1

Flux of a Vector Field

Flux of the Electric Field

Gauss’ Law

A Charged isolated conductor

Applications of Gauss’ law

Gauss’ Law

2

Flux

The word “flux” comes from the latin word meaning “to flow”

For a vector filed flux is the number of lines passing through a surface

3

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

4

Electric flux

Given a chare distribution we can determine an electric field at a point using coulomb ‘s law

PE

E F= q

Where the total field “E” is the vector sum of all the fields due to all the point charges at point “P”

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

To fined out the charge distribution we need to know the electric flux and Gauss’s law

5

Electric fluxNumber 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

ΦE = EA Nm2 /C

6

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

7

+δE

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

8

+q

E

+2q

E

Electric flux

Electric flux through a surface is directly proportional to the magnitude of charges enclosed by that surface

9

The electric flux increases: with A with E

A

E ΦE = EA

Electric flux

10

If area “A” is not exactly perpendicular to the electric field E

ΦE = EAcos

AEE Or

Electric flux

Flux will be maximum when surface area is perpendicular to the electric field

ΦE = EA

11

Flux through an Arbitrary Shape

E is not uniform

Divide the arbitrary shape into small squares of area ΔA

The direction of ΔA 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

12

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

a.

b.

c.

Net Flux a + b + c = 0

2180cos REEdAdAE

θdAE cos

090cos dAEΦ

2)0cos( REEdAdAE

AdEΦ

13

Gauss’s law

Simplify electric field calculation

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

Gives a relation between the electric filed at all the points on the surface and the charge enclosed within the surface

Gauss’s law is used to analyze experiments that test the validity of Coulomb’s law

It is an alternative to Coulomb’s law for expressing the relationship between electric charge and electric field

14

The total electric flux through any closed surface is proportional to the total electric

charge inside the surface

Gauss’s law

enclEq

enclqAdE 0

enclEq0

AEE where

Relates net electric flux to the net enclosed electric charge

0enclq

AdE

15

Gauss’ Law & Coulomb’s LawLet 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

qAdE0

qAEd0

16

+

dA

E

The E field is uniform and thus constant everywhere

Where “ 4πr2” is the area of circular surface

qAEd0

220

20

0

4

)4(

r

kq

r

qE

qrE

qAdE

17

Infinite Line of Charges

++

++

++

++

++

++

++

++

++

++

++

++

+ Let us consider an infinite line of positive

charge 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 will have only the radial components and there is no flux at the ends

18

Infinite Line of Charges

++

++

++

++

++

++

++

++

++

++

++

++

+

EdA

h

Now according the Gauss law

Gauss’s law for electric field determination due to a charge distribution is the simplest of all

rE

hrhE

qAdE

0

0

0

2

)2(

19

Infinite Sheet of Charge

Let us now consider portion of nonconducting (Insulator) sheet of charge having a charge density (charge per unit area)

Consider an imaginary cylindrical Gaussian surface inserted into sheet

The charge enclosed by the surface is q = A

20

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

AEA

AEAEA

qAdE

21

Gauss’ Law & Conductors

Conductors are materials that are electrically neutral There is no net charge inside an isolated metal 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?

+

22

Gauss’ Law & Conductors

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

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

And there would then be no net charges inside conductor If there were any…we would see current inside…which

is never observed The excess charges do not disappear from the scene These excess charges appear as electrostatic charges at

the surface

+

+

+

++

+

23

Property of Conductors

An excess charge placed on or inside an isolated conductor moves entirely to the outer surface of the conductor. None of the excess charge is found within the

body of the conductor

24

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 spread

over both the surfaces Each surface will have a charge equal to

q/2 And now we try to apply Gauss’ Law

Applications of Gauss’ Law

E

++

++

++

++

++

++

++

++

++

++

++

++

+

++

++

++

++

++

++

++

++

++

++

++

++

+

E

E E

25

A Thin Conducting Plate

02

E

E=0We can think of the situation as two noncouducting charged plates connected back-to-back each resulting in

0

E

The total E field of a thin conducting plate would then be

E

++

++

++

++

++

++

++

++

++

++

++

++

+

++

++

++

++

++

++

++

++

++

++

++

++

+

E

E E

26

Two Thin Conducting Plates with Opposite Charge

E = 0

E

+++++++++

---------

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

27

E = 0

E

+++++++++

---------

E = 0

The charges on both plates will move towards the inner surfaces due to force of attraction

02

E

Each surface will set up an E field

The net E field can thus be given as

0

E

28

0

E

E = 0

E

+++++++++

---------

E = 0Putting =q/A

0A

qE

This is the electrical field of a parallel plate capacitor

29

Homework Exercises

Course book, Volume 2

Edition 4th Chapter # 29

Problem # :1, 3, 5, 9, 13, 14, 19, 23, 25, 27, 31, 39 and 42

Edition 5th : Chapter 27

Exercise # 1,3, 5, 9, 15, 17, 19, 23, 26, 29

Problems # 2, 3, 13