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

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.

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:

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

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.

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).

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

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

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.

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.

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

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

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

Method of Images Equivalent systems through method of images

Method of Images Equivalent systems through method of images

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.  

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

A more math approach to the method of images

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

ELEC 3105 Basic EM and Power Engineering Forces in Electrostatics

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

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

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

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

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.

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

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.

ELEC 3105 Basic EM and Power Engineering Principle of virtual work

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

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

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

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

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

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

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

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 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.

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

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:

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

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

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

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

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

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

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.