = |𝜂|𝐻0
𝑬 = −1767.5 𝑒 −100𝑥 cos (2𝜋 × 109 𝑡 − 200𝑥 +
Problem 3
26.57
𝜋) 𝜶𝑧
180
mV/m
Problem 4
Problem 5
Problem 1: External Reflection
An EM-wave is propagating in free space (n1=1) and is incident at an angle θi on a material with
index of refraction n2=1.5. Calculate for both TE and TM components of the wave the
1. Angle of transmission, 𝜃𝑡
2. Reflection and transmission coefficients
3. Reflectance and Transmittance
Snell’s Law to find 𝜃𝑡
𝑛1 sin 𝜃𝑖 = 𝑛2 sin 𝜃𝑡
TE
TM (H-field component)
𝑛1 cos 𝜃𝑖 − 𝑛2 cos 𝜃𝑡
𝑛2 cos 𝜃𝑖 − 𝑛1 cos 𝜃𝑡
𝑟𝑇𝐸 =
𝑟𝑇𝑀 =
𝑛1 cos 𝜃𝑖 + 𝑛2 cos 𝜃𝑡
𝑛2 cos 𝜃𝑖 + 𝑛1 cos 𝜃𝑡
𝑡𝑇𝐸 = 1 + 𝑟𝑇𝐸
θi ()
10
20
30
40
50
𝑡𝑇𝑀 = 1 + 𝑟𝑇𝑀
𝑅𝑇𝐸 = 𝑟𝑇𝐸 2 ,
𝑇𝑇𝐸 = 1 − 𝑅𝑇𝐸
θt ()
6.65
13.18
19.47
25.37
30.71
tTE
0.796
0.783
0.760
0.722
0.665
rTE
-0.204
-0.217
-0.240
-0.278
-0.335
𝑅𝑇𝑀 = 𝑟𝑇𝑀 2 ,
rTM
0.196
0.183
0.159
0.120
0.057
tTM
1.196
1.183
1.159
1.120
1.057
RTE
0.042
0.047
0.058
0.077
0.112
𝑇𝑇𝑀 = 1 − 𝑅𝑇𝑀
TTE
0.958
0.953
0.942
0.923
0.888
RTM
0.038
0.033
0.025
0.014
0.003
TTM
0.962
0.967
0.975
0.986
0.997
Problem 2: Internal Reflection
An EM-wave is propagating in free space (n1=1.5) and is incident at an angle θi on a material with
index of refraction n2=1.
Calculate for both TE and TM components of the wave the
1. Angle of transmission
2. Reflection and transmission coefficients
3. Reflectance and Transmittance
Use the same formulas as above. The only thing that changes is the value of n1 and n2
θi ()
0
20
30
40
50
θτ ()
0.00
30.87
48.59
74.62
rTE
0.200
0.243
0.325
0.625
tTE
1.200
1.243
1.325
1.625
rTM
tTM
-0.200
0.800
-0.156
0.844
-0.068
0.932
0.316
1.316
No transmission
RTE
0.040
0.059
0.106
0.391
TTE
0.960
0.941
0.894
0.609
RTM
0.040
0.024
0.005
0.100
TTM
0.960
0.976
0.995
0.900
1
1 + |𝑟| 1 + 3 4
𝑆𝑊𝑅 =
=
= =2
1 − |𝑟| 1 − 1 2
3
𝑆𝑊𝑅𝑑𝐵 = 20 log 2 ≈ 6 𝑑𝐵
6 dB