 Overview of static electricity( On Electric
Force , field and Gauss’ law, Electric
potential, capacitance and dielectrics).
 http://www.fayoum.edu.eg/nmh
E-Mail: nmh00@fayoum.edu.eg
22-10-2014 FCI 1

2014/2015 ىعماجلا ماعلا ةدنجا
22-10-2014
.
2014/10/11 قفاوملا تبسلا يساردلا ماعلا ءدب •
25/10/2014 ةيرجهلا ةنسلا سار ةزاجا •
. لاالا 22/11/2014 تبااسلا فوتا لااساردلا فاالالا فاااماا ااااحتما •
27/11/2014
ىلا 3/1/2015 تبسلا نم لولأا يساردلا لصفلا تاناحتما •
.
22/1/2015
FCI 2

Course No Course Title
Hrs / week
Lect Tut Lab Final
Marks
Y.Work
oral Lab Total
Exam hours
GEN 110
English Language (1)
) 1 ( ةيزيلجنإ ةغل
2 40 10 50 2
GEN 113
Computational Linguistics
ةيباسح تايوغل
2 -
GEN 114
Humanities
تايناسنا
2
GEN 120
GEN 125
CSC160
Mathematics (1)
) 1 ( تايضاير
Physics (1)
) 1 ( ءايزيف
Basics of computer science
بساحلا مولع تايساسأ
4
4
4
3
3
-
-
-
-
2
4
40
40
115
90
105
10 50 2
10
20
25
50
15 150
10 25 150
2
3
3
15 10 20 150 3
22-10-2014 FCI 3

Physics 1 Gen 125 : Electricity and Magnetism
REF: Physics For Scientists And Engineers 6E By Serway And Jewett
 Ch. 1 : Overview of static electricity( on Electric field and Gauss’ law,
Electric potential, capacitance
and dielectrics).

Ch. 3 - Direct current and circuit
3 -1 Electromotive Force
3 -2 Resistors in Series and Parallel
3 3 Kirchhoff’s Rules
3 -4 RC Circuits
3 -5 Electrical Meters
Ch. 2 : Resistance and current:
2-1 Electric Current
2-2 Resistance
2-3 A Model for Electrical
Conduction
Inductance: Self learning
2-4 Electrical Power
22-10-2014 FCI 4

Physics 1 Gen 125 : Electricity and Magnetism
Ref: Physics For Scientists And Engineers 6E By Serway And Jewett
Ch. 4 : AC circuits:
4.1 AC Sources
4.2 Resistors in an AC
Circuit
4.3 Inductors in an AC
Circuit
4.4 Capacitors in an AC
Circuit
4.5 The RLC Series Circuit
4.6 Power in an AC Circuit
4.7 Resonance in a Series
RLC Circuit
4.8 The Transformer and
Power Transmission
4.9 Rectifiers and Filters
22-10-2014 FCI 5

The aim of the course
•
The graduates have to develop a clear understanding of the basic concepts of electricity .
•
The graduates have to develop a clear understanding of the basic concepts of electronics .
•
Developing the graduate's skills and creative thought needed to meet new trends in science.
•
Supplying graduates with basic attacks and strategies for solving problems.
22-10-2014 FCI 6

- Intended Learning Outcomes
A- Knowledge and
Understanding:
1) Concept of electric field, electric flux.
2) Concept dc circuits and Ac circuits.
3) Investigate the characteristics of simple series circuits that contain resistors, inductors, and capacitors and that are driven by a sinusoidal voltage.
4) Strategies for solving problems related to the previous topics.
B- Intellectual Skills: 1) Utilizations of theories, rules and basic sciences to interpret physical events.
2) Collecting information from its relevant sources and use it in discussion .
C- Professional and
Practical Skills:
D- General and transferable Skills
1) Being able to solve problem sheets related to the material course.
2) Collect and record data and information from libraries and summarize it in suitable forms.
3) The student would be able to apply some experiments related to the course contents.
1) Graduate should be able to access data and information from the Internet related to the course subjects.
2) Graduate should develop self professional, scientific, and personal attitude towards continuous education.
3) Graduate should be able to cooperate in teams
22-10-2014 FCI 7

Chapter 1
Electric Charge and Electric Fields
 What is a field?
 What causes fields?
Field Type gravity electric magnetic
Caused By mass charge moving charge
22-10-2014 FCI 8

Electric Charge
Types: Two types of charges
Positive: Lack of electrons
Negative: surplus of electrons
Unit : Coulomb C
+ + - -
22-10-2014 FCI 9

22-10-2014
Insulators : do not allow charges to move through them(wood , paper…… )
Conductors: allow charges to move through them ( copper……
Semiconductors : partially allow charges to move through them ( Si , Ge …….
FCI 10

22-10-2014 FCI 11

Charging:
+ + +
+ + +
+ + + + + +
2
22-10-2014
1
FCI 12

Charging:
+ + + + + +
22-10-2014 FCI
3
13

Charging:
Ebonite
Wool
Cloth
22-10-2014 FCI 14

22-10-2014
1
FCI
2
3
15

+ r
+
K= 8.89 x10 9 Nm 2 /C 2
22-10-2014 FCI 16

(N)
Ɛ o
- Another constant called permittivity of vacuum
22-10-2014 FCI 17

y
+q
1
F
13
F
13
F
+q
2
X
+q
3
F
23
FCI 22-10-2014
F
23
18

Force acting on a unit positive charge at the point.
+ + + r
22-10-2014 FCI q
(N/C)
19

Force acting on a unit positive charge at the point.
+ + + r
22-10-2014 FCI q
( N/C)
20

+
22-10-2014 positive
FCI
negative
21

22-10-2014 FCI 22

22-10-2014 FCI 23

+q
-q
22-10-2014 FCI 24

-q +q
22-10-2014 FCI 25

F
12
 r
12 2 r
12
ˆr
12
ˆr
12
 r
12 r
12
+q
1 r
12
+q
2
F
12
Coulomb’s constant: permittivity of free space: r
12 k

8.988x10
9
  o
1
2

9.0x10
9
C
2

8.85x10

12
C
2
C
2
2
2

1
4
 o
Charge polarity:
22-10-2014
Same sign
Opposite sign
Force is right
Force is Left
Electrostatics --- Charges must be at rest!
FCI 26

F
0

F
10

F
20

F
30

....
+Q
1
+Q
2 r
20 r
10
+Q
0
+Q
3 r
30
F
30
F
20
F
10
F
0
 kq q
0 1
2 r
10 r
ˆ
10
 kq q
0 2
2 r
20 r
ˆ
20
 kq q
0 3
2 r
30
ˆ r

....
30
F
0
 kq
0

 q
1
2 r
10
22-10-2014 r
ˆ
10
 q
2
2 r
20
FCI r
ˆ
20
 q
3
2 r
30
ˆ r
30

....


 kq
0
N  q i r i0
2 r
ˆ i0
27

 Where is the resultant force equal to zero?
 The magnitudes of the individual forces will be equal
 Directions will be opposite


Will result in a quadratic
Choose the root that gives the forces in opposite directions q
1
= 15.0 m
C q
2
= 6.0 m
C
22-10-2014 FCI 28

 The spheres are in equilibrium
 Since they are separated, they exert a repulsive force on each other
 Charges are like charges
 Proceed as usual with equilibrium problems, noting one force is an electrical force
22-10-2014 FCI 29

 The free body diagram includes the components of the tension, the electrical force, and the weight
 Solve for | q |
 You cannot determine the sign of q , only that they both have same sign
22-10-2014 FCI 30

Surface Charge Density:
22-10-2014
Volume Charge Density:
FCI
 
Q
 dq
A dA
 
Q
 dq
V dV
31

E
P

E
1P

E
2P

E
3P

....
+q
1
+q
2 r
20 r
10
P
+q
3 r
30
E
30
E
20
E
10
E
P
 kq
1
2 r
10 r
ˆ
10
 kq
2
2 r
20 r
ˆ
20
 kq
3
2 r
30
ˆ r

....
30
E
P
 k

 q
2 r
10
1 r
ˆ
10
22-10-2014 FCI
 q
2
2 r
20 r
ˆ
20
 q
3
2 r
30
ˆ r
30

....


 k
N  q i r i0
2 r
ˆ i0
32




Find the electric field due to q
1
, E
1
Find the electric field due to q
2
, E
2
E = E
1

+ E
2
Remember, the fields add as vectors
 The direction of the individual fields is the direction of the force on a positive test charge
22-10-2014 FCI 33

Gauss’s Law
 The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity.
Gauss's Law is a general law applying to any closed surface.
22-10-2014 FCI 34

 Gauss' law permits the evaluation of the electric field in many practical situations by forming a symmetric Gaussian surface surrounding a charge distribution and evaluating the electric flux through that surface.
22-10-2014 FCI 35