Engineering Electromagnetics
Lecture 9
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
2 0 1 3
President University
Erwin Sitompul
EEM 9/1
Chapter 6
Dielectrics and Capacitance
Capacitance
 Now let us consider two conductors
embedded in a homogenous dielectric.
 Conductor M2 carries a total positive
charge Q, and M1 carries an equal
negative charge –Q.
 No other charges present  the total
charge of the system is zero.
• The charge is carried on the surface as
a surface charge density.
• The electric field is normal to the
conductor surface.
• Each conductor is an equipotential
surface
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Erwin Sitompul
EEM 9/2
Chapter 6
Dielectrics and Capacitance
Capacitance
 The electric flux is directed from M2
to M1, thus M2 is at the more positive
potential.
 Works must be done to carry a
positive charge from M1 to M2.
 Let us assign V0 as the potential
difference between M2 and M1.
 We may now define the capacitance
of this two-conductor system as the
ratio of the magnitude of the total
charge on either conductor to the
magnitude of the potential difference
between the conductors.
Q
C
V0
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Erwin Sitompul
C

S
 E  dS

  E  dL

EEM 9/3
Chapter 6
Dielectrics and Capacitance
Capacitance
 The capacitance is independent of the potential
and total charge for their ratio is constant.
 If the charge density is increased by a factor,
Gauss's law indicates that the electric flux
density or electric field intensity also increases
by the same factor, as does the potential
difference.
C

S
 E  dS

  E  dL

 Capacitance is a function only of the physical dimensions of
the system of conductors and of the permittivity of the
homogenous dielectric.
 Capacitance is measured in farads (F), 1 F = 1 C/V.
President University
Erwin Sitompul
EEM 9/4
Chapter 6
Dielectrics and Capacitance
Capacitance
 We will now apply the definition of capacitance to a simple twoconductor system, where the conductors are identical, infinite
parallel planes, and separated a distance d to each other.
S
E
az

D  S a z
 The charge on the lower plane is positive, since D is upward.
DN  Dz   S
 The charge on the upper plane is negative,
DN   Dz   S
President University
Erwin Sitompul
EEM 9/5
Chapter 6
Dielectrics and Capacitance
Capacitance
 The potential difference between lower and upper planes is:
V0   
lower
upper
E  dL   
0
d
S
S
 dz 
d


 The total charge for an area S of either plane, both with linear
dimensions much greater than their separation d, is:
Q  S S
 The capacitance of a portion of the infinite-plane arrangement,
far from the edges, is:
C
Q S

V0
d
President University
Erwin Sitompul
EEM 9/6
Chapter 6
Dielectrics and Capacitance
Capacitance
 Example
Calculate the capacitance of a parallel-plate capacitor having a
mica dielectric, εr = 6, a plate area of 10 in2, and a separation
of 0.01 in.
S  10 in 2
 10 in 2  (2.54 102 m in) 2
 6.452 103 m2
C
d
(6)(8.854 1012 )(6.452 10 3 )

2.54 104
 1.349 nF
d  0.01 in
 0.01in  (2.54 102 m in)
 2.54 104 m
President University
S
Erwin Sitompul
EEM 9/7
Chapter 6
Dielectrics and Capacitance
Capacitance
 The total energy stored in the capacitor is:
WE  12   E 2 dv
vol
2
 S 
1
 2     dv
vol
  
2
1 S d S
 2 
dzdS
0 0 
2

 12 S Sd
 2
1  S S
2
2
d
d 2
C
V0 
d
S
d

Q
C
V0
2
Q
WE  12 CV02  12 QV0  12
C
President University
S
Erwin Sitompul
EEM 9/8
Chapter 6
Dielectrics and Capacitance
Several Capacitance Examples
 As first example, consider a coaxial cable or coaxial capacitor
of inner radius a, outer radius b, and length L.
L
a
 The capacitance is given by:
V 
ln
2
ab
Q
2 L
C

Vab ln(b a)
b
Q  L L
 Next, consider a spherical capacitor formed of two concentric
spherical conducting shells of radius a and b, b>a.
Q
4
C

Vab 1  1
a b
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Q
Vab 
4
Erwin Sitompul
EEM 9/9
1 1
  
a b
Chapter 6
Dielectrics and Capacitance
Several Capacitance Examples
 If we allow the outer sphere to become infinitely large, we
obtain the capacitance of an isolated spherical conductor:
C  4 a
 A sphere about the size of a marble, with a diameter of 1 cm,
will have:
C  0.556 pF
 Coating this sphere with a different dielectric layer, for which
ε = ε1, extending from r = a to r = r1,
Q
Dr 
4 r 2
Q
Er 
(a  r  r1 )
2
41r
Q

(r  r1 )
4 0 r 2
President University
Erwin Sitompul
EEM 9/10
Chapter 6
Dielectrics and Capacitance
Several Capacitance Examples
 While the potential difference is:
a Qdr
r1 Qdr
Va  V   

r1 4 r 2
 4 r 2
1
0
Q

4
1 1 1 1 
   


a
r

r
1 
0 1
 1
 Therefore,
4
C
1 1 1 1
  
1  a r1   0 r1
President University
Erwin Sitompul
EEM 9/11
Chapter 6
Dielectrics and Capacitance
Several Capacitance Examples
 A capacitor can be made up of several dielectrics.
 Consider a parallel-plate capacitor of area S and spacing d,
d << linear dimension of S.
 The capacitance is ε1S/d, using a dielectric of permittivity ε1.
 Now, let us replace a part of this dielectric by another of
permittivity ε2, placing the boundary between the two dielectrics
parallel to the plates.
• Assuming a charge Q on
one plate, ρS = Q/S, while
DN1 = DN2, since D is only
normal to the boundary.
• E1 = D1/ε1 = Q/(ε1S),
E2 = D2/ε2 = Q/(ε2S).
• V1 = E1d1,
V2 = E2d2.
Q
1
Q
1


C

d1
d
1
1
V1  V2
V0
 2

1S  2 S C1 C2
President University
Erwin Sitompul
EEM 9/12
Chapter 6
Dielectrics and Capacitance
Several Capacitance Examples
 Another configuration is when the
dielectric boundary were placed
normal to the two conducting plates
and the dielectrics occupied areas
of S1 and S2.
• Assuming a charge Q on one plate,
Q = ρS1S1 + ρS2S2.
• ρS1 = D1 = ε1E1,
ρS2 = D2 = ε2E2.
• V0 = E1d = E2d.
Q 1S1   2 S 2

C
 C1  C2
d
V0
President University
Erwin Sitompul
EEM 9/13
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 The configuration of the two-wire line consists of two parallel
conducting cylinders, each of circular cross section.
 We shall be able to find complete information about the electric
field intensity, the potential field, the surface charge density
distribution, and the capacitance.
 This arrangement is an important type of transmission line.
• Schematics of a
transmission line
President University
Erwin Sitompul
EEM 9/14
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 The capacitance, together with conductance, forms a shunt
admittance of a transmission line.
 The line capacitance is proportional to the length of the
transmission line.
 When an alternating voltage is applied to the line, the line
capacitance draws a leading sinusoidal current, called the
charging current.
 The charging current is negligible for lines
less than 100 km long. For longer lines, the
capacitance becomes increasingly
important and has to be accounted for.
 The value of such capacitance is
significantly higher with underground
cables than with overhead lines, due to the
close proximity of the individual conductors.
President University
Erwin Sitompul
EEM 9/15
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 The potential field of two
infinite line charges, with
a positive line charge in
the xz plane at x = a and
a negative line at x = –a,
is shown below.
 The potential of a single
line charge with zero
reference at a radius of
R0 is:
R0
L
V
ln
2
R
 The combined potential field can be written as:
R10 R2
L
R20 
 L  R10
ln
V
 ln
 ln

2  R1
R2  2 R20 R1
President University
Erwin Sitompul
EEM 9/16
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 We choose R10 = R20, thus placing the zero reference at equal
distances from each line.
 Expressing R1 and R2 in terms of x and y,
 L ( x  a)2  y 2
L
( x  a) 2  y 2

ln
V
ln
2
2
4 ( x  a)2  y 2
2
( x  a)  y
 To recognize the equipotential surfaces, some algebraic
manipulations are necessary.
 Choosing an equipotential surface V = V1, we define a
dimensionless parameter K1 as:
K1  e 4V1
L
so that
( x  a) 2  y 2
K1 
( x  a) 2  y 2
President University
Erwin Sitompul
EEM 9/17
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 After some multiplications and algebra, we obtain:
K1  1
2
x  2ax
 y2  a2  0
K1  1
2
2
 2a K1 

K1  1 
2

xa
  y  

K1  1 
K

1

1


 The last equation shows that the
V = V1 equipotential surface is
independent of z and intersects
the xy plane in a circle of radius b,
2a K1
b
K1  1
 The center of the circle is x = h, y = 0, where:
K1  1
ha
K1  1
President University
Erwin Sitompul
EEM 9/18
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 Let us now consider a zero-potential conducting plane located
at x = 0, and a conducting cylinder of radius b and potential V0
with its axis located a distance h from the plane.
 Solving the last two equations for a and K1 in terms of b and h,
a  h2  b2
h  h2  b2
K1 
b
 The potential of the cylinder is V0, so that:
K1  e2V0
L
 Therefore,
4V0 2V0
L 

ln K1 ln K1
President University
Erwin Sitompul
EEM 9/19
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 Given h, b, and V0, we may determine a, K1, and ρL.
 The capacitance between the cylinder and the plane is now
available. For a length L in the z direction,
C
C
L L
V0

4 L 2 L


ln K1 ln K1
2 L
ln  h  h2  b 2


2 L

1
cosh
( h b)

b

• Prove the equity by solving
quadratic equation in eα,
where cosh(α)=h/b.
• cosh(α) = (eα+e–α )/2
President University
Erwin Sitompul
EEM 9/20
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 Example
The black circle shows the
cross section of a cylinder of
5 m radius at a potential of
100 V in free space. Its axis is
13 m away from a plane at
zero potential.
b  5, h  13, V0  100
a  h2  b2  132  52  12
h  h2  b2 13  12
K1 

 5  K1  25
b
5
4 V0 4 (8.854 1012 )(100)
L 

 3.46 nC m
ln K1
ln 25
2 (8.854 1012 )
2

 34.6 pF m
C
1
1
cosh (13 5)
cosh (h b)
President University
Erwin Sitompul
EEM 9/21
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 We may also identify the
cylinder representing the 50 V
equipotential surface by
finding new values for K1, b,
and h.
K1  e 4V1  L
4 8.8541012 50 3.46109
e
5
2a K1 2  12 5
 13.42 m
b

5 1
K1  1
ha
K1  1
5 1
 18 m
 12
K1  1
5 1
President University
Erwin Sitompul
EEM 9/22
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 L
( x  a)2  y 2 
E   
ln
2
2
2

(
x

a
)

y


 L  2( x  a)a x  2 ya y 2( x  a)a x  2 ya y 




2
2
2  ( x  a)  y
( x  a) 2  y 2 

D  E =  L
2
 2( x  a)a x  2 ya y 2( x  a)a x  2 ya y 



2
2
2
2
(
x

a
)

y
(
x

a
)

y


 S ,max   Dx , x  h b , y 0 =
 S ,max
3.46 109

2
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L
2
 hb a
hba 

 (h  b  a) 2 (h  b  a) 2 


 13  5  12
13  5  12 
2


0.165
nC
m
 (13  5  12) 2 (13  5  12) 2 


Erwin Sitompul
EEM 9/23
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
L
2
 S ,min  Dx, x hb, y 0 =
 13  5  12
13  5  12 
2


0.073
nC
m
 (13  5  12) 2 (13  5  12) 2 


+
+
+
+
+
+
+
+
-
-
 S ,min
3.46 109

2
 hba
hba 

 (h  b  a) 2 (h  b  a) 2 


-
-
-
 S ,max   Dx , x  h b , y 0
 S ,min  Dx , x  h b, y 0
 S ,max  2.25 S ,min
President University
Erwin Sitompul
EEM 9/24
Chapter 6
Dielectrics and Capacitance
Capacitance of a Two-Wire Line
 For the case of a conductor with b << h, then:

ln  h  h 2  b 2

2 L
C
ln(2h b)
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
b

ln  h  h  b 
ln  2h b 
(b  h)
Erwin Sitompul
EEM 9/25
Chapter 6
Dielectrics and Capacitance
Homework 8
 D6.4.
 D6.5.
 D6.6.
 All homework problems from Hayt and Buck, 7th Edition.
 Due: Monday, 17 June 2013.
President University
Erwin Sitompul
EEM 9/26