electric flux density - Erwin Sitompul

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Engineering Electromagnetics
Lecture 3
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
President University
Erwin Sitompul
EEM 3/1
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Electric Flux Density
 About 1837, the Director of the Royal Society in London,
Michael Faraday, was interested in static electric fields and the
effect of various insulating materials on these fields.
 This is the lead to his famous invention, the electric motor.
 He found that if he moved a magnet through a loop of wire, an
electric current flowed in the wire. The current also flowed if the
loop was moved over a stationary magnet.
►Changing magnetic field produces an electric field.
President University
Erwin Sitompul
EEM 3/2
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Electric Flux Density
 In his experiments, Faraday had a pair of concentric metallic
spheres constructed, the outer one consisting of two
hemispheres that could be firmly clamed together.
 He also prepared shells of insulating material (or dielectric
material), which would occupy the entire volume between the
concentric spheres.
President University
Erwin Sitompul
EEM 3/3
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Electric Flux Density
 Faraday found out, that there was a sort of “charge
displacement” from the inner sphere to the outer sphere, which
was independent of the medium.
 We refer to this flow as displacement, displacement flux, or
simply electric flux.
ψQ
 Where ψ is the electric flux, measured in coulombs, and Q is
the total charge on the inner sphere, also in coulombs.
President University
Erwin Sitompul
EEM 3/4
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Electric Flux Density
 At the surface of the inner sphere, ψ
coulombs of electric flux are produced
by the given charge Q coulombs, and
distributed uniformly over a surface
having an area of 4πa2 m2.
 The density of the flux at this surface
is ψ/4πa2 or Q/4πa2 C/m2.
 The new quantity, electric flux density, is measured in C/m2 and
denoted with D.
 The direction of D at a point is the direction of the flux lines at
that point.
 The magnitude of D is given by the number of flux lines
crossing a surface normal to the lines divided by the surface
area.
President University
Erwin Sitompul
EEM 3/5
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Electric Flux Density
 Referring again to the concentric
spheres, the electric flux density is in
the radial direction :
Q
a
2 r
4 a
Q

a
2 r
4 b
D r a 
(inner sphere)
D r b
(outer sphere)
 At a distance r, where a ≤ r ≤ b,
Q
D
a
2 r
4 r
 If we make the inner sphere smaller and smaller, it becomes a
point charge while still retaining a charge of Q. The electrix flux
density at a point r meters away is still given by:
D
Q
a
2 r
4 r
President University
Erwin Sitompul
EEM 3/6
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Electric Flux Density
 Comparing with the previous chapter, the radial electric field
intensity of a point charge in free space is:
E
Q
4 0 r
2
ar
 Therefore, in free space, the following relation applies:
D   0E
 For a general volume charge distribution in free space:
v dv
E
a
vol 4 R 2 R
0
v dv
a
vol 4 R 2 R
D
President University
Erwin Sitompul
EEM 3/7
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Electric Flux Density
 Example
Find the electric flux density at a point having a distance 3 m
from a uniform line charge of 8 nC/m lying along the z axis in
free space.
L
L
8 109
1.273 109
E
a  D 
a 
a 
a  C m2
2 0 
2
2

For the value ρ = 3 m,
1.273 109
D
 4.244 1010 a C m2  0.424a nC m2
3
• Can you determine the
electric flux density at
(1,7,7) and (3,4,5)?
President University
Erwin Sitompul
EEM 3/8
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Electric Flux Density
 Example
Calculate D at point P(6,8,–10) produced by a uniform surface
charge density with ρs = 57.2 μC/m2 on the plane x = 9.
6
s
s
57.2

10
E
aN  D  aN 
a N  28.6aN C m2
2
2 0
2
At P(6,8,–10),
a N = a x
 D  28.6a x  C m2
• Can you determine D at
(1,8,2) and (12,–2,7)?
President University
Erwin Sitompul
EEM 3/9
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Gauss’s Law
 The results of Faraday’s experiments with the concentric
spheres could be summed up as an experimental law by stating
that the electric flux passing through any imaginary spherical
surface lying between the two conducting spheres is equal to
the charge enclosed within that imaginary surface.
ψQ
 Faraday’s experiment can be generalized to the following
statement, which is known as Gauss’s Law:
“The electric flux passing through any closed surface is
equal to the total charge enclosed by that surface.”
President University
Erwin Sitompul
EEM 3/10
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Gauss’s Law
 Imagine a distribution of charge, shown as a cloud of point
charges, surrounded by a closed surface of any shape.
 If the total charge is Q, the Q coulombs of electric flux will pass
through the enclosing surface.
 At every point on the surface the electric-flux-density vector D
will have some value DS (subscript S means that D must be
evaluated at the surface).
President University
Erwin Sitompul
EEM 3/11
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Gauss’s Law
 ΔS defines an incremental element of area with magnitude of
ΔS and the direction normal to the plane, or tangent to the
surface at the point in question.
 At any point P, where DS makes an angle θ with ΔS, then the
flux crossing ΔS is the product of the normal components of DS
and ΔS.
ψ  flux crossing S  DS cos  S  DS  S
ψ   dψ 

closed
surface
President University
DS  dS
Erwin Sitompul
EEM 3/12
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Gauss’s Law
 The resultant integral is a closed surface integral, with dS
always involves the differentials of two coordinates
► The integral is a double integral.
 We can formulate the Gauss’s law mathematically as:
ψ

S
DS  dS  charge enclosed  Q
 The charge enclosed meant by the formula above might be
several point charges, a line charge, a surface charge, or a
volume charge distribution.
Q  Qn
Q    L dL
President University
Q    S dS
S
Erwin Sitompul
Q    v dv
vol
EEM 3/13
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Gauss’s Law
 We now take the last form, written in terms of the charge
distribution, to represent the other forms:

S
DS  dS   v dv
vol
 Illustration. Let a point charge Q be
placed at the origin of a spherical
coordinate system, and choose a
closed surface as a sphere of radius a.
 The electric field intensity due to the
point charge has been found to be:
E
Q
4 0 r
2
ar
D   0E  D 
President University
Q
a
2 r
4 r
Erwin Sitompul
EEM 3/14
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Gauss’s Law
 At the surface, r = a,
Q
DS 
a
2 r
4 a
dS  a2 sin  d d ar
Q 2
Q
DS  dS 
a
sin

d

d

a

a

sin  d d
r
r
2
4 a
4
 D  dS
 
Q
 
sin  d d


4
ψ
S
S
2
0
0
r a

Q
2

cos    0
4
 0
Q
President University
Erwin Sitompul
EEM 3/15
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Application of Gauss’s Law: Some Symmetrical Charge Distributions
 Let us now consider how to use the Gauss’s law to calculate
the electric field intensity DS:
Q

S
D S  dS
 The solution will be easy if we are able to choose a closed
surface which satisfies two conditions:
1. DS is everywhere either normal or tangential to the closed
surface, so that DSdS becomes either DSdS or zero,
respectively.
2. On that portion of the closed surface for which DSdS is not
zero, DS is constant.
 For point charge ► The surface of a sphere.
 For line charge ► The surface of a cylinder.
President University
Erwin Sitompul
EEM 3/16
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Application of Gauss’s Law: Some Symmetrical Charge Distributions
 From the previous discussion of the
uniform line charge, only the radial
component of D is present:
D  D a
 The choice of a surface that fulfill the
requirement is simple: a cylindrical
surface.
 Dρ is every normal to the surface of a
cylinder. It may then be closed by two
plane surfaces normal to the z axis.
President University
Erwin Sitompul
EEM 3/17
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Application of Gauss’s Law: Some Symmetrical Charge Distributions
Q

S
D S  dS
 D 
sides
 D 
L
dS

2
z 0  0
 
 Dz  dSz
top
z L
 Dz 
bottom
dSz
z 0
 d dz
 D 2 L
Q
 D 
2 L
 We know that the charge enclosed is ρLL,
L
D 
2
E 
L
2 0 
President University
Erwin Sitompul
EEM 3/18
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Application of Gauss’s Law: Some Symmetrical Charge Distributions
 The problem of a coaxial cable is almost
identical with that of the line charge.
 Suppose that we have two coaxial
cylindrical conductors, the inner of radius
a and the outer of radius b, both with
infinite length.
 We shall assume a charge distribution of
ρS on the outer surface of the inner
conductor.
 Choosing a circular cylinder of length L and radius ρ, a < ρ < b,
as the gaussian surface, we find:
Q  DS 2 L
 The total charge on a length L of the inner conductor is:
Q
L

2
z 0  0
S ad dz  2 aLS  DS 
President University
Erwin Sitompul
aS

EEM 3/19
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Application of Gauss’s Law: Some Symmetrical Charge Distributions
 For one meter length, the inner conductor
has 2πaρS coulombs, hence ρL = 2πaρS,
L
D
a
2
 Every line of electric flux starting from the
inner cylinder must terminate on the inner
surface of the outer cylinder:
Qouter cyl  2 aLS ,inner cyl
2 bLS ,outer cyl  2 aLS ,inner cyl
a
 S ,outer cyl    S ,inner cyl
b
 If we use a cylinder of radius ρ > b,
then the total charge enclosed will be zero.
► There is no external field,
DS  0
President University
Erwin Sitompul
• Due to simplicity,
noise immunity and
broad bandwidth,
coaxial cable is still
the most common
means of data
transmission over
short distances.
EEM 3/20
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Application of Gauss’s Law: Some Symmetrical Charge Distributions
 Example
A 50-cm length of coaxial cable has an inner radius of 1 mm
and an outer radius of 4 mm. The space between conductors is
assumed to be filled with air. The total charge on the inner
conductor is 30 nC. Find the charge density on each conductor
and the expressions for E and D fields.
Qinner cyl  2 aLS ,inner cyl
Qinner cyl
  S ,inner cyl 
2 aL
30 109

2 (103 )(0.5)
Qouter cyl  2 bLS ,outer cyl  Qinner cyl
Qinner cyl
  S ,outer cyl 
2 bL
30 109

2 (4 103 )(0.5)
 2.39  C m 2
 9.55  C m 2
President University
Erwin Sitompul
EEM 3/21
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Application of Gauss’s Law: Some Symmetrical Charge Distributions
 S ,inner cyl
D  a

 10

3
(9.55 106 )
9.55


nC m 2
President University
E 
D
0
9.55 109

8.854 1012 
1079

V m

Erwin Sitompul
EEM 3/22
Chapter 3
Electric Flux Density, Gauss’s Law, and Divergence
Homework 3
 D3.3.
 D3.4.
 D3.5. All homework problems from Hayt and Buck, 7th Edition.
 Deadline: 1 May 2012, at 08:00.
President University
Erwin Sitompul
EEM 3/23

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