Electrostatic Lecturer

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

qq

4

1WU 21

0


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End ofsession 5

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Electrostatic Lecturer