chapter22 parti

Chapter 22:Chapter 22:Coulomb’s Law and Electric Fields Part ICoulomb’s Law and Electric Fields Part I

ObjectivesObjectives•To study quantitatively the nature of forces between electric chargesTo study quantitatively the nature of forces between electric charges

•To develop the concept of the electric field as a transmitter of electric forceTo develop the concept of the electric field as a transmitter of electric force

•To learn how work is done and potential energy is stored in an electric fieldTo learn how work is done and potential energy is stored in an electric field

•To learn how to use electric fields to control and direct the motion of electric chargesTo learn how to use electric fields to control and direct the motion of electric charges

Physics is LifeRye High School: Physics

Coulomb’s Law

Experiment shows that the electric force between two charges is proportional to the product of the charges and inversely proportional to the distance between them.

Rye High School: Physics

2

Coulomb’s Law

Coulomb’s law:

Where Q1 and Q2 are the amount of charge and k is a proportionality constant

Charges produced by rubbing ordinary objects (such as a comb or a plastic ruler) are typically around a microcoulomb or less:


3

4

Coulomb’s Law

The magnitude of the charge of an electron, on the other hand, has been determined to be about 1.602 x 10-19 C, and its sign is negative. This is the smallest known charge, and because of its fundamental nature, it is given the symbol e and is often referred to as the elementary charge:

Example:

How many electrons make up a charge of -30.0 micro coulombs (C)?

N = Q/e = (-30 x 10-6 C)/ (-1.60 x 10-19 C/electrons) = 1.88 x 1014 electrons

What is the mass of 1.88 x 1014 electrons?

Mass = (9.11 x 10-31 kg)(1.88 x 1014 electrons) = 1.71 x 10-16 kg


The charges carried by the proton and electron are equal in size. However, the mass of the proton is 2000 times the mass of the electron.

Coulomb’s Law


5

• Double one of the chargesDouble one of the charges– force doubles

• Change sign of one of the chargesChange sign of one of the charges– force changes direction

• Change sign of Change sign of bothboth charges charges– force stays the same

• Double the distance between chargesDouble the distance between charges– force four times weaker

• Double Double bothboth charges charges– force four times stronger

Coulomb’s LawCoulomb’s Law


6

Coulomb’s Law vs. Law of Universal Coulomb’s Law vs. Law of Universal GravitationGravitation

• F = kQF = kQ11QQ22/r/r22 vs. F=GM vs. F=GM11MM22/r/r22

• Both are inverse square laws FBoth are inverse square laws F1/r1/r22

• Both have a proportionality to a product of each Both have a proportionality to a product of each body-mass for gravity, electric charge for electricity.body-mass for gravity, electric charge for electricity.

• A major difference is that gravity is always an A major difference is that gravity is always an attractive force, whereas the electric force can be attractive force, whereas the electric force can be wither attractive or repulsive.wither attractive or repulsive.

• Electrical Force is stronger than gravitational forceElectrical Force is stronger than gravitational force


Comparison of electrical force vs. Gravitational force #1

Comparison of electrical force vs. Gravitational Force #2

7

http://homepages.wmich.edu/~kaldon/classes/ph207-6-forces.htm

8

Sample problem

Find the force between two positive 1.0 C charges when they are 1000m apart?

Solution

q1=q2 = 1.0C r = 1000m

F = kq1q2/r2 where k = 9.0 x 109 Nm2/C2

After substitution, F = 9.0 x 103 N

Solving Problems involving Coulomb’s Law



Sample problem

What is the magnitude of the electric force of attraction between an iron nucleus (q = +26e) and its innermost electron if the distance between them is 1.5 x 10-12 m?

Solution


F = (9.0 x 109 Nm2/C2)(26)(1.6 x 10-19 C)(-1.6 x 10-19 C)/ (1.5 x 10-12 m)2 = -2.7 x 10-3 N


9


Sample problem

What is the repulsive electrical force between two protons in a nucleus that are 5.0 x 10-15 m apart from each other?

Solution


F = (9.0 x 109 Nm2/C2)(1.6 x 10-19 C)(1.6 x 10-19 C)/ (5.0 x 10-15 m)2 = 9.2 N


10


Sample problem

Two charged balls are 20.0 cm apart. They are moved, and the force on each of them is found to have been tripled. How far apart are they now?

Solution

Let F1 = kq1q2/r12 and F2 = kq1q2/r2

2 where F2 = 3 F1

F2/F1 = r12

/r22

3= [(20.0cm)/r2]2, which gives r2 = 11.5 cm


11

ADVANCEDADVANCED

12

In physics, the space surrounding an electric charge has a property called an electric field. This electric field exerts a force on other electrically charged objects. The concept of an electric field was introduced by Michael Faraday.

The electric field is a vector field with SI units of newtons per coulomb (N C−1) or, equivalently, volts per meter. The strength of the field at a given point is defined as the force that would be exerted on a positive test charge of +1 coulomb placed at that point; the direction of the field is given by the direction of that force.

The Electric Field

Michael Faraday (1791-1867)


http://en.wikipedia.org/wiki/Physics

http://en.wikipedia.org/wiki/Electric_charge

http://en.wikipedia.org/wiki/Force_(physics)

http://en.wikipedia.org/wiki/Michael_Faraday

http://en.wikipedia.org/wiki/Vector_field

http://en.wikipedia.org/wiki/SI

http://en.wikipedia.org/wiki/Newton

http://en.wikipedia.org/wiki/Coulomb

http://en.wikipedia.org/wiki/Volt

http://en.wikipedia.org/wiki/Test_charge

http://upload.wikimedia.org/wikipedia/commons/5/5c/Faraday_Cochran_Pickersgill.jpg

13

• One can think of electric force as establishing a “field” telling One can think of electric force as establishing a “field” telling particles which way to move and how fastparticles which way to move and how fast

+

Electric “field lines” tell a positivecharge which way to move.

For example, a positive charge itselfhas field lines pointing away from it,because this is how a positively-charged“test-particle” would respond if placedin the vicinity (repulsive force).

+

Run Away!

The Electric Field


The Electric Field

+ +

++

The direction of the electric field is always directed in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding the source charge


14

15

Measuring an electric field is a quite simple process involving a test charge. To measure the strength of an electric field, first a test charge must be placed in its vicinity, then calculate the force the test charge “feels”. The resulting number is the strength of the electric field. This process is simplified into the following equation

In this equation,F is the magnitude of the force, as found by using Coulomb's Law, q is the magnitude of the test charge. The resulting electric strength is measured in Newton’s per a Coulomb.

The Electric Field


16

Electric Field vs. Gravitational FieldElectric Field vs. Gravitational Field

• Right now you are experiencing a uniform gravitational field: it has a magnitude of 9.8 Right now you are experiencing a uniform gravitational field: it has a magnitude of 9.8 m/sm/s22 and points straight down. If you threw a mass through the air, you know it would and points straight down. If you threw a mass through the air, you know it would follow a parabolic path because of gravity. You could determine when and where the follow a parabolic path because of gravity. You could determine when and where the object would land by doing a projectile motion analysis, separating everything into x object would land by doing a projectile motion analysis, separating everything into x and y components. The horizontal acceleration is zero, and the vertical acceleration is and y components. The horizontal acceleration is zero, and the vertical acceleration is g. We know this because a free-body diagram shows only mg, acting vertically, and g. We know this because a free-body diagram shows only mg, acting vertically, and applying Newton's second law tells us that mg = ma, so a = g. applying Newton's second law tells us that mg = ma, so a = g.

• You can do the same thing with charges in a uniform electric field. If you throw a You can do the same thing with charges in a uniform electric field. If you throw a charge into a uniform electric field (same magnitude and direction everywhere), it charge into a uniform electric field (same magnitude and direction everywhere), it would also follow a parabolic path. We're going to neglect gravity; the parabola comes would also follow a parabolic path. We're going to neglect gravity; the parabola comes from the constant force experienced by the charge in the electric field. Again, you from the constant force experienced by the charge in the electric field. Again, you could determine when and where the charge would land by doing a projectile motion could determine when and where the charge would land by doing a projectile motion analysis. The acceleration is again zero in one direction and constant in the other. The analysis. The acceleration is again zero in one direction and constant in the other. The value of the acceleration can be found by drawing a free-body diagram (one force, F = value of the acceleration can be found by drawing a free-body diagram (one force, F = qE) and applying Newton's second law. This says: qE = ma, so the acceleration is a = qE) and applying Newton's second law. This says: qE = ma, so the acceleration is a = qE / m. qE / m.

• The one big difference between gravity and electricity is that m, the mass, is always The one big difference between gravity and electricity is that m, the mass, is always positive, while q, the charge, can be positive, zero, or negative. positive, while q, the charge, can be positive, zero, or negative.


17

The Electric FieldThe Electric Field

Sample Problem

A positive charge of 1.0 x 10-5C experiences a force of 0.30N when located at a certain point. What is the electric field intensity at that point?

Solution

E=F/q = 0.30N / 1.0 x 10-5 C = 3.0 x 104 N/C


Sample Problem

A test charge experiences a force of 0.20 N on it when it is placed in an electric field intensity of 4.5 x 105 N/C. What is the magnitude of the charge?

Solution

q=F/E = 0.20N / 4.5 x 105 N/C = 4.4 x 10-7 C



18

Sample Problem

A positive charge of 10-5 C experiences a force of 0.2N when located at a certain point in an electric field. What is the electric field strength at that point?

Solution

F= 0.2N q=10-5C

E= F/q = 0.2N/10-5C = 2 x 104 N/C



19


Measuring Potential Difference Measuring Potential Difference (Voltage)(Voltage)

Electric Strength can also be defined in units of work and energy. This alternative way of measuring the strength of an electric field entails finding the potential difference, which exits between any two points. To visualizing this difference picture an electric field, then trying to force a test charge in between two points, if the test charge is repelled, we have to do work to push it into place. This process is also simplified into the formula

In this formula, W is work, q is the magnitude of charge, and V is voltage or potential difference.

20


Measuring Potential Difference Measuring Potential Difference (Voltage)(Voltage)

Sample Problem

It takes 5.0 x 10-3 J of work to move a positive charge of 2.5 x 10-4C from point X to point Y on an electric field. What is the difference of Potential between X and Y?

Solution

W= 5.0 x 10-3 J q = +2.5 x 10-4C

V = W/q = 5.0 x 10-3J/2.5 x 10-4C = 20 J/C = 20 volts

21


ElectronvoltElectronvolt

The joule is too large a unit for measuring the work done on moving elementary charges, such as electrons, protons, or the small charges in ions. For this purpose, the electron volt is a more convenient unit of energy or work. An electron volt is the work done in moving an electron or other body having a unit of elementary charge through a difference of potential of one volt. Thus,

1eV = 1.60 x 10-19 J

22


ElectronvoltElectronvoltSample Problem

The difference of potential between point X and point Y of an electric field is 100 volts. (a) How much work done in electronvolts is done by the electric field in moving a free electron from point X and Y. (b) What happens to this work? (c) What is this work in Joules

(a) V = 100 V q = +1 electron charge

W= 100volts and 1 electron’s charge = 100 eV

(b) This work is used to accelerate the proton. It is converted into the KE gained by the proton on moving from point X to Y.

© 1 ev = 1.60 x 10-19J, so…100eV = 1.6 x 10-17J

23


Electrical Potential EnergyElectrical Potential Energy

In expressing the gravitational potential energy of a body, we learned that a base level such as the surface of the earth must be arbitrarily selected as the level corresponding to zero PE. Similarly, in expressing the PE of a charge in a an electric field, a position of charge corresponding to the base or zero potential energy must first be selected.

A charge in the electric field of a point charge is said to have zero PE when it is at an infinite distance from the point charge. This is expressed as:

U = K (q1 q2)/r

24


Electrical Potential EnergyElectrical Potential Energy

Sample Problem

Assuming that a hydrogen atom consists of one electron and one proton separated by a distance of 5.3 x 10-11 m, what is the PE of the electron in the field of the proton?

Solution

U = K (qprotonqelectron)/r = -4.3 x 10-18 Nm or J

Note that the PE is negative because the charges are opposite and attract each other.

25


The quantity of charge (q) on an object is related to the (unbalanced) number of electrons that have been either gained or lost by the object.

The unit of charge is the coulomb (C).

The charge on one electron is the smallest charge known to exist independently and has the value of 1.60 x 10-19 coulomb. This value of the electron charge is known as the fundamental charge, e = 1.60 x 10-

19 C. Therefore every electron has a charge of -e and every proton (or positive charge due to the loss of one electron) has a charge of +e.

A rubber balloon becomes negatively charged after you rub the balloon with a wool cloth. The quantity of charge due to the excess electrons on the balloon can be found according to the following general relationship:

Quantity of (positive or negative) charge, q = (number of electrons, Ne) x (fundamental (electron) charge, +e or - e)

So, q = (Ne)(e)

Measuring Electric ChargeMeasuring Electric Charge

26


Measuring Electric ChargeMeasuring Electric Charge

Sample Problem

What is the charge of 1000 electrons?

27

Q = (Ne)(e) = 1000 (-1.6 x 10-19 C) = - 1.6 x 10-16 C

28

Field Lines

The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge.

At locations where electric field lines meet the surface of an object, the lines are perpendicular to the surface.


29

Field Lines

Electric dipole: two equal charges, opposite in sign:

•Field lines indicate the direction of the field; the field is tangent to the line.


30

Field Lines

The number of field lines starting (ending) on a positive (negative) charge is proportional to the magnitude of the charge.

Electric field lines never cross each other.

The electric field is stronger where the field lines are closer together.


Field Lines

The electric field between two closely spaced, oppositely charged parallel plates is constant.


31

32

Field Lines

Summary of Field lines Around Charges

• The magnitude of the field is proportional to the density of the lines.

• Field lines start on positive charges and end on negative charges

• Field lines indicate the direction of the field; the field is tangent to the line.

• The electric field between two closely spaced, oppositely charged parallel plates is constant.

• At locations where electric field lines meet the surface of an object, the lines are perpendicular to the surface.

• Electric field lines never cross each other.


Millikan’s Oil Drop Experiment

One important application of the uniform electric field between two parallel plates is the measurement of charge of an electron. This was determined by American physicist Robert A Millikan in 1909

Millikan’s experiment showed that charge is quantized. This means that an object can only have a charge with a magnitude that is some integral multiple of the charge of the electron (1.6 x 10-19 C).


33

Millikan’s Oil Drop ExperimentSample Problem

In a Millikan oil drop experiment, a drop has been found to weigh 1.9 x 10-14 N. When the electric field is 4.0 x 104 N/C, the drop is suspended motionless. (a) what is the charge on the drop? (b) If the upper plate is positive, how many excess electrons does the oil drop have?

Solution

(a) When balanced, Felectric = F gravity Thus, qE=mg solving for q, the charge will be

q=mg/E = 1.9 x 10-14 N/4.0 x 104N/C = 4.8 x 10-19 C

(b) Determine the number of electrons by n=q/e

n=4.4 x 10-19 C/1.6 x 10-19 C/electron = 3 electrons


34

Sample Problem

A positively charged oil drop weighs 6.4 x 10-13 N. An electric field of 4.0 x 106 N/C suspends the drop. (a) What is the charge on the drop? (b) How many electrons is the drop missing?

Solution

(a)Q=F/E = 6.4 x 10-13 N/ 4.5 x 106 N/C = 1.6 x 10-19 C

(b) N = Q/e = 1.6 x 10-19 C/1.6 x 10-19 C/electron =

1 electron

Millikan’s Oil Drop Experiment


35


Electric Fields and Conductors

We have previously shown that any charged object - positive or negative, conductor or insulator - creates an electric field which permeates the space surrounding it. In the case of conductors there are a variety of unusual characteristics about which we could elaborate. Recall that a conductor is material which allows electrons to move relatively freely from atom to atom. It was emphasized that when a conductor acquires an excess charge, the excess charge moves about and distributes itself about the conductor in such a manner as to reduce the total amount of repulsive forces within the conductor. Electrostatic equilibrium is the condition established by charged conductors in which the excess charge has optimally distanced itself so as to reduce the total amount of repulsive forces. Once a charged conductor has reached the state of electrostatic equilibrium, there is no further motion of charge about the surface.

36

http://www.physicsclassroom.com/Class/estatics/u8l1d.cfm



Charged conductors which have reached electrostatic equilibrium share three particular characteristics.

One characteristic of a conductor at electrostatic equilibrium is that the electric field anywhere beneath the surface of a charged conductor is zero. If an electric field did exist beneath the surface of a conductor (and inside of it), then the electric field would exert a force on all electrons that were present there. This net force would begin to accelerate and move these electrons. But objects at electrostatic equilibrium have no further motion of charge about the surface. So if this were to occur, then the original claim that the object was at electrostatic equilibrium would be a false claim. If the electrons within a conductor have assumed an equilibrium state, then the net force upon those electrons is zero. The electric field lines either begin or end upon a charge and in the case of a conductor, the charge exists solely upon its outer surface. The lines extend from this surface outward, not inward.

37


This concept of the electric field being zero inside of a closed conducting surface was first demonstrated by Michael Faraday. Faraday constructed a room within a room, covering the inner room with a metal foil. He sat inside the inner room with an electroscope and charged the surfaces of the outer and inner room using an electrostatic generator. While sparks were seen flying between the walls of the two rooms, there was no detection of an electric field within the inner room. The excess charge on the walls of the inner room resided entirely upon the outer surface of the room.

The inner room with the conducting frame which protected Faraday from the static charge is now referred to as a Faraday's cage. The cage serves to shield whomever and whatever is on the inside from the influence of electric fields. Any closed, conducting surface can serve as a Faraday's cage, shielding whatever it surrounds from the potentially damaging affects of electric fields. This principle of shielding is commonly utilized today as we protect delicate electrical equipment by enclosing them in metal cases. Even delicate computer chips and other components are shipped inside of conducting plastic packaging which shields the chips from potentially damaging affects of electric fields.


38

http://en.wikipedia.org/wiki/Faraday_cage

39


The excess charges arrange themselves in a the conductor surface precisely in the manner needed to make the electric field zero within the material.


40

We can also see that a relatively safe place to be during a lightning storm is inside a car, surrounded by metal..



http://images.google.com/imgres?imgurl=http://www.uberreview.com/wp-content/uploads/car-lightning.jpg&imgrefurl=http://www.uberreview.com/2007/03/what-happens-when-lightning-strikes-your-car.htm&usg=__AwdvcAwRHP__epVqrrUUMc5RG7w=&h=337&w=450&sz=21&hl=en&start=



A second characteristic of conductors at electrostatic equilibrium is that the electric field upon the surface of the conductor is directed entirely perpendicular to the surface. There cannot be a component of electric field (or electric force) that is parallel to the surface. If the conducting object is spherical, then this means that the perpendicular electric field vector are aligned with the center of the sphere. If the object is irregularly shaped, then the electric field vector at any location is perpendicular to a tangent line drawn to the surface at that location.

41


Electric Fields and ConductorsA third characteristic of conducting objects at electrostatic equilibrium is that the electric fields are strongest at locations along the surface where the object is most curved. The curvature of a surface can range from absolute flatness on one extreme to being curved to a blunt point on the other extreme. A flat location has no curvature and is characterized by relatively weak electric fields. On the other hand, a blunt point has a high degree of curvature and is characterized by relatively strong electric fields. A sphere is uniformly shaped with the same curvature at every location along its surface. As such, the electric field strength on the surface of a sphere is everywhere the same. But on an irregularly shaped object, excess electrons would tend to accumulate in greater density along locations of greatest curvature.

42



Summary of Field lines Around Conductors

• The electric field anywhere beneath the surface of a charged conductor in static equilibrium is zero; excess charge of a conductor exists solely on its surface.

• The electric field of the surface of the conductor at electrostatic equilibrium is directed entirely perpendicular to the surface.

• Conductors at electrostatic equilibrium exert strong electric fields along any curvature or sharp bend at its surface.

43


Electron GunElectron GunAn electron gun is a component that produces an electron beam that has a precise kinetic energy, being used in televisions and monitors which use cathode ray tube technology, and in other instruments, as electron microscopes and particle accelerators.

An electron gun is formed of several parts: a hot cathode, which is heated to create a stream of electrons via thermionic emission, electrodes generating an electric field which focus the beam - such as a Wehnelt cylinder - and one or more anode electrodes which accelerate and further focus the electrons. A large voltage between cathode and anode accelerates the electrons. A repulsive ring placed between them focuses the electrons onto a small spot on the anode on the expense of a lower extraction field strength on the cathode surface. Often at this spot is a hole, so that the electrons pass through the anode forming a collimated beam and are finally reach a second anode called a collector. This device can be used to find the mass of an electron (9.1 x 10-31 kg)

44


Electron Beams; Cathode Ray Tubes Electron Beams; Cathode Ray Tubes (CRTs)(CRTs)

• Televisions, Oscilloscopes, Monitors, etc. use an Televisions, Oscilloscopes, Monitors, etc. use an electron beam steered by electric fields to light up the electron beam steered by electric fields to light up the (phosphorescent) screen at specified points(phosphorescent) screen at specified points

E-field

metal plates

- - - - - - -

+ + + + + + +

electron beam

screen

cathode emitter

45


SummarySummary• Coulomb’s Law: F = kQCoulomb’s Law: F = kQ11QQ22/r/r22

• Units of Coulomb’s Law is the Newton.Units of Coulomb’s Law is the Newton.• Electric Field: E = F/QElectric Field: E = F/Q• Units of Electric Field are Newtons per CoulombUnits of Electric Field are Newtons per Coulomb• Potential Difference: V = W/QPotential Difference: V = W/Q• Units of Potential difference is Joules per CoulombUnits of Potential difference is Joules per Coulomb• Work (W) = QVWork (W) = QV• The work done in moving an electric charge can be The work done in moving an electric charge can be

expressed in joules or electron volts (eV).expressed in joules or electron volts (eV).• One electron volt is equal to 1.60 x 10One electron volt is equal to 1.60 x 10-19-19 joule joule

• Potential Energy of a charge: U = kQPotential Energy of a charge: U = kQ11QQ22/r/r

46

47

Summary of Chapter 22

• Electric field can be represented by electric field lines

• Static electric field inside conductor is zero; surface E-field is perpendicular to surface; Higher density of field lines signify a stronger E-field.

MIT LECTURE ON ELECTRIC FIELD


http://www.stmary.ws/highschool/physics/home/notes/electricity/staticElectricity/default.htm

chapter22 parti

Documents