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CHAPTER 23
ELECTRIC FIELD
ELECTRIC CHARGE
There are two kinds of electric charges:
positive(protons) and negative(electrons).
The negative charge of an electron and the positive charge of a proton have the
same magnitude given as
e 1.6 10 19 C
Thus, an atom is electrically neutral when the number of electrons equals the
number of protons.
It is found, experimentally, that like charges repel each other and unlike charges
attract each other.
Charges have the following properties:


Electric charge is conserved.
Electric charge is quantized, i.e., q=ne
With n=0,1,2,
and e is the charge of the electron (proton).
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CONDUCTORS AND INSULATORS
A conductor is the substance that permits electric charge to move freely, e.g,
Silver, copper, and aluminum
An insulator is the substance that does not permit electric charges to move. e.g.,
glass, plastic, and rubber
There are substances that fall in an intermediate position between conductors and
insulators. These substances are called semiconductors such as silicon and
germanium.
With the properties of conductors and insulators in mind, one can charge a
conductor by a technique called charging by induction, (see the Figure 1.1).
he center of the circle.
(a)
(b)
(c)
Figure 1.1
Charging by induction. (a) The charge on a
conducting sphere is redistributed when a charged rubber rod is
brought near the sphere. (b) The access negative charges leave the
sphere as it is grounded. (c) The rubber and the ground connection
are removed, and the sphere has a uniform positive charge.
3
COULOMB’S LAW
It is found that:
(i)
The force is directly proportional to the magnitude of the charges of each
particle.
(ii)
The force is inversely proportional to the square of the distance between
the two particles.
(iii)
The direction of the force is along the line joining the two particles.
(iv) The force is attractive if the charges are of opposite signs and repulsive if the
charges have the same signs.
From these observations the magnitude of the force acting on a point charge q1
due to another point charge q2 can be expressed as
F21 k
q1 q2
2
r21
where r21 is the distance between the two charges, and k is the proportionality
constant known as Coulomb’s constant.
The direction of the force is determined according to the sign of the two charges.
It is repulsion if the two charges are alike and attractive if they are unlike.
The proportionality constant, k, has a numerical value depends on the choice
of units. The SI unit of charge is Coulomb (C), which is considered as the fourth
primary unit beside meter, kilogram, and second. In this system k has the value
k  9.0  109 N.m2/C2
Coulomb’s constant is usually written as
k
1
4o
with
 o 8.8542 1012 C2/N.m2,
is called the permittivity of free space.
4
The cgs unit of charge is statcoulomb or electrostatic unit (esu). In this system, the
constant k is defined to be unity with no units.
In vector form Coulomb's law can be written as

qq
F21k 12 2 r̂21
r21
where r̂21 is a unit vector directed from q2 to q1. It should be noted that the electric
force is of the same form as the gravitational force. As the gravitational force is
conservative, we conclude that the electrostatic force is also conservative.
q1
r̂21
q2
If there are more than two charges, the total force acting on one charge due to the
others is the vector sum of the forces due to the individual charges. For example, if
there are three charges, then the net force on particle 1 due to the other two
particles is given as
  
F1F21F31
Always remember that the electric force obeys Newton’s third law, that is


F21 F12
Remark: Coulomb’s law applies only for point charges.
5
Example 23.1
Find the magnitude of the electrical force and the gravitational
force between the electron and the proton in the H-atom. The average distance
between the two particles is 0.53 Ao.
Solution
q1 q2
Fe k
 (9 10 )
9
r122
(1.6 1019 )2
(0.53 10
10 2
)
 8.2 108 N
To find the gravitational force we have
Fg G
m1m2

r122
 6.67 101011
)(1.67 10
(9.11(10
0.53 10 )
31
27
)
10 2
 3.6 10 47 N
F
Note that e  21039
Fg
Example 23.2
Three
point
charges of 5 C, -4 C, and 8 C
are located at the corners of an
equilateral triangle of side 0.1 m
as shown in the Figure. Find the
resultant electric force exerted on
q3.
Solution Here the charge q3 are
affected by two forces, F13 due to
q1 and F23 due to q2. Now
F13 k
q1 q3
r132

F13
q2= - 2.0 C
0.1 m
q1=5.0 C

0.1 m F23
q3=5.0 C
2 0.1 m
6
(9 10 )
9
(5.0 106 )(5.0 106 )
( 2 0.1)2
and F23 k
(9 109 )
 11 N
q2 q3
r132
(2.0 106 )(5.0 106 )
(2)2
 9.0 N In vector notation we have



F13 11cos iˆ 11sin  ˆj  7.9 iˆ  7.9 ˆj N
And



F23   9.0 iˆ N
Now the resultant force acting on q3 is the sum of the two forces, i.e.,


  
F3F13F23  7.9  9.0iˆ  7.9 ˆj   1.1iˆ  7.9 ˆj N
Example 23.2 Three point charges, q1 = 15C at the origin, q2 = 6.0 C at x=2.0
m, and q3 at x. If the resultant electric force exerted on q3 is zero, find x.
2.0 m
2.0 - x
x
q1=15C
F13
q3
F23
q2= 6.0 C
Solution The two forces acting on q3 are both attractive and so opposite. For the
net force to be zero their magnitudes must, therefore, be equal. i.e.,

F13 F23
k
q1 q3
r132
k
q2 q3
2
r23

7
k
15 10
6
2  x 
q3
2
k
6.0  10
x
6
q3
2

5 x 2  8  4 x  2 x 2  3x 2  8 x  8  0 
x=0.78 m
and x=-3.4 m
The second solution is not acceptable since at a point to the left of q2 the two
forces will be in the same direction, and so their sum can't be zero at all even if
their magnitudes are equal.
Example 23.4
Two identical small charged spheres, each of mass 30 g and
charge q are hanged in equilibrium as shown in the Figure. The length of each
string is 0.15 m, and the angle
 = 5.0o . Find q.
Tcos

0.15 m
0.15 m
T
Fe
Tsin
2a
mg
(a)
(b)
Solution It is enough to study the equilibrium state of one of the spheres. The
forces acting on such a sphere is shown in the free-body diagram shown in Figure
(b). The equilibrium conditions imply that
 Fx 0
and
 Fy 0
From the first condition we write
Fe T sin 
(1)
Here Fe is the electric force acting on the sphere by the other. The second
condition gives
8
mg T cos 
(2)
Dividing Equation (1) by Equation (2), we obtain
tan  
Fe
mg
(3)
Substituting the values of m, g, and  in Equation (3) we get
Fe 30 102 9.8 tan 5.0  2.6 102 N
The magnitude of the electric force Fe is obtained from Coulomb’s law as
Fe k
q
2
r2
Solving for q and knowing that r  2a  20.15sin   0.013 m we get
q
Fe r 2
2.6 102 (0.013) 2

 4.4 108 C
9
k
9 10
9
THE ELECTRIC FIELD
Any charge has its own electric field within a region surrounding the
charge.
The electric field E at a point in space is a vector quantity defined as the electric
force acting on a positive small test charge qo placed at that point divided by the
magnitude of the test charge, i.e.,

 F
E
qo
The test charge should be small enough to ensure that its presence does not affect
the charge distribution that produces E.
Since qo is always positive, the direction of E is the direction of F. The SI unit of
E is Newton per Coulomb (N/C).
If a charge q is placed at a point with an electric field E, this charge experiences
an electric force given by
 
F qE
ELECTRIC FIELD OF POINT
CHARGES:
Let us calculate the electric field at a point p
arising from an isolated point charge q, a
distance r from it. To do so we assume the
existence of a test charge qo at point p, (see the
figure). From Coulomb’s law the force upon the
test charge is
 qq
F k 2o rˆ
r
where the unit vector directed from q to the point p.
Using Eq. 1.5 
qo
EP
P
r
q
10

q
E k 2 rˆ
r
It is clear from the definition of r̂ that
if q is positive the direction of E is outward from the charge,
if the charge is negative E it is directed toward q.
The electric field at a point due to a group of point charges is the vector sum of the
electric fields at that point due to each charge individually, i.e., if we have n
charges the net electric field E is
  

EE1E2     En
where E1, E2, and En are the electric fields due to the charges q1, q2, and qn,
respectively.
Example 23.5
Two point charges of 7.0 C, -5.0 C, are located as shown in
the figure. Find the resultant electric field at point P.
E1
p

E2
0.4 m
0.5 m

q1=7.0 C
0.3 m
q2= - 5.0 C
11
Solution a) Let us first calculate the magnitudes of the electric fields for each
charge individually, E1 due to the 7.0-C charge and E2 due to the -5.0-C.
E1k
q1
r12
and
E2 k
q2
r22

 9 109

 9 10
9
7.(00.410)
6
 3.9 105
2

5.0 106
(0.5)
2
 1.8  106 N/C
In vector notation we have

E1  3.9 105 ˆj N/C
And

E2 1.8105 cos iˆ 1.8 105 sin  ˆj
1.11059 iˆ 1.4 105 ˆj
Now the resultant electric field at point P is


  
E p E1E2 1.1105 iˆ  3.9 105  1.4 105 ˆj


 1.1105 iˆ  2.5 105 ˆj N/C
Example 23.6 A positive point charge q and a negative point charge –q are
separated by a distance 2a (electric dipole). Find E at point P a distance y>>a
from the origin.
E+


Ey
r
q


a
a
-q
12
 

Solution: The resultant electric field at P is E E   E  . Since P is equidistant from
the two charges the magnitude of E+ and E- are equal and are given by
E E  k
q
r2
As it is clear from the directions of the two fields, the y-components of E+ and Ecancel each other. The net electric field at P is, therefore

E  E cos  E cos  i
From the figure we see that cos  d 2r , so we find
E  2k
Now

r3  a2  y2

3
q
r
2
cos   k
q2a 
r3
i
2
If ya we can neglect the first term in the bracket so that r3 is approximated to
r 3  y3 . 
Ek
q2a 
y
3
i=k
p
y3

where p  q2ai is called the electric dipole moment. Its direction is taken to be
from the negative charge to the positive charge.
ELECTRIC FIELD OF A ONTINUOUS CHARGE DISTRIBUTION
To evaluate the electric field of such a configuration, the following
procedure are used:
1- We divide the distribution into small elements such that the electric field for
each element can be calculated easily.
13
2- Each element is assumed a charge dq
3- Next, we calculate the electric field due to one of these elements at the point in
question
4- Finally, we integrate over the charge distribution to evaluate the total electric
field due to the whole charge distribution.
Remark: The integration should be performed for each component of the electric
field, why?
Charge Density When dealing with continuous charge distribution it is
convenient to use the concept of charge density. If the charge is distributed along a
line we define the linear charge density  as

dq
dl
If the charge is distributed over a surface we define the surface charge density  as

dq
dA
If the charge is distributed within a volume we define the volume charge density 
as

dq
dV
where dq is the charge of a small line, surface, or volume element.
The SI units of the charge densities, , , and  are C/m, C/m2, and C/m3,
respectively.
If the charge is uniformly distributed along a line, over a surface, or within a
volume, the charge densities are constants. In this case the total charge Q can be
obtained by integrating both sides of the above Equations
Ql ,
QA ,
Or
Q  V
14
where l, A, and V are the length of the line, the area of the surface, and the
magnitude of the volume, respectively.
Example 23.7
A rod of length L lying along the x-axis is uniformly charged
with a total charge Q. Calculate the electric field at a point p along the axis of the
rod, a distance D from one end, as shown in the Figure.
L
D
dE
x
dx
dq
Solution We divide the rod into small elements each of length dx and charge dq.
Each element can be considered as a point charge, so the electric filed dE due to
one of these elements a distance x from p is
dEk
dq
x2
Note that each element produces a field in the negative x-axis as shown in Figure
1.-, and so we can integrate simply to get
Ek 
dq
x2
Expressing dq in term of the charge density  as dq = dx we get
L D
E k 
D
L D
 1

k

 x 
x2
D
dx
k
L
D L  D 
k
Q
DL  D 
where we have substituted for L by Q. From this result we conclude that if p is
far away from the rod (DL), then we recover the result of the point charge as
expected, that is.
15
E k
L
D2
Example 23.8
A ring of radius R has a uniform charge distribution of
magnitude Q. Calculate the electric field along the axis of the ring at a point p
lying a distance D from the center of the ring.
dq
r
R

D
x
dE
Solution: Again we divide the ring into small elements each charge dq.
The e.f due to one element is
dEk
dq
r
2
k
R
dq
2
 D2

Because of the symmetry about the x-axis we should note that there exist a similar
element, which is also located a distance r from p. This element produces an equal
field dE but differing in direction. As shown in the figure, the vertical components
of these two fields cancel one another and this will be hold for each such pair of
elements. Hence, the total field is the sum of the x-components of dE which is
given as
dE x  dE cos   k
R
Ddq
2
 D2

3
2
Integrating this expression we get
16
Ex k
R
D
2
 D2

3
2
 dq
Ex k
R
DQ
2
 D2

3
2
Again for the limiting case DR we recover the result of the point charge. At the
center of the ring D=0 and E must be zero, comment.
Example 23.9
A disk of radius R has a uniform
charge density . Calculate the electric field along
the axis of the disk at a point p lying a distance D
from the center of the disk.
Solution: In this problem we divide the disk into
thin rings each of thickness dr. From the result of
the previous example, the electric filed due to one
element with radius r is
dE  k
r
Ddq
2
 D2

3
R
r
p
D
dr
2
Integrating the last result and noting that
dqdA   2 rdr 
we get
R
E kD 
0
r
2rdr
2
 D2

3

2
 2
D  r2
 kD 

 12


 12 
R

D
  2k 1 


D2  R2

0




The electric field for a large charged plane can be obtained from the last result by
letting R (or x0). This gives
E 2k 

2 o
17
THE ELECTRIC FIELD LINES
The electric filed in a region is represented by imaginary lines known as
electric field lines (lines of force) introduced by Michael Faraday (1791-1867).
These lines have the following properties:
1- The direction of the lines at any point is the direction of the electric field at that
point.
2- The lines must begin on positive charges and terminate at negative charges.
3- The number of lines per unit area, perpendicular to E, is proportional to the
magnitude of E in that region.
4- No two field lines can cross.
Motion of Charged Particle in a Uniform E.F.
The e. force is a force that obey Newton's laws, i.e., this force will creates an
acceleration on the particle 
If a particle of mass m and charge q exists in a uniform e.f E the e. force acting on
the charge is
18



Fe  qE  ma




 qE
a
m
Since E is uniform  a is constant and the formulas govern the motion with
constant acceleration are still hold.
Example 23.10 A positive point charge q of mass m is

released from rest in a uniform e.f E  E iˆ . Describe its
motion.
E
v
Solution The acceleration of the charge is

 qE qE ˆ
a

i
m
m
Since vo  0 and the acceleration is along the x-axis  the motion is linear.
After a time t the particle is travelled a distance given by
x  vo t  12 at 2 
qE 2
t
2m
v  vo  at 
The speed of the charge at this time is
Note that
qE
 qE  qE 2 
t
v 2  vo2  2ax  2
t  v 

m
 m  2m 
Applying the conservation of energy principle 
Wnet  K  qEx  12 mv2
v
2qEx

m
 qE 2 
2qE 
t 
 2m   qE t
m
m

qE
t
m
19
Example 23.11 An
electron
is
projected, horizontally with initial speed
of 3.00  106 m/s into a region of uniform
electric field between two plates of 200
N/C and directed vertically upward. The
horizontal length of the plates is 0.1 m.
a) Find the acceleration of the electron
while it is in the field.
b) How long is it in the field?
c) What is the vertical displacement of the
electron inside the field?
____________________
E
y
++++++++++++++++++++++
l
Solution: a) The acceleration of the electron is

 F qE
a 
m m
 1.6  1019 200 j

 3.51 1013 j m/s2
 31
9.11 10


This means that ax = 0.
b) For the time we have
t
x
0.1

 3.33  10 8 s.
6
v x 3.00  10
b) The vertical displacement can be calculated from the equation
y  v0 y t  12 a y t 2


 12  3.511013 3.33  108

2
 1.95 cm.