97.315 Basic EM and power engineering

ELEC 3105 Basic EM and Power EngineeringLecture 5Method of imagesEnergy stored in an electric fieldPrinciple of virtual work1

Method of Images2

Consider the following problemWe 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

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 Images4From Gausss Law

No 2 since E only on top side of conducting plate

Method of ImagesJust above the surface

E = 0

5

Rigorous solution to problem:Solve Poissons 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 Images6

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).Method of Images

7

Consider the following charge distribution, electric field lines and equipotential lines.+Q-QEquipotential lineElectric field lineddSimilar to assignment 1 questionMethod of Images8

Consider the following charge distribution, electric field lines and equipotential lines.Equipotential line: in 3-D it is an equipotential surfaceddThe bisecting plane is at zero potential because it is half way between the two charges. V = 0 Method of Images9

Consider the following charge distribution, electric field lines and equipotential lines.Equipotential line: in 3-D it is an equipotential surfaceddThis 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 insertedMethod of Images10The field distribution and location of charges above and on the conducting plane are the same in both figures. Top portions of systems.

dd

Equivalent regionsMethod of Images

11Solve 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.

ddThe result you get is valid for the electric field above the ground planeMethod of Images12 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 helpsFlat, curved, .One or more charges, distributions, Usually around original chargesMethod of Images13

ddP

P

Electric field at point P obtained from two point charges. Charge density from field at surface of metalMethod of Images

14Equivalent systems through method of images

Method of Images15Equivalent systems through method of images

Method of Images16Equivalent systems through method of imagesmetalchargeHere the image charges produce their own images and we get an infinite number of image charges.Method of Images17Equivalent systems through method of imagesMetal spherechargebQaa2/b from center of sphere(a b) in this caseQ= - a / b QMethod of ImagesQ18

A more math approach to the method of imagesMethod of Images

19Equivalent systems through method of images

For currents in conductorsMethod of ImagesELEC 3105 Basic EM and Power EngineeringForces in Electrostatics20

Forces in Electrostatics

+ + + + + + + + + + + +This charge distribution dq is in a field of value:

dadqForce on dq is thus:

With Then

Force on the charges on a metal surfaceProof of: see slides at endGeneral expressionSideView

Forces try to pull conductor apart.- - - - - - - - - -- - - - - -+ + + + + + + + + ++ + + + + +- - - - - - - - - -+ + + + + + + + + +

conductorElectrostatic forces are usually very smallFor E = 106 volts/m, F/a 4.4 N/m2Forces in ElectrostaticsConsider a conducting sphere uniformly charged +Q

Total charge :

Then

asRadial outwardForces in Electrostatics

as

Forces distributed over surface of sphereForce per unit area

Using dq

Forces in Electrostatics

as

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 ElectrostaticsConsider a conducting sphere placed in a uniform electric field.

as

Induced surface charge given by

Forces in ElectrostaticsNet 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 ElectrostaticsELEC 3105 Basic EM and Power EngineeringPrinciple of virtual work

29Energy stored in electric field+ -+Q -Q

VConsider 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:

DA

Easy way to get expression

Conductor caries a surface charge of density , find force on plates of a parallel plate capacitor. Plate area APrinciple of virtual work: Find the work W required to increase plate separation by S

s

Field between plates+Q-QRecall

Principle of virtual workW = change in energy stored in the system = U

Principle of virtual work

s

Field between plates+Q-QPlate area A

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-Qd--------+++++++--------+++++++yLWhat force is pulling metal insert into the capacitor?Metal insertCan apply principle of virtual workPrinciple of virtual workxdyLTotal energy stored in systemMetal insertCapacitor energy in region without insertTop view

Capacitor energy in region with insertPrinciple of virtual workCapacitor plates carry a uniform charge of density , find force on the metal insert introduced between the capacitor plates. Capacitor Plate area A = xLMetal insertxdyL

s

dL

Principle of virtual workCapacitor plates carry a uniform charge of density , find force on the metal insert introduced between the capacitor plates. Capacitor Plate area A = xLxdyL

S

dL

Force pulling metal insert into capacitor

Principle of virtual workCapacitor plates carry a uniform charge of density , find force on the metal insert introduced between the capacitor plates. Capacitor Plate area A = xLyFor 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 workNote 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. Forces in ElectrostaticsELEC 3105 Basic EM and Power EngineeringNext few slides: Proof of Field by other charges on a metal surfaceForces in ElectrostaticsTop view

Side viewConductor caries a surface charge of density

Charge dq experiences electric field of all other charges in the system.Therefore electric force is:

Forces in ElectrostaticsSide viewIn 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

Forces in Electrostatics

Electric field produced by charge dq only.conductor+ + + + + + + + + + + +

Enlarged view of conducting surface near dq

Gaussian surfacedaElectric field produced by all charges on conductorForces in Electrostatics

From Gausss 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 ElectrostaticsFirst calculate electric field produced by charge dq only.conductor+ + + + + +

Gaussian surfacedaThe 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:

Forces in ElectrostaticsFirst calculate electric field produced by charge dq only.conductor+ + + + + +

Gaussian surfacedaThe 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:

Forces in Electrostatics

conductor+ + + + + +

Same magnitudeBy Gausss lawThese add to give:These cancel exactly

Two equationsand two unknownsForces 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.

Gauss Law Uniform surface charge distribution Flat infinite surface
Charged surface
Gaussian surface (imaginary)
We want to obtain the electric field, magnitude and direction, at a point situated on the Gaussian surface
P
Flux
x
y
Charge surface extends to infinity in y-z plane
Charge density on surface

Gauss Law Uniform surface charge distribution Flat infinite surface
x
y
Area A
Direction of E determined from symmetry

Electric field (charge distribution)
Two point charges

Conductors in Electrostatics
Electric Fields inside conductor add vectorially. Notice opposite directions of electric fields.
++++
eeee
Actual electric field lines

Question (5)

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

Forces in Electrostatics
Same magnitudeBy Gausss law
These add to give:
These cancel exactly
Two equationsand two unknowns