Lecture Notes 1

EPF 0024: Physics II 1

1.0 Electric Charge and Electric Field


Outline

1.1 Electric Charge1.2 Conductors and Insulators1.3 Charging by contact1.4 Charging by induction1.6 Coulomb’s Law1.7 The Electric Field


Topics for today's lecture:

Electric Charge

Conductors and Insulators

Charging by contact and by induction


Objectives of this lecture:

Explain the phenomenon of Electric Charge.

Explain conductors, semiconductors and insulators using the atomic model.

Analyze the charging of objects by contact and by induction.

Discuss problems related to these concepts.


1.1 Electric Charge

Rubbing two objects together causes them to become electrically charged. e.g:

i. glass rod rubbed with silk can attract paper piece. ii. plastic rod rubbed with fur.

There are only two types of electric charges:

i. positive charges (glass rod rubed with silk) ii. negative charges (plastic rod rubbed with fur).


A basic law of charges states: like charges repel and unlike charges attract as indicated in Fig. 1.1.

Fig. 1.1


1.1.1 The Atomic Theory of Matter

Matter is made up of atoms as building blocks (quantization of matter).

An atom consists of (Fig. 1.2):

1. a nucleus (consists of +vely charged protons and neutrons.

2. Surrounding the nucleus are vely charged particles called electrons (charge is also quantized)


Fig 1.2: The Structure of the Atom

An atom consists of equal numbers of electrons & protons (electrically neutral). It becomes charged by loosing or receiving electrons.

When plastic rod is negatively charged, it implies it received excess negative charges. When a glass rod is positively charged, it implies negative charges are removed from it.


1.1.2 Conservation of Electric Charge

Electric charge is conserved. It can neither be created nor destroyed. This principle is stated as: The net amount of electric charge produced in any process is zero.

For instance

Glass rod is rubbed with silk: 5 negative charges move to the silk. The glass rod now has less 5 negative charges.

However, the algebraic sum of charges on the glass rod and the silk is zero (i.e. charge is conserved).


1.2 Conductors and Insulators

Using the atomic model, it can be deduced that the charges that can move in a material are the outer (valence) electrons. The protons in the nucleus remain in relatively fixed positions.

Electronic properties of materials can be explained using the atomic model. In a conductor the outer (valence) Electrons of the atoms are loosely bound to the atom and can move freely inside the material when an electric field is applied. Most metals are good conductors.


Conductors Semiconductors Insulators

Valance electrons free to move

The bond of electrons to atoms can be broken when sufficient energy is supplied

Electrons tightly bound to atom

Metals Silicon, Germanium

Plastics, wood, glass

1.2 Conductors and Insulators


Fig. 1.3: Charging by contact

Insulated stand

Metal sphere

-

-

-

-

Plastic rod

-- -

-- --

-

- ---

Electrons

Insulators are charged by rubbing. A conductor can also be charged (Fig. 1.3) - charging by contact.


Fig. 1.4 Charging by induction

+ +

+ +

Metal sphere

Insulated stand

Plastic rod

-- -

-- --

-

Ground

+ --

-

-+

+

+

Conductors can also be charged without contact as shown in Fig. 1.4.


Conceptual questions

1. Explain why a comb that has been rubbed through your hair attracts small bits of paper, even though the paper is uncharged.

2. Small bits of paper are attracted to an electrically charged comb, but as soon as they touch the comb they are strongly repelled. Explain this behavior.

3. A charged rod is brought near a suspended object, which is repelled by the rod. Can we conclude that the suspended object is charged? Explain.

4. A charged rod is brought near a suspended object, which is attracted to the rod. Can we conclude that the suspended object is charged? Explain.


Solution

2. Initially, the bits of paper are uncharged and are attracted to the comb by polarization effects. When one of the bits of paper comes into contact with the comb, it acquires charge from the comb. Now the piece of paper and the comb have charge of the same sign, and hence we expect a repulsive force between them.

4. No. Even uncharged objects will be attracted to a charged rod, due to polarization effects.


Topics for today's lecture:

Coulomb’s Law

The Electric Field


Objectives of this lecture:

To describe and deduce Coulomb’s Law.

To define the electric field.

To solve problems related to these concepts.


1.6 Coulomb’s Law Electric charge exerts force on other charges.

Coulomb’s Law gives the magnitude of the force F between two point charges and is written:

SI unit of electric charge coulomb (symbol C).

k = 1/(40) = 8.99 109 N.m2/C2 = Coulomb’s Law constant and C2/(N.m2)permittivity of free space.

221

02

21

4

1

r

qq

r

qqkF

(1.1)


The smallest unit of charge is the charge of an electron, e (= 1.6 1019 C). The charge of a proton is e (= 1.6 1019 C).

The direction of the forces that two charges exert on each other is dependant on their signs (Fig. 1.6).


Fig. 1.6: Forces acting on like and unlike charges

F21

F12r12

q1q2

F43F34

q3q4

r34

F65F56q6q5

r56

x

y

The two forces are always equal in magnitude and

opposite in direction.


Any charge q can be written, according to the quantization principle as:

e is called elementary charge = 1.6 1019 C. n is known as the charge quantum number.

,...3 ,2 ,1 , nneq (1.2)


Table 1.1: The charges on three particles

Particle Symbol Charge

Electron e or e e

Proton p + e

Neutron n 0

Table 1.1 shows the symbols of and charges on the three atomic particles: the electron, proton and neutron.


Example 1

How many electrons must be removed from an electrically neutral metal to give it a charge of 2.4 C.


Solution

1319

6

105.1C101.6

C104.2

e

qn

neq


Example 2

Two point charges of 1.0 C and.0 C are separated by a distance of 1000 m. Find the magnitude of the attractive force that either charge exerts on the other.


Solution

N100.9

m101.0

C 0.1C 0.1CN.m 1099.8

3

23

229

221

r

qqkF


Example 3

Three charged particles are arranged in a line as shown in the figure below. Calculate the net electric force on q3 due to the charges q1 and q2.

q1= 310 – 6 C q2= + 510 – 6 C q3 = 410 – 6 C

0.3 m 0.2 m

F32F31q1

q2 q3

x

y


Solution Net force on q3 is vector sum of force F31

exerted by particle 1 and F32 exerted by 2.

Magnitude Fnet = F31 (F32) = F31 F32.

N 43.0)m 5.0(

)C100.3)(C100.4)(/CN.m1099.8(2

66229

31

F

N 5.4)m 2.0(

)C100.5)(C100.4)(/CN.m1099.8(2

66229

32

F

N 1.4N 5.4N 4.03231 FFFnet

Net force:

Magnitude is 4.1 N and -ve sign indicates net force points in the –ve x-direction.


Example 4

Calculate the magnitude and direction of the net force on charge q3 (6.5 x 105 C) shown in figure below due to the charges q1 (8.6 105 C) and q2 ( 5.0 x 105 C).

F32

q3

F31F31y

F31x

x

y

60 cm30 cm

q1

90o

q2

30o

52 cm


Solution

Magnitudes of F31 and F32 are

N 140)m 6.0(

)C106.8)(C105.6)(/CN.m1099.8(2

55229

31F

N 330)m 3.0(

)C100.5)(C105.6)(/CN.m1099.8(2

55229

32

F


N 7030sin

N 12030cos0

3131

03131

FF

FF

y

x

N 260 70N 330

N 120

3132

31

NFFF

FF

yy

xx

Resolve F31 into its components along the x and y axes

The force F32 has only a y component. So the net force Fnet on q3 has components


N 290)N 260()N 120( 2222 yx FFF

o65

2.2N 120

N 260tan

x

y

F

F

Magnitude of net force

Direction of net force

That is 65o above the positive x-axis

Fy

Fx

65o

+x

+ y

0


1.7 The Electric Field

The electric field, E, may be conceptualized as surrounding a charge q just as the gravitational field, g, surrounds a mass m.

Magnitude of E at a point is defined as the ratio of the electric force experienced by a tiny (point) positive test charge q0 and the magnitude of the charge.


The magnitude E at any point in space is therefore given by the equation

or

Using equation 1.1 and 1.3 and considering q2 as ve test charge q0, E at a distance r from a point charge q would have magnitude

2

02

0

20

0 4

1

r

q

r

qk

q

r

qqk

q

FE

(1.4)

0q

FE (1.3)EqF 0(N/C)


The direction of E is defined as the direction of the force on a very tiny ve test charge placed at a point in the field and is represented by field lines that start on positive charges and end on negative charges.

The field lines of E are as shown in Fig. 1.7 (a) and (b) for a positive and a negative charge, respectively. Fig. 1.8 and 1.9 shows electric-field lines due to other charge configurations.


Fig. 1.7: The Electric Field lines surrounding an isolated positive and negative charge.


Fig. 1.8: Electric Field Lines due to two point charges close together


Fig 1.9: Field lines due to a parallel sheet of positive and negative charges


Example 1

Find the magnitude and direction of the electric field E due to a point charge q of magnitude 3.00 106 C at a point P 30 cm from the charge. (Take k = 8.99 109 N.m2).


Solution

N/C100.3 )m30.0(

)C100.3)(C/N.m1099.8( 5

2

6229

2

r

qkE

Eq = 3.0 106 C

P

30 cm

x

y

-ve x direction

Magnitude


Example 2

Two charges q1 (25 C) and q2 (50 C) are separated by a distance of 10.0 cm. (a) What is the direction and magnitude of the net electric field at a point P, 2.0 cm from the first charge. (b) If an electron is placed at P, what will its acceleration (direction and magnitude) be initially. (Take the mass of the electron to be 9.1 1031 kg).

ENG 1 & 2 START HERE


Solution

E1 E2

q1 q2

25C 50 C

P

d1 = 2.0 cm d2 = 8.0 cm

10 cm

x

y

(a)


(a) Direction of Enet: ve x-direction

Magnitude of Enet :

N/C 106.3

m100.8

C1050

m100.2

C1025/CN.m1099.8

8

2

6

22

6229

22

22

1

122

22

1

121net

d

q

d

qk

d

qk

d

qkEEE

(b) The electron will experience a force in +ve x- direction. Therefore: Direction of Acceleration: +ve x direction.

Magnitude of acceleration:

220

31

819

m/s101.1

kg101.9

N/C103.6)C1060.1(

em

eE

m

qE

m

Fa


Example 3

Two point charges are lying on the y axis with q1 = 4.00 C and q2 = 4.00 C. They are equidistant from a point P, which lies on the x-axis (see figure). (a) What is the net electric field at P? (b) A small object of charge q0 = 8.00 C and mass m = 1.20 g is placed at P. When it is released, what is its acceleration (magnitude and direction)?


Solution

Sketch diagrams

q2

P31.0o

q1

31.0o

Enet

E1 E2

+y

+xP

0.700 m

0.700 m

31.0o

31.0o

q1

q2


N/C1034.7m700.0

C1000.4)/CN.m1099.8( 42

629

21

11

r

qkE

N/C1034.7m700.0

C1000.4)/CN.m1099.8( 42

629

22

22

r

qkE

N/C1078.30.31sin 41 oE

N/C1078.30.31sin 42 oE

N/C 56.70.31sin0.31sin 21net oo EEE

(a)

Therefore Enet has a magnitude of 7.57 N/C and directed along the y axis.


(b) According to Newton’s second law:

223

46

0

m/s1004.5kg101.20

N/C1056.7C1000.8

m

Eq

m

Fa

Thus the magnitude of the acceleration is 5.04 102 m/s2 and is directed along the +y axis.

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