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Flux of a Vector Field

Flux of the Electric Field

Gauss Law

A Charged isolated conductor

Applications of Gauss law

Gauss Law

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Flux

The word flux comes from the latin

word meaning to flow

For a vector field flux is the number of

lines passing through a surface

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Flux of a vector field Vector field

Velocity field of a flowing fluid

The velocity field is a representation of a fluid flow

Field itself is not flowing but is a fixed representation of the

flow

Water flowP

In the velocity field of a water flow point p represents the

flow of water (fluid)

Velocity field

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Electric flux

Given a charge distribution we can determine an electric

field at a point using coulomb s law

P

E

E F= q

Where the total field E is the vectorsum of all the fields due to all the point

charges at point P

Alternatively if an electric field E is given we can determinethe charge distribution

To find out the charge distribution we need to know the

electric flux and Gausss law

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Electric flux

Number of electric lines of force passing through a surface

of area A perpendicular to the electric field E

A

E

Mathematically it is the product of the surface area A and

the component of the electric field E perpendicular to the

surface

E = EA Nm2 /C

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Electric flux

Empty enclosed surface no electric field

+q

EElectric field

directed outward

Electric field

directed inward

Positive charge

enclosed

Negative charge

enclosed

-q

E

Outward flux

No flux

Inward flux

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+E

In this case the box is placed inside an electric field of

some out side charge distribution

Again here the net flux is zero, because the number of lines

entering the box is exactly the same as leaving the box

Electric flux

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+q

E

+2q

E

Electric flux

Electric flux through a surface is directly proportional to the

magnitude of charges enclosed by that surface

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The electric flux increases: with A

with E

A

E E = EA

Electric flux

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If area A is not exactly perpendicular

to the electric field E

E = EAcos

AEE

Or

Electric flux

Flux will be maximum when surface

area is perpendicular to the electricfield

E = EA

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Flux through an Arbitrary Shape

E is not uniform

Divide the arbitrary shapeinto small squares of area A

The direction ofA is

drawn outward Calculate the electric flux at

each square and sum all these

This is a surface integral, i.e. an integral

over a closed surface, enclosing a

volume

AdE

AE D

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Sample problem 2: Find the electric flux through a cylindrical surface in a

uniform electric field E

a.

b.

c.

Net Fluxa + b + c = 0

2180cos REEdAdAE

dAEcos

090cos dAE2

)0cos( REEdAdAE

AdE

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Gausss law

Simplify electric field calculation

Gives an in site about the electric charge distribution overthe conducting body

Gives a relation between the electric filed at all the points

on the surface and the charge enclosed within the surface

Gausss law is used to analyze experiments that test the

validity of Coulombs law

It is an alternative to Coulombs law for expressing therelationship between electric charge and electric field

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The total electric flux through any closed

surface is proportional to the total electric

charge inside the surface

Gausss law

enclEq

enclqAdE 0e

enclEq0e

AEE

where

Relates net electricfluxto the net enclosed

electric charge

0e

enclqAdE

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Gauss Law & Coulombs Law

Let us consider a

positive point charge q

Surround the charge with

an imaginary surface

the Gaussian Surface

+

dA

E

The angel between vector area and

E field is zero everywhere

qAdE0e

qAEd0e

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+

dA

E

The E field is uniform and

thus constant everywhere

Where 4r2 is the area of

circular surface

qAEd0e

22

0

2

0

0

4

)4(

r

kq

r

qE

qrE

qAdE

e

e

e

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Infinite Line of Charges

+++++++++++++++

+++++

+++++

Let us consider an infinite line of positivecharge with a linear charge density = q/h

We wish to find the E field at a distance r

from the line

Applications of Gauss Law

EdA

hLet us now enclose this line with a

cylindrical Gaussian surface

The symmetry indicates that E field willhave only the radial components and there is

no flux at the ends

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Infinite Line of Charges

+++++++++++++++

+++++

+++++

EdA

h

Now according the Gauss law

Gausss law for electric field determination

due to a charge distribution is the simplest of

all

rE

hrhE

qAdE

0

0

0

2

)2(

e

e

e

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Infinite Sheet of Charge

Let us now consider portion

of nonconducting (Insulator)sheet of charge having a

charge density (charge perunit area)

Consider an imaginary

cylindrical Gaussian surface

inserted into sheet

The charge enclosed by thesurface is q =A

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Due to symmetry we can conclude that E field is right

angles to the end caps

There is no flux from the curved surface of the cylindricalThere is equal flux out of both caps

Infinite Sheet of Charge

A very useful result that can be directly applied

on similar applications of Gauss Law

0

0

0

0

2

2

)(

e

e

e

e

+

E

AEA

AEAEA

qAdE

G & C

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Gauss Law & Conductors

Conductors are materials that are electrically neutral

There is no net charge inside an isolatedmetal ball

And therefore the E field inside an isolated conductor is

zero

Suppose we are able to inject some charge into the

center of the metal ball

What would then happen?+

G L & C d

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Gauss Law & Conductors

Initially there would be an E field that would cause allthe charges to redistribute

Within nanoseconds the charges would settle and stopmoving, which is called an electrostatic equilibrium

And there would then be no net charges inside conductor

If there were anywe would see current insidewhichis never observed

The excess charges do not disappear from the scene

These excess charges appear as electrostatic charges at

the surface

+

+

+

++

+

P t f C d t

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Property of Conductors

An excess charge placed on or inside an

isolated conductor moves entirely to the

outer surface of the conductor. None ofthe excess charge is found within the

body of the conductor

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A Thin Conducting Plate

Suppose we take a thin conducting plate

This plate has definitely two surfaces

And spray a charge q on any surface

This charge q will move and will spreadover both the surfaces

Each surface will have a charge equal to

q/2And now we try to apply Gauss Law

Applications of Gauss Law

E

+++

+++++++++++

+

+++++

+++++

+++

+++++++++++

+

+++++

+++++

E

E E

A Thi C d ti Pl t

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A Thin Conducting Plate

02e

E

E=0

We can think of the situation

as two noncouducting chargedsheets connected back-to-back

each resulting in

0e

E

The total E field of a thinconducting plate would then be

E

+++

++++++++++++

+++++

+++++

+++

++++++++++++

+++++

+++++

E

E E

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Two Thin Conducting Plates with Opposite

Charge

E = 0E

+

+

+

+

+

+

+

+

+

-

-

-

-

-

-

--

-

E = 0

Let we bring closer

two thin conducting

sheets each with an

equal and opposite

charge of magnitude of

q

Both conductors now cannot be considered

as isolated conductors

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E = 0E

+

+

+

+

+

+

+

+

+

-

-

-

-

-

--

-

-

E = 0

The charges on both plates will

move towards the inner

surfaces due to force ofattraction

02e

E

Each surface will set

up an E field

The net E field canthus be given as

0e

E

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0e

EE = 0

E

+

++

+

+

++

+

+

-

-

-

-

-

--

-

-

E = 0Putting =q/A

0eAqE

This is the electrical field of a parallel plate capacitor