chapter 2
DESCRIPTION
ELECTROMAGNETIC FIELDS THEORY. CHAPTER 2. Electrostatic Fields: Electric Field Intensity. In this chapter you will learn. Electrostatic Fields -Charge, charge density -Coulomb's law - Electric field intensity - Electric flux and electric flux density -Gauss's law - PowerPoint PPT PresentationTRANSCRIPT
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Electrostatic Fields:Electric Field Intensity
ELECTROMAGNETIC FIELDS THEORY
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Electrostatic Fields -Charge, charge density-Coulomb's law-Electric field intensity- Electric flux and electric flux density-Gauss's law-Divergence and Divergence Theorem-Energy exchange, Potential difference, gradient-Ohm's law-Conductor, resistance, dielectric and capacitance- Uniqueness theorem, solution of Laplace and Poisson equation
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 2
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Two like charges repel one another, whereas two charges of opposite polarity attract
The force acts along the line joining the charges
Its strength is proportional to the product of the magnitudes of the two charges and inversely propotional to the square distance between them
NrR tt
121
ˆ0
t1t 4
QQ F
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 3
Charles Agustin de Coulomb
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 4
Q1
Qt
F1t
F1t
F1t
This proof the existence of electric force field which radiated from Q1
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Consider a charge fixed in a position, Q1.
Let’s say we have another charge, say Qt which is a test charge.
When Qt is moved slowly around Q1, there exist everywhere a force on this second charge.
Thus proof the existence of electric force field.
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 5
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Electric field intensity, E is the vector force per unit charge when placed in the electric field
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 6
Qt1
Qt2
Qt3
Q1
Experience E due to Q1
Experience the most E due to Q1
Experience less E due to Q1
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According to Coulomb’s law, force on Qt is
We can also write
Which is force per unit charge
NrR tt
1210
t1t ˆ
4QQ F
NrR tt
1210
1
t
t ˆ4
Q Q
F
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 7
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If we write E as
Thus we can rewrite the Coulomb’s law as
Which gives the electric field intensity for Qt at r due to a point charge Q located at r
t
t
Q F
E
3
1
1
0
1
4 rrrrE
t
t
πεQ
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 8
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For N points charges, the electric field intensity for at a point r is obtained by
Thus
N
k k
k
rrrr
13
0 ||)(
NQ4
1E
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 9
30
32
2
0
23
1
1
0
1
||4...
||4||4 N
nN
rrrr
πεQ
rrrr
πεQ
rrrr
πεQ
E
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 10
The behavior of the fields can be visualized using field lines:
Field vectors plotted within a regular grid in 2D space surrounding a point charge.
Field Lines
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 11
Some of these field vectors can easily be joined by field lines that emanate from the positive point charge.
The direction of the arrow indicates the direction of electric fields
The magnitude is given by density of the lines
Field Lines
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 12
The field lines terminated at
a negative point charge
The field lines for a pair of
opposite charges
Field Lines
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 13
RQ a
RE 2
04
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 14
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q q
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 15
q
q
mCddqq
L /lim0
Imagine charges distributed along a line
q
We can find the charge density over the line as:
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We can find the total charge by applying this equation:
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 16
L LL dlQdldQ
Charge density Length of the line
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 17
S Ss dSQdSdQ
2
0/lim mC
dsdq
sq
Ss
Similarly, for surface charge, we can imagine charges distributed over a surface
q
We can find the charge density over the surface as:
qq qqq
q q
q
And the total charge is
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 18
v vv dvQdvdQ
q qq
q qq
3
0/lim mC
dvdq
vq
vv
We can find the volume charge density as:
And the total charge is
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for line charge distribution:
for surface charge distribution:
for volume charge distribution
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 19
S Ss dSQdSdQ
L LL dlQdldQ
v vv dvQdvdQ
++ + + +ρL
+ ++ +
ρS
+ + ++++
+
+
ρv
++++
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RS
RπεdSρ a E 204
RL
Rπεdlρ a E 204
RRdv a
4 E
0
v 2
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 20
Surface charge
Volume charge
Line charge++ + + +ρL
+ ++ +
ρS
+ + ++++
+
+
ρv
++++
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 21
E field due to line charge
E field due to surface charge
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FIELDS THEORY 22
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 23
q qq
q qq
q
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 24
Infinite Length of Line Charge:
To derive the electric field intensity at any point in
space resulting from an infinite length line of
charge placed conveniently along the z-axis
LINE CHARGE
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 25
Place an amount of charge in coulombs along the z axis.
The linear charge density is coulombs of charge per meter length,
Choose an arbitrary point P where we want to find the electric field intensity.
mCL
zP ,,
LINE CHARGE (Cont’d)
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 26
LINE CHARGE (Cont’d)
The electric field intensity is:
zzEEE aaaE But, the field is only vary with the radial distance from the line.
There is no segment of charge dQ anywhere on the z-axis that will give us . So,E
zzEE aaE
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 27
LINE CHARGE (Cont’d)Consider a dQ segment a distance z above radial axis, which will add the field components for the second charge element dQ.
The components cancel each other (by symmetry) , and the adds, will give:
zE
E
aE E
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 28
LINE CHARGE (Cont’d)
Recall for point charge,
RQ a
RE 2
04
For continuous charge distribution, the summation of vector field for each charges becomes an integral,
RdQ a
RE 2
04
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 29
LINE CHARGE (Cont’d)
The differential charge,
dzdldQ
L
L
The vector from source to test point P,
z
R
zR
aa aR
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 30
LINE CHARGE (Cont’d)
Which has magnitude, and a unit vector, 22 z R
22 z
z zR
aaa
So, the equation for integral of continuous charge distribution becomes:
2222204 z
z
z
dz zL
aaE
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 31
LINE CHARGE (Cont’d)
Since there is no component,
a
aE
23
220
23
220
4
4
z
dz
z
dz
L
L
za
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 32
IMPORTANT!!
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 33
IMPORTANT!!
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 34
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 35
Hence, the electric field intensity at any point ρ away from an infinite length is:
aE
02L
For any finite length, use the limits on the integral.
LINE CHARGE (Cont’d)
ρ is the perpendicular distance from the line to the point of interest
aρ is a unit vector along the distance directed from the line charge to the field point
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 36
aE
02L
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A uniform line charge of 2 µC/m is located on the z axis. Find E in cartesian coordinates at P(1, 2, 3) if the charge extends from
a) −∞ < z < ∞: b) −4 ≤ z ≤ 4:
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 37
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 38
dEz
4
-4
dl
r
ρdEdEρ
z
x
y
P(1, 2, 3)
z
r’
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With the infinite line, we know that the field will have only a radial component in cylindrical coordinates (or x and y components in cartesian).
The field from an infinite line on the z axis is generally
Therefore, at point P:
aE
02L
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 39
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Here we use the general equation :
Where the total charge Q is
RdQ a
RE 2
04
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 40
dzdldQ LL
B
A
z
z LdzQ
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Or, we can also write E as:
Whereand
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 41
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So E would be:
Using integral tables, we obtain:
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 42
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FIELDS THEORY 43
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 44
dl
R ρ
(0, y’, 0) (0, y, 0)
z
x
y
P(x, y, z)
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 45
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A uniform line charge of 16 nC/m is located along the line defined by y = −2, z = 5. If ε= ε0. Find E at P(1, 2, 3)
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 46
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1. Draw the line y = −2, z = 5.2. Find aρ 3. Find E
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FIELDS THEORY 47
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 48
ρ
(1, -2, 5)
z
x
y
P(1, 2, 3)
-2
5
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2
),,(),,(
P
P
linelineP zyxzyx
RRa
R
zy
P
P
P
P
aaRRa
R
R
2420
)2,4,0(
20)2(4
)2,4,0()5,2,1()3,2,1(
2
2222
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 49
To find aρ , we must draw a “projection” line from point P to our line. Then find out what is the x-axis value at that projection point
Then aρ can be calculated as:
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aE
02L
mVzy
zy
/8.285.5720
242
1016
0
9
aa
aaE
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 50
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 51
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FIELDS THEORY 55
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 56
EXAMPLE
Use Coulomb’s Law to find
electric field intensity at
(0,0,h) for the ring of
charge, of charge density,
centered at the origin in
the x-y plane.
L
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Step 1: derive what are dl and R, R2 and aR
Step 2: Replace dl, R2 and aR into line charge equation:R
L
Rπεdlρ a E 204
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FIELDS THEORY 57
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 58
By inspection, the ring
charges delivers only
and contribution to
the field.
component will be
cancelled by symmetry.
zdEEd
Ed
Solution
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 59
RdQ a
RE 2
04
Each term need to be determined:
The differential charge,
addldQ
L
L
Solution
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 60
The vector from source to test point,
z
R
haR
aa aR
Which has magnitude, and a unit vector, 22 ha R
22 ha
ha zR
aaa
Solution
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 61
The integral of continuous charge distribution becomes:
2222204 ha
ha
ha
ad zL aaE
zL h
ha
ad aE2
322
04
Solution
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 62
Rearranging,
2
023
2204
zL dha
ah aE
Easily solved,
zL
ha
ah aE2
322
02
Solution
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 63
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FIELDS THEORY 64
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Static charges resides on conductor surfaces and not in their interior, thus the charges on
the surface are known as surface charge density or ρs
Consider a sheet of charge in the xy plane shown below
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FIELDS THEORY 65
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 66
Φ
z’
y
x
z
R
ρ
P(0,0,z’)
1
2
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The field does not vary with x and y Only Ez is present The charge associated with an elemental
area dS is
dddQ s
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 67
dSdQ S
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The total charge is
From figure,
RRdQd aE 2
04
dSQ S
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 68
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From figure ,
,
||,)(2/122
RRa
hRRhaaR
R
z
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 69
h
R
-ρ
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Replace in
Thus becomes
RRdSd a
4 E
0
S2
2/3220
'
][4][
hhdd
d zS
aaE
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 70
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 71
Φ
z’
y
x
z
R
ρ
P(0,0,z’)
1
2
For an infinite surface, due to symmetry of charge distribution, for every element 1, there is a corresponding element 2 which cancels element 1.
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Thus the total of Eρ equals to zero. E only has z-component.
For infinite surface, 0 < ρ < ∞
zzdE aE
zS
zddh aE 2/322
0
2
00 ]'[4
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 72
Infinite Surface charge
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Thus in general, for an infinite sheet of charge,
Where an is a vector normal to the sheet.E is normal to the sheet and independent of
the distance between the sheet and the point of observation, P .
nS aE02
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 74
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In parallel plate capacitor, the electric field existing between the two plates having equal and opposite charges is given by
nS
nS
nS aaaE
000
)(22
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 75
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 80
EXAMPLE
An infinite extent sheet of charge
exists at the plane y = -2m. Find the electric field
intensity at point P (0, 2m, 1m).
210mnC
S
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 81
SOLUTIONStep 1: Sketch the figure
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 82
Step 2: find an.
The unit vector directed away from the sheet and toward the point P is
Thus
ya
yS
nS aaE
00 22
SOLUTION
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 83
Step 3: solve the problem!
mV
y
y
yS
a
a
aE
565
10854.821010
2
12
9
0
SOLUTION
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Given the surface charge density, ρs = 2µC/m2, in the region ρ < 0.2 m, z = 0, and is zero elsewhere, find E at PA(ρ = 0, z = 0.5):
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 84
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Sketch the figure:
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 85
2
2
(0,0,0.5)
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First, we recognize from symmetry that only a z component of E will be present. Considering a general point z on the z axis, we have r = zaz. Then, with r = ρaρ, we obtain r − r = zaz − ρaρ. The superposition integral for the z component of E will be:
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 86
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With z = 0.5 m,
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 87
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Planes x = 2, and y = -3, respectively, carry charges 10nC/2 and 15nC/m2. if the line x = 0, z = 2 carries charge 10π nC/m, calculate E at (1, 1, -1) due to three charge distributions.
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 88
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 89
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Consider the volume charge distribution with uniform charge density, in the figure below
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 90
Φ’
θ’
y
x
Rρv
P(0,0,z’)=P(z’,0,0)
r’
dv at (r,θ’,Φ’)
Z’
α
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The charge dQ is
The total charge in a sphere with radius a is
34 3a
dv
dvQ
v
v
v
dvdQ v
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 91
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The electric field dE at point P(0,0,z) due to the elementary volume charge is
Where
Rv
Rdvd aE 204
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 92
aaa sincos zR
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For a point P2 at (r,θ,Φ) E is
Which is identical to the electric field at the same point due to a point charge Q at the
origin .
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 99
rrQ aE 2
04
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 100
EXAMPLE 6
Find the total charge over
the volume with volume
charge density,
3105
5m
Ce zV
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 101
SOLUTION
V
VdVQ The total charge, with volume:
dzdddV
Thus,
C
dzdde
dVQ
z
z
VV
14
01.0
0
2
0
04.0
02.0
10
10854.7
55
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 102
To find the electric field intensity resulting from a volume charge, we use:
RV
RdVdQ aR
aR
E 20
20 44
Since the vector R and will vary over the volume, this triple integral can be difficult. It can be much simpler to determine E using Gauss’s Law.
V
SOLUTION
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Total E field at a point is the vector sum of the fields caused by all the individual
charges.
Consider the figure below. Positions of the charges q1, q2, q3, …… qn, (source points )
be denoted by position vectors R1’, R2’, R3’, R4’, ….. Rn’. The position of the field
point at which the electric field intensity is to be calculated is denoted by R
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 103
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Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 104
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Thus
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 105
...)()()(
41 3'
3
33'
2
23'
1
1
0 RR
q
RR
q
RR
q '3
'2
'1 RRRRRRE
n
k k
k
RR
q
1
3'0
)(4
1 'kRRE
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Three infinite lines of charge, all parallel to the z-axis, are located at the three corners of the kite-shaped arrangement
shown in the figure below. If the two right triangles are symmetrical and of equal corresponding sides, show that the
electric field is zero at the origin .
Chapter 2BEE 3113ELECTROMAGNETIC
FIELDS THEORY 106