1. ELEC 3105 Basic EM and Power Engineering
Lecture 5
Method of images
Energy stored in an electric field
Principle of virtual work

2. Consider the following problem
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.

3. Method of Images
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.
We reason that:

4. Method of Images From Gauss’s Law Just above the surface E = 0
No 2 since E only on top side of conducting plate
Clarify on the board

5. Method of Images Solving this way is a formidable task.
Rigorous solution to problem:
Solve Poisson’s equation subject to the boundary conditions V = 0 at x = 0.
The electric field is then obtained from.
And the induced charge distribution from.
Solving this way is a formidable task.

6. Method of Images Alternate solution technique:
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).

7. Method of Images
Consider the following charge distribution, electric field lines and equipotential lines. +Q d
Equipotential line d
Electric field line -Q
Similar to assignment 1 question

8. Method of Images
Consider the following charge distribution, electric field lines and equipotential lines. d
Equipotential line: in 3-D it is an equipotential surface d
The bisecting plane is at zero potential because it is half way between the two charges. V = 0
Clarify on the board

9. Method of Images
Consider the following charge distribution, electric field lines and equipotential lines.
Thin metal sheet inserted d
Equipotential line: in 3-D it is an equipotential surface d
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.

10. Method of Images d
Equivalent regions
The field distribution and location of charges above and on the conducting plane are the same in both figures. Top portions of systems.

11. Method of Images 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

12. Method of Images
Recognize that problem can be solved by method of images.
Place an imaginary thin metal sheet along an equipotential surface.
Image charges through the metal sheet such that equipotential surface of thin metal sheet is unchanged.
Remove the imaginary metal sheet you put in.
Obtain the electric field, potential, …. in the region of interest.
Experience helps
Flat, curved, ….
One or more charges, distributions, …
Usually around original charges

13. Method of Images P d P
Electric field at point P obtained from two point charges. Charge density from field at surface of metal

14. Method of Images
Equivalent systems through method of images

15. Method of Images
Equivalent systems through method of images

16. 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.  

17. Method of Images Equivalent systems through method of images charge
Q’= - a / b Q a Q Q b
a2/b from center of sphere
(a  b) in this case
Metal sphere

18. A more math approach to the method of images

19. For currents in conductors
Method of Images
Equivalent systems through method of images
For currents in conductors

20. ELEC 3105 Basic EM and Power Engineering
Forces in Electrostatics

21. Forces in Electrostatics
Force on the charges on a metal surface
This charge distribution dq is in a field of value: da dq
Proof of: see slides at end
Force on dq is thus:
General expression
With
Then

22. Forces in Electrostatics
Forces try to pull conductor apart.
Side
View
conductor
These types of electrostatic forces are usually very small
For E = 106 volts/m,
F/a 4.4 N/m2

23. Forces in Electrostatics
Consider a conducting sphere uniformly charged +Q a
Radial outward s
Total charge :
Then

24. Forces in Electrostatics
Force per unit area dq a
Using s
Forces distributed over surface of sphere

25. Forces in Electrostatics
Total radial force a s
Forces distributed over surface of sphere, but, there is no translational force on the sphere since all force elements cancel in pairs.

26. Forces in Electrostatics
Consider a conducting sphere placed in a uniform electric field. a s
Induced surface charge given by

27. Forces in Electrostatics
Net outward force trying to pull sphere apart.
Forces distributed over surface of sphere, but, there is no translational force on the sphere since all force elements cancel in pairs.

28. ELEC 3105 Basic EM and Power Engineering
Principle of virtual work

29. Energy stored in electric field
Easy way to get expression
Consider a capacitor at potential difference V and of charge +Q ,- Q on the plates. Area of plates (A) and spacing (D)
+Q Q
Energy stored in the capacitor:
But: A D
+ - V

30. Principle of virtual work
Conductor caries a surface charge of density , find force on plates of a parallel plate capacitor. Plate area A +Q s
Field between plates -Q
Principle of virtual work:
Find the work W required to increase (or decrease) plate separation by S
Recall

31. Principle of virtual work
W = change in energy stored in the system = U
Plate area A +Q s
Field between plates -Q

32. Principle of virtual work Can apply 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 s d
Metal insert -Q y L
What force is pulling metal insert into the capacitor?
Can apply principle of virtual work

33. 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
Metal insert
Metal insert y L
Total energy stored in system
Capacitor energy in region without insert
Capacitor energy in region with insert

34. 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 d L y s d L

35. 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 d L y S d L y
Force pulling metal insert into capacitor

36. 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:.

37. 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).

38. Forces in Electrostatics
           
                  
                      

39. Forces in Electrostatics

40. Forces in Electrostatics

41. Forces in Electrostatics
The micromechanical angular rate sensor has a butterfly-shaped 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.

42. ELEC 3105 Basic EM and Power Engineering
Next few slides: Proof of Field by other charges on a metal surface

43. Forces in Electrostatics
Conductor caries a surface charge of density 
Side view
Top view
Charge dq experiences electric field of all other charges in the system.
Therefore electric force is:

44. Forces in Electrostatics
In electrostatics is perpendicular to the surface at dq. If not, the charges would move along the surface and  change value.
Side view
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

45. Forces in Electrostatics
Electric field produced by charge dq only.
Gaussian surface da
conductor
Enlarged view of conducting surface near dq
Electric field produced by all charges on conductor

46. Forces in Electrostatics
From Gauss’s law on surface element da
Force on dq is not:
since dq on da contributes to electric field.
Must get
Due to all other charges on conductor which act on dq. Then

47. Forces in Electrostatics
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:
conductor

48. Forces in Electrostatics
First calculate electric field produced by charge dq only.
Gaussian surface da
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

49. Forces in Electrostatics
These add to give:
conductor
Same magnitude
By Gauss’s law
Two equations
and two unknowns
These cancel exactly

50. Forces in Electrostatics
Solving two equations gives:
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.