1. Lecture 5 Method of images Energy stored in an electric field Principle of virtual work 1

2. Method of Images 2 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. 3 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. Method of Images

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

5. 5 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. Method of Images

6. 6 Alternate solution technique: Method of images ?????????? Caution: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). Method of Images

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

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

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

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

11. 11 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. d d The result you get is valid for the electric field above the ground plane Method of Images

12. 12 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 Method of Images

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

14. 14 Equivalent systems through method of images Method of Images

15. 15 Equivalent systems through method of images Method of Images

16. 16 Equivalent systems through method of images metal charge Here the image charges produce their own images and we get an infinite number of image charges. Method of Images  

17. 17 Equivalent systems through method of images Metal sphere charge b Q a a 2 /b from center of sphere (a  b) in this case Q’= - a / b Q Method of Images Q

18. 18 A more math approach to the method of images Method of Images

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

20. Forces in Electrostatics 20

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

22. Side View Forces try to pull conductor apart. - - - - - - - - + + + + + + + + - - - - - + + + + + conductor Electrostatic forces are usually very small For E = 10 6 volts/m, F/a  4.4 N/m 2 Forces in Electrostatics

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

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

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

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

27. 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. Forces in Electrostatics

28. Principle of virtual work

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

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

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

32. Capacitor Plate area A = xL Capacitor plates carry a uniform charge of density , find force on the metal insert introduced between the capacitor plates. Capacitor Plate area A = xL s +Q -Q d -------- +++++++ -------- +++++++ y L What force is pulling metal insert into the capacitor? Metal insert Can apply principle of virtual work Principle of virtual work

33. x d y L Total energy stored in system Metal insert Capacitor energy in region without insert Top view Capacitor energy in region with insert Principle of virtual work Capacitor Plate area A = xL Capacitor plates carry a uniform charge of density , find force on the metal insert introduced between the capacitor plates. Capacitor Plate area A = xL Metal insert

34. x d y L s d L Principle of virtual work Capacitor Plate area A = xL Capacitor plates carry a uniform charge of density , find force on the metal insert introduced between the capacitor plates. Capacitor Plate area A = xL

35. x d y L S d L Force pulling metal insert into capacitor Principle of virtual work Capacitor Plate area A = xL Capacitor plates carry a uniform charge of density , find force on the metal insert introduced between the capacitor plates. Capacitor Plate area A = xL y

36. 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). Electrostatic actuators 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:. Principle of virtual work

37. 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

38. Forces in Electrostatics

41. 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. Forces in Electrostatics

42. Next few slides: Proof of Field by other charges on a metal surface

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

44. Forces in Electrostatics Side view In electrostatics is perpendicular to the surface at dq. If not, the charges would move along the surface and  change value. 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. conductor + + + + + + Enlarged view of conducting surface near dq Gaussian surface da 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. conductor + + + 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:

48. Forces in Electrostatics First calculate electric field produced by charge dq only. conductor + + + 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:

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

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.