X. Things We Skipped
XA. Capacitance Revisited
Two Component Capacitors
• In general, we defined capacitance (chapter 1) as the proportionality constant
between the voltage and the charge for a group of conductors
Qi   CijV j
• For a general combination of conductors, Cij is defined by:
j
• A very common actual situation is two conductors
+Q
with equal and opposite charge on them
• In this case we commonly define the capacitance as
the difference between the voltages on the two conductors Q  C V
–Q
C11C22  C21C12
• With some work, you can show that
C
C11  C22  C21  C12
• I don’t think this formula is ever useful
• Though we discussed them before dielectrics, they work well with dielectrics too
Energy in Capacitors
• In general, the energy in a capacitor can
U  12    r    r  d 3r
be most easily computed from the formula:
• For a two-component capacitor, the charge is on the two components
of the capacitor, and on each of them the potential is constant, so
U  12 QV
U  1 QV  1 Q V  1 QV  1 QV  1 QV
2
1 1
2
2 2
2
1
2
2
2
• This can be rewritten in terms of the capacitance in two ways:
U  12 C  V 
2
Q2

2C
Q  C V
Sample Problem X.1
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•
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•
•
•
•
•
A parallel plate capacitor has dimensions l  w and separation d much
smaller than the size. If a dielectric with dielectric constant  fills the
space to a distance x along the length l, what is the capacitance?
l
We need to find electric field
Away from edges, it will point straight across the gap
Since E|| is continuous, it will have the same value
in the dielectric and in vacuum
l–x
x
We therefore have E  Ezˆ
0
V  Ed
The potential difference is therefore b  t  d Edz
The displacement is
  Ezˆ on the left
D
on the left
 0 Ezˆ on the right
 V d
The surface charge on either surface is
  nˆ  D  
 0 V d on the right
Multiplying by the area, the charge is
w
V
V
C    x   0l   0 x 
 wx 
 0w l  x 
Q    da 
d
d
d
Sample Problem X.2
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•
For the capacitor problem in problem X.1, write the total energy for (a) fixed
potential difference V, and (b) fixed charge Q. In each of these cases, if the
dielectric is free to move, will it be pulled into or expelled from the gap?
l
w
C    x   0l   0 x 
2
d
Q
2
U  12 C  V  
2C
The first part is trivial:
Since for most materials,  > 0,
l–x
w
x
capacitance increases as x increases C  d     0  x   0l 
If V is fixed, increasing x increases the energy
If Q is fixed, increasing x decreases the energy
Systems tend to go to lower energy
This suggests, incorrectly, that dielectric is (a) expelled and (b) pulled in
In order to maintain constant V, there must be an external source of power
– Like a battery
In reality, in both cases the dielectric is drawn in
Sample Problem X.3 (’11 test) (1)
A parallel plate capacitor has length l, width w, and separation d, and is filled with
two triangular wedges of dielectric constants 1 and 2. Find the capacitance.
Assume d is small compared to the other dimensions.
l
• This is tricky because the boundary is at a weird angle
• Fortunately, if d is small, then the boundary is almost parallel
to the conducting plates
• Keeping in mind that D is continuous, this suggests a constant D  D  x  zˆ
D pointing straight across the boundary, depending only on x
• The electric field will depend on both x and z:
11 D  x  zˆ in material 1
E   1
• The vertical height of each of these
 2 D  x  zˆ in material 2
regions is given, as a function of x, by
x
 x
d1  d , d 2  d 1  
• The potential difference, is
l
 l
D  x  dx D  x  d  l  x 
d  x lx
V   Edz 

 D  x  

1l
 2l
l  1  2 
Sample Problem X.3 (’11 test) (2)
A parallel plate capacitor has length l, width w, and separation d, and is filled with
two triangular wedges of dielectric constants 1 and 2. Find the capacitance.
Assume d is small compared to the other dimensions
l
l

V
d  x lx
D  x 
V  D  x   

d  x 1   l  x   2 
l  1  2 
• The potential is a constant, so we now solve this for D
  nˆ  D  D  x 
• The surface charge density on the conductor is
• We now integrate this over the surface to get the charge:
w
l
dx
l 1 du
lwV
lwV 1 2  l 1 
Q    da  l V dy


ln


0
0
d
x 1   l  x   2 d 1 1  1  2  l  2 u d  2  1   l  2 
• Make a “u substitution” u  x 1   l  x   2
• So capacitance is
 2 
lw1 2
C
ln  
d   2  1   1 
XB. Conductivity
Conductivity
• We have mostly ignored the effects of conductivity, treating objects as perfect
conductors or perfect insulators
– Perfect conductors: E = 0; Perfect insulators: J = 0
• Realistic materials have finite conductivity
• Most materials, at reasonable electric fields, follow Ohm’s law: J   E
–  is the (microscopic) conductivity; unit: m-1ohm-1 = m-1-1
• Consider a macroscopic material in the shape of a cylinder
with conductivity  with length L and cross-sectional area A
L
• Apply a potential difference V across it
V
• Electric field is
 AV
E
ˆ
I   J  n da   EA 
• The current will be
L
L
• Define the resistance as
L
– Unit is ohm ( = V/A) R   A
V  IR
• Then we have macroscopic Ohm’s law
Sample Problem X.4 (’14 and ’15 exam) (1)
Two very large metal plates are held at a distance d apart. One is at potential 0,
and the other at potential V0. A metal sphere of radius a (a << d) is sliced in two,
and one hemisphere is placed on the grounded plate so that it’s potential is also
zero. If the region between the plates is filled with a material with conductivity
, determine an expression for the current that flows to the hemisphere.
  V0
• We need a picture
• Since the sphere is small, it will only have a local effect
– Background field from just the capacitor
0
• Because the plates are large,
V0
E   zˆ
we get a constant field
d
• But it won’t be constant near the hemisphere
• Goal – find a potential that
– Vanishes on the surface z = 0 and r = a
V0
– Has asymptotic constant E-field
 z
d
• Because E = – , we must have asymptotically
Sample Problem X.4 (’14 and ’15 exam) (2)
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•
Two very large metal plates are held at a distance d apart. One is at potential 0,
and the other at potential V0. A metal sphere of radius a (a << d) is sliced in two,
and one hemisphere is placed on the grounded plate so that it’s potential is also
zero. If the region between the plates is filled with a material with conductivity
, determine an expression for the current that flows to the hemisphere.
  V0
We’ll work in spherical coordinates
V0 z

at the center of the hemisphere
d
General solution
l
 l 1


A
r

B
r
 lm lm  Ylm  ,    0
in spherical

l ,m
coordinates is
Y10  ,    cos
Somehow, you have to remember that
Since z = rcos, we can therefore get right asymptotic properties just with A10
This suggests a solution of the form
   Ar  Br 2  cos 
Need it to vanish at r = a
V0
Solution is easy to find:
   r  a 3r 2  cos 
d
Sample Problem X.4 (’14 and ’15 exam) (3)
Two very large metal plates are held at a distance d apart. One is at potential 0,
and the other at potential V0. A metal sphere of radius a (a << d) is sliced in two,
and one hemisphere is placed on the grounded plate so that it’s potential is also
zero. If the region between the plates is filled with a material with conductivity
, determine an expression for the current that flows to the hemisphere.
V0
  V0
   r  a 3r 2  cos 
d
• We can get the current if we knew E: J   E
0
• The electric field is:


V0
V0 ˆ
3 3
ˆ
ˆ
ˆ
E    r   θ    r 1  2a r  cos   θ 1  a 3r 3  sin 
r

d
d
• We want it on the surface, where it has to be radial anyway E a  rˆ 3V0 cos 
 
• We integrate the current over the surface
d
2
1
I   J  rˆ dA   3V0 cos  dA   3V0 a 2
d  cos  d  cos  


0
0
d
d
3 V0 a 2
I 
d
Sample Problem X.2½ (1)
For the capacitor problem in problem X.1, the dielectric is slid into the capacitor while
a battery maintains the capacitor at constant voltage V. (a) How much does the
energy increase in the capacitor? (b) How much work does the battery do?
l
w
• The capacitance increases
C    x   0l   0 x 
wl
wl
d
C
 0  C  
d
d
• Therefore the charge
Q  C V  Q  C V
l–x
x
in the capacitor increases
wl
2
• Therefore the energy increases
U 




V


0 
1
1
2d
U  2 QV  U   2 QV
• The battery is continuously moving the charge from one side to the other
• The work required to move small charge Q across potential V: W   Q  V 
• The total work done is therefore
wl
2
W





V
W  W  V  Q  V  CV  CV 


0 
d
Sample Problem X.2½ (2)
For the capacitor problem in problem X.1, the dielectric is slid into the capacitor while
a battery maintains the capacitor at constant voltage V. (a) How much does the
energy increase in the capacitor? (b) How much work does the battery do?
l
wl
2
U 
   0  V 
2d
wl
2
W
    0  V 
d
x
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•
Why aren’t these numbers equal?
We are doing more work than the increase in the capacitor energy
There is also a force pulling the dielectric into the capacitor
This force is doing mechanical work on the dielectric
l–x
Finite Frequency Conductivity
• In the steady state, there is no particular reason for conductivity to be
independent of frequency
• We will, as always, assume all quantities are proportional to e –it
– We always take the real part at the end
J     E
• Ohm’s Law will now be frequency dependent:
• Assume we have a linear medium as well:
• Take a look at Ampere’s Law:
• This becomes
D     E

H  J  D
t
 H   E  i E
• One way to think of this is an effective complex susceptibility
 eff   
i

Free or Bound?
• What is the difference between free and bound charges?
• In a steady electric field, free charges move an unlimited distance, but bound
charges move only a finite distance
• In the presence of an oscillating field, even a free charge will only move a finite
distance
• It is therefore an artificial choice to distinguish
i
between free and bound charges
 eff   
• All that matters is the effective dielectric constant

• If you treat all charges as free, then we would have
i
D   0E
 eff   0 

• If you treat all charges as bound, then effectively the bound charges are just
included in 
• The presence of actual free electrons shows up as a divergence in Im() at  = 0
Energy Absorption
• Let’s, for the moment, treat all charge as
i  
 eff   0 
if it is explicit, rather than bound

J     E
• Then we have
• We recall that electric fields do work on charges:
dx
dW
 F
P
 q  E  v  B   v  qv  E
dt
dt
• The power per unit volume will, therefore, be 1 P  nqv  E  J  E   E  E
V
• Recall that actual fields are found by
multiplying by e–it and taking the real part, so
1
P  Re  Ee it   Re  Ee it   14  Eeit   *E*eit    Eeit  E*eit 
V
 14    *  E  E*  14  E  Ee 2it  14  *E*  E*e2it
1
2i

t
P  12 Re   E  E*
• If we do a time average, the terms with e
go away, so
V
• The real part of  causes energy loss/absorption
• If we work with eff, absorption given you information about Im(eff)
Sample Problem X.5
Suppose we have a thin gas of charged particles of charge q and mass m with number
density n, so thin they don’t collide. Find the conductivity as a function of .
 it
• Assume we have an electric field E  E0 e
• This will cause the charged particles to accelerate due to
• Their positions will then
d2
 it
m
x

q
E
e
0
satisfy the equation
dt 2
• Integrating, we find a velocity
v
• The current is therefore given by
• Conductivity is defined by Ohm’s Law
• The conductivity is therefore:
F  qE  qE0 e  it
d
iq
x
E0 e  it
dt
m
inq 2
J   v  nqv 
E0 e  it
m
J E
inq 2

m
Plasma Frequency
•
•
•
•
•
i
inq 2
 eff   


m
ne e 2
For a plasma of electrons, the effective dielectric constant is  eff   0 
2
m

e
2
Define the plasma frequency as
n
e
2
e


Then we have
p
  p2 
 0 me
 eff   0 1  2 
  
Consider an EM wave passing through a thin plasma. We have  2   k 2
The plasma contributes little to , so this is
k   0 eff
2
2
2

 1

p
2
  0 0 1  2   2  2   p2 
   c


 2  c2 k 2   p2
Plasma Frequency (2)
2
n
e
 p2  e
 0 me
 eff
  p2 
  0 1  2 
  


• As written, this applies only to free electrons at low density
– We ignored any binding of the electrons
– We ignored any collisions
• At high frequency, electrons move only a tiny distance
– Any restoring force is negligible
• Collisions are also negligible
• At sufficiently high frequency, all electrons are effectively free
• What about the nuclei? Why did we ignore them?
• Electron mass is 103 or more times lighter than the nuclei
 2  c2 k 2   p2
XC. The Kramers-Kronig Relations
Are All Dielectric Constants Possible?
• Let’s work in a notation where we incorporate everything into :
• We can find the polarization using the formula D   E  P
0
D  E
P     0  E
Are all possible complex functions () possible?
The reaction of P (the motion of the charges) must come after the stimulus E
Let G() be the polarization that results a time  after a stimulus E

Then we have
P  t    E  t    G   d
– Note that G must be real
0
• Let both P and E be periodic
P  t   Peit , E  t   Eeit
•
•
•
•
• Substituting, we have
• We therefore have
Pe
 it

  Eeit i G   d
0

     0   G   ei d
0
Analyticity and Properties of ()

     0   G   ei d
0
• Take the complex conjugate  *       G   ei d        0
0
0
of this expression
*





   
• We therefore have
• If we are in an insulator, G must fall off to zero as  goes to infinity
– Because the charges have to return to their starting point eventually
• This leads to a smooth, differentiable function for all real 
• We can also consider this formula for complex 
• We note that if Im() > 0, this expression will converge, and we will get a
smooth, differentiable function
• There are no “poles” or other problems for Im()  0
• We can use contour integration anywhere in the upper half of the complex plane
Using Contour Integration
• Consider the
      
0
I


P
d 
   
following integral:
 
• The “P” means principal value
• Integral has a pole at ' = 
• The “P” means to symmetrically skip the singularity

     0   G   ei d
0
Im 
      
         0

0
I    lim  
d   
d 

 
 0
  
  


• To find this integral, close contour in the uhp
– Justified since  – 0 small when Im() > 0
 Re 
• We now distort this contour into a tiny half-circle around 
• We now write       ei
 1

i
i




I    lim 




e



ie
d


i
lim




e
  0  d




0
i






 0 0  e
 0 0
      
1
0
     0  P 
d 
I     i      0 

i
  
Kramers-Kronig Relations
      0
1
     0  P 
d 
i
  
• To get the Kramers-Kronig relations, take real and imaginary parts of this equation
Re      0 
1
Im   
d 
Im     
1
P

Re      0
d 
  

  
*







 
• Then substitute   –  for the negative values of , and use
   Im     
1 
 Im      Im      

2

 d 

Re       0  P  

d




P

0
 0     
    
 0 2   2
1 
 Re        0 Re        0 

Im       P  

 d 
0

  
  






P

2


P

0
Re     0
  
2
2
d 

Comments on Kramers-Kronig Relations
Re      0 
2

P

0
 Im  
2  Re      0
d Im     
P
d 
2
2
2
2
0
  

  
• We note that if we want dispersion (  0), we must have an imaginary part to 
– Though not necessarily at the same frequency
• It is often easier to measure absorption than it is the dielectric constant
• Then Kramers-Kronig can then be used to find the real part
Sample Problem X.6 (1)
Suppose that absorption measurements indicated that the
A
Im     
imaginary part of the permittivity is as given at right.
2
2 2
  

What is the real part of the permittivity?
• We use one of the
Kramers-Kronig relations:
Re       0 
2A

 0 
 0 
P
 2 d  

0
 
2A
   
2
A
   
2
Re      0 
2


2 2

2 2
2

2

2

P

0
 Im  
d 
2
2
  

2 2
2
2
2 

2
2










 d 
P
 2

2
2
2
2
0    
  
2   2  





   
P  ln 

   
    
  

1    

tan   

2
2

   
 

0
2
2
2
2

Sample Problem X.6 (2)
Suppose that absorption measurements indicated that the
A
Im     
imaginary part of the permittivity is as given at right.
2
2 2
  

What is the real part of the permittivity?
Re       0 
A
   
2

2 2

   
P  ln 

   
    
  

1    

tan   

2
2

   
 

0
2
2
2
2

• The principal part only affects the logarithmic term
 

2
2

    
     
A
   

Re       0 
lim 
  ln 
  ln 

 
2
2 2   0



2
    
    0
       



 2   2         
A
Re       0 
lim 
 ln 
  ln 

2
2 2   0
2


2




2







    
Re       0 
A  2   2 
2   
2

2 2
XD. Waves with Complex Numbers
Poynting Vector and Complex Notation
S t   E t   H t 
• Recall for real waves we had
• For complex waves, E and H
have implied time-dependence
E  t   Re  Ee  it 
H  t   Re  He  it 
• The time-averaged Poynting vector is then
S  E t   H t  

1
4
 Ee
 it
 E*eit    Heit  H*eit 
E  H*  E*  H  E  He 2it  E*  H*e2it 
• We therefore have
• Also recall
1
4
S  12 Re  E  H* 
I S
1
4
E  H
*
 E*  H 
Complex Wave Numbers
S  12 Re  E  H* 
•
•
•
•
Suppose we had a complex permittivity 
Frequency and wave number k are related by: k   
Because  is complex, k will have an imaginary part k  k R  ik I
Our electric field
ik R x  k I x
ik x

E
e
e
E
x

E
e
  0
0
will look like
• And our magnetic field will look like
H B 
1
1

k E
• Poynting vector is
1
1
2
1
*
*
*
*
S 
Re E   k  E   
k R E0 e2k i x
Re  E  E  k  
2
2
2
• We note that intensity is falling exponentially, due to absorption
Square Roots of Complex Numbers
z  x  iy
• How do we take the square root of a complex number?
• First, write the complex number in the form
z  rei , r  z ,   arg  z 
 Im  z  
  tan 

 Re  z  
1
• The argument can be found from
• Then to find the square root, we simply write
z
ze
i

Sample Problem X.7
What’s the square root of i?
i  02  12  1
  arctan 1 0   12 
ie
1 i
2
i e
1 i
2
e
1 i
4
 cos  14    i sin  14  
1 i
i
2
ze
1 i
2
Good Conductors
• The effective dielectric constant is given by
 eff   
i

• Many materials have high conductivity, so that
 
– Free electrons move much farther than bound electrons
– Especially at low frequency
• Conductivity tends to also be real, at low frequencies
i
• In this case, we approximate
 eff 

• As frequencies increase, the current starts to lag the electric field
• This implies that  becomes complex
• Indeed, at high frequencies, we already talked about the plasma frequency
ine e2

m
Sample Problem X.8
A wave with frequency  is attempting to propagate through a good conductor
with conductivity . At what distance is the intensity down by a factor of 103?
i
• The effective permittivity is
 eff 

• We therefore have
i
k    eff  
 12  1  i 

• We define the skin depth

2

kI  1 
• The Poynting vector will have space dependence
• The intensity will have fallen by 103 when
x
3
 k i  x  12 ln 10   3.454

• So the answer is
x  3.454
I
1
2
e2k i x  103
kR E0 e2k i x
2
Index of Refraction and So On
• The index of refraction of a material was defined as

n
• If  is complex, we can still use
 0 0
all our formulas from before
• For example, we can find reflection and transmission probabilities using
Perpendicular
Parallel
E
2n cos 

,
2
2
2
E n cos   n  n sin 
E n cos   n2  n2 sin 2 

E n cos   n2  n2 sin 2 
E
2nn cos 

,
2
2
2
2
E n cos   n n  n sin 
E n2 cos   n n2  n2 sin 2 

E n2 cos   n n2  n2 sin 2 
• The complex results can result in changes in expected behavior
• The presence of complex numbers can lead to a phase shift at the boundary
• Define the phase shift as the argument
 Im  E  E  
  arctan 

  arg  E E 

 Re  E E  
Sample Problem X.9 (1)
A wave with frequency  is normally incident from vacuum on a good
conductor with conductivity  and susceptibility 0. What fraction of the
power is reflected, and what is the phase shift for that reflected light?
• The conductor has effective permittivity
• The index of refraction is then n 
 eff  i 
i


 0
 0 0
• For normal incidence, E  n  n
i  0  1
i   0



both formulas have
E n  n
i  0  1
i   0
• For good conductors,  >> 0


i 1   0 i
E 

E
i 1   0 i
  1 2

 0
2 0
 1  1  i 
i

Sample Problem X.9 (2)
A wave with frequency  is normally incident from vacuum on a good
conductor with conductivity  and susceptibility 0. What fraction of the
power is reflected, and what is the phase shift for that reflected light?
• The intensity is
2 0
E
 1  1  i 
E

I  E 

 1  1  i  2 0   1  1  i  2 0    1  2 0 
I
E
2

• Now work on the phase
2 0
I 
 1 2
I


 E  
  arg    tan 1  2 0 
 1  2  
 E 
0


  tan 1



2 0 

 
 
2
2 0

2 0 

2