Chapter 6 Gauss’ law

In this chapter we will introduce the following new concepts:

The flux (symbol Φ ) of the electric field. Symmetry Gauss’ law

We will then apply Gauss’ law and determine the electric field generated by: An infinite, uniformly charged insulating plane. An infinite, uniformly charged insulating rod. A uniformly charged spherical shell. A uniform spherical charge distribution.

We will also apply Gauss’ law to determine the electric field inside and outside charged conductors.

Gauss’s Law Gauss proposed that by summing electric field lines that penetrate a closed surface which surrounds a group of charges, one can determine the net charge enclosed in the surface.

Q

++ + +

+ +

++ +

++ +

++ +

++ +

++ +

++ +

This is called the divergence theorem in mathematics.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

Electric Flux

n

A

Φ=0 Φ = EA

A A

Φ = EAcosθ

n θ E E

E

Electric Flux Φ =!E ⋅!A

What is the net electric flux through

the cube given!E = E0 j ?

1)The x and z surfacesides do not contribute

to the flux where!E ⊥!A ( perpendicular).

2) At the surface!Ay=0 = −a2 j

Φ y=0 =!E ⋅!Ay=0 = (E0 j) i (−a2 j) = −a2E0

3) At the surface!Ay=a = +a2 j

Φ y=a =!E ⋅!Ay=a = (E0 j) i (a2 j) = +a2E0

4) Φnet = Φ y=0 +Φ y=a = 0

x

z

y

!E

x

a a

a

dA E dA

E dA

n

nn

Flux of the electric field. Consider the closed surface shown in the figure. In the vicinity of the surface assume that we have a known electric field

!E. The flux Φ of the

electric field through the surface is defined as follows: 1. Divide the surface into small "elements" of area ΔA2. For each element calculate the term

!E ⋅ Δ!A = EAcosθ

3. Form the sum Φ =!E ⋅ Δ!A∑

4. Take the limit of the sum as the area ΔA→ 0The limit of the sum becomes the integral:

Φ =!E ⋅d!A"∫ Flux SI unit: N ⋅m2 / C

Note 1 : The circle on the intergral sign indicates thatthe closed surface is closed. When we apply Gauss'law the surface is known as "Gaussian" Note 2 : Φ is proportinal to the net number of electric field lines that pass through the surface

Φ =!E ⋅d!A

S"∫

n

nn

Applying Gauss' Law

!E ⋅d!A =

q inc

εo"∫

o encqε Φ =

Note 1 : Gauss' law holds for any closed surface. Usually one particular surface makes the problem ofdetermining the electric field very simple. Note 2 : When calculating the net charge inside a closed surface we take into account the algebraic sign of each charge.Note 3 : When applying Gauss' law for a closed surface we ignore the charges outside the surface no matter how large they are.Example : Surface S1 : εoΦ1 = +q , Surface S2 : εoΦ2 = −q

Surface S3 : εoΦ3 = 0 , Surface S4 : εoΦ4 = −q + q = 0

Note : We refer to S1, S2, S3, S4 as "Gaussian surfaces"

S Q 1 ? 2 -2q 3 +q 4 -4q 5 -2q 6 +3q

What is the net charge in surface S1 ? A) +q B) 0 C) -2q D) +4q E) -4q

Recipe for applying Gauss’ Law 1. Make a sketch of the charge distribution.

2. Identify the symmetry of the distribution and its effect on the electric field.

3. Gauss’ law is true for any closed surface S. Choose one that makes the calculation of the flux Φ as easy as possible.

4. Use Gauss’ law to determine the electric field vector.

Φ =

qinc

εo

Surfaces of Symmetry Gauss’s Law is best implemented when we pick a surface over which the electric field takes a constant value so Can be easily evaluated.

!E id!A"∫

Gaussian pillbox or cylinder

Gaussian cylinder

+Q

Gaussian sphere

ndA

Gauss' law and Coulomb's lawHere we will derive Coulomb's law from Gauss' law.Consider a point charge q. We will use Gauss' lawto determine the electric field

!E generated at a point

P at a distance r from q. We choose a Gaussian surface which is a sphere of radius r and has its center at q.

Φ =!E i d!A = E"∫

Coulomb's law Gauss' law↔

We divide the Gaussian surface into elements of area dA.

The flux is: Φ =!E i d!A = E"∫ dA ="∫ E 4πr 2( ) (

!E # d

!A on the surface!)

From Gauss' law we have: Φ =qinc

εo

= qεo

→ E(4πr 2 ) = qεo

→ E = q4πεor

2

This is the same answer we got in Chapter 5 using Coulomb's law

1n

2n

3n S1

S2 S3

Electric field generated by a thin, infinite, non - conducting uniformly charged sheet.We assume that the sheet has a positive charge ofsurface density σ . From symmetry, the electric fieldvector

!E is perpenducular to the sheet and has a

constant magnitude. Furthermore it points away fromthe sheet. We choose a cylindrical Gaussian surface Swith the caps of area A on either side of the sheet asshown in the figure. We devide S into three sections:S1 is the cap to the right of the sheet. S2 is the curved

surface of the cylinder. S3 is the cap to the left of the

sheet. The net flux through S is Φ = Φ1 +Φ2 +Φ3

Φ1 = Φ2 = EAcos0 = EA. Φ3 = 0 (θ=90°)

→Φ = 2EA. From Gauss's law we have: Φ =qenc

εo

= σ Aεo

→ 2EA = σ Aεo

→ E = σ2εo

2 o

Eσε

=

S1 1n

2nS2

S3 3n

Consider the long rod shown in the figure. It is uniformly charged with linera charge density . Using symmetry arguments we can show that th

λ

Electric field generated by a long, uniformly charged rod

e electric field vector points radially outwards and has the same magnitude for points atthe same distance from the rod. We use a Gaussian surfaceS that has the same symmetry. It is a cylinder of r

radius

and height whose axis coincides with the charged rod. r

h

We divide S into three sections: Top flat section S1. Middle curved section S2.

Bottom flat section S3. The net flux through S is Φ = Φ1 +Φ2 +Φ3.

Fluxes Φ1 = Φ3 = 0 because the electric field is at right angles with the

normal to the surface. Φ2 = EAcosθ = E(2πrh)cos0 = 2πrhE

Φ = 2πrhE From Gauss's law we have: Φ =qinc

εo

= λhεo

If we compare these two equations we get: 2πrhE = λhεo

→ E = λ2πεor

2 o

Er

λπε

=

The Electric Field inside a Conductor1)Any free charge q inside a conductor will be forcedto the surface by the other charges until staticequlibrium is established. σ = q / A may not be uniform!2) All charges must sit on the surface such that theaction-reaction

forcesare balanced Fi, ji, j∑ =0 (charges i,j). (no motion)

3) The flux through a Gaussian surface placed just inside the

conductor's wall will be !E ⋅d!A∫ = 0 ! qinside = 0

4) Since!E " d

!A and d

!A ≠ 0 then

!Einside = 0

1)The electric field !E inside a conductor is always zero.

2)No charges can accumulate on any interior cavity. 3) The elecric field vector

!E is perpendicular to the conductor surface.

If it were not then !E would have a component parallel E" to the

conductor surface and chargeswould not be static.

+

E!

F!

+ + + +

+ + +

+ +

+ + + +

+ +

+ +

+ +

+ +

+

S

S'

A

A'

Let the surface charge densities σ 1 and -σ 1. When the two plates are moved close

to each the charges one plate attract those on the other moving to the inner faces of each plate. We apply Gauss' law for the cylindrical surface S which has caps of area A.

The net flux Φ = Ei A =qinc

εo

=2σ 1Aεo

→ Ei =2σ 1

εo

To find the field Eo outside

the plates we apply Gauss' law for the cylindrical surface ′S which has caps of area A.

The net flux Φ = Eo A =qinc

εo

=σ 1 −σ 1

εo

= 0→ Eo = 0

Parallel Plates of Infinite Size(far apart)

(together)

Spherical Shell of charge q of radius R :

Inside the shell : Consider a Gaussian surface S1

which is a sphere with radius r < R and whose center coincides with that of the charge shell. S1 :Inside the shell : The electric field flux

Φ = 4πr 2Ei =qinc

εo

= 0 Thus Ei = 0 r < R

S2 :Outside the shell : Consider a Gaussian surface S2

which is a sphere with radius r > R and whose center coincides with that of the charge shell.

The electric field flux Φ = 4πr 2Eo =qinc

εo

= qεo

Thus Eo =q

4πεor2 r < R

Note : Outside the shell the electric field is the same as if all the charge of the shell were concentrated at the shell center.

1n iE!

2n oE!

0iE =

24oo

qE

rπε=

S2

S1

iE!

2n

oE!

1n

Electric field generated by a uniformly charged sphere of radius R and charge q Outside the sphere : Consider a Gaussian surface S1

which is a sphere with radius r > R and whose center coincides with that of the charge shell.

The electric field flux Φ = 4πr 2Eo = qinc / εo = q / εo

Thus Eo =q

4πεor2

Inside the sphere : Consider a Gaussian surface S2

which is a sphere with radius r < R and whose center coincides with that of the charge shell.

The flux Φ = 4πr 2Ei =qinc

εo

qinc =4πr3

4πR3 q = r3

R3 q

4πr 2Ei

Φ!"#

= r3

R3

qεo

qinc /εo

$→ Ei =

q4πεoR3

⎛

⎝⎜⎞

⎠⎟r

R

34 o

qRπε

r

E

O

PRACTICE

x

y

z E=(0, 4000, 3000) N/C

dA=(0 ,0, 12.6)m2

Φ=E�dA=(0, 4000, 3000) �(0, 0, 12.6)= 0x0+4000x0+3000x12.6 N-m2/C Φ=3.7x104 N-m2/C

dA=πr2=12.6 m2

ι r

Thefluxonlypenetratestheoutcansurface.

Φ = Er idA= (λ

2πε01r)

Er

!"# $#(2πrℓ)

dA!"$ = λℓ

ε0

Er dA

Electric Flux Through a Cube The edge of a Gaussian cube is placed at a distance y=2a from the origin. The cube sits in an electric field E = 5i + 10yj - 2k. Find the net electric flux through the cube and the net charge inside.

Φ = Σ E�dA (6 sides)E�dA1 = (5,10y,-2)�(0,-a2,0)= -20a3 y=2aE�dA2 = (5,10y,-2)�(0,+a2,0)= +30a3 y=3aE�dA3 = (5,10y,-2)�(+a2,0,0)= +5a2 y=2a to 3aE�dA4 = (5,10y,-2)�(-a2,0,0)= -5a2 y=2a to 3aE�dA3 = (5,10y,-2)�(0,0,+a2)= -5a2 y=2a to 3aE�dA4 = (5,10y,-2)�(0,0,-a2-)= +5a2 y=2a to 3a

Φ = Σ E�dA = 10a3=q/εoΦ =10a3 q=10a3εo

1)InsidethepipethereisnochargesobyGauss'sLawΦ =0and E=0.2)OutsidethepipeΦ = Er dA= Er (2πrℓ)

Φ =qincε0

Gauss 's Law

Er (2πrℓ)=σ (2πaℓ)

ε0

Er =σε0

ar

Er

dA

75) (a) Calculate the electric flux through the openhemispherical surface due to the electric field E = E0k(see below). (b) If the hemisphere is rotated by 90° around the x-axis, what is the flux through it?

Φ =!E id!A

S"∫ = Eok i d

!A∫ = Eok i( !Adisk +

!Adome )

!Adisk = −(πR2)k

!Adome = +(2πR2)

1/2 x4πR2#$%k

(a) Φ =EoπR2

(b)Φ =0becauseallfieldlinesentering alsoexitresultinginzeroflux.

85. A non-conducting spherical shell of inner radius a1 and outer radius b1 is uniformly charged with charged density ρ1 inside another non-conducting spherical shell of inner radius a2 and outer radius b2 that is also uniformly charged with charge density ρ2 . See below. Find the electric field at space point P at a distance r from the common center such that (a) r > b2, (b) a2 < r < b2, (c) b1 < r < a2, (d) a1 < r < b1, and (e) r < a1 .

(a)r>b2 E4πr2= ρ1 +q2)ε0

(b)a2<r<b2 E=k(q1 +Δq2)r2

Δq2 = q2(4 3πr2 −4 3πa224 3πb22 −4 3πa22

)

E =k[q1 +q2(

r2 −a22

b22 −a2

2 )]

r2

88. Concentric conducting spherical shells carry charges Q and –Q, respectively (see below). The inner shell has negligible thickness. Determine the electric field for (a) r < a; (b) a < r < b; (c) b < r < c; and (d) r > c.

(a) r <a E(4πr2)= 1εo(0) ⇒ E =0

(b)a< r <b E(4πr2)= 1εo(+Q) E = k Q

r2

(c)b< r < c E(4πr2)= 1εo(+Q−Q) E =0

−Q sits on inner surface

(d) r > c E(4πr2)= 1εo(+Q−Q) E =0

Net inside charge is zero

86. Two non-conducting spheres of radii R1 and R2are uniformly charged with charge densities ρ1 and ρ2, respectively. They are separated at center-to-center distance a (see below). Find the electric field at point P located at a distance r from the center of sphere 1 and is in the direction θ from the line joining the two spheres assuming their charge densities are not affected by the presence of the other sphere. (Hint: Work one sphere at a time and use thesuperposition principle.)

!E1 =|

kq1r2|(cosθ ,sinθ )

r" #$ %$q1 = ρ1V1

!E2 =|

kq2r '2 |(−cosϕ ,sinϕ) q2 = ρ2V2

E =|kq1r2|(cosθ ,sinθ )+|kq2

r '2 |(−cosϕ ,sinϕ)

Let q1 = q2 r = a/2i (center point)

E =|kq1r2|(1,0)+|kq2

r '2 |(−1,0)=0

φ r’

STOP HERE

24oo

qE

rπε=

34io

qE r

Rπε⎛ ⎞

= ⎜ ⎟⎝ ⎠

R

34 o

qRπε

r

E

O

S2

S1

iE!

2n

oE!

1nElectric field generated by a uniformly charged sphere of radius R and charge q. Summary

e E!

v!F!

We shall prove the the electric field inside a conductor vanishesConsider the conductor shown in the figure to the left. It is an experimental fact that such an o

The electric field inside a conductor

bject contains negatively charged electrons which are free to move inside the conductor. Lets assume for a moment that the electric field is not equal to zero.

In such a case an non-vanishing force F!

is exerted by thefield on each electron. This force would result in a non-zero velocity and the moving electrons would constitute an electriccurrent. We will see in subserquent chapters tha

eE

v

= −!

!

t electric currents manifest themselves in a variety of ways: a) they heat the conductorb) they generate magnetic fields around the concuctor No such effects have ever been observed. Thus the original assumption that there exists a non-zero electric field inside the conductor. Thus we conclude that :

The electrostatic electric field inside a conductor is equal to zeroE!

A charged isolated conductor :Consider the conductor shown in the figure that has a total charge q. In this section we will ask the questionwhere is this charge located. To answer the question we will apply Gauss' law to the Gaussian surface shown in the figure which is located just below the conductor surface. Inside the conductor the electric field

!E = 0

Thus Φ =!E ⋅ Δ!A"∫ = 0 (eqs.1)

From Gauss's law we have: Φ =qenc

εo

(eqs.2)

If we compare eqs.1 with eqs.2 we get: qenc = 0

Thus no charge exists inside the conductor. Yet we know that the conductorhas a non-zero charge . Where is this charge located? There is only one placefor it to be: On the surface of the conductor

q.

electrostatic charges can exist inside a conductor. charges reside on the conductor surface

NoAll

An isolated charged conductor with a cavity :Consider the conductor shown in the figure that has a total charge q. This conductor differs from the one shown in the previous page in one aspect: It has a cavity. We ask the question whether charges can reside on the walls of the cavity. As before Gauss's law provides the answer. We will apply Gauss' law to the Gaussian surface shown in the figure which is located just below the conductor surface. Inside the conductor the electric field

!E = 0

Thus Φ =!E ⋅ Δ!A"∫ = 0 (eqs.1)

From Gauss's law we have: Φ =qenc

εo

(eqs.2)

If we compare eqs.1 with eqs.2 we get: qenc = 0

Conclusion :There is no charge on the cavity walls. All the excess charge remains on the outer surface of the conductor

q

Symmetry: We say that an object is symmetric under a particular mathematical operation (e.g. rotation, translation, …) if to an observer the object looks the same before and after the operation. Note: Symmetry is a primitive notion and as such is very powerful

featureless sphere

rotation axis

observer Rotational symmetry Example of spherical symmetry

Consider a featureless beach ball that can be rotated about a vertical axis that passes through its center. The observer closes his eyes and we rotate the sphere. Then, when the observer opens his eyes, he cannot tell whether the sphere has been rotated or not. We conclude that the sphere has rotational symmetry about the rotation axis.

Featureless cylinder

rotation axis

observer

Rotational symmetry A second example of rotational symmetry

Consider a featureless cylinder that can rotate about its central axis as shown in the figure. The observer closes his eyes and we rotate the cylinder. Then, when he opens his eyes, he cannot tell whether the cylinder has been rotated or not. We conclude that the cylinder has rotational symmetry about the rotation axis

observer

magic carpet

infinite featureless plane

Translational symmetry

Example of translational symmetry:

Consider an infinite featureless plane. An observer takes a trip on a magic carpet that flies above the plane. The observer closes his eyes and we move the carpet around. When he opens his eyes the observer cannot tell whether he has moved or not. We conclude that the plane has translational symmetry