1 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 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 filed 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 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
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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
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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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+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
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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 ΦE = EA
Electric flux
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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
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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
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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 Flux a + b + c = 0
2180cos REEdAdAE
θdAE cos
090cos dAEΦ
2)0cos( REEdAdAE
AdEΦ
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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
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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
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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
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+
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
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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
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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(
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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
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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
AEA
AEAEA
qAdE
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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?
+
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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
+
+
+
++
+
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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 of the 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 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
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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
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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
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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
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0
E
E = 0
E
+++++++++
---------
E = 0Putting =q/A
0A
qE
This is the electrical field of a parallel plate capacitor
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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