General physics 2 electricity magnetism lecture 1

 All materials acquire an electric charge Neutral object: Total positive charge Q+= Total negative charge Q-.. Defining the electric field16 The electric field vector at a point in spa

GENERAL PHYSICS 2

Electricity & Magnetism

1

 Text book:

Fundamentals of Physics, David Halliday et al., 8th Edition.

Physics for Scientists and Engineers, Raymond A Serway and John W.

Jewett, 6th Edition.

 Instructor: Dr Ngac An Bang

Faculty of Physics, Hanoi University of Science ngacanbang@hus.edu.vn

 Homework: will be assigned and may be collected.

 Quizzes and Exams:

 There will be at least two (02) 15-minute quizzes.

 There will be a mid-term exam and a final exam.

Fundamentals of Physics, David Halliday et al., 8th Edition.

Physics for Scientists and Engineers, Raymond A Serway and John W.

Jewett, 6th Edition.

 Instructor: Dr Ngac An Bang

Faculty of Physics, Hanoi University of Science ngacanbang@hus.edu.vn

 Homework: will be assigned and may be collected.

 Quizzes and Exams:

 There will be at least two (02) 15-minute quizzes.

 There will be a mid-term exam and a final exam.

 Grading policy:

 Homework and Quizzes: 20 %

 Midterm exam: 20 %

 Final exam: 60 %

Electric Charge and Field

Lecture 1

 Electric Charges

 Coulomb’s Law

 Electric Fields

 Electric Field of a Continuous Charge Distribution

 Motion of Charged Particles in a Uniform Electric Field

3

Mother and daughter are both enjoying the effects of electrically charging their bodies Each individual hair

on their heads becomes charged and exerts a repulsive force on the other hairs, resulting in the

“stand-up’’ hairdos that you

see here (Courtesy of Resonance Research Corporation)

4

Mother and daughter are both enjoying the effects of electrically charging their bodies Each individual hair

on their heads becomes charged and exerts a repulsive force on the other hairs, resulting in the

“stand-up’’ hairdos that you

see here (Courtesy of Resonance Research Corporation)

Electric charge Some simple experiments demonstrate the existence of

electric forces and charges

Quantization of Charge

 The smallest unit of “ free ” charge known in nature is the charge

of an electron or proton, which has a magnitude of

e = 1.602 x 10-19 C

 Charge of any ordinary matter is quantized in integral multiples

of the elementary charge e, Q = ± Ne.

 An electron carries one unit of negative charge, -e,

 While a proton carries one unit of positive charge, +e.

 Note that although quarks (u, d, c, s, t, b) have smaller charge in

comparison to electron or proton, they are not free particles.

Charge is quantised

6

 The smallest unit of “ free ” charge known in nature is the charge

of an electron or proton, which has a magnitude of

e = 1.602 x 10-19 C

 Charge of any ordinary matter is quantized in integral multiples

of the elementary charge e, Q = ± Ne.

 An electron carries one unit of negative charge, -e,

 While a proton carries one unit of positive charge, +e.

 Note that although quarks (u, d, c, s, t, b) have smaller charge in

comparison to electron or proton, they are not free particles.

Charge conservation

• In a closed system, the total amount of charge is conserved since charge

can neither be created nor destroyed.

• A charge can, however, be transferred from one body to another.

A universal conservation law

 All materials acquire an electric charge

 Neutral object: Total positive charge Q+= Total negative charge Q-.

 Positively charged object: Q+ > Q-,

 Negatively charged object: Q+ < Q

- In this part, we consider only two types of materials

• Conductors: Electrical conductors are materials in which some of the electronsare free electrons that are not bound to atoms and can move relatively freely throughthe material;

• Insulators: are materials in which electrons are bound to atoms and can not movefreely through the material.

Some basic concepts

8

 All materials acquire an electric charge

 Neutral object: Total positive charge Q+= Total negative charge Q-.

 Positively charged object: Q+ > Q-,

 Negatively charged object: Q+ < Q

- In this part, we consider only two types of materials

• Conductors: Electrical conductors are materials in which some of the electronsare free electrons that are not bound to atoms and can move relatively freely throughthe material;

• Insulators: are materials in which electrons are bound to atoms and can not movefreely through the material.

Charge transfer by contact

Charging Objects By Induction

9

Coulomb’s Law

vacuum

 The force F12 exerted by q1on q2is given by Coulomb's law

 The force F21 exerted by q2on q1is given by

 The Coulomb constant k in SI units has the value

 The constant ε0 is known as the permittivity of free space and has the value

r

r r

q q k

r r

q q k F





2

2 1 2

2 1

10

vacuum

 The force F12 exerted by q1on q2is given by Coulomb's law

 The force F21 exerted by q2on q1is given by

 The Coulomb constant k in SI units has the value

 The constant ε0 is known as the permittivity of free space and has the value

Nm10

9875

84

0

-Nm

C10

28.854 





Electric force

11

 The electric force between charges q1and q2 is

(a) repulsive if charges have same signs (b) attractive if charges have opposite signs

 The electric force is a radial force, thus, a conservative force.

 More than one force, Superposition principle is applied.

Example 1

The electron and proton of a hydrogen atom are separated (on the average) by a

electric force and the gravitational force between the two particles

• From Coulomb’s law, we find that the magnitude of the electric force is

• Using Newton’s law of universal gravitation we find that the magnitude of the

gravitational force is

• The ratio of them is

N10

2

8)

m10

3.5(

)C106

.1(C

Nm10

8975

2 19 2

2 9

F E

12

• From Coulomb’s law, we find that the magnitude of the electric force is

• Using Newton’s law of universal gravitation we find that the magnitude of the

gravitational force is

• The ratio of them is

N10

2

8)

m10

3.5(

)C106

.1(C

Nm10

8975

2 19 2

2 9

F E

N10

6

3)

m10

3.5(

)kg10

67.1)(

kg10

1.9(kg

Nm10

67

27 31

2

2 11

3910

F



1 Does the ratio γ depend on the distance r between the electron and the proton?.

2 What is the fundamental difference between the two forces?

Questions

Example 2

q1 = -q2 = 6.0 μC, q3 = 3.0 μC, a = 2.0 x10-2 m

 The total force F3 acting on the charge q3 is

 The electric force F13 can be calculated as

 The electric force F23 can be calculated as

 Finally,

23 13

i a

q q F

r r

q q r

r r

q q F

1

sin

cos 2

4 1

ˆ

4

1 4

1

2 3 1 0 2

3 1 0 13

13 2 13

3 1 0 13

13 2 13

3 1 0 13

 The total force F3 acting on the charge q3 is

 The electric force F13 can be calculated as

 The electric force F23 can be calculated as

 Finally,

a

q q j

i a

q q F

r r

q q r

r r

q q F

1

sin

cos 2

4 1

ˆ

4

1 4

1

2 3 1 0 2

3 1 0 13

13 2 13

3 1 0 13

13 2 13

3 1 0 13

q q r

r

q q r

r r

q q

23

3 2 0 23

23 2

23

3 2 0

F

4

2 1

4

2 4

1

2

3 1 0

23 13

3



4

2 1

4

2 4

1

2

3 1 0

23 13

3 4

2 1

4

2 4

1

2 / 1 2 2

2

3 1 0

F

4

2 1

4

2 4

1

2

3 1 0

23 13

3



0 3

3

3 151 1

4 2

4 /

Defining the electric field

 What is the mechanism by which one particle can exert a force on another

across the empty space between particles?

 Suppose a charge is suddenly moved Does the force exerted on a second

particle some distance r away change instantaneously?

15

 A charge produces an electric field everywhere in space.

 The force is exerted by the field at the position of the second charge.

 The field propagates through space at the speed of light.

 It’s a vector field.

Defining the electric field

16

The electric field vector at a point in space is defined as the electric

force acting on a positive test charge q0 placed at that point divided

by the test charge:

Electric field of a point charge

 An electric charge q produces an electric field

everywhere.

 If we put a positive test charge q0 at any point P a

distance r away from the point charge q, the

electrostatic force exerts on a test charge is

 The electric field created by the charge q at point P is

 If we put a positive test charge q0 at any point P a

distance r away from the point charge q, the

electrostatic force exerts on a test charge is

 The electric field created by the charge q at point P is

F 

r

r r

qq F





2 0

04

q q

F E







2 0

Field lines

1.The electric field vector is tangent to the electric field line at each point

2.Field lines point away from positive charges and terminate on negative charge

3.Field lines never cross each other

4 The number of lines per unit area through a surface perpendicular to the lines is proportional to the magnitude of the electric field in a given region.

18

Superposition principle

At any point P, the total electric field due to a group of source charges equals the vector sum of the electric fields of all the charges.

 If we place a positive test charge q0 near n point charges q1, q2, q3…, qn, then the

net force F0 from n point charges acting on the test charge is

 By definition, the electric field E at the position of the test charge is

19

 If we place a positive test charge q0 near n point charges q1, q2, q3…, qn, then the

net force F0 from n point charges acting on the test charge is

 By definition, the electric field E at the position of the test charge is

F F

F

1

0 0

30 20

F q

F q

F E

1

1 0

0 0

1 0

Electric dipole

An electric dipole is defined as a positive charge +q and

a negative charge -q separated by a distance d For the

dipole shown in this figure,

1 Find the electric field E at P due to the dipole,

where P is a distance y from the origin.

2 Find the electric field E at Q due to the dipole,

where Q is a distance x from the origin.

20

An electric dipole is defined as a positive charge +q and

a negative charge -q separated by a distance d For the

dipole shown in this figure,

1 Find the electric field E at P due to the dipole,

where P is a distance y from the origin.

2 Find the electric field E at Q due to the dipole,

where Q is a distance x from the origin.

Electric dipole

1 Find the electric field E at P due to the dipole,

where P is a distance y from the origin.

Answer

 The electric field E+ at P due to the charge +q

 The electric field E- at P due to the charge –q

 The electric field E at P due to the dipole

  cos i sin j2

4

1 4

1

2

2 0

2 0

q r

r r

q E

21

1 Find the electric field E at P due to the dipole,

where P is a distance y from the origin.

Answer

 The electric field E+ at P due to the charge +q

 The electric field E- at P due to the charge –q

 The electric field E at P due to the dipole

  cos i sin j2

4

1 4

1

2

2 0

2 0

q r

r r

q E

  cos i sin j2

4

1 '

' ' 4

1

2

2 0

2 0

q r

r r

q E

 

y d

d y

d

q E

i y

d

q E

2 0

2 2 0

2

2 2

2 4

1

cos 2 2

4 1

2 0

24

Electric dipole moment

 Definition of electric dipole moment:

 The electric field E at P due to the dipole

 In case of y >> d

d q

P e 



22

 Definition of electric dipole moment:

 The electric field E at P due to the dipole

 In case of y >> d

2

2 0

2 / 3 2

2 0

2 4

1

2 4

P i

y d

Superposition principle

Continuous charge distribution

• Volume charge density

• Surface charge density

• Linear charge density

23

• Volume charge density

• Surface charge density

• Linear charge density

i Q q q

i Q q q

r

r r

dq E

d r

r r

q E

i

i i

i i

2

14

Electric Field of a Rod

A non-conducting rod of length l with a uniform positive charge density λ and a total charge q is lying along the x-axis, as illustrated in figure.

1 Calculate the electric field at a point P(x0,0) located along the axis of the rod

2 Calculate the electric field at a point Q(0,y0) located along its perpendicular bisector

25

Electric Field of a Rod

A non-conducting rod of length l with a uniform positive charge density λ and a total charge q is lying along the x-axis, as illustrated in figure.

1 Calculate the electric field at a point P(x0,0) located along the axis of the rod

dx

dq  

i x

x

dq E

2 0

x

dq E

2 0

x

q i

l x

l E

i x x

dx E

4

1)

4(

4

1

)(

4

2 2

0 0

2 2

0 0

2 /

2 /

2 0

q E

Electric Field of a Rod

A non-conducting rod of length l with a uniform positive charge density λ and a total charge q is lying along the x-axis, as illustrated in figure.

2 Calculate the electric field at a point Q(0,y0) located along its perpendicular bisector

 sin i cos j 

) (

4

1

2 2

0 0

dx E

d

27

 sin i cos j 

) (

4

1

2 2

0 0

dx E

d

 sin i cos j 

) (

4

1

0 0

dx E

j x

y

dx dE

E d

) (

cos 4

j

Electric Field of a Rod

j x

y

dx

) (

1 (

cos

tan

2 2

0

2 0 2

2

0 0

x

d

y dx

y x

28

) tan

1 (

cos

tan

2 2

0

2 0 2

2

0 0

x

d

y dx

y x

j y

j y

d y

max 0

0

2 2

2 0 0

4

1 cos

) tan

1 (

cos 4

max max

y y

l j

l y

l y

4 4

1 4

2

/

2 4

1

2 2

0 0

0 2

2 0 0

y y

q

4 4

1

2 2

0 0

Electric Field of a Circular Arc

29

Electric Field on the Axis of a Ring

A non-conducting ring of radius R with a uniform charge

density λ and a total charge Q is lying in the xy-plane, as

shown in figure Compute the electric field at a point P,

located at a distance z from the center of the ring along its

axis of symmetry.

 Let’s consider a small length element dl on the ring The

amount of charge contained within this element is

dq = λdl

 The electric field dE created by the charge dq at point

P is

30

 Let’s consider a small length element dl on the ring The

amount of charge contained within this element is

dq E

d





2 0

1

z R

dq dE







Electric Field on the Axis of a Ring

 Using the symmetry argument illustrated in this

figure, we see that the electric field at P must point

R

z E

n z

R

z z

R

dq E

n dE

E d E

ring ring

( 4

1

) (

) (

4 1

cos

2 / 3 2 2

0

2 / 1 2 2

2 2

R

Qz

2 / 3 2 2

Electric Field on the Axis of a Ring

 The electric field at point P a distance z from the center of the ring along its axis

• At the center O: z = 0, E = 0

• In the limit z >> R :

 Graphical representation

n z

R

Qz

2 / 3 2 2

Q z

2 0

4

1)

(



2 0

(





R z

R z E

z

E

-10 -8 -6 -4 -2 0 2 4 6 8 10

-0.4 -0.2 0.0 0.2

Electric Field due to a Charged Disk

A circular plastic disk of radius R that has a positive surface charge of

uniform density σ on its upper surface is shown in the figure on the

right What is the electric field at point P, a distance z from the disk

along its central axis?

33

k R

A point charge in an electric field

 A particle of charge q and mass m is placed in an electric field E, the electric

force exerted on the charge is

 If this is the only force exerted on the particle, it must be the net force and causes theparticle to accelerate according to Newton’s second law

 If the particle has a positive charge, its acceleration is in the direction of the

electric field

 If the particle has a negative charge, its acceleration is in the direction

opposite the electric field

E q

F  



a m E

 A particle of charge q and mass m is placed in an electric field E, the electric

force exerted on the charge is

 If this is the only force exerted on the particle, it must be the net force and causes theparticle to accelerate according to Newton’s second law

 If the particle has a positive charge, its acceleration is in the direction of the

electric field

 If the particle has a negative charge, its acceleration is in the direction

opposite the electric field

a m E

q



