ch21a electromagnetism

8/11/2019 Ch21a Electromagnetism

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2005 Pearson Education South Asia Pte Ltd

ELECTROMAGNETISM

FE1001 Physics I NTU - College of Engineering

21. Electric Charge and

Electric Field

22. Gausss Law

23. Electric Potential

24. Capacitance and

Dielectrics

25. Current, Resistance, and ElectromotiveForce

26. Direct-Current Circuits


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ELECTROMAGNETISM

FE1001 Physics I NTU - College of Engineering

27. Magnetic Field and

Magnetic Forces

28. Sources of Magnetic Field

29. Electromagnetic Induction

30. Inductance

31. Alternating Current

32. Electromagnetic Waves


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21. Electric Charge and Electric Field


Chapter Objectives

The nature of electric charge Interactions of electric charges

Coulombs law

The concept of electric field


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Chapter Outline

1. Electric Charge

2. Conductors, Insulators, and Induced Charges

3. Coulombs Law

4. Electric Field and Electric Forces

5. Electric-Field Calculations6. Electric Field Lines

7. Electric Dipoles


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21.1 Electric Charge

Electric chargeis a fundamental attribute of

particles.

Electrostatics are defined as the interactions

between electric charges that are at rest (or nearlyso).

The figure shows some experiments used todemonstrate electrostatics.


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Fig. 21.1


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Electrostatics experiments show that there are

exactly two kinds of electric charge, negativeand

positive.

Two positive charges or two negative chargesrepel each other. A positive charge and a

negative charge attract each other.


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CAUTION

Like charges only mean that two charges have

the same algebraic sign(both positive or both

negative).

Opposite charges means that the electriccharges on both objects have different signs

(one positive and the other negative).

A technological application is in a laser printer; the

figure shows a schematic diagram of such a printer

in operation.


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2005 Pearson Education South Asia Pte Ltd Fig. 21.2


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Electric Charge and the Structure of Matter

The atomic structure consists of three particles: the

negatively charged electron, the positively charged

proton, and the uncharged neutron.

Protons and neutrons make up the nucleuswhile

electrons orbit it from a distance.

The figure shows how changes in the atomic

structure of lithium determines its net electric

charge.


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Fig. 21.4


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Atomic numberis defined as the number of

protons or electrons in a neutral atom of an

element.

A positive ionis formed by removing one or more

electrons from an atom; a negative ionis one that

has gainedone or more electrons. This process is

called ionization.


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When the total number of protons equals the total

number of electrons in a macroscopic body, its total

charge is zero and the body as a whole is

electrically neutral.

When we speak of the charge of a body, we always

mean its netcharge.


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The Principle of Conservation of Charge: The

algebraic sum of all the electric charges in any

closed system is constant.

In any charging process, charge is not created or

destroyed but merely transferredfrom one body to

another.


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The magnitude of charge of the electron or

proton is a natural unit of charge.

Every observable amount of electric charge on any

macroscopic body is always either zero or an

integer multiple (positive or negative) of this basic

unit, the electron chargequantization of charge.


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21.2 Conductors, Insulators, and Induced Charges

Fig. 21.5


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Charging by inductionis the process in which a

charged body can give another body a charge of

oppositesign without losing any of its own charge.

The figure shows the charging of a metal sphere byinduction.


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Fig. 21.6


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Excess charges that develop in the region of a

body during electrical induction are called induced

charges.

The earth is a conductor, and it is so large that itcan act as an infinite source of extra electrons or

sink of unwanted electrons. The charge it acquires

via induction will be equal and opposite to the

charge remaining on the electrically induced body.


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In a metallic conductor, the mobile charges are

always negative electrons.

In ionic solutions and ionized gases, both positive

and negative charges are mobile.


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A charged body can exert forces even on objects

that are notcharged themselvesinduced-charge

effect.

This is due topolarization, in which a chargedobject of eithersign exerts an attractive force on an

uncharged (neutral) insulator.


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Coulombs Lawstates that:

The magnitude of the electric force between

two point charges is directly proportional to the

product of the charges and inverselyproportional to the square of the distance

between them.

The directions of the forces the two charges exert

on each other are always along the line joining

them, as shown in the figure.

21.3 Coulombs Law


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21.3 Coulombs Law

Fig. 21.9


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21.3 Coulombs Law

Coulombs Law is usually written as:

where

229

0

22120

/100.94

1

/10854.8

CmN

mNC

21 C


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21.3 Coulombs Law

The most fundamental unit of charge is the

magnitude of the charge of an electron or proton,

denoted by e, where

e= 1.602176462(63) x 10-19

C

21 El t i Ch d El t i Fi ld


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Example 21.1 Electric force versus gravitational force

An particle (alpha) is the nucleus of a helium

atom. It has a mass of m= 6.66 x 10-27kg and a

charge q= +2e= 3.2 x 10-19C. Compare the force of

the electric repulsion between two particles with the

force of gravitational attraction between them.



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Example 21.1 (SOLN)

Identify and Set Up

The magnitudeFeof the electric force is given by

Eq. (21.2),

The magnitudeFgof the gravitational force is givenby Eq. (12.1),

We compare these two magnitudes by calculating

their ratio.

2

2

04

1

r

qFe

2

2

r

mGFg



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Example 21.1 (SOLN)

Execute

The ratio of the electric force to the gravitational force

is

35

227

219

2211

229

2

2

0

101.3

)1066.6(

)102.3(

/1067.6

/100.9

4

1

kg

C

kgmN

CmN

m

q

GF

F

g

e



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Example 21.1 (SOLN)

Evaluate

This astonishingly large number shows that the

gravitational force in this situation is completely

negligible in comparison to the electric force. This is

always true for interactions of atomic and subatomicparticles. (Notice that this result doesnt depend on

the distance rbetween the two particles.) But within

objects the size of a person or a planet, the positive

and negative charges are nearly equal in magnitude,and the net electric force is usually much smallerthan

the gravitational force.


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Example 21.2 (SOLN)

Identify and Set Up

We use Coulombs law, Eq. (21.2), to calculate the

magnitude of the force that each particle exerts on

the other. The problem asks us for the force on

each particle due to the other particle, so we useNewtons third law.

21 Electric Charge and Electric Field


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Example 21.2 (SOLN)

Execute

a) Converting charge to coulombs and distance to

meters, the magnitude of the force that q1exerts on q2 is

N

m

CCCmN

r

qq

F on

019.0

)030.0(

|)1075)(1025(|)/100.9(

||

4

1

2

99229

221

021



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Example 21.2 (SOLN)

Since the two charges

have opposite signs, the

force is attractive; that is,

the force that acts on q2is

directed toward q1alongthe line joining the two

charges, as shown in Fig.

21.10b.

Fig. 21.10



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Example 21.2 (SOLN)

b) Remember that Newtons third law applies to the

electric force. Even though the charges have

different magnitudes, the magnitude of the force that

q2exerts on q1is the same as the magnitude of the

force that q1exerts on q2:

Newtons third law also states that the direction of the

force that q2exerts on q1is exactly opposite the

direction of the force that q1exerts on q2 ; this is shown

in Fig. 21.10c.

NF on 019.012



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Example 21.2 (SOLN)

Evaluate

Note that the force on q1is directed toward q2, as it

must be, since charges of opposite sign attract each

other.



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Example 21.3 Vector addition of electric forces on a line

Two point charges are located on the positivex-axis

of a coordinate system (Fig. 21.11a). Charge q1= 1.0

nC is 2.0 cm from the origin, and charge q2= -3.0 nC

is 4.0 cm from the origin. What is the total force

exerted by these two charges on a charge q3= 5.0 nClocated at the origin? Gravitational forces are

negligible.



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Example 21.3 (SOLN)

Identify

Here there are twoelectric forces acting on the

charge q3, and we must add these forces to find the

total force.



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Example 21.3 (SOLN)

Set Up

Figure 21.11a shows the

coordinate system. Our

target variable is the net

electric force exerted oncharge q3by the other

two charges. This is the

vector sum if the forces

due to q1and q2individually. Fig. 21.11



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Example 21.3 (SOLN)

Execute

Figure 21.11b is a free-body diagram for charge q3.

Note that q3is repelled by q1(which has the same

sign) and attracted to q2(which has the opposite

sign).



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Example 21.3 (SOLN)

Execute

Converting charge to coulombs and distance to

meters, we use Eq. (21.2) to find the magnitudeF1 on 3

of the force of q1on q3:

This force has a negativex-component because q3is

repelled (that is, pushed in the negativex-direction)

by q1.

NN

m

CCCmN

rqqF on

1121012.1

)020.0(

|)100.5)(100.1(|)/100.9(

||4

1

4

2

99229

231

031



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Example 21.3 (SOLN)

Execute

The magnitudeF2 on 3of the force of q2on q3is

This force has a positivex-component because q3isattracted (that is, pulled in the positivex-direction) by

q2.

NN

m

CCCmN

r

qqF on

84104.8

)040.0(

|)100.5)(100.3(|)/100.9(

||

4

1

5

2

99229

232

032



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Example 21.3 (SOLN)

Execute

The sum of thex-components is

There are noy-orz-components. Thus the total

force on q3 is directed to the left, with magnitude 28N = 2.8 x 10-5N.

NNNFx 2884112



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Example 21.3 (SOLN)

Evaluate

To check the magnitudes of the individual forces,

note that q2 has three times as much charge (in

magnitude) as q1but is twice as far from q3. From

Eq. (21.2) this means that F2 on 3must be 3/22= aslarge as F1 on 3. Indeed, our results show that this ratio

is (84 N)/(112 N) = 0.75. The direction of the net

force also makes sense: is opposite to and has a

larger magnitude than , so the net force is in the

direction of .

4

3

31onF

32onF

31onF



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Example 21.4 Vector addition of electric forces in a plane

In Fig. 21.12, two equal

positive point charges q1

= q2= 2.0 C interact with

a third point charge Q=

4.0 C. Find themagnitude and direction

of the total (net) force on

Q.

Fig. 21.12



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Example 21.4 (SOLN)

Identify and Set Up

As in Example 21.3, we have to compute the force

each charge exerts on Q and then find the vector sum

of the forces. The easiest way to do this is to use

components.



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Example 21.4 (SOLN)

Execute

Figure 21.12 shows the force on Qdue to the upper

charge q1. From Coulombs law the magnitudeFof

this force is

N

mCCCmNF Qon

29.0

50.0)100.2)(100.4()/100.9(

2

66

2291



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Example 21.4 (SOLN)

Execute

The angle is below thex-axis, so the components

of this force are given by

Nm

mNFF

N

m

mNFF

QonyQon

QonxQon

17.050.0

30.0)29.0(sin)()(

23.0

50.0

40.0)29.0(cos)()(

11

11



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Example 21.4 (SOLN)

Execute

The lower charge q2exerts a force with the same

magnitude but at an angle abovethex-axis. From

symmetry we see that itsxcomponent is the same as

that due to the upper charge, but itsycomponent hasthe opposite sign. So the components of the total

force on Qare

The total force on Qis in the +x-direction, with

magnitude 0.46 N.

F

017.017.0

46.023.023.0

NNF

NNNF

y

x



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Example 21.4 (SOLN)

Evaluate

The total force on Qis in a direction that points

neither directly away from q1nor directly away from

q2. Rather, this direction is a compromise that points

away from the systemof charges q1 and q2. Can yousee that the total force would notbe in the +x-

direction if q1 and q2were not equal or if the

geometrical arrangement of the charges were not so

symmetrical?

ch21a electromagnetism

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