Transcript CHAPTER 23 : ELECTRIC FIELDS 23.1 : PROPERTIES OF ELECTRIC CHARGES •Existence of electric forces and charges : •Running a comb through your.

CHAPTER 23 : ELECTRIC FIELDS
23.1 : PROPERTIES OF ELECTRIC CHARGES
•Existence of electric forces and charges :
•Running a comb through your hair on a dry day
•Glass or rubber - rubbed with silk or fur
•Inflated balloon rubbed with wool – adheres to a wall
•Rubbing shoes on a wool rug
• Electrified / Electrically charged
• Electric charges - positive
- negative
• Like charges – repel (+ve with +ve / -ve with –ve)
Unlike charges – attract (+ve with –ve)
• Electric charge is always conserved
• Electric charge is quantized
Figure (23.1)
FIGURE (23.1)
Rubber
Rubber
F
F
F
Rubber
Glass
(a)
A negatively charged rubber rod suspended by
a thread is attracted to a positively charged
glass rod
F
(b)
A negatively charged rubber rod is repelled by
another negatively charged rubber rod.
Quiz (23.1) : If you rub an inflated balloon against your hair, the two materials
attract each other (Fig. 23.2). Is the amount of charge present in the balloon and
your hair after rubbing (a) less than, (b) the same as, or (c) more than the
amount of charge present before rubbing?
23.2 : INSULATORS AND CONDUCTORS
• Electrical conductors – materials in which electric charges move freely.
• Electrical insulators – materials in which electric charges cannot move freely.
• Conductor – e.g.: copper, aluminum, silver
• Insulator – e.g.: glass, rubber, wood
• Semiconductors - their electrical properties are somewhere between those of
insulators and those of conductors.
- Silicon and germanium.
- Electrical properties of semiconductors can be changed.
• Grounded – when a conductor is connected to the earth by means of a
conducting wire or pipe.
• Induction – process of charging a conductor (Fig. 23.3).
•A process similar to induction in conductors takes place in insulator.
Insulator
Quiz 23.2 : Object A is attracted to object
B. If object B is known to be positively
charged, what can we say about object A?
(a) It is positively charged, (b) It is
negatively charged, (c) It is electrically
neutral, or (d) Not enough informatin to
answer.
Charged object
Induced charges
(a)
FIGURE (23.4)
23.3 : COULOMB’S LAW
• Coulomb’s experiment showed the electric force between two stationary charge
of particles :
i)
F
1
r2
ii)
F  q1q 2
iii) Opposite sign - attractive
Same sign - repulsive
• Coulomb’s Law
Fe  k e
q1 q 2
r2
(23.1)
1
ke 
4o
Coulomb constant, ke= 8.9875 x 109 Nm2/C2
Permittivity of free space, o = 8.8542 x 10-12 C2/Nm2
Table 23.1
Particle
Charge (C)
Mass (kg)
Electron (e)
- 1.6021917 x 10-19
9.1095 x 10-31
Proton (p)
+ 1.6021917 x 10-19
1.67261 x 10-27
Neutron (n)
0
1.67492 x 10-27
Example (23.1) : The electron and proton of a hydrogen atom are
separated (on the average) by a distance of approximately 5.3 x 10-11 m.
Find the magnitudes of the electric force and the gravitational force
between the two particles.
e
r
5.3 x 10-11 m
p
•Coulomb’s Law – vector quantity
q1q 2
F12  k e 2 rˆ
r
(23.2)
F21  F12
Quiz (23.2) : Object A has a charge of +2C, and object B has a charge of +6
C. Which statement is true?
(a)FAB = -3FBA
(b)FAB = -FBA
(c) 3FAB = -FBA
F1  F21  F31  F41
Example (23.2) : Find the Resultant Force
Consider three point charges located at the corners of a right triangle (Fig. 23.7),
where q1 = q3 = 5.0 C, q2 = -2.0 C, and a = 0.10 m. Find the resultant force
exerted on q3.
y
a
q2 a
q1 +
F13
F 23
+
q3
FIGURE (23.7)
2a
x
Example (23.3) : Where is the Resultant Force Zero?
Three point charges lie along the x-axis as shown in Figure (23.8). The
positive charge q1 = 15.0 C is at x = 2.00 m, the positive charge q2 = 6.00 C
is at the origin, and the resultant force acting on q3 is zero. What is the x
coordinate of q3?
2.00 m
FIGURE (23.8)
2.00 - x
x
+
q2
+
-
F 23
q3
F13
q1
x
Example (23.4) : Find the Charge on the Spheres
Two identical small charged spheres, each having a mass of 3.0 x 10-2 kg, hang
in equilibrium as shown in Fig. (23.9a). The length of each string is 0.15 m,
and the angle  is 5.0o. Find the magnitude of the charge on each sphere.
 
L
q
T cos 
L
Fe
T sin 

mg
L = 0.15 m
 = 5.0o
(a)


q
a
T
FIGURE (23.9)
(b)
23.4 : THE ELECTRIC FIELD
Q
+
+
+
+ +
+ +
+ +
+ +
- direction
qo
+
E
FIGURE (23.10)
•The strength (magnitude) of the electric
field, E at a point in space is defined as
the electric force Fe acting on a positive
test charge, qo placed at that point
divided by the magnitude of the test
charge :
Fe
E
qo
•Newtons per coulomb (N/C)
• E is a vector - magnitude
(23.3)
• An electric field exists at a
point if a test charge at rest at
that point experiences an
electric force.
•Charge qo is small enough that
it does not disturb the charge
distribution.
'
q
+ o  q o
+ qo
- - - -- - -
- -- - - - - - --
(a)
(b)
FIGURE 23.11
• To calculate electric field E at a point
P due to a group of point charges :
qq o
F e  k e 2 rˆ
r
1) Calculate the E vectors at P
individually using Equation (23.4).
2) Add them vectorially
Fe
E
qo
q
E  k e 2 rˆ
r
At any point P, the total electric field
due to a group of charges equals the
vector sum of the electric fields of the
individual charges.
(23.4)
qi
E  k e  2 rˆi
i ri
(23.5)
Quiz 23.4 : A charge of +3 C is at a point P where the electric field is directed
to the right and has a magnitude of 4 x 106 N/C. If the charge is replaced with a
-3 C charge, what happens to the electric field at P?
Example (23.5) : Electric Field Due to Two Charges
A charge q1=7.0C is located at the origin, and a second charge q2=-5.0 C is
located on the x-axis, 0.30 m from the origin (Fig. 23.13). Find the electric field at
the point P, which has coordinates (0, 0.40)m.

E1
y

P


E
FIGURE (23.13)

E2
Exercise : Find the electric
force exerted on a charge
of 2.0x10-8C located at P.
0.50m
0.40m

+
q1
x
0.30m
q2
Example (23.6) : Electric Field of a Dipole
An electric dipole is define as a positive charge q and a negative carge –q
separated by some
distance. For the dipole shown in Figure 23.14, find the

electric field E at P due to the charges, where P is a distance y >> a from the
origin.
23.5 : ELECTRIC FIELD OF A CONTINUOUS CHARGE DISTRIBUTION
Procedure :
rˆ
q
2) Use Eq. (23.4) to calculate the electric
field due to one of these elements at a
point P.
r

E
FIGURE (23.15)
P
1) Devide the charge distribution into
small elements, each of which contains
a small charge q (Fig. 23.15)
3) Evaluate the total field at P due to the
charge distribution by summing the
contributions of all the charge elements
(applying the superposition principle).
The volume charge density, 

q
E  k e 2 rˆ
r
Q

V
Unit : C/m3

q i
E  k e  2 rˆi
i ri
The surface charge density, 
q i
dq
E  k e lim0  2 rˆi  k e  2 rˆ
i ri
r
Unit : C/m2
q i
(23.6)
Q

A
The linear charge density, 
Q


Unit : C/m
dQ

dV
dQ

dA
dQ

d
Example (23.7) : The Electric Field Due to a
Charged Rod
A rod of length  has a uniform positive charge
per unit length  and a total charge Q. Calculate
the electric field at a point P thant is located along
the long axis of the rod and a distance a from one
end (Fig. 23.16).
Example (23.8) : The Electric Field of a
Uniform Ring of Charge
A ring of radius a carries a uniformly distributed
positive total charge Q. Calculate the electric
field due to the ring at a point P lying a distance
x from its center along the central axis
perpendicular to the plane of the ring (Fig.
23.17a).
Exercise (Example (23.8)) : Show that at great distances from the ring
(x>>a) the electric field along the axis shown in Figure (23.17) approaches
that of a point charge of magnitude Q.
Example (23.9) : The Electric Field of a Uniformly Charged Disk
A disk of radius R has a uniform surface charge density . Calculate the
electric field at a point P that lies along the central perpendicular axis of the
disk and a distance x from the center of the disk (Fig. 23.18).
23.6 : ELECTRIC FIELD LINES
•
Drawing lines that follow the same direction as the electric field vector at
any point
•
Electric field lines.
•
Electric filed in any region of space :
1)

The electric field vector E is tangent to the electric field line at each point.
2) The magnitude of the electric field – depends on the number of lines per
unit area through a surface perpendicular to the lines.
3) Nonuniform
Positive Point Charge
+
q
Negative Point Charge
-
(a)
-q
(b)
REPELLED positive test charge
ATTRACT positive test charge
Outward
Inward
Rules for drawing electric field lines :
1) The lines must begin on a positive charge and terminate on a negative charge.
2) The number of lines drawn leaving a positive charge or approaching a
negative charge is proportional to the magnitude of the charge.
3) No two field lines can cross.
B
A
C
FIGURE (23.22(a))
+
FIGURE (23.21(a))
Quiz (23.5) : Rank the
magnitude of the
electric field at points
A, B and C shown in
Figure (23.22(a))
(greatest magnitude
first).
23.7) : MOTION OF CHARGED
PARTICLES IN A UNIFORM
ELECTRIC FIELD
+
+2q
-
-q
•When a particle of charge q and mass m is
placed in an electric field  , the electric
force exerted on the charge isEq  .
E
•The only force exerted on particle – it must
be the net force and cause the particle to
accelerate.
•Newton’s second law applied to the particle,
gives :



Fe  qE  ma
FIGURE (23.23)
•The acceleration of the particle is therefore

 qE
a
m
(23.7)

• E is uniform (constant in magnitude and direction) – acceleration is constant.
• Positive charge - its acceleration is in the direction of the electric field.
• Negative charge – its acceleration is in the direction opposite the electric field.
Example (23.10) : An Acceleration Positive Charge
A positive
point charge q of mass m is released from rest in a uniform electric

field E directed along the x-axis, as shown in Figure (23.24). Describe its
motion.

+
E
-
+
+ 
v0
+ +
-
+
-
q
+
+

v
x
-
FIGURE (23.24)


i i
-
y
- - - - - - - - x
(0,0)

E
(x,y)
+
+
+
+
+
+
+
+
FIGURE (23.25)
•The electric field,  is in the
E
positive y direction – the
acceleration of the electron is in the
negative y direction.

eE 
a
j
m
(23.8)
•Acceleration
is
constant
–
equations of kinematic in two
dimensions (Chapter 4) with :
 yi  0
 xi  i
-


•After the electron has been in
the electric field for a time t, the
components of its velocity are :
 x  i  constant (23.9)
eE
y  a yt  
t
m
(23.10)
•Its coordinates after a time t in
the field are :
x  i t
(23.11)
1 eE 2
y  a yt  
t
2 m
1
2
2
(23.12)
• Subtituting the value t 
propotional to x2.
x (Eq. 23.11) into Eq. (23.12), we see that y is
i
•Hence, the trajectory is a parabole.
•After the electron
leaves the field, it continues to move in a straight line in the

direction of  in Figure (23.25), obeying Newton’s first law, with a speed   i.
Example (23.11) : An Accelerated Electron
An electron enters the region of a uniform electric field as shown in Figure
(23.25), with i = 3.00 x 106 m/s and E = 200 N/C. The horizontal length of the
plates is  = 0.100 m.
(a) Find the acceleration of the electron while it is in the electric field.
(b) Find the time it takes the electron to travel through the field.
(c) What is the vertical displacement y of the electron while it is in the field?
Exercise : Find the speed of the electron as it emerges from the field (3.22x106m/s).