Transcript electrical field lines.

Chapter 23
Electric field
23.1 Properties of Electric Charges
23.2 Charging Objects By Induction
23.3 Coulomb’s Law
23.4 The Electric Field
23.6 Electric Field Lines
23.7 Motion of Charged Particles in a Uniform Electric Field
1
Norah Ali AL.Moneef
23.1 properties of electric charge
2
Norah Ali AL.Moneef
Objects can be charged by rubbing or friction
(a) Rub a plastic ruler
(b) bring it close to some tiny pieces of paper
Objects charged by this method will attract each other.
3
Norah Ali AL.Moneef
we conclude that charges of the same sign repel one another and
charges with opposite signs attract one another
4
Norah Ali AL.Moneef
In 1909, Robert Millikan discovered that electric charge always occurs as
some integral multiple of a fundamental amount of charge e
When a glass rod is rubbed with silk, electrons are
transferred from the glass to the silk. Because of
conservation of charge, each electron adds negative charge
to the silk, and an equal positive charge is left behind on the
rod. Also, because the charges are transferred in discrete
bundles,
the charges on the two objects are
Note that the electrical charge is measured in coulombs).
and we can write q = N e, where N is some integer
the electric charge q is said to be quantized
the electron has a charge – e and the proton has a charge of equal magnitude
but opposite sign +e . Some particles, such as the neutron, have no charge.
5
Norah Ali AL.Moneef
A coulomb is the charge resulting from the transfer of 6.24 x 1018 of the
charge carried by an electron
6
Norah Ali AL.Moneef
• During any process, the net electric charge of an isolated
system remains constant i.e. is conserved.
How many electrons constitute 1 mC?
N =1x10-6/1.6x10-19
= 6x1012 electrons
7
Norah Ali AL.Moneef
23.2 Charging Objects By Induction
Electrical conductors are materials in which some of the electrons are free
electrons that are not bound to atoms and can move relatively freely through the material;
electrical insulators are materials in which all electrons are bound to atoms and cannot
move freely through the material
.When materials such as copper, aluminum, and silver are good electrical conductors are charged in
some small region, the charge readily distributes itself over the entire surface of the material
When electrical insulators such as glass, rubber are charged by rubbing, only the area
rubbed becomes charged, and the charged particles are unable to move to other regions of
the material
Semiconductors are a third class of materials, and their electrical properties
are somewhere between those of insulators
• Conductor
transfers charge on contact
• Insulator
does not transfer charge on contact
• Semiconductor might transfer charge on contact
8
Norah Ali AL.Moneef
(a) The charged object on the
left induces a charge
distribution on the surface of
an insulator due to realignment
of charges in the molecules.
9
Norah Ali AL.Moneef
Charging a metallic object by induction (that is, the two objects never
touch each other)
(a) A neutral metallic
sphere , with equal
numbers of positive and
negative charges
(b) The electrons on the neutral
sphere are redistributed when a
charged rubber rod is placed near
the sphere.
(d) When the ground connection is removed,
the sphere has excess positive charge that is
nonuniformly distributed
10
Norah Ali AL.Moneef
. (c) When the sphere is
grounded, some of its
electrons leave through the
ground wire.
(e) When the rod is removed, the remaining
electrons redistribute uniformly and there is a net
uniform distribution of positive charge on the
sphere
Conceptual Question
Assume that you have two uncharged, insulated metallic spheres A and B that
are in contact with each other. If you bring a positively charged insulated rod
near sphere A and hold it there while you move sphere B away, what charge
will sphere B have?
(A) Negative charge
(D) No net charge
(B) Positive charge
(E) None of these
(C) Either pos. or neg. charge
11
Norah Ali AL.Moneef
12
Norah Ali AL.Moneef
23.3 Coulomb’s Law
Charles Coulomb (1736–1806) measured the magnitudes of the electric forces between
charged objects using the torsion balance
Fe 
1
r
2
we can express Coulomb’s law as an equation giving the magnitude of the electric force
(sometimes called the Coulomb force) between two point charges
Coulomb’s law
13
Norah Ali AL.Moneef
where ke is a constant called the Coulomb constant = 8.987 5 x 109 N m 2/C2
Ke = 9x 109 N m 2/C2
charges of the same sign repel one another
charges with opposite signs attract one another
14
Norah Ali AL.Moneef
magnitude
direction
where rˆ is a unit vector directed from q1 toward q2,.
the electric force obeys Newton’s third law, the electric force exerted by q2 on
q1 is equal in magnitude to the force exerted by q1 on q2 and in the opposite
direction; that is, F21 = - F12.
15
Norah Ali AL.Moneef
(do not put signs on the charges when you use Coulomb’s law)
 The force between two charges gets
stronger as the charges move closer
together.
 The force also gets stronger if the amount
of charge becomes larger.
 The force between two charges is directed
along the line connecting their centers.
 Electric forces always occur in pairs
according to Newton’s third law, like all
forces.
16
Norah Ali AL.Moneef
Example
• Which charge exerts greater force? Two positive
point charges, Q1=50mC and Q2=1mC, are
separated by a distance L. Which is larger in
magnitude, the force that Q1 exerts on Q2 or the
force that Q2 exerts on Q1?
What is the force that Q1 exerts on Q2?
What is the force that Q2 exerts on Q1?
Q1Q2
F12  k 2
L
Q2Q1
F21  k 2
L
Therefore the magnitudes of the two forces are identical!!
Well then what is different? The direction.
17
Which direction?
Opposite to each other!
What is this law?
Newton’s third law, the law of action and reaction!!
Norah Ali AL.Moneef
Two point charges separated by a distance r
exert a force on each other that is given by Coulomb’s law. The
force F21 exerted by q2 on q1 is equal in
magnitude and opposite in direction to the force F12 exerted
by q1 on q2.
(a) When the charges are of the same sign, the force is
repulsive. (b) When the charges are of
opposite signs, the force is attractive
When more than two charges are present, the resultant force on any one of them equals
the vector sum of the forces exerted by the various individual charges. For example, if
four charges are present, then the resultant force exerted by particles 2, 3, and 4 on
particle 1 is
18
Norah Ali AL.Moneef
• Double one of the charges
– force doubles
• Change sign of one of the charges
– force changes direction
• Change sign of both charges
– force stays the same
• Double the distance between charges
– force four times weaker
• Double both charges
force four times stronger ( the force quadruples )
19
Norah Ali AL.Moneef
The electric force is very much like gravity. Both forces act at a
distance. Charge is like mass, although mass is always positive
and the force of gravity is always attractive. In fact, the general
law of gravitation is very much like the Coulomb's Law. The
force between two bodies of mass m1 and mass m2 a distance r
apart is
Gravitation force is always attractive.
20
Norah Ali AL.Moneef
Example
Two point charges repel each other with a force of 4×10 -5 N at a distance of 1
m. The two charges are:
(a) both positive
(b) both negative
(c) alike
(d) unlike
Example
Two charges of- Q are 1 cm apart. If one of the charges is replaced by a
charge of +Q, the magnitude of the force between them is;
(a) zero (b) smaller
(c) the same
(d) larger
21
Norah Ali AL.Moneef
Example :
Example :
22
Norah Ali AL.Moneef
Example :
find the magnitude of the force exerted between the proton and the
electron in the hydrogen atom. The force between these tiny charges, each
one of size 1.6 x 10-19 C (but opposite in sign, which we ignore here)
separated by a distance of 5.29 x 10-11 m is
23
Norah Ali AL.Moneef
Example :
24
Norah Ali AL.Moneef
Example :
Two protons in a molecule are separated by 3.80 x 10-10 m. Find
the electric force exerted by one proton on the other. (b) How
does the magnitude of this force compare to the magnitude of the
gravitational force between the two protons (mp=1.67 x10-27 Kg )
Thus, the gravitational force between charged atomic particles is negligible when compared with the
electric force.
25
Norah Ali AL.Moneef
Example - Forces between Electrons
• What is relative strength of the electric force compared with the
force of gravity for two electrons?(me=9.11x10-31 Kg)
(e) 2
Felectric  k 2
r
me2
Fgravity  G 2
r
Felectric
ke2
42


(do
the
calculatio
n)

4
.
2

10
Fgravity Gm2
26
Norah Ali AL.Moneef
Example :
27
Norah Ali AL.Moneef
Example :
 What is the magnitude of the electric force of attraction
between an iron nucleus (q=+26e) and its innermost
electron if the distance between them is 1.5 x 10-12 m
3.The magnitude of the Coulomb force is
F = kq1q2/r2
F=(9 x109N·m2/C2)(26)(1.6x10–19 C)(1.6 x10–19C)/(1.5 x 10–12
m)2
= 2.7 x 10–3 N.
28
Norah Ali AL.Moneef
Zero Resultant Force, Example
 Where is the resultant force equal to zero?
 The magnitudes of the individual forces will be equal
 Directions will be opposite
This is another location at which the magnitudes of the forces on q3 are
equal, but both forces are in the same direction at this location.
29
Norah Ali AL.Moneef
Electric Force of 2 charges
Ex1) Find the total electric force on q1.
You must use Newton’s Laws for these
problems
Step 1) Draw all forces
Step 2) Find the magnitudes of the forces individually




q1 q2
9  109 N  m 2 C 2 3.0  10 6 C
F12  k

2
r
0.20m 2


F12  2.7 N ( i )




q1 q3
9 109 N  m 2 C 2 3.0 10 6 C 7.0 10 6 C
F13  k

r2
0.15m 2


F13  8.4 N ( i )
30
Norah Ali AL.Moneef
4.0 10

6
C

 2.7 N
 8.4 N
 F  2.7N  8.4N
Example
Calculate the net electrostatic force on charge Q3 shown in the
figure due to the charges Q1 and Q2.
. (a) F32 is repulsive (the force on Q3 is in
the direction away from Q2 because Q3
and Q2 are both positive) whereas F31 is
attractive (Q3 and Q1 have opposite
signs), so F31 points toward Q1.
31
Norah Ali AL.Moneef
(b) Adding F32 to F31 to
obtain the net force.
The forces, components, and signs are as shown in the
figure. Result: The magnitude of the force is 290 N, at an
angle of 65° to the x axis.
Conceptual Example Make the force on Q3
zero.
In the figure, where could you place a fourth
charge, Q4 = -50 μC, so that the net force on
Q3 would be zero?
Solution: The force on Q3 due to Q4 must exactly cancel the net force on
Q3 from Q1 and Q2. Therefore, the force must equal 290 N and be
directed opposite to the net force
K Q3 Q 4
9  109  65  106  50  106
r

F
290
r  0.318m  31.8 cm
32
Norah Ali AL.Moneef
23 .4 The Electric Field
 The electric field at any point in space is defined as the
force exerted on a tiny positive test charge divide by the
test charge
F
 Electric force per unit charge
E
q
 What kind of quantity is the electric field?
 Vector quantity.
 What is the unit of the electric field?
 N/C
 The magnitude of the electric field at a distance r from a
single point charge Q is
1 Q
F
kQq r 2
kQ
E

 2 
2
4

q
r
q
r
0
33
Norah Ali AL.Moneef
Example :
Find the electric field
34
Norah Ali AL.Moneef
Example:
Electric field of a single point charge.
Calculate the magnitude and direction of the electric field
at a point P which is 30 cm to the right of a point charge
Q = -3.0 x 10-6 C.
3 10 6
9
E  K 2  9 10 
 3 105 N / C
2
r
30102
Q


. Electric field at point P (a) due to a negative charge Q, and
(b) due to a positive charge Q, each 30 cm from P.
Solution: Substitution gives E = 3.0 x 105 N/C. The field points
away from the positive charge and towards the negative one.
35
Norah Ali AL.Moneef
A map of the electrical field can be made by bringing a positive test charge into
an electrical field.
•When brought near a negative charge the test charge
is attracted to the unlike charge and when brought
near a positive charge the test charge is repelled.
•You can draw vector arrows to indicate the direction of
the electrical field.
•This is represented by drawing lines of force or
electrical field lines.
36
Norah Ali AL.Moneef
Point negative charge
q1
r
E= kq1/r2
q1
37
Norah Ali AL.Moneef
A test chargeq0 at point P is a distance r from apoint charge q.
(a) If q is positive, then the force on the test charge is directed away
from q.
(b) For the positive source charge, the electric field at P points radially
outward from q.
(c) If q is negative, then the force on the test charge is directed toward
q.
(d) For the negative source charge, the electric field at P points radially
inward toward q.
38
Norah Ali AL.Moneef
23.6
Electric Field Lines
Electric field lines penetrating two surfaces. The magnitude of
the field is greater on surface A than on surface B.
 The electric field vector 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 field vector. The direction of the line is
that of the force on a positive test charge placed in the field.
 The number of lines per unit area through a surface perpendicular to
the lines is proportional to the magnitude of the electric field in that
region. Thus, the field lines are close together where the electric field
is strong and far apart where the field is weak.
39
Norah Ali AL.Moneef
23.6 Electric Field Lines (Point Charge)
Electric Field
(vector)
Field Lines
(Lines of force)
Electric field lines (lines of force) are continuous lines
whose direction is everywhere that of the electric field
40
Norah Ali AL.Moneef
The charge on the right is twice the magnitude of the charge on the left
(and opposite in sign), so there are twice as many field lines, and they
point towards the charge rather than away from it.
41
Norah Ali AL.Moneef
Equal charges:
same number density
 Because the charges are of equal magnitude,
 the number of lines that begin at
the positive charge must equal the number
that terminate at the negative charge.
 At points very near the charges, the lines
are nearly radial. The high density of lines
between the charges indicates a region of
strong electric field
Unequal charges
–Ve > + Ve
42
Norah Ali AL.Moneef
Like charges (++)
Opposite charges (+ -)
The number of field lines starting (ending) on a positive (negative)
charge is proportional to the magnitude of the charge
since
N lines  Q
if object 1 has charge Q1 and object 2 has charge Q2, then the ratio of
number of lines is
N2/N1=Q2/Q1.
43
Norah Ali AL.Moneef
Electric Field Lines:.
The rules for drawing electric field lines are :
• 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.
44
Norah Ali AL.Moneef
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.
For discrete point charges, we can use the superposition
principle and sum the fields due to each point charge:
q2
 

E(r)   Ei
q3
q1
45
Norah Ali AL.Moneef
p
i
1)
What are the signs of the
charges whose electric
fields are shown at below?
2)
3)
4)
5) no way to tell
Electric field lines originate on
positive charges and terminate
on negative charges.
46
Norah Ali AL.Moneef
Which of the charges has
the greater magnitude?
1)
2)
3) Both the same
The field lines are denser around
the red charge, so the red one
has the greater magnitude.
Follow-up: What is the red/brown ratio of
magnitudes for the two charges?
47
Norah Ali AL.Moneef
Example :
Which of the following statements about electric field lines associated
with electric charges is false?
(a) Electric field lines can be either straight or curved.
(b) Electric field lines can form closed loops.
(c) Electric field lines begin on positive charges and end on negative
charges.
(d) Electric field lines can never intersect with one another.
Answer: (b). Electric field lines begin and end on charges and
cannot close on themselves to form loops.
48
Norah Ali AL.Moneef
Example :
Find the electric field due to a point charge of 0.5 mC at a distance of
4 cm from it in vacuum
q =0.5×10–3 C,
r = 4×10–2 m,
Recall E =kq/r2
and k=9 x 109 N.m2/C2
E = 9x109 N.m2/C2 0.5 X10-3 C/(4x 10-2 m)2 = 2.82×1010 N/C
49
Norah Ali AL.Moneef
Example :
3. Rank the magnitudes E of the
electric field at points A, B,
and C shown in the figure.
A) EC>EB>EA
B) EB>EC>EA
C) EA>EC>EB
D) EB>EA>EC
E) EA>EB>EC
50
Norah Ali AL.Moneef
.B
.C
.A
Example :
Rank the electric field strength in
order from smallest to largest.
A: E1 < E2 < E3 = E4
B: E3 = E4 < E2 < E1
C: E2 = E3 < E4 < E1
D: E1 < E4 < E2 = E3
51
Norah Ali AL.Moneef
Example :
A test charge of +3 µC is at a point P where an external electric
field is directed to the right and has a magnitude of 4×106
N/C. If the test charge is replaced with another test charge of
–3 µC, what happens to the external electric field at P ?
A. It is unaffected.
B. It reverses direction.
C. It changes in a way that cannot be determined.
52
Norah Ali AL.Moneef
Example :
At the position of the dot, the
electric field points
1. Left.
2. Down.
3. Right.
4. Up.
5. The electric field is zero.
53
Norah Ali AL.Moneef
Example of finding electric field from two charges
We have q =+10 nC at the origin, q = +15 nC at x=4 m.
1
2
What is E at y=3 m and x=0? point P
y
P
3m
q1=10 nc
Find electric field due to both charges at
point p
54
Norah Ali AL.Moneef
4m
x
q2 =15 nc
Example continued
E =kq/r2
k=9 x 109 N.m2/C2
Field due to q1
E=
9x109
N.m2/C2
X10-9 C/(3m)2
10
(in the y direction).
= 10 N/C
E

 j
 
f
5
3
q1=10 nc
Ex= 0
Ey= 10 N/C
y
f
4
x
q2 =15 nc
Field due to q2
E = 9 x 109 N.m2/C2 15 X10-9 C/(5m)2 =5.4 N/C
at some angle f Resolve into x and y components
Now add all components
Ey=E sin f  5.4 3/5  3.4N/C
Ex=E cos f  5.4  4/5   4.3 N/C
E
55
E E
2
x
Norah Ali AL.Moneef
2
y
magnitude
Ey= 10 + 3.4 = 13.4 N/C
Ex=
- 4.3 N/C
Example continued
E
Ey= 10 + 3.4 = 13.4 N/C

Ex= - 4.3 N/C
3
4
q1=10 nc
x
q2 =15 nc
Magnitude of electric field
E
Using unit vector notation we can
also write the electric field vector as:
E x2  E y2
E  13.4  4.3  14.1N / C
2
2
  tan -1 Ey/Ex= tan -1 (13.4/-4.3)= 72.8⁰
56
Norah Ali AL.Moneef



E   4.3 i  13.4 j
Example :
Three point charges are aligned
along the x axis as shown in
the Figure. Find the electric
field at
(a) the position (2.00, 0) and
57
Norah Ali AL.Moneef
Example :
From the figure find the electric field at (0,0)
E1 
E2 
ke q1
r12
ke q2
r22
8.99  10  3.00  10 

ˆ
j 
9
9
 
 0.100
(a)
 ˆj    2.70  10
3
2
8.99  10  6.00  10 

ˆ
ˆ   5.99  10
i

 i 


 0.300
9
9
2
2

 

ˆ
N C i

ˆ 2.70  103 N C ˆ
E  E 2  E1   5.99  102 N C i
j

F  qE  5.
00  109 C

ˆ 2 700ˆ
j
  599i

ˆ 13.5  106 ˆ
F  3.00  106 i
j N 
58
Norah Ali AL.Moneef

N C ˆ
j
N C
 3.00iˆ 13.5ˆj mN
23.7 Motion of a Charged Particle in a Uniform Electric Field
If we know the electric field, we can calculate
the force on any charge:
The direction of the force depends on the
sign of the charge – in the direction of the
field for a positive charge, opposite to it for
a negative one.
59
Norah Ali AL.Moneef
Electric Field lines
Uniform Field
From infinity
To infinity
Equispaced parallel straight lines
60
Norah Ali AL.Moneef
23.7 Motion of a Charged Particle in a Uniform Electric
Field
When a particle of charge q and mass m is placed in an electric field E, the
electric force exerted on the charge is q E.
If this is the only force exerted on the particle, it must be the net force and
causes the particle to accelerate according to Newton’s second law


 If the electric field E is uniform (magnitude
F  qE
and direction), the electric force F on the
particle is constant.



F  qE  ma

 qE
a
m
61
Norah Ali AL.Moneef

If the particle has a positive charge, its
acceleration a and electric force F are in the
direction of the electric field E.

If the particle has a negative charge, its
acceleration a and electric force F are in the
direction opposite the electric field E.
An Accelerating 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,. Describe its motion.
The acceleration is constant
a = qE/m.
The motion is simple linear motion along the x axis.
We apply the equations of kinematics in one dimension
A positive point charge q in a uniform electric field E
undergoes constant acceleration in the
direction of the field.
Choosing the initial position of the charge as xi=0 and assigning vi = 0 because the particle
starts from rest, the position of the particle as a function of time is
The speed of the particle is
62
Norah Ali AL.Moneef
the kinetic energy of the charge after it has moved a distance ∆x = xf - xi :
from the work–kinetic energy
theorem because the work done by the electric force is
Fe ∆ x = qE ∆ x
and
W = ∆ K.
63
Norah Ali AL.Moneef
The electric field in the region between two oppositely charged flat
metallic plates
is approximately uniform .
Suppose an electron of charge "e is projected horizontally into this field
from the origin with an initial velocity v i at time t = 0.
Because the electric field E in the Figure is in the positive y direction,
the acceleration of the electron is in the negative y direction.That is,
i
64
Norah Ali AL.Moneef
An electron is projected horizontally into a uniform electric field produced
by two charged plates. The electron undergoes a downward acceleration
(opposite E), and its motion is parabolic while it is between the plates.
Because the electric field E in the Figure is
in the positive y direction, the acceleration of
the electron is in the negative y direction.
That is,
the acceleration is constant, we can apply the equations
of kinematics in two Dimensions with vxi = vi and vyi = 0.
65
Norah Ali AL.Moneef
After the electron has been in the electric field for a time interval, the
components of its velocity at time t are
Its position coordinates at time t are
t = xf / vi
After the electron leaves the field, the electric force vanishes
and the electron continues to move in a straight line in the
direction of v in with a speed v > vi
66
Norah Ali AL.Moneef
Electron Beams; Cathode Ray Tubes
• Televisions, Oscilloscopes, Monitors, etc. use an
electron beam steered by electric fields to light up
the (phosphorescent) screen at specified points
screen
+ + + + + + +
cathode emitter
E-field
electron beam
- - - - - - -
metal plates
67
Norah Ali AL.Moneef
Which electric field is responsible for
the trajectory of the proton?
68
Norah Ali AL.Moneef
An electron (mass m = 9.11×10-31kg) is accelerated in the
uniform field E (E = 1.33×104 N/C) between two parallel
charged plates. The separation of the plates is 1.25 cm. The
electron is accelerated from rest near the negative plate and
passes through a tiny hole in the positive plate, as seen in the
figure. With what speed does it leave the hole?
F = qE = ma
Vf 2 = vi2 + 2a (∆)
a = qE/m
Vf
 Vf
69
2=
2ad = 2(qE/m) ∆
2a ∆
= 2(qE/m) ∆
=2(1.9 x 10 -19C) (1.33×104 N/C) (1.25m) l 9.11×10-31kg
Vf = 8.3 x 10 6 m/s
2=
Norah Ali AL.Moneef
Example : Electron moving perpendicular to E .
Suppose an electron traveling with speed v0 = 1.0 x 107 m/s
enters a uniform electric field E =2x104N/C , which is at right
angles to v0 as shown. Describe its motion by giving the
equation of its path while in the electric field. Ignore gravity.
Solution: The acceleration is in the
vertical direction (perpendicular to
the motion in -y -direction)
a = –eE/m
=1.6x1019x2x104/9.11x10-31.
y = ½ ay2
x = v0t;
y = -(eE/2mv02)x2
70
Norah Ali AL.Moneef
Example: An electron is projected perpendicularly to a
downward electric field of E= 2000 N/C with a horizontal
velocity v=106 m/s. How much is the electron deflected
after traveling 1 cm.
V
•e
d
E
E
Since velocity in x direction does not change, t=d/v =10-2/106 = 10-6 sec, so the
distance the electron falls upward is
y =1/2at2 = 0.5*eE/m*t2 = 0.5*1.6*10-19*2*103/10 - 30*(10-8)2 = 0.016m
71
Norah Ali AL.Moneef
Example :
; The electrons in a particle beam each have a kinetic
energy of 1.60 x 10-17 J. What are the magnitude and
direction of the electric field that stops these electrons in a
distance of 10.0 cm?
72
Norah Ali AL.Moneef
Example :
An electron and a proton are each placed at rest in an electric field of 520 N/C.
Calculate the speed of each particle 48.0 ns after being released.
73
Norah Ali AL.Moneef
Example :
An object having a net charge of 24.0 C is placed in a
uniform electric field of 610 N/C that is directed vertically.
What is the mass of this object if it “floats” in the field?
74
Norah Ali AL.Moneef
Example :
Three point charges are located at the corners of an equilateral
triangle. Calculate the net electric force on the 7.00 μC charge.
75
Norah Ali AL.Moneef
Electric charges have the following important properties:
• Unlike charges attract one another, and like charges repel one another.
• Charge is conserved.
• Charge is quantized—that is, it exists in discrete packets that are some integral
multiple of the electronic charge.
Conductors are materials in which charges move freely. Insulators are materials
in which charges do not move freely.
76
Norah Ali AL.Moneef
where ˆr is a unit vector directed from the charge to the point in question.
The electric field is directed radially outward from a positive charge and
radially inward toward a negative charge.
The electric field due to a group of point charges can be obtained by using
the superposition principle. That is, the total electric field at some point
equals the vector sum of the electric fields of all the charges:
77
Norah Ali AL.Moneef
Electric field lines describe an electric field in any region of space. The number
of lines per unit area through a surface perpendicular to the lines is proportional
to the magnitude of E in that region.
A charged particle of mass m and charge q moving in an electric field E has an
acceleration
78
Norah Ali AL.Moneef
Unit Modifiers for Reference
Smaller
 Centi
=
 Milli ( m ) =
 Micro (m) =
 Nano ( n ) =
 Pico ( p ) =
Examples:
5mC = .005C
10k = 10000 
79
Norah Ali AL.Moneef
10-2
10-3
10-6
10-9
10-12
Larger
o Kilo (K )=
o Mega =
o Giga
=
o Tera
=
103
106
109
1015