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ELECTRICITY and
MAGNETISM
NỘI DUNG MÔN HỌC
Chương 1. Trường tĩnh điện
Chương 2. Vật dẫn trong điện trường
Chương 3. Dòng điện không đổi
Chương 4. Từ trường tĩnh
Chương 5. Cảm ứng điện từ
TÍNH ĐIỂM MÔN HỌC
Điểm thường kỳ: là tổ hợp của:
Bài kiểm tra trên lớp.
Chuyên cần.
Thái độ học tập.
Điểm chấp nhận phải từ 3 trở lên.
Lưu ý: không nghỉ học buổi nào, thái độ học tập tốt,
không vi phạm qui chế sẽ được cộng thêm từ 1 đến
1,5 điểm TK và nếu điểm giữa kỳ nhỏ hơn 4, lớn hơn
hoặc bằng 3 vẫn được dự thi cuối kỳ.
Thi giữa kỳ:
Nội dung thi: Chương1,2.
Hình thức thi: Trắc nghiệm và tự luận, thời gian 45’.
Thời gian thi: Dự kiến vào tuần thứ 7 (thực học).
Điểm đạt 4 trở lên (thi một lần duy nhất)
Thi cuối kỳ:
Nội dung thi: Chương 3,4
Hình thức thi: Trắc nghiệm và tự luận, thời gian 60’
Điểm đạt 4 trở lên (thi một lần duy nhất)
Điều kiện để được thi cuối kỳ:
Mắc phải một trong những điều sau đây sẽ không được
dự thi cuối kỳ:
Nghỉ quá 20% tổng số tiết học ( quá 6 tiết).
Điểm thi giữa kỳ nhỏ hơn 3.
Điểm thường kỳ dưới 3.
Chapter 1
ELECTRIC FIELDS
1 – Properties of electric charges
2 – Coulomb’s Law
3 – The electric field
4 – Electric field lines
1 – Properties of electric charges
-There are two kinds of charges in nature : charge
positive and charge negative.
- Charges of the same sign repel one another and
charges with opposite signs attract one another.
- Total charge in an isolated sytems is conserved
- Electric charge has the smallest value is call
element charge.
q = Ne
where N is some interger
e = -1,6.10-19C
me = 9,1.10-31kg
2 – Coulomb’s Law
Electric force between two stationary charged particles:
Fk
q1q 2
2
r
F12   F21
Electric force between two stationary charged particles:
Fk
1
k
4π ε 0
q1q 2
r
2
k = 9.109 (Nm2/C2) Coulomb constant
r : space between two point charges (m)
ε 0  8,85.10
12
2
2
C m /N
is known as the permitivity of free space
Exercise
There are two point electric charges: q1 = -3.10-8C
at A, q2 = 5.10-8C at B, for from AB = 10cm. Find
the electric force exerted by q1 on q2?
The electric force exerted by q1 on q2 is:
Fk
q1q 2
r
2
 ...
Fk
q1q 2
r
2
vector

q1q 2 
Fk 3 r
r
To calculate the electric force due to a group of
point charges, the vector sum of the electric force of
all the charges exerted by q is:
  


F  F1  F2  ...  Fn   Fi
i
Plus two vector:
  
c ab
 


2
2
c  a  b  2ab.cos a, b
Exercise Consider three point charges located at the
corners of a right triangle as shown in figure, where
q1 = q3 = 5μC, q2 = -2μC, and a = 0,1m. Find
resultant force exerted on q3.
a
q2

F23

F3

F13
The resultant force
exerted on q3 is:
 

F3  F13  F23
q3
a
F3  F  F  2F13F23cos135
2
13
q1
 q1q 3
  k
2
2a

2
  q 2q 3
  k 2
  a
 
2
2
23
0

q1q 3
q 2q 3
0
  2k
.k
cos135
2
2

2a
a

Exercise
Three point charges lie along the x axis. The
positive charge q1 = 15μC is at x = 2m, the positive
charge q2 = 6μC is at the origin, and the net force
acting on q3 is zero. What is the x coordinate of q3.
The x coordinate of q3 is:
O
x = 2m
+
q2
+
x
q1
k


F31  F32
q1q 3
2  x 
2
k
x ?
q 2q 3
x
2
3 – The electric field
Electric fields is the region of space around a
charged object the source charge. When another
charged obejct the test charge enters this electric
field, an electric force acts on it.

The electric field vector EV/m  at a point in space is
defined as the electric force F acting on a positive test
charge q0 placed at that point divided by the test
charge:


F
E
q0


F  q0 E
* The electric field created by q is positive :
If the source charge q positive, the situation with the test
charge removed the source charge sets up an electric field at
point P, directed away from q.


+
P
E
r
q0


q0  0  E  F
The force is in the same direction as the field
* The electric field created by q is negative:
If q is negative , the force on the test charge is toward the
sourse charge, so the electric field at P is directed toward the
sourse charge.
P

r

E
-
q <0


q0  0  E  F
The force and the field are in opposite directions.

P

+
E
r
q0
Ek
q.q 0
F
E
k
2
q0
q0r
q
r
2
vector

q 
Ek 3 r
r
r : is the space from point P to the source charge q (m)
* The electric field at point due to a group of sourse
charges can be expressed as the vetor sum:

qi 
E  k  3 ri
ri
i
  


or E  E1  E 2  ...  E n   E i
i
ri is the distance from the i th sourse charge qi to
the point P (m).
* The charge distribution is modeled as continuous,
the total field at P in the lim it Δqi → 0 :

E  k lim
Δq i 0

i
qi 
dq 
r  k 3 r
3 i
ri
r
. If a charge q is unifromly distributed throughout a
volume V, the volume charge density ρ is defined
by: ρ = q/V (C/m3)
. If a charge q is unifromly distributed throughout
on a surface of area A, the surface charge density
σ is defined by: σ = q/A (C/m2)
. If a charge q is unifromly distributed a long a line
of length ℓ, the linear charge density σ is defined
by: ℓ = q/λ (C/m2)
. If the charge q is nonuniformly distributed over a
volume, surface, or line, throughout a volume V,
the of charge dq in a small volume, surface, or
length element are:
dq  ρ.dV
dq  σ.dA
dq  λ.d
Exercise
In the air, there are a point electric charges Q = 3.10-6 C. Caculate the electric field at the point P
to electric charge Q is 30cm ?
The electric field at the point P to electric charge
Q is 30cm:
Ek
q
r
2
 ...
Exercise
There are two point electric charges: q1 = -3.10-8C
at A, q2 = 5.10-8C at B, for from AB = 10cm. At P,
with PA = PB = 5cm. Find the electric field at the
point P? 
A q1
E2 P

E1
+ B
r1 = r2 = 0,05m
q2

 
E P  E1  E 2
EP  E1  E2  k
q1
2
1
r
k
q2
2
2
r
 ...
Exercise: There are two point electric charges: q1 =
8µC at A, q2 = -6µC at B, for from AB = 10cm. At
P, with PA = 8cm, PB = 6cm. Find the electric field
at P ?

 
2
2


E

E

E
E P  E1  E 2
P
1
2
E1
2
 q1   q 2 
  k 2    k 2 
 r1   r2 
2
P
r1
+
q1 A

EP

E2
r2
B
-
q2
Exercise
A point charge of -4nC is located at (0, 1)m. What
is the x component of the electric field due to the
point charge at (4,-2)m?
Exercise
1. Put a point electric charge Q = - 5μC is at the
origin. Caculate the electric field at point P
located (4,-3)m?
2. There are two point electric charges: q1 = -10-8
C at A, q2 = 3.10 -8C at B, for from AB = 20cm. At
point P is a centre of AB. Caculate :
a. The electric field at point P. Draw a figure.
b. The electric force affecting on electron put at P.
Exercise
Four charged particles are at the corners of a square
of side a. Determine:
a. The total electric field at the centre of square.
b.The total electric force exerted on electron at the
centre of square.
2q
3q
+
+
q
+
+
4q
4 – Electric field lines
The electric field vector E
is tangent to the electric
field line at each point.
The line has a direction,
indicated by an arrowhead,
that is the same as that of
the electric vector.
The number of lines per unit area through a surface
perpendicular to the lines is proportional to the
magnitude of the electric field in the region. Thus, the
field lines are close together where the electric field is
strong and far apart where the field is weak.
B
A
EA > EB
The rules for drawing electric field lines are as
follows:
. The lines must begin on a positive charge and
terminate on a negative charge. In the case of an
excess of one type of charge, some lines will
begin or end infinitely far away.
. The number of lines drawn leaving a positive
charge or approaching a negative charge is
proportional to the magnitude of the charge.
. No two field lines can cross.
Electric field lines
Chapter 2
GAUSS’S LAW
1 – Electric Flux
2 – Gauss’s Law
3 – Application of Gauss’s Law to various
charge distributions
4 – Conductors in electrostatic equilibrium
1. Electric Flux ΦE (Vm) is proportional to the
number of electric field lines that penetrate a surface.
1. Electric Flux ΦE (Vm) is proportional to the
number of electric field lines that penetrate a
 
surface.
Φ E   E.dA   E.dA.cosθ


E
n
θ
dA
If the electric field is uniform
and makes an angle θ with the
normal to a surface of area A,
the electric flux
through
the

surface is: E  const
 


ΦE  EA.cosθ θ  n, E
A is area (m2)
Exercise. In the electric field is uniform E =
2 is area of suface
4.10-5V/m has
A
=
20cm
 
and θ  n, E   1200 . Find the electric flux
send to a area A.
Exercise. A flat surface of area 3,2m2
is rotated in a unifrom electric field of
magnitude E = 6,2.105V/m. Determine
the electric flux through this area :
a. When the electric field is
perpendicular to the surface.
b. When the electric field is parallel to
the surface.
2. Gauss’s Law
Electric flux ΦE through any closed gaussian
surface is equal to the net charge qin inside the
surface divided by ɛ0 :
 
 E   E.dA 
q
i
ε0
in
Exercise. There are three point electric
charges: q1 = -10-6C, q2 = 2.10-6C, q3 =
3.10-6C put inside the sphere, radius 30cm.
Find the electric flux send to plane of
sphere it.
The electric flux send to plane of sphere is:
E 
q
i
ε0
in
q1  q 2  q 3

 ...
ε0
Exercise. There are three point electric
charges: q1 = -10-6C at A, q2 = 2.10-6C at B,
q3 = 3.10-6C at C. For AB = 20cm and BC =
30cm. Caculate the electric flux send to
plane of sphere has centre B, radius 25cm.
q1
A
q2
+
B
q3
+
C
The electric flux send to plane of sphere has the
radius 25cm is:
q1  q 2
ΦE 
 ...
ε0
Exercise. Give a right triangle ABC, AB =
BC = 10cm, put electric charge q1 = 10-6C at
A and electric charge q2 = - 2.10-6C at C.
Caculate :
a. The electric field at point B.
b. The electric force affecting on electron put
at B.
c. The electric flux send to plane of sphere
has centre C, radius 5cm
Chapter 3
ELECTRIC POTENTIAL
1 – Potential difference and electric potential
2 – Potential difference in a uniform electric
field
3 – Electric potential and potential energy due
to point charge
4 – Obtaining the value of the electric field form
the electric potential
1 – Potential difference and electric potential
When a positive test charge q is moved between
points A and B in an electric field E of electric charge
Q, the potential energy (J) of the charge field system
is:
rB
B
 
dr
W   F.d r  kQq  2
r
r
A
A
A
rrMA
Q
r
+
rB
rN
+
B
q
 1
1
W  kQq  
 rA rB 
Fk
Qq
εr
2
 qE
 1
1
W  kQq  
 rA rB 
rA is the space from point A to the source charge Q (m)
rB is the space from point B to the source charge Q (m)
A
rrMA
Q
r
+
rB
rN
+
B
q
Fk
Qq
εr
2
 qE
On the other side:
 
Q
W   F.d r  q  E.dr  q  k 2 dr
r
rA
r
rB
rB
A
rB
Q
Q
W
Q
= VA - VB
-k
  k 2 dr  k
rA
rB
q
r
rA
W
 U AB  VA  VB
q
UAB : is the potential difference between points A & B (V)
VA & VB : is the electric potential at point A & B (V)
Comment
The electric energy a charge – field sytems do
not depend on the shape way, only depend on
the first point and the final point.
W
 U AB  VA  VB
q
If the first point A coincide the final point B
(A≡ B) then W = 0 and UAB = 0 → VA = VB
Exercise there are two point electric charges: Q
= -10-6 C at A, q = 2.10-8C at B, for from AB =
90cm. Caculate the potential energy of the
charge – field system when charge q is moved
from B to C ? BC = 10cm
1
1
W  kQq    kQq
 rA rB 
A
-
Q
B
+
q
1
1
    ...
 rB rC 
rB = AB, rC = AC
C
* The electric potential at a point due to a point
charge Q:
If we choose V = 0 at r = ∞ (V∞ = 0 ), the electric
potential due to a point charge at any distance r from
the charge is:
Q0V0
Q
Notice
Vk
Q<0V<0
r
r: is the space from point caculate the electric
potential to the source charge Q (m)
* The electric potential at a point due to several
point charges:
The total electric potential at some point due to
several point charges is the sum of the potential
due to the individual charges. For a group of point
charges, we can write the total electric potential :
n
Qi
V  k
i 1 ri
ri : is the distance from the point caculate the
electric potential to the source charge Qi (m)
Exercise. There are three point electric charges:
q1 = -10-6C, q2 = 2.10-6C, q3 = 3.10-6C put at three
the tops of regular triangle, side a = 20cm. Choose
V∞ = 0. Caculate:
a. Electric potental at centre triangle.
b. The potential energy of the charge – field system
when a electron is moved from centre triangle far
to endless.
q1 = -10-6C, q2 = 2.10-6C, q3 = 3.10-6C , a = 20cm.
a. Electric potental at centre triangle is:
q1  q 2  q 3
Qi
 ...
V  k  k
r
i 1 ri
n
r
O
r
r
3
2
3
2
a
r h  a
3
3 2
3
b. The potential energy of the charge – field
system when a electron is moved from centre
triangle far to endless is:
W V V
 O

q
r
O
r
W  qVo  eVo  ...
r
Exercise. Put two point electric charges q1 = 10-6C at A and q2 = 2.10-6C at B, for AB =
10cm. Point C consistant with AB a right
triangle at C. Choose V∞ = 0. Caculate
1. The electric potential at centre point of AB.
2. The potential difference UMC = ?
Problem. Put two point electric charges q1 = 3.10-6C
at A and q2 = - 4.10-6C at B, for AB = 10cm. Point C
is centre of AB. Choose V∞ = 0. Caculate :
a. The electric force exerted by q1 on q2 ?
b. The electric field at C.
c. The electric flux send to plane of sphere has centre
B, radius 15cm.
d. The electric potential at point C.
e. The potential energy of the charge – field system
when a electric charges q0 = 2.10-6C moved from
point C far to endless.
2 – Potential difference in a uniform electric field
W
 U AB
q
  B 
 VA  VB   F.d r   E.d r
A
B
U  E  dr  E.r
A
Because E = constant
where r = |s| is space of the electric charge q moved
between points A and B (m)
VA > VB ; EA > EB
VA < VB ; EA > EB
B
A
A
B
A
C
B
VA = VB > VC
EA = EB = EC
Electric field lines always
point in the direction of
decreasing electric
potential
Exercise
BC = 3cm, AC = 4cm
B
UAB = E.AC = - 240V
UCB =VC – VB = 0
UBA = -UAC = 240 V

E
A
C
MỘT SỐ ỨNG DỤNG CỦA TĨNH ĐIỆN
1 – Sơn tĩnh điện:
Fine mist of negatively
charged gold
particles adhere to
positively charged
protein on fingerprint.
Negatively charged
paint adheres
to positively charged
metal.
71
MỘT SỐ ỨNG DỤNG CỦA TĨNH ĐIỆN
2 – Làm sạch không khí:
72
MỘT SỐ ỨNG DỤNG CỦA TĨNH ĐIỆN
3 – Băng dính:
73
The drum
is anSỐ
aluminum
coated
with
a thin ĐIỆN
layer of
MỘT
ỨNGcylinder
DỤNG
CỦA
TĨNH
Selenium.
4 – Kỹisthuật
photocopy:
Aluminum
a conductor.
Selenium is a photoconductor, it is an insulator in the dark and
a conductor when exposed to lightlight.
So, a positive charge deposited on the Selenium layer will stay
there.
However, when the drum is esposed to light, electrons from the
aluminum will pass through the conducting selenium and
neutralize the positive charge.
1. Charging the drum
2. Imaging the document on the drum
3. Fixing the toner
4. Transferring the toner to the paper.
74
MỘT SỐ ỨNG DỤNG CỦA TĨNH ĐIỆN
5 – Kỹ thuật in phun:
75
MỘT SỐ ỨNG DỤNG CỦA TĨNH ĐIỆN
6 – Kỹ thuật in Laser:
76
MỘT SỐ ỨNG DỤNG CỦA TĨNH ĐIỆN
7 – Đèn hình TV:
77
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