Lecture 14
Capacitance and Conductance
Sections: 6.3, 6.4, 6.5
Homework: See homework file
Definition of Capacitance
capacitance is a measure of the ability of the physical structure to
accumulate electrical free charge under certain voltage
Q
C  , F=C/V
V
C
D

  E  ds
S
N
Gauss law
V
S
, F
P
 E  dL
P
E
LECTURE 14
E
Q
r
Q
E
N
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2
Capacitance of Parallel-Plate Capacitor
s  an  D  0
an
s
E  az

an
s  an  D  0
C 

 E  ds
S
N
 E  dL

S
Ez  A
A
 C  , F
Ez  d
d
P
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3
1. A capacitor whose insulator has relative permittivity εr1 = 1
has capacitance C1 = 1 μF. What is going to be its capacitance
if the insulator is replaced by another one with εr2 = 30?
C2 
2. If the two capacitors (of C1 and of C2) are biased with voltage V
= 1 kV, what would be their respective charges (Q1 and Q2)?
Q1 
Q2 
3. What is the nature of the charges Q1 and Q2?
(a) free charge deposited on electrode surface
(b) bound charge deposited on the insulator surface at the electrode
(c) total charge at the electrode-insulator surface
LECTURE 14
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4
4. In the previous example:
(a) Find the ratios of the free-to-total charge at the
insulator-electrode interface.
(b) Find the ratios of bound-to-total charge.
(c) Compare the total (free + bound) charge values.
 
     
 sf
 sb
       
LECTURE 14
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5
Capacitance and Stored Energy – 1
• general energy expression (Lecture 9)
1
We   ( vf  V )dv
2 v
• there are two electrodes: one at a potential V1 and the other at V2
• charge is distributed on the surface of the electrodes
1
1
(1)
(1)
We  V1 
 sf ds  V1Q1

2 S
2
1
1
(2)
We(2)  V2 

ds

V2Q2
sf

2 S
2
• the capacitor is assumed charge neutral as a whole both before
and after voltage is applied (charge conservation)
Q1  Q2  0  Q1  Q2  Q
LECTURE 14
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6
Capacitance and Stored Energy – 2
• total energy of the two electrodes
We  We(1)
 We(2)
1
 Q (V1  V2 )
2
1
 We  QV12
2
1
1
 We  (C V12 )V12  CV122
2 
2
Q
1 Q 1 Q2
 We  Q 
C 2 C
2 
V12
2We
C 2
V
LECTURE 14
Q2
C
2We
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7
Capacitance Example 1: Double-Layer Plate Capacitor
• voltage between plates V  E1d1  E2 d 2
• at the dielectric interface
 E1 
C
C
Dn1  Dn 2
why
1
 E2  E1
2
V
V
  s  D1  1E1 
d1  d 2 (1 /  2 )
( d1 / 1 )  ( d 2 /  2 )
Q | s | S

V
V
1
d1
d2

1S  2 S

s
1
1
1

C1 C2
s
2 capacitors in series
LECTURE 14
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8
Equivalence of Metallic Structures
• principle: placing a PEC sheet at an equipotential surface does not
change the field distribution → capacitance does not change
• follows from the uniqueness theorem: potential values at the
boundary surfaces remain the same
V  10 V                     

V  6 V        

E
V 0 V 
• the structure is effectively split into two
capacitors in series
V 10
V 6 V
LECTURE 14
V 0 V
9
Capacitance Example 2: Spherical Capacitor

 D  ds  Q   | E | 4 r
2
Q
Q
r
S
1 Q
E
ar
2
4 r
b
V   E  dL 
a
Q
4
S
 1 1  Q (b  a )
  
 a b  4 ab
Q
a
b
E

Q
ab
C   4
, F
V
(b  a )
ab
single sphere capacitance: Ca  lim C  lim 4
 4 a, F
b 
b 
(b  a )
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10
Capacitance per Unit Length: Parallel-Plate Line
A
wl
C  
, F
h
h
 C 
h
w
C
w
  , F/m
l
h
l
LECTURE 14
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11
Capacitance per Unit Length: Coaxial Cable
cross-section
2b
2a
• apply Gauss’ law to find E field
l
1 Q
E 

 , V/m
2 2 l
V0
• find voltage from E
V0  
b
a
b
E d  
ln   , V
2 l  a 
Q
V 0
• find capacitance from voltage
Q
2 l

C
,F
V0 ln(b / a )
• find PUL capacitance
C
2
C  
, F/m
l ln(b / a )
LECTURE 14
E
l
S

slide
12
Capacitance per Unit Length: Twin-Lead Cable – 1 (optional)
Step 1: Find equation of equipotential lines of two line charges at
x = −s and x = s.
l
 l
s
s
ln , V 
ln
• at observation point P: V 
2  2
2 1
l  1 
 V ( P )  V  V 
ln  
2   2 
y
h
1
s
s
0
V 0
r
 l
h
P
2
r
 l
x
equipotentials
LECTURE 14
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13
Capacitance per Unit Length: Twin-Lead Cable – 2 (optional)
• at an equipotential line V = Vc we have
 2Vc 
 1  2Vc
1
ln   

 K  exp 





2 
2
l
l



K
• at P(x,y)
( s  x) 2  y 2
1
K

2
( s  x) 2  y 2
• squaring and re-arranging we obtain the equation of a circle
2
2


K 1
 2 Ks 
2
 x s 2   y   2 
1
K
1



 K
2
h
r
radius
x-coordinate of center
LECTURE 14
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14
Capacitance per Unit Length: Twin-Lead Cable – 3 (optional)
• the equation of the equipotential line V = Vc
2
2

K 1
2
Ks


2
x

s

y



 2 
2
K 1 
 K 1 

is a circle of radius
2 Ks
r 2
2
h
K 1
h
K     1
and a center on the x-axis at a distance
r
r
from the origin  K 2  1 
2
2
s
h
r


h  s 2

 K 1
Step 2: Construct an equivalent problem of wires of finite radius r
by placing the wires so that their surfaces coincide with the equipotential lines of the ideal line charges (a distance h from origin).
2
LECTURE 14
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15
Capacitance per Unit Length: Twin-Lead Cable – 4
• potential at the positive wire
• potential at the negative wire
l
V1 
ln K
2

V2   l ln K
2
• potential difference between wires
2


l
h
h

V12  V1  V2 
ln      1 
  r

r

K
C 
l
V12


2
h

h

ln      1 
r
 r

, F/m
LECTURE 14
C 

 2h 
ln  
 r 
, hr
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16
Analogy between Capacitance and Conductance
Q
C 
V
  E  ds

S
N
 E  dL
, F
I
G 
V
P
  E  ds

S
N
, S
 E  dL
P
for a given geometry the expressions for capacitance and
conductance are identical except for the material constant
Examples:
1) coaxial capacitor/resistor
2 l
2 l
ln(b / a )
C
, F  G
, S; R 
,
ln(b / a )
ln(b / a )
2 l
2) homework: conductance G of a parallel-plate resistor
LECTURE 14
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17
Conductance per Unit Length – 1
3) coaxial cable with lossy (dissipative) insulator (σd ≠ 0)
2 d
2

, F/m  G 
, S/m
ln(b / a )
ln(b / a )
• note that a cable in general suffers loss not only due to the
conducting wires (described by Rꞌ) but also due to its non-ideal
insulator (current flows through the insulator, described by Gꞌ)
C 
• the loss in the metallic leads of a coaxial
cable Rꞌ was obtained in Lecture 10:
1  1
1 

R 
 2
 ,  /m
2
2
 m  a c  b 
I metal
no skin effect taken into account!
I metal
NOTE: G   1 / R
LECTURE 14
I diel
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Conductance per Unit Length – 2
4) twin-lead cable with lossy insulator (σd ≠ 0)
C 

2
h

h

ln      1 
r
r



, F/m  G 
 d
2
h

h

ln      1 
r
r



the loss in the metallic wires of a twin-lead
cable was obtained in Lecture 10
1

,  /m
R  2
mA
, S/m
G   1 / R
5) homework: derive the conductance per unit length Gꞌ of a
parallel-plate line
LECTURE 14
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Parameters per Unit Length in Circuit Models of TLs
iN
iN 1
Rl
vN
Ll
Gl
l
LECTURE 14
C l vN 1
z
slide
20
You have learned:
what capacitance is and what capacitance per unit length is
how to calculate capacitance from the field distribution
how capacitance relates to the stored electric energy
how to calculate the capacitance per unit length and the
conductance per unit length of a parallel-plate line, coaxial cable
and twin-lead cable
LECTURE 14
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