Lecture 5
Method of images
Energy stored in an electric field
Principle of virtual work
1
Method of Images
Consider the following problem
We place a charge +Q
above an infinite
conducting plane and
wish to find the field
above the plane as well
as the charge
distribution on the
plane.
2
Method of Images
We reason that:
• On the conducting plane the
induced surface charge is
negative.
• The magnitude of the surface
charge is largest just below
the charge +Q and tapers off
to zero far from the +Q charge.
• The E field lines terminate
normal to the surface of the
conductor.
• Conducting surface remains
an equipotential surface.
3
Method of Images
From Gauss’s Law
Just above the surface
s
E 
o
E=0
No 2 since E only on top side of conducting plate
Gauss’ Law Uniform surface charge
distribution Flat infinite surface
y
y
Flux
Charged surface
S
Gauss’ Law Uniform surface charge
distribution Flat infinite surface
Gaussian surface (imaginary)
Charge surface extends to
infinity in y-z plane
E
x
P

E
We want to obtain the electric field, magnitude
and direction, at a point situated on the
Gaussian surface
S
Charge density on surface
E 
 E  2EA
x
S A

S
2
Direction of E determined
from symmetry
Area A
4
Method of Images
Rigorous solution to problem:
•Solve Poisson’s equation
subject to the boundary
conditions V = 0 at x = 0.
2

 V 
o
•The electric field is then
obtained from.


E  V
• And the induced charge
distribution from.

V
 s   o E   o
x
Solving this way is a formidable task.
x 0
5
Method of Images
Alternate solution technique:
•Method of images ??????????
•Caution: The method of images is
not always the best way to proceed
in solving EM problems. You are
in the process of acquiring many
different techniques for solving
EM problems and it is up to you to
chose the best technique given the
problem.
Solving this way is much simpler
(for this geometry at least).
6
Method of Images
Consider the following charge
distribution, electric field lines
and equipotential lines.
d
+Q
Equipotential line
d
Electric field line
-Q
Similar to assignment 1 question
7
Method of Images
Consider the following charge
distribution, electric field lines
and equipotential lines.
d
d
Equipotential line: in 3-D it is
an equipotential surface
The bisecting plane is at zero potential because it is half way
between the two charges. V = 0
8
Method of Images
Consider the following charge
distribution, electric field lines
and equipotential lines.
Thin metal sheet inserted
d
d
Equipotential line: in 3-D it is
an equipotential surface
This means, we could insert a thin metallic sheet along this plane and it would
not disturb the electric field lines or the equipotential lines.
9
Method of Images
Equivalent regions
d
d
The field distribution and location
of charges above and on the
conducting plane are the same in
both figures. Top portions of
systems.
10
Method of Images
d
d
Solve this problem for two point
charges as shown and focus on
the electric field lines in the
upper portion of the figure only.
You remove the metal sheet at V
= 0 since it has no effect on the
field lines.
The result you get is valid for the
electric field above the ground plane
11
Method of Images
• Recognize that problem can be solved
by method of images.
Experience helps
• Place an imaginary thin metal sheet
along an equipotential surface.
Flat, curved, ….
• Image charges through the metal
sheet such that equipotential surface of
thin metal sheet is unchanged.
One or more charges,
distributions, …
• Remove the imaginary metal sheet
you put in.
• Obtain the electric field, potential, ….
in the region of interest.
Usually around
original charges
12
Method of Images
P
P
d
d
Electric field (charge distribution)
Electric field at point P
obtained from two point
charges. Charge density
from field at surface of
metal
q1
z

r1
 q2
r2
 
r  r1
 
r  r2

r
y
P
Two point charges
x

kq
E   1 2
r  r1
 r  r1
 
 r  r1

kq 2
   2
 r  r2
 r  r2
 
 r  r2



13
Method of Images
Equivalent systems through method of images
14
Method of Images
Equivalent systems through method of images
15
Method of Images
Equivalent systems through method of images
charge
metal

Here the image charges produce their own images and we
get an infinite number of image charges.

16
Method of Images
Equivalent systems through method of images
charge
Q’= - a / b Q
Q
a
Q
b
a2/b from center of sphere
(a  b) in this case
Metal sphere
17
Method of Images
A more math
approach to the
method of
images
18
Method of Images
Equivalent systems through method of images
For currents in conductors
19
Forces in Electrostatics
20
Forces in Electrostatics
Force on the charges on a
metal surface
da
dq
+ + + + + + + + + + + +

Einside  0
This charge distribution dq is in a
field of value:


Eothers 
nˆ
2 o
Proof of: see slides at end



Force on dq is thus: dF  dqEothers  dq
nˆ
2 o
General expression
With
Then
dq  da
 2
dF 
danˆ
2 o
dF  o E

da
2
2
Forces in Electrostatics

F
Side
View
Forces try to pull conductor apart.
- - - - - - - - - dq  da
+ + + + + + + + + +

F
For E = 106 volts/m,
- - - - - - - - - - - - - - conductor
+ + + + + + + + + + + + + + + +
Electrostatic forces are usually very small
F/a 4.4 N/m2
Forces in Electrostatics
Force per unit area

dF
dq

dF

dF
a
s
 s
E
o

Einside  0

dF
Using
dF  o E

da
2
2

dF

dF
Forces distributed over surface of sphere

 s2
dF  o  s2

rˆ 
rˆ
2
da
2 o
2 o
Principle of virtual work
Energy stored in electric field
Easy way to get expression
+Q
Consider a capacitor at potential difference V and of
charge +Q ,- Q on the plates. Area of plates (A) and
spacing (D)
-Q
 
Q
E  s 
o o A
Energy stored in the capacitor:
QV CV 2
U

2
2
But:
A
 o AV
 o AD  V   o E 2
CV
 AD 
U


  
2
2D
2 D
2
2
D
+V-
U
oE 2
2
volume
2
between plates 
2
U
o
2

E 2 dvol
Volume
29
Principle of virtual work
Conductor caries a surface charge of density , find force on plates of a
parallel plate capacitor. Plate area A

F
s

F
+Q

Eo
Field between plates
-Q
Principle of virtual work:
Find the work W required to increase (or decrease) plate
separation by S
Recall
 FS  W
W U
F

S
S
Principle of virtual work
W = change in energy stored in the system = U
Plate area A
+Q

F
s

E

F
Field between plates
-Q
U
 Q 
 AS   AS  
2
2
 o A 
oE
o
2
F
W U

S
S
2
U
Q2
Q Q
QE
F



S 2 o A 2  o A
2
Principle of virtual work
Capacitor plates carry a uniform charge of density , find force on the metal insert
introduced between the capacitor plates. Capacitor Plate area A = xL
+Q
+++++++

E
s

F
-------+++++++
-------L
d
Metal insert
-Q
y
What force is pulling metal insert into the capacitor?
W U
F

S
S
Can apply principle of virtual
work
Principle of virtual work
Capacitor plates carry a uniform charge of density , find force on the metal insert
introduced between the capacitor plates. Capacitor Plate area A = xL
Top view
x
d
y
L
Total energy stored in system
Metal
Metal
insert
insert
U  U L y  U y
Capacitor energy in region without insert
Capacitor energy in region with insert
Principle of virtual work
Capacitor plates carry a uniform charge of density , find force on the metal insert
introduced between the capacitor plates. Capacitor Plate area A = xL
x
U L y 
d
L
s
o E 2
2
x( L  y ) S
y

E
Uy 

F
d
L
U  U L y  U y 
o E
2
2
oE
2
2
xy( S  d )
x( L  y ) S  yS  d 
Principle of virtual work
Capacitor plates carry a uniform charge of density , find force on the metal insert
introduced between the capacitor plates. Capacitor Plate area A = xL
x
U
d
L
y
E
o
2
2
xLS  yd 
U  E xd
F

y
2
2
o

E
S
L

F
d
y
Force pulling metal insert into capacitor
F
 E xd
2
o
2
Principle of virtual work
Electrostatic actuators
For a parallel plate capacitor, the energy stored, U, is given in the
equation (where C is the Capacitance, and V is the voltage across
the capacitor).
When the plates of the capacitor move towards each other, the work done by the
attractive force between them can be computed as the change in U with distance (x).
The force can be computed by:.
Forces in Electrostatics
Note that only attractive forces can be generated in this instance. Also, to
generate large forces (which will do the useful work of the device), a large
change of capacitance with distance is required. This leads to the development
of electrostatic comb drives .
Comb Drives. These are particularly popular with surface micromachined devices. They consist of
many interdigitated fingers (a). When a voltage is applied an attractive force is developed between
the fingers, which move together. The increase in capacitance is proportional to the number of
fingers; so to generate large forces, large numbers of fingers are required. One potential problem
with this device is that if the lateral gaps between the fingers are not the same on both sides (or if
the device is jogged), then it is possible for the fingers to move at right angles to the intended
direction of motion and stick together until the voltage is switched off (and in the worst scenario,
they will remain stuck even then).
Forces in Electrostatics
Forces in Electrostatics
Forces in Electrostatics
Forces in Electrostatics
The micromechanical angular
rate sensor has a butterflyshaped polysilicon rotor
suspended above the
substrate, free to oscillate
about the center tether [top].
The rotor's perforations are a
necessary evil, needed to allow
etching beneath the rotor
during manufacturing. Four
interdigitated combs on the
outer edge of the rotor drive it
into resonant oscillation
[bottom]. Electrical leads carry
the driving signal to the combs
and the measurement signals
from the detection electrodes
below the rotor.
Next few slides: Proof of
Field by other charges on a
metal surface
Forces in Electrostatics
Conductor caries a surface charge of density 

dF
dq  da
Side view
Top view

Charge dq experiences electric field Eothers of all other charges in the syste


•
Therefore electric force is:
dF  dqEothers
Forces in Electrostatics

dF is
In electrostatics
perpendicular to the surface at dq.
If not, the charges would move
along the surface and  change
value.

dF
dq  da
Side view
Question (5)
Since dq is bound to the conductor
by internal forces, the forces acting
on charge dq are transmitted to the
conductor itself. Recall question 5 in
assignment.
Assignment 1 slide 6


dF  dqEothers
Forces in Electrostatics

nˆ
2 o
dq  da

nˆ
2 o
Gaussian surface
da

E
+ + + + + + + + + + + +

Einside  0
Electric field produced by charge
dq only.
conductor
Enlarged view of conducting
surface near dq
Electric field produced by all charges on conductor
Forces in Electrostatics
Forces in Electrostatics

2 o

2 o
Gaussian surface
da
nˆ
dq  da
nˆ
Electric field produced by
charge dq only.

E
  qenclosed
 E  da 
conductor
+ + + + + + + + + + + +

E in sid e  0
From Gauss’s law on surface element da
Enlarged view of
conducting surface near dq
o
s
 
E  nˆ
o
Electric field produced by all charges on conductor
Force on dq is not:


F  dqE
since dq on da contributes to electric field.
Must get

Eothers
Due to all other charges on conductor which act on dq. Then


F  dqEothers
Forces in Electrostatics

Edq up
First calculate electric field
produced by charge dq only.
Gaussian surface
da
+ + + + + +
The charge dq will
produce and electric
field out of the
conductor which will
add to the field
produced by all other
charges giving external
electric field:

Edq down
conductor
 
E  nˆ
o
Forces in Electrostatics
First calculate electric field
produced by charge dq only.
Gaussian surface
da

Edq up
+ + + + + +
The charge dq will also
produce and electric
field into the conductor
which must cancel
exactly with the field
produced by all other
charges giving internal
electric field:
conductor

Edq down

Einside  0
Forces in Electrostatics

Edq up
Same magnitude
By Gauss’s law

Eothers
These add to give:



Edqu p  Eother  nˆ
o
+ + + + + +

Edq down
conductor
These cancel exactly

Eothers


Edq down  Eother  0
Two equations
and two unknowns
Forces in Electrostatics
Forces in Electrostatics

Edq up
Same magnitude
By Gauss’s law

Eothers
These add to give:



Edqup  Eother  nˆ
o
+ + + + + +
conductor

Edq down

Eothers

Solving two equations gives:
Two equations
and two unknowns

These cancel exactly Edq down  Eother


Eother 
nˆ
2 o
Thus the segment da containing charge dq produces 1/2 “half” of the electric field
close to its charge distribution.
All other charges produce the other 1/2 “half” of the electric field.