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Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
CHAPTER I
THE ELECTRIC FIELD AND
POTENTIAL DIFFERENCE
1
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
After completing this chapter, the students should know:
 The different types of electric charges.
 The concept of Coulomb’s law.
 The relation between electric force and electric field.
 The units of electric charge, electric force and electric field.
 The concept of Gauss’s law.
 The relation between potential energy and potential difference.
 The motion of charged particle.
2
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
CHAPTER I
THE ELECTRIC FIELD AND POTENTIAL DIFFERENCE
The Electrical Force
Coulomb’s Law:
Coulomb measured the magnitudes of the electric forces between charged
objects using the torsion balance, and he showed that:
1- The electric force is inversely proportional to the square of the separation r
between the particles and directed along the line joining them, i.e.

FC 
1
r2
(1.1)
2- The electric force is directly proportional to the product of the charges q1
and q2 on the particles.

FC  q1 q2
(1.2)
Therefore, from these observations, we can express Coulomb’s law as follow:

q1 x q2
r2

q xq
FC  kC 1 2 2
r
FC 
(1.3)
Where kc is a constant called the Coulomb’s constant.
The value of the
Coulomb’s constant depends on the choice of units. The Coulomb’s constant kc in SI
units has the value of
kC 
1
4o
 9 x 109 N .m 2 / C 2
Where 0 is known as the permittivity of free space and has the value of
8.85x10-12 C2/N m2. The smallest unit of charge known in nature is the charge on the
electron or proton, which has an absolute value of
3
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
q  e 1.6 x1019 C
Note that:
The electric force is attractive if the charges are of opposite sign and repulsive
if the charges have the same sign.
              (repulsion force)
              (repulsion force)
             (attraction force)
Example (1.1):
The electron and proton of a hydrogen atom are separated by a distance of
approximately 5.3x10-11 m.
Find the magnitudes of the electric force and the
gravitational force between the two particles, knowing that, me= 9.11x10-31 kg, mp=
1.67x10-27, kc= 9x109 N.m2/m2 and G= 6.7x10-11 N.m2/kg2.
Solution:
-e
Firstly, the Coulomb force is

FC  kC
P
q1 x q2
r2
 9 x109 x
1.6 x10 x 1.6 x10 
5.3x10 
19
19
11 2
 8.2 x10 8 N
Secondly, the gravitational force is

FG  Gm
m1 x m2
r2
 6.7 x10 11 x
9.11x10  x 1.67 x10 
5.3x10 
31
27
11 2
 3.6 x10 47 N
Example (1.2):
Two charges, one of is +8x10−6 C and the other is −5x10−6 C, attract each other with a
force of −40 N. How far apart are they?
Solution:
4
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
According to Coulomb’s law, we have

FC  kC
q1 x q2
r2
q xq
r  kC 1 2
F
r
q1 x q2
F
r 2  kC
r
9 x10 3 m2 
9 x109 x 8 x106 x(5 x106 )
(40)
9 x10 3 x106 mm2  30 10 mm
Example (1.3):
Consider three point charges located at the corners of a right triangle as shown in
figure, where q1=q3= 5 C and q2 = -2 C and a= 0.1 m.
F13y
Find the resultant force exerted on q3.
q2
Solution:
θ
F23
q3
0.1 m
Note that, the force F23 exerted
0.1 m
by q2 on q3 is attractive
F13x
F13
Therefore, the magnitude of F23 is

F23  kC
q1
q2 x q3
a2
 9 x10 x
9
2 x10  x 5 x10 
6
6
0.12
9 N
Also, the force F13 exerted by q1 on q3 is repulsive
Therefore, the magnitude of F13 is

F13  kC
q1 x q3
( 2 a) 2
 9 x109 x
5x10  x 5 x10 
6

6

2 x 0.1
2
9 N
But F13 must be analyzed in two components in x-axis (horizontal component) and yaxis (vertical component), as follow:
5
Electromagnetism I
Chapter I

F13 x  kC
q1 x q3
( 2 a) 2
 9 x10 x
9
Prof. Dr T. Fahmy
cos  ,
  450
5 x10  x 5 x10 
6

6

2 x 0 .1
2
x
1
2
 7.9 N

F13 y  kC
q1 x q3
( 2 a) 2
 9 x109 x
sin  ,
  450
5x10  x 5x10 
6

6

2 x 0.1
2
x
1
2
 7.9 N
Therefore, the total force in x-axis is
Fx= 7.9 – 9 = -1.1 N
and the total force in y-axis is
Fy= 7.9 N
Then, the total force acting on the charge q3 is



F total  F x  Fy

F total  (1.1) x  (7.9) y
The value of this force is
F  ( Fx ) 2  ( Fy ) 2
 (1.1) 2  (7.9) 2
 7.97 N
Example (1.4):
Consider three point charges located at the corners of a right triangle as shown in
figure, where q1=2 C, q2= 5 C and q3 = 4 C. Find the resultant force exerted on
q3.
6
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
F23y
Solution:
Note that, the force F13 exerted by q1 on q3 is repulsive
F13
q1
Therefore, the magnitude of F13 is

F13  kC
q1 x q3
2
r13
 9 x10 x
9
0.03 m
F23x
0.05 m
0.04 m
F23
2 x10  x 4 x10 
6
q3
6
0.032
q2
 80 N
Also, the force F23 exerted by q2 on q3 is repulsive
Therefore, the magnitude of F23 is

F23  kC
q2 x q3
2
r23
 9 x109 x
5 x10  x 4 x10 
6
6
0.052
 72 N
But F23 must be analyzed in two components in x-axis (horizontal component) and yaxis (vertical component), as follow:

F23x  F 23 cos  ,
 72 x
cos  
3
 43.2 N
5

F23 y  F23 x sin  ,
 72 x
3
5
sin  
4
 57.6 N
5
Therefore, the total force in x-axis is
Fx= 43.2 + 80 = 123.2 N
and the total force in y-axis is
7
4
5
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
Fy= 57.6 N
Then, the total force acting on the charge q3 is



F total  F x  Fy

F total  (123.2) x  (57.6) y
The value of this force is
F  ( Fx ) 2  ( Fy ) 2
 (123.2) 2  (57.6) 2
 136 N
The Electric Field
The electric field E 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 magnitude of the test
E
charge’’
Therefore,
P
E
q


E F
q
(1.4)
0
q
To define the direction of an electric field, consider a point charge q located at a
distance r from a test charge q0 located at a point p.
According to Coulomb’s law, the electric force exerted
by q on q0 is

q x q0
r2
Therefore
F  kC

E

r

F
q0
 kC
q
r2
(1.5)

r
Also, note that:
8
(1.6)
P
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
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.
q3
Then:
E3




E  E1  E2  E3  E4          En

q2


r3

E
n 1
r2
n

qn
n 1
r
 kC 
E2
q4

r
2
(1.7)
q1
r4
r1
n
E4
rn
qn En
Example (1.5):
Two point charges have the same charge q with opposite sign located at a distance r


between them. Calculate the electric force F and electric field E on a third charge q3
if this charge is located at different positions such as a, b, c and d. Knowing that,
q1 and q2 = 5 C, q3 = 4 C and r= 0.12 m.
d
c
a
Solution:
a
At the point (a):


3/4 r

F a  F13  F23
 kc x
q1 q3
2
 kc x
q2 q3
2
3 
1 
 r
 r
4 
4 
6
6
5 x10 x 4 x10 6
x 4 x10 6
9 5 x10
 9 x109

9
x
10
2
2
3

1

 x0.12 
 x 0.12 
4

4

 22.22  200  222.22 N


F
222.22
Ea  a 
 5.55 x10 7 N / C
6
q3
4 x10
9
¼r
b
E1
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
r
At the point (b):
r
b
There are two forces against to each other, therefore, we will obtain the total force as a
difference between them as follow



F b  F23  F13
 kc x
q2 q3
qq
 kc x 1 32
2
r 
2r 
 9 x109
6
5 x10 6 x 4 x10 6
x 4 x10 6
9 5 x10

9
x
10
0.122
2 x 0.122
 12.5  6.25  6.25 N


F
6.25
Eb  b 
 1.56 x106 N / C
6
q3 4 x10
C
½r
r
At the point (c):
There are two forces against to each other, therefore, we will obtain the total force as a
difference between them as follow



F c  F13  F23
 kc x
q1 q3
2
 kc x
q2 q3
2
1 
3 
 r
 r
2 
2 
6
6
x 4 x10 6
x 4 x10 6
9 5 x10
9 5 x10
 9 x10
 9 x10
2
2
1

3

 x 0.12 
 x 0.12 
2

2

 50  5.55  44.45 N


F
6.25
Ec  c 
 1.11 x 107 N / C
q3 4 x10 6
10
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
At the point (d)
d



F d  F13  F23
 kc x
q1 q3
q q
 kc x 2 3 2
2
r 
r 2


Then

F13  kc x
q1 q3
5 x10 6 x 4 x10 6
 9 x109 x
 12.5 N
2
r 
 0.122

F23  kc x
q2 q3
r 2 
2
 9 x109 x
5 x10 6 x 4 x10 6
 0.12 2 
2
 6.27 N
Therefore


1
 4.43 N
2


1
F23 y  F23 x sin   6.27 x
 4.43 N
2
Then, the total force is
F23 x  F23 x cos   6.27 x



Fd  Fx  Fy  4.43 x
F 

F
2
x
Fy 
2
 12.5  4.43 y
4.432
 8.07   9.2
2

F
9.2
Ec  c 
 2.3 x 106 N / C
q3
4 x10 6
Example (1.6):
The electric field (E) in a certain neon sign is 5000 V/m. What force does this field (F)
exert on a neon ion of mass 3.3x10−26 kg and charge +e?
Solution:
The force (F) on the neon ion is

F q E

F 1.6 x1019 x5000  8 x1016 N
Example (1.7):
In the following figure calculate
+Q1=2C
+Q2=4C
r12=2cm
11
+Q3=1C
r32=3cm
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy

1- The total force ( F ) acting on Q2, due to the effect Q1 and Q3

2- The electric field ( E ) at the position of Q2.
Solution:
1- The total force is



F = F 12 + F32

F 12 = K c x
Q1 x Q2
r12 2
9
= 9 x10 x
2 x106 x 4 x106
2x10 
2 2
8 x1012
= 9 x10 x
180 N (to the right)
4 x10 4
9

F 13 = K c x
Q3 x Q2
r32 2
= 9 x109 x
= 9 x109 x
5 x106 x 1x106
3x10 
2 2
5 x1012
 50 N (to the left)
9 x10 4
Therefore, the total force is



F = F 12 + F32 = 180 - 50 =130 N (to the right)

2- The electric field E is


F
130
= 3.25x107 N/m
E = total 
Q2
4 x10 6
Example (1.8):
In the following figure calculate
-Q1=2C
+Q2=4C
r12=2cm
+Q3=5C
r32=3cm

3- The total force ( F ) acting on Q2, due to the effect Q1 and Q3

4- The electric field ( E ) at the position of Q2.
Solution:
3- The total force is
12
Electromagnetism I

Chapter I
Prof. Dr T. Fahmy


F = F 12 + F32

F 12 = K c x
Q1 x Q2
r12 2
= 9 x109 x

F 13 = K c x
(to the left)
= 9 x10 x
2 x106 x 4 x106
2x10 
2 2
8 x1012
180 N
4 x10 4
Q3 x Q2
r32 2
= 9 x109 x
9
= 9 x109 x
5x106 x 4 x106
3x10 
2 2
20 x1012
 200 N
9 x10 4
Therefore, the total force is



F = F 12 + F32 = 180 + 200 =380 N

4- The electric field E is


F
200
= 5x107 N/m
E = total 
Q2
4 x10 6
Example (1.9):
Find the force on the charge q2 in the diagram below due to the charges q1 and q3.
q1 = 1 µC
q2 = -2 µC
0.1 m
q3 = 3 µC
0.15 m
Solution:
F12  k e
| q1 || q 2 |
F32  k e
| q3 || q 2 |
r12 2
r32 2
 ( 8.99 x10 9 N  m 2 / C 2 )
( 1x10 6 C )( 2 x10 6 C )
 ( 8.99 x10 9 N  m 2 / C 2 )
( 0.1m )2
 1.8 N ( to the left )
( 3 x10  6 C )( 2 x10  6 C )
( 0.15m )2
F2   F12  F32  1.8 N  2.4 N  0.6 N ( to the right )
13
 2.4 N ( to the right )
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
Example (1.11):
q3 = 3 µC
Find the force on q2 in the diagram to the
right.
0.15 m
Fx   F12  1.8 N
q1 = 1 µC
F y  F32  2.4 N
q2 = -2 µC
F2  Fx2  F y2  ( 1.8 ) 2  ( 2.4 ) 2  3.0 N
tan 
Fy
Fx

0.1 m
2.4
 1.33
 1.8
   126.9 o ( ccw from pos . x  axis )
q3 = 3 µC
F2
q1 = 1 µC
F32

F12
q2 = -2 µC
Example (1.12):
q2 = -1.5 µC
Find the electric field at P due to charges q1 and q2.
0.3 m
q1 = 2 µC
14
0.4 m
P
Electromagnetism I
Chapter I
E1  k e
6
| q1 |
9 ( 2 x10 )

(
8
.
99
x
10
)
 1.13 x105 N / C
2
2
r1
(0.4)
E2  k e
6
| q2 |
9 (1.5 x10 )

(
8
.
99
x
10
)
 5.4 x10 4 N / C
r22
(0.5) 2
Prof. Dr T. Fahmy
E x  E1  E2 cos   1.13 x105  5.4 x10 4 cos(36.9)
 6.9 x10 4 N / C
q2 = -1.5 µC
E y  E2 sin   5.4 x10 4 sin( 36.9)  3.24 x10 4 N / C
E  E x2  E y2  (6.9 x10 4 ) 2  (3.24 x10 4 ) 2
0.3 m
0.5 m
E2
 7.6 x10 4 N / C
E
 = 36.9
o
  tan 1 ( E y / E x )  tan 1 (3.24 / 6.9)  25o
q1 = 2 µC
15
0.4 m

P
E1
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
Gauss's Law
This law is relating the distribution of electric charge to the resulting electric
field. Gauss's law states that:
The electric flux through any closed surface is proportional to the enclosed electric
charge.
The law was formulated by C. F. Gauss in 1835, but was not published until
1867. It is one of four of Maxwell's equations which form the basis of classical
electrodynamics, the other three being Gauss's law for magnetism, Faraday's law of
induction, and Ampère's law with Maxwell's correction. Gauss's law can be used to
derive Coulomb's law, and vice versa.
S
E
As shown in the figure, if a surface with an
Area of A is placed in an electric field of intensity
E, therefore, the total normal electric flux can be expressed (ɸ) as follow:
  E. S
(1.8)
So, the vertical flux (dɸ) through an area of (dS) can be
written as follow:
dS
E
r
d  E. dS
(1.9)
q
where E is the electric field at any point of sphere surface and equals to
E
1
4  o
q
r2
16
Electromagnetism I
Chapter I
d 
1
q
r2
4  o
Prof. Dr T. Fahmy
dS
Then by using the integration, the total flux can be determined:

   d 
0
1
4  o
1

q
r2
4  o
q
r2
4r 2
 dS
0
4 r 2
q

(1.10)
o
But if the surface of the sphere is non-homogeneous, the flux will be
  E S cos 
   E S cos 
Or in other form as follow:
   E S cos 
(1.12)
Therefore, from equations (1.11) and (1.12), we can conclude that:
 E S cos 

1
0
q
(1.13)
Example (1.13):
Calculate the electric flux (φ) through a circle has a radius of 5 cm and a charge
of 4 C at its center.
Solution:
r = 5 cm
φ= E. A
r= 5x10-2 m
E  kc
Q=4x10-6 C
and
6
Q
9 4 x10

9
x
10
 72 x106 N / C
2 2
r2
5 x10


17
Q=4C
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
A  4  r 2  4 x3.14 x(5x102 )2  3.14 x102 m2
  72 x106 x3.14 x102  2.26 x108 N .m2 / C
The Potential Energy (U) and Potential Difference (V)
If a test charge q0 is placed in an electric field E, created by some other charged
objects, the electric force (F) acting on the test charge is q0 E. If the charge is moved
a distance d  , then the total work done can be written as follow:
dw  F cos  d
(1.14)
Where θ is the angle between the force (F) and the displacement (d  ).
Therefore if the charge is moved from the point A to the point B, the total work done
can be written as follow:
wAB 

B
A
F cos  d
(1.15)
Also, in terms of the potential energy (U), the equation (1.15) can be written as follow
B
U AB    F cos  d
A
Where
F  qo E
Then
B
U AB    q0 E cos  d
(1.16)
A
The negative sign mean that, the motion against the electric field direction.
Note that,

at θ= 00
 at θ= 900
→ WAB and UAB have a maximum value
→ WAB and UAB have a minimum value, i.e., (WAB and UAB = 0).
18
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
The Potential Difference (V)
The potential difference (V) is defined as the work done to transport a positive charge
against the electric field.
Also,
The potential difference (V) is defined as the ratio between the potential energy
(U) to a test charge (q0).
Therefore,
VAB 
U AB
q0
(1.17)
B
1
V AB  
q 0 E cos  d
q 0 A
B
V AB    E cos  d
A
Or
V AB    E cos  d
If θ= 00
V AB    E d
Or in a differentiation form, we can write
dV   E d
E 
dV
d
(1.18)
Note that,
The potential difference is a scalar quantity.
Example (1.14):
19
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
Calculate the potential difference at a point located
R
at a distance r from the centre of a charged sphere
with appositive charge?
r
Solution:
As known,
E
q
1
40 r 2
(1)
Also,
dV
dr
dV   E dr
E
(2)
From 1 and 2, we can write
dV  
q
1
dr
4o r 2
The value of V can be estimated using the integration process, as follow:
V
V   dV  
0
V 
q
40
r
1
r
2
dr
0
q
1
 K1
40 r
Where K1 is the integration constant. To calculate the value of this constant, the
boundary condition must be used as follow;
at r → α
→V=0 , therefore, K1= 0 ,
Then
V 
q
4o
V  k
1
r
q
r
(3)
20
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
Also, at the sphere surface, the potential difference is
V  k
q
R
(4)
c
Example (1.15):
In the following figure,
1- Calculate the potential at the points (a, b and c).
0.1 m
0.1 m
2- Calculate the potential energy of a positive
charge of 4x10-9 C at the points (a, b and c).
b
0.04 m
a
0.06 m
3- Calculate the potential difference Vab, Vba and Vcb.
4- Calculate the work to transfer a charge of 4x10-9 C
From a to b and from c to a.
Solution:
Va  k
1-
q1
q
 k 2,
r1
r2
q1  q 2  q
1
1
Va  kq     9 x10 9 x12 x10 6
r2 
 r1
Va   9 x10 5 V
1 
 1
 0.06  0.04 


1 
 1
Vb  9 x10 9 x12 x10 6 

 .25 x10 4 V

 0.04 0.14 
1 
 1
Vc  9 x10 9 x12 x10 6 
   0 V
 0.1 0.1
U a  qVa   9 x10 5 x 4 x10 9   36 x10 4 J
2-
U b  qVb  225 x10 4 x 4 x10 9  9 x10 3 J
U c  qVc  0 x 4 x10 9  0 J
21
0.04 m
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
-


Vab  Vb  V a  225 x10 4   9 x0 5  22.5 x10 5  9 x10 5  31.5 x10 5 V
3-




Vba  Va  Vb   9 x0  225 x10   9 x0 5  (22.5 x10 5 )   31.5 x10 5 V
5
4
Vcb  Vb  Vc  (225 x10 ) V
4
Wab  Wb  Wa  qVab
 4 x10 9 x31.5 x10 5  126 x10  4 J
4-
U ca  Wa  Wc  qVca


 4 x10 9 x  9 x10 5   36 x10  4 J
Example (1.16):
Consider two charges as shown below. Compute the potential at point P.
Solution:
Q1=- 6 C
As known:
Vp  k 
qi
ri
q q 
 2 X 10
Vp  k  1  2  = 9 X 109 
 4
 r1 r2 
V p   6.3 X 103 Volt
6

6 X 10 

5 
6
3m
m
Q2= 2 C
P
4m
m
Note that:
We knew that, objects have potential energy because of their positions. In this case
charge in an electric field has also potential energy because of its positions. Since there
is a force on the charge and it does work against to this force we can say that it must
have energy for doing work. In other words, we can say that Energy required
increasing the distance between two charges to infinity or vice verse. Electric potential
energy (U) is a scalar quantity and Joule is the unit of it. The following formula is
used to calculate the magnitude of U;
22
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
Q1 x Q2
r
U  Kc x
Be careful:
 In this formula if the charges have opposite sign then, (U) becomes negative, if
they are same type of charge then, (U) becomes positive.
 If (U) is positive then, electric potential energy is inversely proportional to the
distance (r).
 If U is negative then, electric potential energy is directly proportional to the
distance (r).
(a)
(b)
(c )
In Figure a and Figure b, charges repel each other, thus external forces does work
for decreasing the distance between them. On the contrary, in Figure c, charges
attract each other, distance between them is decreased by electric forces, and there is
no need for other external forces.
Q1= 10q
Example (1.17):
System given below is composed of the charges
10q, 8q and -5q.
3d
Find the total electric potential energy of the system
Solution:
U12  Kc

10q x 5q   Kc  50 q 2
Q1 x Q2
 Kc
r12
3d
3d


Q1 x Q3
10q x8q 
20 q 2
U13  Kc
 Kc
 Kc
r13
4d
d
U 23  Kc
Q2 x Q3
 Kc
r23
 5q x8q   K
5d


 8 q 
2
c
d
23
Q2= -5q
Q3= 8q
4d
5d
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
U total  U12  U13  U 23
 50 q   K 20 q   K  8 q 
2
 Kc
 Kc
3d
 14 q 2
3d

2
c

d
2
c
d
Joule
Note that:
Electric Potential
Electric potential is the electric potential energy per unit charge. It is known as
voltage in general, represented by V and has unit volt (Joule/C).
Electric Potential and Potential Energy due to Point Charge
The potential difference V can be calculated between the points A and B as follow:
rB
VB  V A    E dr
rB
(1.19)
rA
But , as known
q
1
4o r 2
E
rA
Therefore,
VB  V A  
q
4o
rB
2
dr
rA
q
1

4o r

1
r
rB
rA
q 1
1
  
4o  rB rA 
If rA  
VB 
q
4o rB
or ingeneral form
24
1
Electromagnetism I
Chapter I
V
Prof. Dr T. Fahmy
1
q
4o r
(1.20)
Example (1.18):
Two charges of 2mC and -6mC are located at positions (0,0) m and (0,3) m,
respectively as shown in figure 5.13. (i) Find the total electric potential due to these
charges at point (4,0) m. (ii) How much work is required to bring a 3mC charge from
 to the point P? (iii) What is the potential energy for the three charges?
q= -6 mC
(0,3)
Solution:
(i)
(i) Vp = V1 + V2
q= 2 mC
V
(4,0)
(0,0)
1  q1 q2 

40  r1 r2 
p
 2 x106 6 x106 
3
V  9 x109 

   6.3x10 volt
4
5


(ii) The work required is given by
W = q3 Vp = 3  10-6  (-6.3  103) = -18.9  10-3 J
The -ve sign means that work is done by the charge for the movement from  to P.
(iii) The potential energy is given by
U = U12 + U13 + U23
q q q q q q 
U k  1 2  1 3  2 3 
r13
r23 
 r12
25
Electromagnetism I

Chapter I

 

 
Prof. Dr T. Fahmy


 2 x106  6 x106
2 x106 3x106
 6 x106 3x106 
U  9 x109 



3
4
5


U   5.5 x102 J
Example (1.19):
A particle having a charge q=3x10-9C moves from point a to point b along a straight line,
a total distance d=0.5m. The electric field is uniform along this line, in the direction
from a to b, with magnitude E=200 N/C. Determine the force on q, the work done on it
by the electric field, and the potential difference Va-Vb.
Solution:
The force is in the same direction as the electric field since the charge is positive; the
magnitude of the force is given by
a
F =qE = 3x10-9 x 200 = 600x10-9 N
b
d= 0.5 m
The work done by this force is
W =Fd = 600x10-9 x 0.5 = 300x10-9 J
The potential difference is the work per unit charge, which is
Va-Vb = W/q = 100 V
Or
Va-Vb = Ed = 200x 0.5 = 100 V
Example (1.20):
A charge q is distributed throughout a non-conducting spherical volume of radius R.
(a) Show that the potential at a distance r from the center where r < R, is given by
26
Electromagnetism I
Chapter I
V
Prof. Dr T. Fahmy

q 3R 2  r 2
80 R 3

Solution:
To calculate the voltage inside the non-conducting spherical at a point A, we will
calculate the voltage between a point at infinity and the point A as follow:


VA V    E. d
Since, the field has two different values outside and inside the sphere as we know
Eout 
q
4o r
2
Ein 
qr
4o R 3


VA  V  VA  VB   VB  V 




VA  V    E . d    E
.
in
out d
Note that:
The angle between E and d  is 1800, i.e., cos 180= -1
Also d  =-dr
VA  V   
qr
q
dr  
dr
3
4o R
4o r 2
r2 
q 1 
q 3R 2  r 2


 
4o R 3  2  4o  r 
4o R 3
q
But if the point A at the surface of the sphere, the voltage will be
V
q
4o R
Example (1.21):
For the charge configuration shown in the following figure, Show that V(r) for the
points on the vertical axis, assuming r >> a, is given by
27
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
p
r
a
+q
+q
a
Solution:
Vp = V1 + V2 + V3
V
V
V
q
q
q


40 (r  a) 40 r 40 (r  a)
q (r  a)  q(r  a)
q

2
2
40 (r  a )
40 r
2aq
q

2
 a  40 r
40 r 2 1  2 
 r 
when r>>a then a2/r2 <<1
2aq
q

 a 2  40 r
40 r 2 1  2 
 r 
V
Hence, using the following formula:
(1 + x)n = 1 + nx
when x<<1
We can get:
V
2a q
40 r 2
 a2 
q
1 2  
 r  40 r
a2
can be neglected compared to 1, Then
r2
V
 q 2aq 
  2 
40  r
r 
1
28
-q
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
Example (1.22):
Find the potential difference between points A and B, VAB in terms of kcq/r?
A
Solution:
W
VAB  VB  VA  AB
q
K xq
K xq
VA  c
VB  c
3r
2r
Kc x q  3 2 
VAB  VB  VA 
  
r 6 6 q
K xq
VAB  c
6r
3r
q
2r
B
The Motion of Charged Particle Under the Effect of Electric Field:
If a negatively charged particle (the electron) moves under the effect of an electric field
as shown in the figure, there are two forces acting
on the electron as follow:
+++++++++++++
e
F1  q E
and
F2  m a
--------------------
(1.21)
Therefore
q E ma
Hence, the acceleration
(a)  q E
(1.22)
m
The velocity of the charged particle can be estimated using Eq. (1.22) as follow
qE
m
but , as known
a
29
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
dv
dt
Then
a
dv q E

dt m

v
qE
v   dv 
m
0
t
 dt
0
qE
dt
m
qE
 v
t  K1
m
dv 
Where K1 is the integration constant, and to calculate, the boundary conditions must be
used as follow
At t=0 → y=0, then K2=0
then
v 
q E
t
m
(1.23)
The displacement of the charged particle through the electric field can be deduced as
follow
qE
t
m
but , as known
v
dy
dt
Then
v
dy q E

t
dt m
y
y   dv 
0

qE
m
dy 
qE
t dt
m
t
 t dt
 y
0
qE 2
t  K2
2m
Where K2 is the integration constant, and to calculate, the boundary conditions must be
used as follow
At t=0 → v=0, then K2=0
then
y 
q E
t2
2m
(1.24)
30
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
In addition the kinetic energy of the charged particle can be expressed as follow:
K .E 
1
m v2
2
But, using equation (1.24), we can get:
1
1  qE 
K .E  m v 2  m 
t
2
2  m 
2
1 q2E 2 2
K .E 
t
2 m
(1.25)
Example (1.22):
An electron is displaced a distance of 1 cm in an electric field with intensity of
104 N/C . Calculate the velocity of the electron, its kinetic energy and the required time
to move this distance. Knowing that, e= 1.6x10-19 C and m= 9.11x10-31 Kg
Solution:
 The acceleration (a) is:
a
F e E 1.6 x1019 x 104


 1.8 x1015 m / sec 2
 31
m
m
9.11x10
 The velocity (v) can be calculated as follow:
v
2 a y  2 x 1.8 x1015 x 0.01  6 x106 m / sec
 The kinetic energy (K.E) is:
K .E 

1
1
m v 2  x9.11x10 31 x 6 x106
2
2
 The required time (t)
v 6 x106
t 
 3.3x10 9 sec
15
a 1.8 x10
Example (1.23):
31

2
 16 x1018 J
Electromagnetism I
Chapter I
Prof. Dr T. Fahmy
The strength of the electric field in the region between two parallel deflecting plates of a
cathode-ray oscilloscope is 25 kN/C. Determine the force exerted on an electron passing
between these plates. What acceleration will the electron experience in this region?
Solution:
d
As indicated in the figure, one of the plates is positively
charged, while the other is negatively charged. The electric
field between parallel charged plates is constant (as long as we
stay in the middle away from the edges of the plates), and is
+
a
-
A
-
+
-
+
+
+
b
-
directed from the + plate toward the - plate. The electric force
on a charge q placed in a field E is:
Fon q = q E
Since q is negative for an electron, then the force (F) on the
electron is opposite to the direction of (E), i.e., to the left
(toward the + plate). The magnitude of this force is:
|Fon e-| = e E = (1.6 x 10-19)x(25 x 103) = 4 x 10-15 N
While this is an extremely small force, we find the acceleration is quite large:
a = F/m = (4 x 10-15)/(9.11 x 10-31) = 4.4 x 1015 m/sec2.
The direction of the acceleration is, of course, the same as that of the force acting;
toward the + plate.
32