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7/28/2019 Electrostatic Lecturer
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Chapter 3 - Electrostatics
Understanding electrical charges
of its characters and phenomena
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Electrical charge3.1
Explain electrical charge, positive and negative charges,conductors and insulators, charge and discharge, conservation of
charges and quantization of charges.
Two plastic rods being rubbed by fur and two glass rods beingrubbed by silk
silk
fur
plastic
rod A
plasticrod A
glass
rod C
glass
rod D
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When two plastic rods are brought closer to each other, the result of will
be
Both rods repel each other.
rod A rod B
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As two glass rods are brought closer to each other,
both rods repel each other.
rod C
rod D
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However, as a plastic rod is brought closer to a glass rod,
both rods attract each other.
rod C
rod B
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When the fur is brought closer to the hanging glass rod,
both items attract each other.
furThe rod initial
position
The rod final position
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As the silk is brought closer to the hanging plastic rod,
Both items attract each other.
silkThe rod initial
position
The rod final position
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The rubbing of the items cause the electrical charges to transfer from
one material to another.
What is an electrical charge?
In nature, it can be either a positive charge or a negative charge.
A crude representation of an atom, showing the positively charged
nucleus at its center and the negatively charged electrons orbiting about
it.
The structure of atoms can be
described in terms of three particles;
electron, proton and neutron.
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A unit of an electrical charge is coulomb, C
An electron has a magnitude of e = -1.6x10-19 C.
Likewise, a proton has a magnitude of e = +1.6x10-19 C.
It would take 1/e = 6.3 x 1018 protons to create a total charge of +1C.
Likewise, it would take 6.3 x 1018 electrons to create a total charge of
-1C.
The identical charges repel each other.
However, the opposite charges attract each other.
initialfinal
finalinitial
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The two plastic rods acquired negative charges.
Since both rods have similar charges, they repel each other.
Rod A
Rod B initialposition
Rod B finalposition
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Therefore, the glass rod acquires ________ charges.
The fur acquires ________ charges.
The silk acquires _________ charges.
Negative or positive charges?
positive
positive
negative
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4F
F
r
2r
The force between two charges is inversely proportional to the distance
between two charges.
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The experiment conducted by BenjaminFranklin (1706 1790) showed that the
electrical charges were conserved.
Principle of conservation of charge:
The sum of all the electriccharges in any closed system is
constant.
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End ofsession 1
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Quantization of Electrical charge3.1.1
In 1909, Robert Millikan discovered the existence of electrical charge is
in discreet amount or a packet. The electric charge cannot be divided
into amounts smaller than the charge of one electron or proton.
N = 3
N = ???
Number of charges = 3
Number of charges = ????
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Free electrons in conductor or insulator3.1.2
Charges or free electrons can move freely in conductor. When certain
amount of charges are transferred to the conductor, the charges are
uniformly distributed to the other parts of conductor.
Metals and alloys are normally good conductor.Certain chemical solution like NaCldisplays good conductor.
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Substances like glass, Perspex, silk and rubber are good insulators.
When an insulator is rubbed, the rubbed part contains charges.
Charges cannot be transferred to the other parts of the insulator.
This part is still neutral
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Charging by induction3.1.3
In the induction process of charging, a charged object requires nocontact with the object inducing the charge.
Grounding
the charge
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Coulombs Law3.2
State Coulombs Law and use equation F=Qq/40r2 for a pointcharge and system.
In 1785, Charles Coulomb (1736 -1806) established thefundamental law of electric force between two stationary chargedparticles.
It applies only to point charges and to spherical distributions of
charges. The vector F can be written as:
rr
qqkF
2
2121
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Suppose there are two charges, q1 and q2 at a distance, r. If the two
charges have the same sign, both will repel each other with a force,
F.
r
FORCE
q1 q2
Electrical charge in focusThis repulsive electrical force is a vector. It is written as F21.
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Likewise, if q1 has negative sign and q2 has positive sign, both will
attract each other with a force, F.
r
FORCE
q1 q2
Electrical charge in focusThis attractive electrical force is a vector. It is written as F21.
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Suppose there are three charges, q1, q2 and q3.
The resultant force due to q1 and
q2 on q3 will be FT.
q2
q1
q3F31
F32
FT
It can be said that FT is a vectoraddition of F31 and F32.
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The Coulomb force is a field force a force exerted by oneobject on another even though there is no physical contact
between them.
The magnitude of force on charge q2 can be written as:
2
2121
r
qqkF F21
q1 q2
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End ofsession 2
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Electric Field3.3
Define the electric field strength E=F/q and describe the electricfield lines for isolated point charge, dipole charges and plate of
uniform charges..
An electric field exists in the
region of space around a
charged object. When another
charged object enters this
electric field, the field is what
exerts a force on the second
charged object.
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+
Qq0
Etest charge
+ + + + +
+ + + + + + +
+ + + + + + +
+ + + + + +
+ +
0
q
FE
We define the electric field at the location of the small test charge
to be the electric force acting on it divided by the charge q0 of the
test charge.
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Suppose there are two charges, q1 and q2.
The resultant electric field due to
q1 and q2 at point C will be ET.
q2
q1
CE2
ET
It can be said that ET is a vectoraddition of E1 and E2.
E1
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Michael Faraday introduced the electric field lines in the following
manner:
The electric field vector E is tangent to the electric field
lines at each point.
The number of lines per unit area through a surface
perpendicular to the lines is proportional to the strength
of the electric field in a given region.
For a positive point charge, the lines radiate outward and for anegative point charge, the lines converge inward.
E is large when the field lines are close together and small when they
are far apart.
Electric Field Lines3.3.1
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Electric field of positive
and negative point
charges
Two dimensional drawing contains only the field lines that lie in the
plane containing the point charge.
Electric field of positive
point charges.
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Moving charge in a uniform electric field3.3.2
Describe quantitatively the motion of a charge in a uniformelectric field.
A charged particle in the electric field experiences electrical
force, F = q E. The charged particle also accelerates as
dictated by second Newtons Law, F = ma.
Thus, the acceleration
of charged particle is
written:
m
qEa
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Gauss Law3.4
Consider a uniform electric field passing through an area that
is perpendicular to the field.
State and use Gauss Law to determine the electric field of acharged body.
Electric flux exists as
electric field flows
through the area
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Electric flux, is defined as
A.E
If the area A is parallel to the field lines, E = 0; thus, = 0.
No electric flux exists as
no electric field pierce
the area.
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If the E lines pierce at the area A at an angle away from thenormal line, the flux is written:
cosA.E
normal line E
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End ofsession 3
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Gauss Theorem3.4.1
Consider a point charge q and aspherical surface of radius r from
centre on the charge. The constant
magnitude of electric field on the
surface of the sphere is written:
2r
qkE
Since the electric field is everywhere
perpendicular to the spherical surface,the electric flux will be :
0
2
2
0
Qr4
r4
QA.E
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For any type of surface, the general
flux equation through a close surface
can be written :
0
QdA.E
Therefore, it is equally true for any
surface that encloses the charge q, the
flux would simply be the charge divided
by the permittivity of free space :
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3.4.2 The equivalence of Gauss Law and Coulombs Law
Consider a point charge q with a spherical Gaussian surface of radius rfrom centre on the charge. The Gauss law states that the flux througha close surface:
0
QdA.E
The close surface area, dA = 4r2.
The electric field, E for the charge:
0
2 Q)r4.(E
2
0r4
QE
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The positive test charge exerts electrical force due to the electric field.
The magnitude of electrical force is written:
E.qF 0Finally, transformation of Gauss law to Coulombs law is simply writtenas:
20
0r4
Q.qF
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End ofsession 4
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3.4.3Sphere of concentrated charges
Consider positive electric charge Q is distributed uniformly throughout
the volume of an insulating sphere with radius R.
Since the charge density is
constant,
33R
3
4
q
r3
4
'q
The equation is simplified into:
3
3
R
rq'q
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The electric field enclosed by the surface is considered as if that
enclosed charge were concentrated at the center.
20r4
'qE
From substitution of q, we can get the electric field enclosed by thesurface.
rR4
q
E 30
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3.4.4Electric field of charged thin spherical shell
Consider a charged spherical shell of total charge
q and two concentric spherical surfaces, S1 and
S2.
For r R, the electric field is:
20r4
qE
For r < R, the electric field is:
0E
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3.4.5Electric field of infinite charged line
Consider = Q/L for uniformly distributed charge along an infinitely long,thin wire. Gaussian surface of a cylinder with arbitrary radius r and
arbitrary length L is used. No flux through the ends because E lies in the
plane of the surface. E has the same value everywhere on the cylinder
walls. The area, A = 2rL. The electric field for the cylinder is:
+ + + + + + + + + + + + +r
L
r2
1E
0
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3.4.6Electrical field of infinite charged sheet
Consider is the density of charge perunit area. Therefore Q = A. Thecharged sheet passes through the
middle of cylinders length, so thecylinders ends are equidistant from thesheet. No flux passes through the
cylinders side walls, therefore, E = 0.The total flux in Gausss law, = 2EAsince EA from each cylinders end. Thus,
the electric field for the infinite planesheet of charge is written:
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
EE
02E
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3.5.1Electric equipotential surface
The potential at various points in an electric field can be visualized byequipotential surfaces. Equipotential surface is three dimensional
surface on which the electric potential V is the same at every point. If
the test charge q0 is moved from a point to point on such a surface,
the electric potential energy q0V remains constant.Field lines and
equipotential surfaces are always mutually perpendicular.
Equipotential
surface
Field line+
+ -
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3.5Electric potential
Consider a system of charges. V1 is thepotential at point P from a charge q1. The
work done to bring charge q2 to point P
from infinity without acceleration is equal
to q2V1. As the two charges are
separated at distance r, the work done isequal to the potential energy. The work
done to bring the charge Q from a to b
is:
W = Q(VaVb) = -U.
Define the electric potential and use equation V=Q/40r. Use relation E = - dV/dr.
Understand the relationship between electric potential andpotential energy.
drr
1Q
4
1V
20
r
4
1WU 21
0
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End ofsession 5