Gauss Law

CHAPTER 23: GAUSS’ LAW

Dr Reem M. Altuwirqi

•

Flux of an electric field

•

Gauss’ law (another way to calculate E)

•

Gauss’ law and Coulomb law

•

Applying Gauss’ law to:

• A charged isolated conductor

• Cylindrical symmetry

• Planar symmetry

• Spherical symmetry

What we will learn

Gauss’ Law

General Concept



In the past chapter, one of our aims was to find E from different charged objects.

How to find E?

1. Find dq 2. Find dE

3. Integrate dE to find E

How to find E?

1. Choose a Gaussian (hypothetical) surface 2. Find electric flux 

3. Find enclosed charge q_enc 4. Calculate E

What is a Gaussian surface?



A hypothetical (imaginary) closed

surface enclosing a charge distribution.



Can take any shape, best mimic the symmetry of the charge distribution to minimize calculation of E.

Gaussian surface imaginary sphere

What is flux?

(a) A uniform airstream of velocity is

perpendicular to the plane of a square loop of area A.

(b) The component of perpendicular to the plane of the loop is v cos q, where q is the angle between v and a normal to the

plane.

(c) The area vector A is perpendicular to the plane of the loop and makes an angle q with v.

(d) The rate of volume flow through the loop is

= (v cos q) A.

This rate of flow through an area is an example of a flux—a volume flux in this situation.

Φ = 𝜈𝐴 𝑐𝑜𝑠𝜃 = 𝜈 ∙ 𝐴

Electric Flux

Φ = 𝐸𝐴 𝑐𝑜𝑠𝜃 = 𝐸 ∙ 𝐴

EA  = 0 EA||  = EA

θ ABcos



B A

What do we do if E

is non- uniform?

Eis uniform

The electric flux  through a Gaussian surface is proportional to the net number of E lines passing

through that surface

Electric Flux

Non-Uniform E

i i

i i

E  E A  E A

 cos

q

^{E } _^lim_{A 0}



^{Ei Ai }



^EdA

i

nˆ

nˆ nˆ

Electric Flux

Electric Flux

Electric Flux

Electric Flux

Electric Flux

Electric Flux

Gauss’ Law



Relates net flux  of E through a (closed) Gaussian surface to the net charge enclosed q

_enc

by that

surface.

Gauss’ Law

Net  through a closed surface

enclosing a charge q is:

 = q / _o

Net  through a closed surface

enclosing no charge is:

 = 0

o enc E

d q



^{ }





 E A

m N

o  8.8510^12C² / 



Gauss’ Law

o in E

d q



^{ }





 E A

o E

S q



¹ )

( 



o E

q S q



) ) (

'

(  ¹  ²



0 )

''

( 

_E S

Algebraic sum From all charges

inside and outside the surface

Gauss’ Law

o in E

d q



^{ }





 E A

Gauss' law holds for closed surface.

Usually one particular surface makes the problem of determining the electric field very simple.

When calculating the net charge inside a c

Note 1 : any

Note 2 : losed

surface we take into account the algebraic sign of each charge.

When applying Gauss' law for a closed surface we ignore the charges outside the surface no matter how large they are.

Note 3 :

Examp

1 0 1 2 0 2

3 0 3 4 0 4

1 2 3 4

Surface : , Surface :

Surface : 0, Surface : 0

We refer to , , , as "Gaussian surfaces."

S q S q

S S q q

S S S S

 

 

     

      

le :

Note :

Gauss’ Law

Gauss’ Law

Gauss’ Law

Gauss’ Law

Gauss’ Law

Gauss’ Law and Coulomb’s Law

Coulomb’s Law Gauss’ Law

Different ways of describing the relation between E and q in static situations

 We divide the Gaussian surface into elements of area dA

 The flux for each element d = E dA cos 0 = E dA

 Total flux Φ = 𝐸 𝑑𝐴 = 𝐸 𝑑𝐴 = 𝐸 (4𝜋𝑟²)

 From Gauss’ law 𝜀_𝑜Φ = 𝑞_𝑒𝑛𝑐 = 𝑞

 𝐸 4𝜋𝑟² 𝜀_𝑜 = 𝑞 → 𝐸 = ¹

4𝜋𝜀_𝑜 𝑞

𝑟² (Coulomb’s Law)

Gauss’ Law and Coulomb’s Law

Application of Gauss’ Law

How to find E?

1. Find dq 2. Find dE

3. Integrate dE to find E

How to find E?

1. Choose a Gaussian (hypothetical) surface 2. Find electric flux 

3. Find enclosed charge q_enc 4. Calculate E

o enc E

d q



^{ }





 E A

Application of Gauss’ Law

A charged isolated conductor

Cylindrical symmetry Line of charge

Planar symmetry Non-conducting

sheet

Planar symmetry Two conducting

plates

Spherical symmetry Charged

conducting sphere

Spherical symmetry Charged non- conducting sphere

Application of Gauss’ Law

A charged isolated conductor

e E v F

1. Is there charge inside the conductor?

2. Is there an electric field inside the conductor?

3. What is external E due to a charged conductor?

If E was inside the conductor:

Free electrons will be influenced by E E  F

F  movement

Movement  current

Current  heat or magnetic field

The electrostatic field E inside a conductor is equal to zero.

All excess charge will move entirely to the conductor surface E_ext

Application of Gauss’ Law

A charged isolated conductor

3. What is external E due to a charged conductor?

o enc E

d q



^{ }





 E A

If conductor not spherical, the charge doesn’t distribute

uniformly.

 (charge/area) varies.

Difficult to find E!

Application of Gauss’ Law

A charged isolated conductor

3. What is external E due to a charged conductor?

o enc E

d q



^{ }





 E A

ˆ1 3 n

nˆ

ˆ2

n

S₁ S₂

S₃

Φ = Φ₁ + Φ₂ + Φ₃ Φ₁=EA cos 0 = EA Φ₂=EA’ cos 90 = 0 Φ₃= 0 (E=0)

Φ = EA = ^{𝑞𝑒𝑛𝑐}

𝜀₀

𝜎 = ^{𝑞𝑒𝑛𝑐}

𝐴

E= ^𝜎

𝜀_𝑜

What happens when we bring charge

near the conductor?

Application of Gauss’ Law

A charged isolated conductor

E= ^𝜎

𝜀_𝑜

Application of Gauss’ Law

A charged

isolated

conductor

Application of Gauss’ Law

3. What is external E due to a non-conducting sheet?

o enc E

d q



^{ }





 E A

Φ = Φ₁ + Φ₂ + Φ₃ Φ₁=Φ₂= EA cos 0 = EA Φ₃=EA’ cos 90 = 0

Φ = 2EA = ^{𝑞𝑒𝑛𝑐}

𝜀₀

𝜎 = ^{𝑞𝑒𝑛𝑐}

𝐴

E=

^𝜎

2𝜀_𝑜

Planar symmetry Non-conducting

sheet

Application of Gauss’ Law

3. What is external E due to two conducting sheets?

E=

^2𝜎1

𝜀_𝑜

=

^𝜎

𝜀_𝑜

Planar symmetry Two conducting

plates

E= ^𝜎1

𝜀_𝑜

E= ^𝜎1

𝜀_𝑜 E= ^𝜎1

𝜀_𝑜

E= ^𝜎1

𝜀_𝑜

S

S' A

A'

E=

^{2𝜎1−2𝜎1}

𝜀_𝑜

= 0

Application of Gauss’ Law

Application of Gauss’ Law

Application of Gauss’ Law

3. What is external E due to a line of charge?

o enc E

d q



^{ }





 E A

Φ = Φ₁ + Φ₂ + Φ₃ Φ₁=Φ₂= EA cos 90 =0 Φ₃=EA’ cos 0 = 2rh E Φ = 2rh E = ^{𝑞𝑒𝑛𝑐}

𝜀₀

= ^{𝑞𝑒𝑛𝑐}

ℎ

E= 

2^𝜀_𝑜^𝑟 Cylindrical

symmetry Line of charge

S₁ ¹

nˆ

ˆ2

n

S₂

S₃

ˆ3

n

Application of Gauss’ Law

Application of Gauss’ Law

Application of Gauss’ Law

3. What is external E due to a charged conducting sphere?

o enc E

d q



^{ }





 E A

Spherical symmetry Charged

conducting sphere

ˆ1

n Ei

ˆ2

n E0

Inside the shell (r<R):

Φ = 4𝜋𝑟²𝐸_𝑖 = 𝑞_𝑒𝑛𝑐

𝜀_𝑜 = 0 Outside the shell (r>R):

Φ = 4𝜋𝑟²𝐸_𝑜 = 𝑞_𝑒𝑛𝑐

𝜀_𝑜 = 𝑞 𝜀_𝑜 𝐸_𝑜 = 𝑞

4𝜋𝜀_𝑜𝑟²

Shell Theorem

Application of Gauss’ Law

Spherical symmetry Charged non- conducting sphere

S₁

Eo

ˆ1

n

Inside the shell (r<R):

Φ = 4𝜋𝑟²𝐸_𝑖 = 𝑞_𝑒𝑛𝑐 𝜀_𝑜 𝑞_𝑒𝑛𝑐

4 𝜋𝑟3 ³ = 𝑞

4 𝜋𝑅3 ³ 𝐸_𝑖 = 𝑞

4𝜋𝜀_𝑜𝑅³ 𝑟 Outside the shell (r>R):

Φ = 4𝜋𝑟²𝐸_𝑜 = 𝑞_𝑒𝑛𝑐

𝜀_𝑜 = 𝑞 𝜀_𝑜 𝐸_𝑜 = 𝑞

4𝜋𝜀_𝑜𝑟²

S₂

Ei

ˆ2

n

R

3

4 0

q

 R

r E

O

Application of Gauss’ Law

A non- conducting line

of charge

Conductor (surface charge

density )

Non- conducting

sheet (surface charge

density )

Conducting sphere

Non- conducting

sphere

Comparisons…

l Q/

   _{Q/ A}  Q/ A

k r E _e 

 2

 0



in

o out

E

E 



o

E 



 2

R r R r k Q E

R r r

k Q E

e e









, ,

3 2

R r E

R r r

k Q E _e







 , 0

2 ,

•

Electric flux 

•

Using Gauss’ law to calculate E

•

Applying Gauss’ law to calculate:

• E for a non-conducting line of charge

• E for a conducting surface and between two conducting surfaces

• E for a non-conducting sheet

• E for a conducting sphere

• E for a non-conducting sphere

What we have learnt…