2. ELECTROSTATICS
2.1 COULUMB’S LAW
2.2 ELECTRIC FIELD INTENSITY
2.3 LINE, SURFACE & VOLUME CHARGES
2.4 ELECTRIC FLUX DENSITY
2.5 GAUSS’S LAW
2.6 ELECTRIC POTENTIAL
2.7 BOUNDARY CONDITIONS
2.8 CAPACITANCE
2
3. INTRODUCTION
Electromagnetics is the study of the effect of
charges at rest and charges in motion.
Some special cases of electromagnetics:
Electrostatics: charges at rest
Magnetostatics: charges in steady motion (DC)
Electromagnetic waves: waves excited by
charges in time-varying motion
3
4. INTRODUCTION (Cont’d)
Fundamental laws of
Maxwell’s
classical electromagnetics
equations
Special Electro- Magneto- Electro- Geometric
cases statics statics magnetic Optics
waves
Statics: 0 Transmission
t Line
Theory
Input from Circuit
other Theory Kirchoff’s
disciplines d
Laws
4
5. INTRODUCTION (Cont’d)
• Electrical phenomena caused by friction
are part of our everyday lives, and can be
understood in terms of electrical charge.
• The effects of electrical charge can be
observed in the attraction/repulsion of
various objects when “charged.”
5
6. INTRODUCTION (Cont’d)
• Charge comes in two varieties called “positive”
and “negative.”
• Objects carrying a net positive charge attract
those carrying a net negative charge and repel
those carrying a net positive charge.
• Objects carrying a net negative charge attract
those carrying a net positive charge and repel
those carrying a net negative charge.
• On an atomic scale, electrons are negatively
charged and nuclei are positively charged.
6
7. INTRODUCTION (Cont’d)
• Electric charge is inherently quantized such
that the charge on any object is an integer
multiple of the smallest unit of charge which
is the magnitude of the electron charge
e = 1.602 10-19 C.
• On the macroscopic level, we can assume
that charge is “continuous.”
7
8. COULUMB’S LAW
In the late 18th century, Colonel Charles
Augustus Coulomb invented a sensitive
torsion balance that he used to
experimentally determine the force
exerted in one charge by another.
8
9. COULUMB’S LAW (Cont’d)
He found that the force is proportional to the
product of two charges, inversely proportional
to the square of the distance between the
charges and acts in a line containing the two
charges.
Q1Q 2
F k 2
R
9
10. COULUMB’S LAW (Cont’d)
The proportional constant, k is:
1
k , r 0
4
Where the free space permittivity with
a value given by:
9
12 F 10 F
0 8 . 85 10
m 36 m
10
11. COULUMB’S LAW (Cont’d)
Charge Q1 exerts a vector force F12 in Newton's (N)
on charge Q2,
Q1Q 2
F12 a 12
4 0 R 12
2
11
12. COULUMB’S LAW (Cont’d)
If more than two charges, use the principle of
superposition to determine the force on a
particular charge.
If there are N charges, Q1,Q2...QN located
respectively at point with position vectors
r1,r2...rN the resultant force F on a charge Q
located at point r is the vector sum of the forces
exerted on Q by each charges Q1,Q2...QN
12
13. COULUMB’S LAW (Cont’d)
F
QQ 1 r r1 QQ 2 r r2
4 0 r r1
2
r r1 4 0 r r2
2
r r2
...
QQ N r r N
4 0 r r N
2
r rN
Or generally,
Q N Q k r rk
F
4 0 k 1 3
r rk
13
14. COULUMB’S LAW (Cont’d)
For example,
FTOTAL F12 F32 ,
Q1Q 2
F12 a 12
4 0 R 12
2
Q 3Q 2
F32 a 32
4 0 R 32
2
F12 F21
14
15. EXAMPLE 1
Suppose 10nC charge Q1 located at (0.0, 0.0, 4.0m)
and a 10nC charge Q2 located at (0.0, 4.0m, 0.0). Find
the force acting on Q2 from Q1.
15
16. SOLUTION TO EXAMPLE 1
To employ Coulomb’s Law, first find vector R12
R12 r2 r1
4a y 4a z
4a y 4a z
Magnitude of R12
R12 42 42
32 4 2
16
17. SOLUTION TO EXAMPLE 1 (Cont’d)
R12 4a y 4a z 1 1
And a12 ay az
R12 4 2 2 2
Then
Q1Q 2
F12 a 12
4 0 R 12
2
10 10 10 10
9 9
1
ay
1
az
4 8 . 854 10 4 2
12 2
2
2
4 . 0 a y 4 . 0 a z nN
17
18. 2.2 ELECTRIC FIELD INTENSITY
If Q1 is fixed to be at origin, a second charge
Q2 will have force acting on Q1 and can be
calculated using Coulomb’s Law. We also
could calculate the force vector that would
act on Q2 at every point in space to generate
a field of such predicted force values.
18
19. ELECTRIC FIELD INTENSITY (Cont’d)
It becomes convenient to define electric field
intensity E1 or force per unit charge as:
F12
E1
Q2
This field from charge Q1 fixed at origin results
from the force vector F12 for any arbitrarily
chosen value of Q2
19
20. ELECTRIC FIELD INTENSITY (Cont’d)
Coulomb’s law can be rewritten as
Q
E aR
4 0 R
2
to find the electric field intensity at any point in
space resulting from a fixed charge Q.
20
21. EXAMPLE 2
Let a point charge Q1 = 25nC be located
at P1 (4,-2,7). If ε = ε0, find electric field
intensity at P2 (1,2,3).
21
22. SOLUTION TO EXAMPLE 2
By using the electric field intensity,
Q
E aR
4 0 R
2
This field will be:
9
25 10
E a12
4 0 R12
22
23. SOLUTION TO EXAMPLE 2 (Cont’d)
Where, R12 r2 r1 3a x 4a y 4a z
and R12 41
Q Q
E a
2 12
R12
4 0 R12 4 0 R12
3
9
25 10
3a x 4a y 4a z
4 8.854 1012 41 3
2
23
24. ELECTRIC FIELD INTENSITY (Cont’d)
If there are N charges, Q1,Q2...QN located
respectively at point with position vectors
r1,r2...rN the electric field intensity at point r is:
E
Q1 r r1 .. QN r r N
4 0 r r1
2
r r1 4 0 r r N
2
r rN
1 N Q k r rk
E
4 0 k 1 3
r rk
24
25. FIELD LINES
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.
25
26. FIELD LINES (Cont’d)
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
26
27. FIELD LINES (Cont’d)
The field lines terminated
at a negative point charge
The field lines for a pair
of opposite charges
27
28. 2.3 LINE,SURFACE &
VOLUME CHARGES
Electric fields due to continuous charge
distributions:
28
29. LINE,SURFACE &
VOLUME CHARGES (Cont’d)
To determine the charge for each distributions:
Line charge: Surface charge:
dQ L dl dQ S dS
Q L dl Q S dS
L S
Volume charge:
dQ V dV
Q V dV
V
29
30. LINE CHARGE
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
30
31. LINE CHARGE (Cont’d)
Place an amount of charge in
coulombs along the z axis.
The linear charge density is
coulombs of charge per meter
length,
L C m
Choose an arbitrary point P
where we want to find the
electric field intensity.
P , , z
31
32. LINE CHARGE (Cont’d)
The electric field intensity is:
E E a E a Ez a z
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 E . So,
E E a E z a z
32
33. 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 E z components cancel
each other (by symmetry) ,
and the E adds, will give:
E E a
33
34. LINE CHARGE (Cont’d)
Recall for point charge,
Q
E aR
4 0 R
2
For continuous charge distribution, the
summation of vector field for each charges
becomes an integral,
dQ
E aR
4 0 R
2
34
35. LINE CHARGE (Cont’d)
The differential charge,
dQ L dl
L dz
The vector from source
to test point P,
R Ra R
a za z
35
36. LINE CHARGE (Cont’d)
Which has magnitude, R 2 z 2 and a
unit vector,
a za z
aR
2 z2
So, the equation for integral of continuous
charge distribution becomes:
L dz a za z
E
4 0 z
2 2
2 2 z2
36
37. LINE CHARGE (Cont’d)
Since there is no az component,
L dz
E a
4 0 2 z 2 3
2
L dz
a
4 0 2
z2 3
2
37
38. LINE CHARGE (Cont’d)
Hence, the electric field intensity at any point ρ
away from an infinite length is:
L
E a
2 0
For any finite length, use the limits on the integral.
38
39. EXAMPLE 3
Use Coulomb’s Law to
find electric field
intensity at (0,0,h) for
the ring of charge, of
charge density, L
centered at the origin in
the x-y plane.
39
40. SOLUTION TO EXAMPLE 3
By inspection, the ring
charges delivers only dE
and dE z contribution
to the field.
dE component will be
cancelled by symmetry.
40
41. SOLUTION TO EXAMPLE 3 (Cont’d)
Each term need to be
determined:
dQ
E aR
4 0 R
2
The differential charge,
dQ L dl
L ad
41
42. SOLUTION TO EXAMPLE 3 (Cont’d)
The vector from source to test point,
R Ra R
aa ha z
Which has magnitude, R a 2 h 2 and a
unit vector,
aa ha z
aR
a 2 h2
42
43. SOLUTION TO EXAMPLE 3 (Cont’d)
The integral of continuous charge distribution
becomes:
L ad aa ha z
E
4 0 a h
2 2
2
a2 h2
L ad
E
3
ha z
4 0
2
a h
2 2
43
44. SOLUTION TO EXAMPLE 3 (Cont’d)
Rearranging,
2
L ah
E da z
4 0 a 2 h 2
3
2 0
Easily solved,
L ah
E az
2 0 a 2 h 2 3
2
44
45. EXAMPLE 4
nC
An infinite length line of charge L 4.0
m
exists at x = 2m and z = 4m. Find the
electric field intensity at the origin.
45
47. SOLUTION TO EXAMPLE 4 (Cont’d)
The vector from line charge to the origin:
R a 2a x 4a z
Which has magnitude, R 20 and a unit
vector,
2 4
aR a ax az
20 20
47
48. SOLUTION TO EXAMPLE 4 (Cont’d)
Inserting into the infinite line charge equation:
L
E a
2 0
9
4 10 2a x 4a z
2 8 . 854 10 12
20 20
V
7 . 2 a x 14 . 4 a z
m
48
49. SURFACE CHARGE
Infinite Sheet of Surface Charge:
To derive the electric field intensity at point
P at a height h above a charge sheet of
infinite area (x-y plane).
The charge distribution, S is in C
m2
49
51. SURFACE CHARGE (Cont’d)
Consider a differential charge,
dQ S dS
S dd
The vector from surface charge to the origin:
R a ha z
51
52. SURFACE CHARGE (Cont’d)
Which has magnitude, R 2 h 2 and a
unit vector,
a ha z
aR
2 h2
Where, for continuous charge distribution:
dQ
E aR
4 0 R
2
52
53. SURFACE CHARGE (Cont’d)
The equation becomes:
S dd a ha z
E
4 0 2 h2
2
a2 h2
Since only z components exists,
S dd
E
3
ha z
4 0 h
2 2 2
53
55. SURFACE CHARGE (Cont’d)
A general expression for the field from a sheet
charge is:
S
E aN
2 0
Where aN is the unit vector normal from the
sheet to the test point.
55
56. EXAMPLE 5
nC
An infinite extent sheet of charge S 10 2
m
exists at the plane y = -2m. Find the electric
field intensity at point P (0, 2m, 1m).
56
58. SOLUTION TO EXAMPLE 5 (Cont’d)
The unit vector directed away from the sheet
and toward the point P is ay
S
E aN
2 0
10 10 9
2 8 . 854 10 12
ay
V
565 a y
m
58
59. VOLUME CHARGE
A volume charge is distributed over a
volume and is characterized by its volume
charge density, V in C 3
m
The total charge in a volume containing a
charge distribution, V is found by
integrating over the volume: Q dV
V
V
59
60. EXAMPLE 6
Find the total charge
over the volume with
volume charge density,
10 5 z
V 5e C 3
m
60
61. SOLUTION TO EXAMPLE 6
The total charge, Q V dV with volume:
V
dV dddz
Thus, Q V dV
V
0.01 2 0.04
10 5 z
5e dddz
0 0 z 0.02
14
7.854 10 C
61
62. VOLUME CHARGE (Cont’d)
To find the electric field intensity resulting from
a volume charge, we use:
dQ V dV
E aR aR
4 0 R 4 0 R
2 2
Since the vector R and V will vary over the
volume, this triple integral can be difficult. It can
be much simpler to determine E using Gauss’s
Law.
62
63. 2.4 ELECTRIC FLUX DENSITY
Consider an amount of charge
+Q is applied to a metallic
sphere of radius a.
Enclosed this charged sphere
using a pair of connecting
hemispheres with bigger radius.
63
64. ELECTRIC FLUX DENSITY (Cont’d)
The outer shell is grounded. Remove the ground
then we could find that –Q of charge has
accumulated on the outer sphere, meaning the
+Q charge of the inner sphere has induced the –Q
charge on the outer sphere.
64
65. ELECTRIC FLUX DENSITY (Cont’d)
Electric flux, psi extends from the positive
charge and casts about for a negative charge. It
begins at the +Q charge and terminates at the
–Q charge.
The electric flux density, D in C 2 is:
m
D
4 0 R
2
a R where D E
65
66. ELECTRIC FLUX DENSITY (Cont’d)
This is the relation between D and E, where is
the material permittivity. The advantage of using
electric flux density rather than using electric
field intensity is that the number of flux lines
emanating from one set of charge and terminating
on the other, independent from the media.
We can find the total flux over a surface as:
D dS
66
67. ELECTRIC FLUX DENSITY (Cont’d)
We could also find the electric flux density, D
for:
Infinite line of charge:
Where
L L
E a So, D a
2 0 2
67
68. ELECTRIC FLUX DENSITY (Cont’d)
Infinite sheet of charge:
Where
S S
E aN So, D aN
2 0 2
Volume charge distribution:
V dV V dV
E aR So, D aR
4 R
2
4 0 R
2
68
69. EXAMPLE 7
Find the amount of electric flux through the
surface at z = 0 with 0 x 5m , 0 y 3m
and
D 3 xya x 4 xa z C m 2
69
70. SOLUTION TO EXAMPLE 7
The differential surface vector is
dS dxdya z
We could have chosen dS dxdy a z but the
positive differential surface vector is pointing in
the same direction as the flux, which give us a
positive answer.
70
71. SOLUTION TO EXAMPLE 7 (Cont’d)
Therefore,
D dS
3 xy a x 4 x a z dxdy a z
5 3
4 xdxdy Why?!
x0 y0
150 C
71
72. EXAMPLE 8
Determine D at (4,0,3) if there is a point
charge 5 mC at (4,0,0) and a line
charge 3 m C m along the y axis.
72
74. SOLUTION TO EXAMPLE 8 (Cont’d)
Let total flux,
DTOTAL DQ D L
Where DQ is flux densities due to point charge
and DL is flux densities due to line charge.
DQ 0 E 0 aR
Thus, Q
4 R 2
0
Q
a
2 R
4 R
74
75. SOLUTION TO EXAMPLE 8 (Cont’d)
Where,
R 4,0,3 4,0,0 0,0,3
3a z
Which has magnitude, R 3 and a unit vector,
3a z
aR az
3
75
76. SOLUTION TO EXAMPLE 8 (Cont’d)
So,
Q
DQ aR
4 R
2
5 103
az
4 9
0.138a m C z
m2
76
77. SOLUTION TO EXAMPLE 8 (Cont’d)
L
And DL a
2
Where, a
4,0,3 0,0,0 4a x 3a z
4,0,3 0,0,0 5
So,
3 4a x 3a z
DL
2 5 5
0.24a 0.18a m C
x z
m2
77
78. SOLUTION TO EXAMPLE 8 (Cont’d)
Therefore, total flux:
DTOTAL DQ D L
0.318a z 0.24a x 0.18a z
240a x 42a z C
m2
78
79. 2.5 GAUSS’S LAW
If a charge is enclosed, the net flux passing
through the enclosing surface must be equal to
the charge enclosed, Qenc.
Gauss’s Law states that:
The net electric flux through any closed surface is
equal to the total charge enclosed by that surface
D dS Qenc
79
80. GAUSS’S LAW (Cont’d)
It can be rearranged so that we have relation
between the Gauss’s Law and the electric flux.
D d S Q enc
Q enc V dV
V
Q D dS V dV
S V
80
81. GAUSS’S LAW (Cont’d)
Gauss’s Law is useful in finding the fields for
problems that have high degree of symmetry.
• Determine variables influence D and what
components D present
• Select an enclosing surface, called Gaussian
Surface, whose differential surface is directed
outward from the enclosed volume and is
everywhere (either tangent or normal to D)
81
82. GAUSS’S LAW APPLICATION (Cont’d)
Use Gauss’s Law to determine electric
field intensity for each cases below:
Point Charge
Infinite length of Line Charge
Infinite extent Sheet of Charge
82
83. POINT CHARGE
• Point Charge:
It has spherical coordinate
symmetry, where the field
is everywhere directed
radially away from the
origin. Thus,
D Dr a r
83
84. POINT CHARGE (Cont’d)
For a gaussian surface, we could find the
differential surface vector is:
dS r sin dda r
2
So,
D d S D r a r r sin d d a r
2
D r r sin d d
2
84
85. POINT CHARGE (Cont’d)
Since the gaussian surface has a fixed radius,
Dr will be constant and can be taken from
integration to yield
2
D dS sin d d
2
Dr r
0 0
4 r 2 D r
85
86. POINT CHARGE (Cont’d)
By using Gauss’s Law, where:
D dS Qenc
Dr 4r Q
2
Q
So, Dr which leads to expected result:
4r 2
Q
E ar See page 16!
4 0 r 2
86
87. INFINITE LENGTH LINE OF CHARGE
• Infinite length line of charge:
Find D and then E at any
point P , , z
A Gaussian surface
containing the point P is
placed around a section
of an infinite length line
of charge density L
occupying the z-axis.
87
88. LINE CHARGE (Cont’d)
An element of charge dQ along the line will give
Dρ and Dz. But second element of dQ will result
in cancellation of Dz. Thus,
D D a
The flux through the closed surface is:
D d S D d S top D d S bottom
D d S side
88
89. LINE CHARGE (Cont’d)
Where,
dStop dda z ,
dSbottom dd a z , dS side ddza
Then, we know that Dρ is constant on the side
of gaussian surface
D d S side D a d dz a
2 h
D d dz 2 h D
0 z0
89
90. LINE CHARGE (Cont’d)
The charge enclosed by the gaussian surface:
h
Qenc L dz L h
0
We know that,
D d S 2 h D L h Q enc
So, L Thus, as L
D E a
2 expected:
2 0
90
91. INFINITE EXTENT SHEET OF CHARGE
• Infinite extent sheet of charge:
Determine the field everywhere resulting
from an infinite extent sheet of charge ρS
placed on the x-y plane at z = 0.
Locate a point at which we want to find the
field along the z axis at height h.
91
92. SHEET OF CHARGE (Cont’d)
Gaussian surface must contain this point
and surround some portion of the charged
sheet.
A rectangular box is
employed as the
Gaussian surface
surrounding a section
of sheet charge with
sides 2x, 2y and 2z
92
93. SHEET OF CHARGE (Cont’d)
Only a DZ component will be present, and the
charge enclosed is simply:
x y
Q S dS S dx dy
x y
4 S xy
No flux through the side of the box, so find
the flux through the top and bottom surface
93
94. SHEET OF CHARGE (Cont’d)
D d S D d S top D d S bottom
D z a z dxdy a z
top
D z a z dxdy a z
bottom
2 4 xy D z
Notice that the answer is independent of the
height of the box.
94
95. SHEET OF CHARGE (Cont’d)
Then we have:
D dS Q
2 4 xy D z 4 S xy
S S
Dz or D az
2 2
And electric field intensity, as expected:
S
E aN
2 0
95
96. GAUSS’S LAW (Cont’d)
Related to Gauss’s Law, where net flux is
evaluated exiting a closed surface, is the
concept of divergence.
Expression for divergence by applying Gauss’s
Law might be too lengthy to derive, but it can be
described as:
D V
96
97. GAUSS’S LAW (Cont’d)
The expression is also called the point form of
Gauss’s Law, since it occurs at some particular
point in space. For instance,
Plunger stationary – no net
movement of molecules
Plunger moves up – net
movement where air molecules
diverging air is expanding
Plunger pushes in – net flux is
negative and molecules diverging
air is compressing
97
98. EXAMPLE 9
Suppose:
D a
2
Find the flux through the surface of a cylinder
with 0 zh and a by evaluating the
left side and the right side of the divergence
theorem.
98
99. SOLUTION TO EXAMPLE 9
Remember the divergence theorem?
D dS DdV
V
We can first evaluate the left side of the
divergence theorem by considering:
D d S D d S top D d S bottom
D d S side
99
100. SOLUTION TO EXAMPLE 9 (Cont’d)
A sketch of this cylinder is shown with
differential vectors.
The integrals over
the top and bottom
surfaces are each
zero, since:
a az 0
100
101. SOLUTION TO EXAMPLE 9 (Cont’d)
Thus,
D d S D d S side
2 h
a d dz
2
0 z0
2 ha 3
101
102. SOLUTION TO EXAMPLE 9 (Cont’d)
For evaluation of the right side of the divergence
theorem, first find the divergence in cylindrical
coordinate:
1
D
D
1 3
3
102
103. SOLUTION TO EXAMPLE 9 (Cont’d)
Performing a volume integration on this divergence,
DdV
V
3 dddz
2 a h
3 2dddz
0 0 z 0
2ha 3
This is the same!
103
104. 2.6 ELECTRIC POTENTIAL
To develop the concept of electric potential and
show its relationship to electric field intensity.
In moving the object from point a to b, the
work can be expressed by:
b
W F dL
a
dL is differential length vector along some
portion of the path between a and b
104
105. ELECTRIC POTENTIAL (Cont’d)
The work done by the field in moving the charge
from a to b is
b
WE field Q E dL
a
If an external force moves the charge against the
field, the work done is negative:
b
W Q E dL
a
105
106. ELECTRIC POTENTIAL (Cont’d)
We can defined the electric potential difference, Vba
as the work done by an external source to move a
charge from point a to point b as:
b
W
Vba E dL
Q a
Where,
Vba Vb Va
106
107. ELECTRIC POTENTIAL (Cont’d)
Consider the potential difference between two
points in space resulting from the field of a
point charge located at origin, where the
electric field intensity is radially directed, then
move from point a to b to have:
b b
Q
Vba E dL a r dra r
a 4 0 r
2
a
107
108. ELECTRIC POTENTIAL (Cont’d)
r b
Q
Thus, Vba
4 0r r a
Q 1 1
Vb Va
4 0 b a
The absolute potential at some finite radius
from a point charge fixed at the origin:
Q
V
4 0 r
108
109. ELECTRIC POTENTIAL (Cont’d)
If the collection of charges becomes a continuous
distribution, we could find:
dQ
V
4 0 r
L dL
Where, V 4 0 r Line charge
S dS
V 4 0 r Surface charge
V dV
V 4 0 r Volume charge
109
110. ELECTRIC POTENTIAL (Cont’d)
The principle of superposition, where applied to
electric field also applies to potential difference.
Q1 Q2
V
4 0 r r1 4 0 r r2
QN
...
4 0 r r N
Or generally, 1 N
Qk
V rr
4 0 k 1 k
110
111. ELECTRIC POTENTIAL (Cont’d)
Three different paths to
calculate work moving
from the origin to point
P against an electric
field.
Based on figure, if a closed path is chosen, the
integral will return zero potential:
E dL 0
111
112. EXAMPLE 10
Two point charges -4 μC and 5 μC are
located at (2,1-,3) and (0,4,-2)
respectively. Find the potential at
(1,0,1).
112
113. SOLUTION TO EXAMPLE 10
Let Q1 4 C and Q2 5C
So,
Q1 Q2
V
4 0 r r1 4 0 r r2
Where,
r r1 1, 0 ,1 2 , 1,3 1,1, 2 6
r r2 1, 0 ,1 0 , 4 , 2 1, 4 , 2 26
113
114. SOLUTION TO EXAMPLE 10 (Cont’d)
Therefore,
Q1 Q2
V 1, 0 ,1
4 0 r r1 4 0 r r2
6 6
4 10 5 10
4 0 6 4 0 26
V 1,0,1 5.872 kV
114
115. ELECTRIC POTENTIAL (Cont’d)
The electrostatic potential
contours from a point charge
form equipotential surfaces
surrounding the point charge.
The surfaces are always
orthogonal to the field lines.
The electric field can be
determined by finding the max.
rate and direction of spatial
change of the potential field.
115
116. ELECTRIC POTENTIAL (Cont’d)
Therefore,
E V
The negative sign indicates that the field is
pointing in the direction of decreasing potential.
By applying to the potential field:
Q Q
E V ar a
r 4 0r 4 0r 2 r
116
117. IMPORTANT!!
Three ways to calculate E:
If sufficient symmetry, employ Gauss’s Law.
Use the Coulomb’s Law approach.
Use the gradient equation.
117
118. EXAMPLE 11
Consider a disk of charge ρS, find the
potential at point (0,0,h) on the z-axis and
then find E at that point.
118
119. SOLUTION TO EXAMPLE 11
Find that,
dQ S dS
and r h2 2
S dd
dQ
With V then,
4 0 r
S a 2 dd
V r
4 0 0 0
119
120. SOLUTION TO EXAMPLE
11 (Cont’d)
How to calculate the integral?
Let u h 2 2 and du 2 d leads to
integral
u 1 2 du then,
a
S
V h
2 2
2 0 0
S
2 0
h 2
a2 h
120
121. SOLUTION TO EXAMPLE
11 (Cont’d)
To find E, need to know how V is changing with
position. In this case E varies along the z-axis, so
simply replace h with z in the answer for V, then
proceed with the gradient equation.
V
E V az
z
S 1 2 z S z
1a z 1 2 a z
2 0 2 z 2 a 2 2 0 z a2
121
122. BOUNDARY CONDITIONS
So far we have considered the existence of electric
field in a region consisting of two different media,
the condition that the field must satisfy at the
interfacing separating the media called “boundary
condition”. Thus, we could see how the fields
behave at the boundary between a pair of
dielectrics or between a dielectric and a conductor.
122
123. BOUNDARY CONDITIONS
First boundary condition can be determined by
performing a line integral of E around a closed
rectangular path,
123
124. BOUNDARY CONDITIONS
Fields are shown in each medium along with
normal and tangential components. For static
fields,
E dL 0
Integrate in the loop clockwise starting from a,
b c d a
E dL E dL E dL E dL 0
a b c d
124
125. BOUNDARY CONDITIONS
Evaluate each segment,
b w
E dL ET 1aT dLaT ET 1w
a 0
c 0 h 2
E dL EN1a N dLa N EN 2a N dLa N
b h 2 0
h
E N 1 E N 2
2
125
126. BOUNDARY CONDITIONS
d 0
E dL ET 2aT dLaT ET 2 w
c w
a 0 h 2
E dL EN 2a N dLa N EN1a N dLa N
d h 2 0
h
E N 1 E N 2
2
Summing for each segment, then we have the first
boundary condition: ET 1 ET 2
126
128. BOUNDARY CONDITIONS
The Gauss’s Law,
D dS Qenc
Where,
D dS D dS D dS D dS
top bottom side
The pillbox is short enough, so the flux
passes through the side is negligible.
128
129. BOUNDARY CONDITIONS
So, only top and bottom where:
D dS DN1a N dSa N DN1S
top
D dS DN 2a N dS a N DN 2S
bottom
Which sums to:
DN1 DN 2 S Qenc
129
130. BOUNDARY CONDITIONS
And the right side of Gauss’s Law,
Qenc S dS S S
Thus, it leads to the second boundary condition:
DN 1 DN 2 S
This is when the normal direction from
medium 2 to medium 1.
130
131. BOUNDARY CONDITIONS
If the normal direction is from medium 1 to
medium 2,
DN 2 DN 1 S
Generally,
a 21 D1 D2 S
131
132. BOUNDARY CONDITIONS
For a boundary conditions between a dielectric
and a good conductor,
ET 0
Because in a good conductor, E = 0. And since
the electric flux density is zero inside the
conductor,
DN S
132
133. EXAMPLE 12
Consider that the field E1 is known as:
E1 3a x 4a y 5a z V m
Find the field E2 in the other dielectrics.
133
135. BOUNDARY CONDITIONS
We can employ Poisson’s and Laplace’s
equations to help find the potential function
when conditions at the boundaries are specified.
From divergence theorem expression,
D V
By considering D E ,
V
E
135
136. BOUNDARY CONDITIONS
From the gradient expression,
E V
Which gives us Poisson’s equation,
V
V
2
In charge free medium in which V 0 , it
becomes Laplace’s equation:
V 0
2
136
137. EXAMPLE 13
Determine the electric
potential in the dielectric
region between a pair of
concentric spheres that
have a potential
difference Vab.
137
138. SOLUTION TO EXAMPLE 13
0 C
The charge distribution is: V 3
r m
This employs Poisson’s equation and the potential
is only a function of r,
V 0
V
2
r
1 2 V (r ) 0
2 r
r r r r r 0
138
139. SOLUTION TO EXAMPLE 13
Multiply with r2 and integrate to obtain,
V (r ) 0r 2
r2 A
r 2 r 0
Dividing both sides with r2 and integrate again,
0r A
V (r ) B
2 r 0 r
139
140. SOLUTION TO EXAMPLE 13 (Cont’d)
Assume that the Vab consists of a voltage Va on
the inner conductor and the outer conductor is
grounded. So,
0a A
V (r a) Va B 1
2 r 0 a
0b A
V (r b) Vb 0 B 2
2 r 0 b
140
141. SOLUTION TO EXAMPLE 13
0 a A 0b A
Vab Va Vb
2 a B 2 b B
r 0 r 0
0 b a
Va b a A (3)
2 r 0 ab
Va ab 0 ab
From (3) you can get: A
a b 2 r 0
Thus, B 0
a b Va ab
2 r 0 a b
141
142. SOLUTION TO EXAMPLE 13
Then, the potential between the spheres is
given by:
0 a Va a
V (r ) r
2 r 0 r r
142
143. CAPACITANCE
The amount of charge that accumulates as a
function of potential difference is called the
capacitance.
Q
C
V
The unit is the farad (F) or coulomb per volt.
143
144. CAPACITANCE (Cont’d)
Two methods for determining capacitance:
Q Method
•Assume a charge +Q on plate a and a charge
–Q on plate b.
• Solve for E using the appropriate method
(Coulomb’s Law, Gauss’s Law, boundary
conditions)
•Solve for the potential difference Vab between
the plates (The assumed Q will divide out)
144
145. CAPACITANCE (Cont’d)
V Method
• Assume Vab between the plates.
• Find E , then D using Laplace’s equation.
•Find ρS, and then Q at each plate using
conductor dielectric boundary condition
(DN = ρS )
• C = Q/Vab (the assumed Vab will divide out)
145
146. EXAMPLE 14
Use Q method to find the capacitance for
the parallel plate capacitor as shown.
146
147. SOLUTION TO EXAMPLE 14
Place charge +Q on the inner surface of the top
plate, and –Q charge on the upper surface of the
bottom plate, where the charge density,
S Q from Q S dS
S
Use conductor dielectric boundary, to obtain:
D Q
S
a z from DN S
147
148. SOLUTION TO EXAMPLE 14
We could find the electric field intensity, E
Q
E az
0 r S
The potential difference across the plates is:
a
Vab E dL
b
d
Q Qd
a z dza z
0 0 r S 0 r S
148
149. SOLUTION TO EXAMPLE 14
Finally, to get the capacitance:
Q Q
C
Vab Qd
0 r S
0 r S
C
d
149
150. EXAMPLE 15
Use V method to find the capacitance for a length L
of coaxial line of inner conductor radius a and
outer radius b, filled with dielectric permittivity as
shown.
150
151. SOLUTION TO EXAMPLE 15
Employ Laplace’s equation to find the potential
field everywhere in the dielectric. Assume that
fringing fields are neglected and the field is only
in function of ρ.
Laplace’s equation becomes:
V
0
151
152. SOLUTION TO EXAMPLE 15
Integrating twice to obtain:
V A ln B
Apply boundary condition to determine A and
B. Let V(b) = 0 and V(a) = Vab to get:
Vab Vab ln b
B A ln b, A So, V
ln b a lnb a
152
153. SOLUTION TO EXAMPLE 15
Apply the gradient to obtain E:
V
E V
Vab
E a
lnb a
and also..
r 0Vab
D a
lnb a
153
154. SOLUTION TO EXAMPLE 15 (Cont’d)
At inner conductor, the flux is directed outward
indicating a positive surface charge density,
r 0Vab
S
a ln b a
We can find Q on the inner conductor with:
r 0Vab 2L r 0Vab
Q S S 2aL lnb a
a lnb a
154
155. SOLUTION TO EXAMPLE 15 (Cont’d)
Thus, we can find the capacitance:
Q 2L r 0
C
V ln b a
155
156. EXAMPLE 16
Conducting spherical shells with radius a =10cm
and b = 30cm are maintained at a potential
difference of 100V, such that V(r = a) = 100V and
V(r = b) = 0V. Determine:
• The V and E in the region between the shells.
• If εr = 2.5 in the shells, determine the total
charges induced on the shells and the capacitance
of the capacitor.
156
158. SOLUTION TO EXAMPLE 16
Employ Laplace’s equation to find the potential
field and the field is only in function of r.
Laplace’s equation becomes:
1 2 V
V 2 r
2
0
r r r
Multiply by r2, 2 V
r 0
r r
158
159. SOLUTION TO EXAMPLE 16 (Cont’d)
Integrating once gives
V V A
r 2
A 2
r r r
Integrating again gives
A
V B
r
159
160. SOLUTION TO EXAMPLE 16 (Cont’d)
To obtain the value of constant A and B, use
boundary conditions, where:
A
When, r b, V 0 0 B
b
A
B
b
Thus,
1 1
V A
b r
160
161. SOLUTION TO EXAMPLE 16 (Cont’d)
When, r a, V V0 :
1 1
V V0 A
b r
1 1
V0 A
b a
V0
A
1 1
b a
161
162. SOLUTION TO EXAMPLE 16 (Cont’d)
Therefore, the potential difference V
1 1
V V0 b r
1 1
b a
162
163. SOLUTION TO EXAMPLE 16 (Cont’d)
Apply the gradient to obtain E:
V A
E V ar 2 ar
r r
So, the electric field intensity E
1 V0
E 2 a r
r 1 1
b a
163
164. SOLUTION TO EXAMPLE 16 (Cont’d)
The total charges Q is:
Q D dS E dS
2
0 rV0 1 2
2 1 1 r sin dd
0 0 r
b a
164
165. SOLUTION TO EXAMPLE 16 (Cont’d)
This yields to:
2
0 rV0
Q d sin d
1 1 0 0
b a
4 0 rV0
1 1
b a
165
166. SOLUTION TO EXAMPLE 16 (Cont’d)
Thus, substituting the values of a, b and V0
to get the total charges, Q
Q
4 8.854 10 2.5100
12
1 1
0.3 0.1
4.1723 nC
166
167. SOLUTION TO EXAMPLE 16 (Cont’d)
And for the potential difference V = V0.
So, the capacitance, C
Q 4.1723 nC
C 41.723 pF
V 100 V
167
171. LASER PRINTER (Cont’d)
• OPC drum : Organic Photoconductive Cartridge –
has a special coating will hold electrostatic charge.
• The surface is photoconductive – will discharge if
surface hit by light.
i. A portion of drum passes under a negative
charged wire large negative charges induces a
positive charge on the drum.
ii. Image to be printed is delivered to this charged
region by laser and spinning mirror combination.
iii. Wherever the laser light strikes the drum, the
photoconductive material is discharged.
171
172. LASER PRINTER (Cont’d)
iv. The drum rolls past a toner The toner is black
powder and positive charge drawn to those
portions of the charge that have been discharged
by the laser.
v. Paper is fed through same speed as drum it
passes over positively charged wire that gives the
paper a strong negative charge.
vi. The positively charged toner drum is transferred to
the stronger negative charged on the paper then
it passes near a negatively charged wire that
removes the negative charge from the paper,
prevents it from statically clinging to the drum
172
173. LASER PRINTER (Cont’d)
vii.The paper and loose toner powder passed through
heated fuser rollers powder melts into the paper
fiber the warm paper exits the printer.
• The drum continues rolling, passing through high
intensity light discharges all the photoconductors
to erase the image from the drum, and ready for
application of positive charge again from corona
wire.
173
174. SUMMARY (1)
•The force exerted on a charge Q1 on charge Q2 in a
medium of permittivity ε is given by Coulomb’s Law:
Q1Q 2
F12 a 12
4 R 12
2
Where R12 R12a12 is a vector from charge Q1 to Q2
•Electric field intensity E1 is related to force F12 by:
F12
E1
Q2
174
175. SUMMARY (2)
The Coulomb’s Law can be rewritten as:
Q
E aR
4 0 R
2
For a continuous charge distribution:
dQ
E 4 0 R
a
2 R
•For a point charge at origin:
Q
E ar
4 0 r 2
175
176. SUMMARY (3)
•For an infinite length line charge ρL on the z axis
L
E a
2
•For an infinite extent sheet of charge ρS
S
E aN
2
•Electric flux density related to field intensity by:
D r 0E
Where εr is the relative permittivity in a linear, isotropic
and homogeneous material.
176
177. SUMMARY (4)
• Electric flux passing through a surface is given by:
D dS
• Gauss’s Law states that the net electric flux through
any closed surface is equal to the total charge enclosed
by that surface:
D dS Qenc
Point form of Gauss’s Law is
D V
177
178. SUMMARY (5)
• The electric potential difference Vab between a pair
of points a and b in an electric field is given by:
b
Vba E dL Vb Va
a
Where Va and Vb are the electrostatics potentials at a
and b respectively.
For a distribution of charge in the vicinity of the
origin, where a zero reference voltage is taken at
infinite radius:
dQ
V
4r
178
179. SUMMARY (6)
• E is related to V by the gradient equation:
E V
Which for Cartesian coordinates is:
V V V
V ax ay az
x y z
• The conditions for the fields at the boundary between
a pair of dielectrics is given by:
ET 1 ET 2 and a 21 D1 D2 S
179
180. SUMMARY (7)
Where ET1 and ET2 are the electric field components
tangential to the boundary, a21 is a unit vector from
medium 2 to 1 and ρS is the surface charge at the
boundary. If no surface charge is present, the
components of D normal to the boundary are equal:
D N1 D N 2
At the boundary between a conductor and a dielectric,
the conditions are:
ET 0 and DN S
180
181. SUMMARY (8)
• Poisson’s equation is:
V
V
2
Where the Laplacian of V in Cartesian coordinates is
given by:
V V V
2 2 2
V 2 2 2
2
x y z
In a charge free medium, Poisson’s equation reduces to
Laplace’s equation
V 0
2
181
182. SUMMARY (9)
• Capacitance is a measure of charge storage capability
and is given by:
Q
C
V
For coaxial cable:
L 2L
Vab ln b a So, C
2 ln b a
For two concentric spheres:
Q 1 1 4
Vab So, C
4 a b 1 a 1 b
182
