Transcript Lecture 6 Slides elec 3105 cOMPLETEDx

ELEC 3105 Basic EM and
Power Engineering
Electric dipole
Force / torque / work on electric dipole
Z
1
The Electric Dipole
z
Consider electric field and
potential produced by 2
charges (+q, -q) separated
by a distance d.

r1
+q
q1

r1
 q2
r2
 
r  r1
 
r  r2

r
y
 r  r1
 
 r  r1
Two point charges

kq 2
   2
 r  r2
 r  r2
 
 r  r2

r2
x
P
x

kq
E   1 2
r  r1


r
d
Electric field (charge distribution)
z
P(x, z)



-q
The Electric Dipole
z
The dipole is represented by

p
a vector of magnitude qd
and pointing from –q to +q.

r1
+q

p
P(x, z)


r

r2
d
x
-q
Note: small letter p
Units
{p} dipole moment; Coulomb meter {Cm}
The Electric Dipole
Suppose (x, z) >>> d
P(x, z)
z

p
+q
d
-q


r

r1

r2
x
V  x, z  
zp
4 o r
3
The Electric Dipole
Suppose (x, z) >>> d
z

p
+q
d
-q
Spherical coordinates (r, , )


r

r1

r2
x
p cos 
V r ,  ,   
2
4 o r
P(r, , )
The Electric Dipole
Now to compute the electric field expression
Cartesian coordinates (x, z)
Spherical coordinates (r, , )
z

p
+q
d
-q
P(x, z))
P(r, , )


r


E  V

r1

r2
x
V  x, z  
zp
4 o
x
2
z
2

3
The Electric Dipole
Now to compute the electric field expression


E  V
V  x, z  

 V
V
V
E  
xˆ 
yˆ 
y
z
 x

E  E x xˆ  E z zˆ
V  x, z 
p 
Ex 

x
4 o 


V  x, z 
p 
Ez 

z
4 o 


zˆ 

 V

 y

  0



5 
x2  z 2 
3 xz
x

3z 2
2
p 
E
4 o 

z

2

 
5

  p 3 cos 2    1
3 
x 2  z 2  4 o
1




 xˆ  p 3 cos 2    1 zˆ
5 
2
2
4 o
x z 
3 xz



zp
4 o
x
2
z
2

3
The Electric Dipole
Spherical coordinates (r, , )


E  V
 
p cos 
pr
V r ,  ,   

2
4 o r
4 o r 3
 

1  pr 
E
 3 
4 o  r 

E
  
1  3 p  r r  
 p
3 
2
4 o r  r

No 𝜙 dependence
The Electric Dipole
p 
E
4 o 



 xˆ  p 3 cos 2    1 zˆ
5 
4 o
x2  z 2 
3 xz



Force on a dipole in a uniform electric field

E
Here consider
dipole as a rigid
charge
distribution

F
+q
d

p
No net translation
since
-q

F


F  F
Fˆ   Fˆ
Opposite
direction
Force on a dipole in a non-uniform electric field

E
Here
consider
dipole as a
rigid charge
distribution

F
+q
d

p
net translation since
-q

F


F  F
And / Or
Fˆ   Fˆ



Fnet  F  F
y
Force on a dipole in a non-uniform electric field



Fnet  F  F

E


F  qE ( x  x, y  y)


F  qE ( x, y)



Fnet  qE ( x  x, y  y )  qE ( x, y )

F
+q
Manipulate
expression to
get simple
useful form
y
d
-q

F

Fnet
x
x
y
Force on a dipole in a non-uniform electric field
After the manipulations end
we get:

E
 
Fx  p  E x
 
Fy  p  E y
 
Fz  p  Ez

  
Fnet   p  E 
-q

F

F
+q
d
y

p

Fnet
x
We will obtain this expression using a different technique.
x

E
Torque on a dipole
Here
consider
dipole as a
rigid charge
distribution

F
+q

p
d/2


d/2

F
-q
The torque components + and
 - act in the same rotational
direction trying to rotate the
dipole in the electric field.
Review of the
concept of torque
Torque on a dipole

Pivot

F

r



Torque: 
  
  r F
 
  r F sin  
  rF sin  
Moment arm length
Force
Angle between vectors r and F
For simplicity consider the
dipole in a uniform electric
field
Torque on a dipole
  rF sin  
d
  F sin  
2
d
  F sin  
2
Act in same
direction

      dF sin  
  pE sin  
d
 
  p E

p

 
  p E
-q


p  qd

F
+q
  (qd ) E sin  


F  qE

E

F

Also valid for small
dipoles in a nonuniform electric field.
Work on a dipole
By definition
dW  d
Consider work dW required
to rotate dipole through an
angle d
When you have rotation
If we integrate over
some angle range then

+q
W   d   pE sin  d
W   pE cos 
d
 
W  pE
 
  p E

F
-q

F


p

Work on a dipole
 
W  pE
W   pE cos 
For  = 90 degrees
W = 0. Thus  = 90
degrees is reference
orientation for the
dipole. It
corresponds to the
zero of the systems
potential energy as
well. U=W


E

p
 
  p E
Work on a dipole
 
W  pE
W   pE cos 
For  = 0 degrees
W = -pE. Thus  = 0
degrees is the
minimum in energy
and corresponds to
having the dipole
moment aligned
with the electric
field.

p
 0

E
 
  p E
Work on a dipole
 
W  pE
W   pE cos 
  180
For  = 180 degrees
W = pE. Thus  =
180 degrees is the
maximum in energy
and corresponds to
having the dipole
moment anti-aligned
with the electric field.

p

E
 
  p E
After the manipulations end
we get:
Force on a dipole

  
Fnet   p  E 

F
+q
d
We will obtain this expression using a different technique.
 
Work W   p  E
-q
Force F  W

F
Recall principle of virtual work and force
y

p
x
Exam question: once upon
a time
+q
Stator dipole
-q
+Q
2R
2r
Rotor dipole
(0,0)
a)
b)
c)
d)
-Q
E on +q
F on +q
 on +q
 on rotor
Exam question: Once upon
a time
+Q
D>>R
2R
e)  on dipole
-Q
Polarization

E
Atom
Negative electron cloud
Positive nucleus

p
No external electric field
With an external electric field
Charge polarization occurs in the presence of electric field
Polarization

E
Negative electron cloud
Positive nucleus
Each atom acquires
 a small
dipole moment p . For low
intensity electric fields the
polarization is expected to be
proportional to the field
intensity:


p  o E


p
With an external electric field
Atomic polarizability of the atom
Atomic Polarizability And
Ionization Potential
Polarization
No external electric field
With an external electric field
If the density of particles per cubic meter is N, the
net polarization is:


P  Np

{P} has units of {C/m2}
Polarization
Some molecules have built in dipole moments due to
ionic bonds.
Negative region
O2H+
𝑝
H+
Positive region
Water
𝑝
Polarization
Negative region
O2H+
𝑝
H+
𝑝
Positive region
For low intensity applied fields this polarization of the material is
again proportional to the field intensity so a more general
expression for polarization is:


P   C o E
With  C the electric susceptibility of the material
Polarization
For low intensity applied fields this polarization of the material is
again proportional to the field intensity so a more general
expression for polarization is:


P   C o E


Materials where P is proportional to E are called
DIELECTRIC materials.
Dielectrics
 medium   r  o
Why is the dielectric
constant in the medium
different than the dielectric
constant in vacuum?
The answer is contained in the nature of the
material being placed in the electric field.
Dielectrics
Consider the slab of
material “dielectric”
immersed
in an external

field Eo .
d
Endface
Area A

Eo
The molecules inside will be polarized due to the presence of
the electric field.
Dielectrics
-
+
dipole
Consider the slab of
material “dielectric”
immersed in an
external field Eo .
+
+ +
+ ++
+ +
+ ++++
+ + +
+ ++++ +
+ ++ + +
+ + + + +
+
+++ + ++++
+
+
+ +++ + ++
+ +
+
+ + +++ +
+ + + + +
+ +++++++
+
+ + +++
+ +
+ + +
+ +
+ +
+
+

Eo
The molecules inside will
be polarized due to the
presence of the electric
field.
Polarization induced surface charge density
 sp
SLAB
Area A
Dielectrics

p
For the single dipole,
the dipole moment is:
-


p  qd
dipole
For the dipoles along
one line between the
endfaces, the dipole
moment is:



p   p  q d
+

p
d


p  q  d
d
o


p  qd
Dielectrics
The total dipole moment
of the slab is then
.


p  Qd

Eo
Q
+
+ +
+ ++
+ +
+ ++++
+ + +
+ ++++ +
+ ++ + +
+ + + + +
+
+++ + ++++
+
+
+ +++ + ++
+ +
+
+ + +++ +
+ + + + +
+ +++++++
+
+ + +++
+ +
+ + +
+ +
+ +
+
+
Q: Total charge on one face of the slab.
Q   sp A
Dielectrics
The polarization or
dipole moment per unit
volume of the slab is
then.

Eo
d
+
+
++
+
+++++
++++
+
++++++++
+++++++
++
++++++
++++
++
++++++++
++++
++++++
+++
++++++
++
+++
++++
+++++
+++
+
End face Area A


 p Qd Q
P 
 dˆ   sp dˆ
v Ad A

P   sp dˆ
Electric flux Density
𝐷 = 𝜀𝐸
FLUX CAPACITOR
MEETS THE
FUSOR
Dielectrics
d
Electric field is shown E
o
normal to the surface.
Endface
Area A
o

Eo

Do
d

Ed

Dd
The electric flux density vectors Do = Dd . We are treating only
normal components here.
o

Eo

Do
Dielectrics
Original slab in
external electric
field
These two charge
sheets will produce
an electric field
directed from the
positive sheet
towards the negative
sheet.
Polarization
bound charge
sheets of each
endface.
In order to determine the magnitude of the electric
field, a parallel plate capacitor analysis can be
applied to this charge configuration.
Dielectrics

P

Ei

Di
Polarization bound
charge sheets of
each end face.
Induced electric field due to
the polarization effect in the
dielectric.
Through Gaussian
analysis:
  sp
Ei 
o


Di   sp   P
Note: The external field Eo and the induced field Ei are in opposite
directions.
If electrons where free in the dielectric, then the magnitudes of Eo and Ei would end up the same,
their directions would be opposite and the net electric field inside the medium would be zero. Such
a material is called a metal due to the free electrons.
Dielectrics
d
Further manipulations of equations
required to obtain desired result.
Recall that desired result is: Why
mediums have a different dielectric
constant from that of vacuum?
o

Eo

Do
Endface
Area A
d

Ed

Ei

Dd

Di

P
 sp
o

Eo

Do
Dielectrics

Eo

Eo

Eo


o
  sp
Vector
diagram
d
Endface
Area A

Ei
o

Ed
d



Ed  Eo  Ei
inside dielectric
  sp
Dielectrics

Eo
Endface
Area A
d

Eo

Eo


o

Ei
o

Ed
d
  sp
Since:


Di   sp   P
and
  sp
  sp
Ei 
o




Then P   E   E  E 
o i
o
d
o
Vector
diagram
Dielectrics

Eo
Endface
Area A
d

Eo

Eo


o

Ei
o

Ed
  sp



And:  E   E  P
o o
o d



Then D   E  P
d
o d
d
  sp


with  o Eo   d Ed
Vector
diagram
Dielectrics

Eo
Endface
Area A
d

Eo

Eo


o

Ei
o

Ed
d
  sp
  sp


 d Ed :
The dielectric constant can now be obtain Ded
d using
P
d  o 
Ed
Vector
diagram
Dielectrics
With  C the electric susceptibility of the material.
P
d  o 
Ed
 d   o 1   


P   C o E
Why is the dielectric
constant in the medium
different than the
dielectric constant in
vacuum?
Answer: Polarization
and orientation of
internal dipoles.
 r  1   
ELEC 3105 Basic EM and
Power Engineering
Next slides: forget me not
Dielectric Materials
Polarization Field
P = electric flux density induced by E
Electric Breakdown
Electric Breakdown
Electric flux
Density
𝐷 = 𝜀𝐸
From other definitions of flux
we can obtain other useful
expressions for electrostatics

 E   E  dA
E 
S

 
 E   E  dA     E dV
S
E 
V
Divergence
theorem
qenclosed
o

 
V
   E dV   dV
V
V
o
V
V
dV
o
qenclosed
o
Divergence theorem
 
V
   E dV   dV
V
S
o
Integrands must be the same for all dV
  V
E 
o
Medium dependence
Point function
Gauss’s law in differential form
  V
E 

Divergence theorem


    o E dV   V dV
V
S
Integrands must be the same for all dV

  D  V
Point function
Gauss’s law in differential form
No dependence on the dielectric constant
ELECTROSTATICS
Boundary conditions Normal Component of
Dn1  Dn 2   s
Dn1  Dn 2  0
Dn1  Dn 2

D
Dn1   s
Air Dielectric
Gaussian Surface
Gaussian surface on
metal interface encloses
a real net charge s.
Gaussian surface on
dielectric interface
encloses a bound
surface charge sp , but
also encloses the other
half of the dipole as
well. As a result
Gaussian surface
encloses no net surface
charge.
Top Ten
1. What is the definition of a shock absorber?
A careless electrician.
2. Do you know how an electrician tells if he's working
with AC or DC power? If it's AC, his teeth chatter
when he grabs the conductors. If it's DC, they just
clamp together.
3. What did the light bulb say to the generator?
I really get a charge out of you.
4. What Thomas Edison's mother might have said to her son: Of course I'm
proud that you invented the electric light bulb.
Now turn it off and get to bed.
5. Sign on the side of the electrician's van: Let Us Get Rid of Your Shorts.
6. Two atoms were walking down the street one day, when one of them
exclaimed, 'Oh no - I've lost an electron!'
'Are you sure?' the other one asked.
'Yes,' replied the first one, 'I'm positive.'
7. Why are electricians always up to date?
Because they are "current" specialists.
8. What would you call a power failure?
A current event.
9. Did you hear about the silly gardener?
He planted a light bulb and thought he
would get a power plant.
10. How do you pick out a dead battery
from a pile of good ones?
It's got no spark.