Download Electric Field - O6U E

Ass.Prof. HALA MOUSTAFA






Definition of the electric field E
Calculating E due to a charged particle
To find E for a group of point charges
Electric field lines
Motion of charge particles in a uniform electric
field
The electric dipole in electric field
‫‪The direction of E‬‬
‫‪‬‬
‫‪‬‬
‫‪‬‬
‫‪‬‬
‫‪If Q is +ve the electric field at point p in space is‬‬
‫‪radially outward from Q‬‬
‫‪If Q is -ve the electric field at point p in space is‬‬
‫‪radially inward toward Q‬‬
‫يكون اتجاة المجال الكهربائي الناتج عن شحنة موجبة في اتجاة‬
‫الخروج من الشحنة‪.‬‬
‫يكون اتجاة المجال الكهربائي الناتج عن شحنة سالبة في اتجاة‬
‫الدخول على الشحنة‪.‬‬
‫قيمة المجال ‪E=Kq/r2‬‬
‫اتجاة المجال في اتجاة الخروج‬
‫‪‬‬
‫‪‬‬




The lines must begin on
positive charges (or infinity)
‫خطوط المجال تبدا من الشحنة الموجبة‬
‫وتنتهي عند السالبة‬
The lines must end on
negative charges (or infinity)
‫خطوط المجال تنتهي عند الشحنة السالبة‬
The no of lines leaving a+ve
charge (approaching a-ve
charge)is proportional to the
magnitude of the charge
Electric Field lines cannot
cross


The e field two
charges (positive)
The Electric Field
two
charges
(negative)
These are imaginary lines pointing outward from positive charge and
inward to the negative charge
Electric Field
It is the space around the electric charge in which
the electrical effects appear.
Electric Field Strength at Apoint
It is the force per unit charge at the point.SI unit is
)C/N(
E=F/Q
)N/C(
It is defined as the work done to move a positive
unit charge from one point to another
If WAB is the work done to move a charge Q from
point A to point B ,then the potential difference
V is given by:
VAB =
(J/C) or Volts
WAB =QVAB (J)
The work done ΔW to move a charge Q against
an electric field of strength E a distance ΔX is
given by:
ΔW=F ΔX
F=QE
ΔW=QE ΔX
ΔV= ΔW = QEΔX=EΔX
Q
Q
E= ΔV (V/m)
ΔX
‫‪The electric dipole is positive charge and‬‬
‫‪negative charge of equal magnitude‬‬
‫‪‬‬
‫المناظر لألبرة المغناطيسية في الكهربائية بمعنى الشحنة الموجبة‬
‫والشحنة السالبة متساويتا‬
‫‪‬‬
We start with special case of a point charge in uniform
electric field E. The electric field will exert a force on a
charged particle is given by
F = qE
The force will produce acceleration
a = F/m
where m is the mass of the particle. Then we can write
F = qE = ma
The acceleration of the particle is therefore given by
a = qE/m
If the charge is positive, the acceleration will be in the
direction of the electric field. If the charge is negative,
the acceleration will be in the direction opposite the
electric field.

F=QE
qE ‫عند وضع شحنة كهربائية كتلتها في مجال كهربي فان المجال الكهربائي يبذل قوة مقدارها‬

A charge of 2μC experiences a force of 34 N
when placed at a point in an electric
field.Calculate the electric field strength at
the point?
Q = 2μC= 2x 10-6 C
F = 34 N
E=F/Q
(N/C)
E=34 N /2x 10-6 C = 68x 10-6 (N/C)
The work done in bringing a charge of 4 C from
one point to another is 10 J .What is the
potential difference between the two points?
Q=4C
W = 10 J
V = W = 10 = 2.5 V
Q
4
The potential difference between two points is
12 V.Find the work done to move a charge of
8 C between the two points?
V = 12 C
Q=8C
W=QV = 8x12 = 96 (J)
If the potential difference between two parallel
plates separated by a distance 1 mm is 400
v,Find the electric field intensity between the
two plates.?
ΔX = 1mm = 1 x 10-3 m
ΔV = 400 (V/m)
E = ΔV = 400 = 4 x 105 (V/m)
ΔX
1 x 10-3
Acharge of 2µC experiences a force of 34 N when
placed at a point in an electric field.calculate
the electric field strength at the point?
F=34 N
q = 2µC
E=F/q
E = 34 = 1.7x107 ( N/C)
2x10-6
Calcullate the magnitude of the force on charge
of 6 µC when placed in an electric field of
strength 4 x10-3 NC-1?
E=F/q
F= Eq
F = (4x10-3) (5x10-6) = 2x10-8 N




Find the electric field due to electric dipole
along x-axis at point p,
which is a distance r from the origin, then
assume r>>a
The electric dipole is positive charge and
negative charge of equal
magnitude placed a distance 2a apart as shown
in figure 3.6

Find the electric field at point p in figure due
to the charges shown.
‫ نرقم الشحنات‬-1
‫نحدد المجال الناتج عن كل شحنة عند النقطة‬-2
E ‫نوجد‬
Eh