Chapter 23Chapter 23

GaussGauss’’s Laws Law

GaussGauss’’s Law:s Law:--It is a new formulation of coulombIt is a new formulation of coulomb’’s law which take s law which take advantage of symmetry situationsadvantage of symmetry situations

Gaussian Surface:Gaussian Surface:--Can be any surface Can be any surface (shape) you wish to (shape) you wish to make, but must be make, but must be useful, means mimics useful, means mimics the symmetry of the the symmetry of the problem.problem.It must be a closed It must be a closed surface (know what is surface (know what is inside, outside and on inside, outside and on the surface).the surface).

Flux ( Flux ( φφ ))

Flowratemsm

VA

==

=

φφ

φ2./

Flux Flux φφIf V If V ┴┴ A A φφ=VA=VAIf VIf V∥∥A A φφ=0=0If V has as angle with A If V has as angle with A φφ=V =V coscosθθ AAWe define an area vector We define an area vector with magnitude equal to the with magnitude equal to the area & direction area & direction ⊥⊥ to A.to A.If If θθ=90 =90 φφ=V cos90 A=0=V cos90 A=0If If θθ=0 =0 φφ=V cos0 A=VA=V cos0 A=VA

Flux Flux φφ

φφ=V =V coscosθθ AA

φφ= V.A or for vector E = V.A or for vector E φφ= E.A = E.A

If AIf A∥∥E E θθ=90 =90 φφ=0=0If A If A ⊥⊥ E E θθ=0 =0 φφ=EA=EAIf A has an angle with E If A has an angle with E φφ=Ecos =Ecos θθAA

PositivePositiveZeroZeroNegativeNegativeSign of Sign of flux (E.flux (E.∆∆A)A)

Out of Out of the the surfacesurface

Parallel to Parallel to the the surfacesurface

In to the In to the surfacesurface

Direction Direction of Eof E

θθ << 9090θθ=90=90θθ >> 9090θθ

∫=

surfaceClosed

dAEE .φ

Example 1 :Example 1 :--

The figure shows a Gaussian surface in the The figure shows a Gaussian surface in the form of a cylinder of radius R immersed in a form of a cylinder of radius R immersed in a uniform E with the cylinder axis parallel to uniform E with the cylinder axis parallel to the field. What is the field. What is φφthrough this closed through this closed surface?surface?

EClosedsurface

E .dAφ = ∫

∫ ∫ ∫++=a

bcdAEdAEdAE ...φ

Solution:Solution:--

2

.

.

180cos.

REEA

dAEdAE

EdAdAE

AEdAE

aa

aa

a a

Π−=−=

−=

−=

=

∫ ∫

∫ ∫

∫ ∫

∫ ∫ ==b b

AEdAE 090cos.

20cos. REEAAEdAEc c

Π===∫ ∫

022 =Π+Π−= REREφ

GaussGauss’’s Laws LawThis law relates the net flux (This law relates the net flux (φφ ) )

of an electric field through a of an electric field through a closed surface closed surface ““ Gaussian Gaussian

SurfaceSurface”” to the net charge that is to the net charge that is enclosed by that surface ( Q enclosed by that surface ( Q encenc ) )

FORMULAFORMULA

E =( I/4E =( I/4πεπεοο)) (q/R(q/R22 ))

φφ == EA = (EA = (I/4I/4πεπεοο)) (q/R(q/R22 ) () (44πεπεοοRR22 ) = ) = QQencenc//εεοο

φφ = = E.dAE.dA ==QQencenc //εεοο

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∫

NOTESNOTES

QQencenc : is the algebraic sum of all the enclosed : is the algebraic sum of all the enclosed positive, negative or zero positive, negative or zero The sign of ( The sign of ( QQencenc ) must be included < ) must be included < because it tell us something about the net because it tell us something about the net flux through the Gaussian Surface flux through the Gaussian Surface The exact form or location of ( The exact form or location of ( QQencenc ) is no ) is no concern concern EE : is the electric field resulting from all : is the electric field resulting from all charges both inside and outside the Gaussian charges both inside and outside the Gaussian SurfaceSurface

Example 1Example 1

The figure shows five charged plastic lump The figure shows five charged plastic lump and an electrically natural coin. the cross and an electrically natural coin. the cross section is Gaussian surface (S) is section is Gaussian surface (S) is indicated. What is the net electric flux (indicated. What is the net electric flux (φφ) ) through the surface if qthrough the surface if q11=q=q33=+3.1 =+3.1 nCnC , , qq22=q=q55=+5.9 =+5.9 nCnC , and q, and q44==--3.1 3.1 ncnc

coin

SOLUTIONSOLUTION

φ φ =q1+q2+q3/=q1+q2+q3/εε00

=q2/=q2/εε00

==--5.9x105.9x10--

99/8.85x10/8.85x10--1212

==--670 N.m670 N.m22/c/c

* E .d Aφ = ∫0* E.dA q /φ ε= =∫

Example : with uniform fieldExample : with uniform field

The figure shows a The figure shows a GaussainGaussain surface in surface in the form of the cylinder of radius R the form of the cylinder of radius R immersed in a uniform E with the immersed in a uniform E with the cylinder axis parallel to the field . What cylinder axis parallel to the field . What is the through this closed surface ?is the through this closed surface ?

Φ

a b c2

a a a a

b b2

c c c2 2

E dA E dA E dA E dA

E dA E cos180 dA EdA E dA EA E

E dA E cos90 dA zero

E dA E cos0 dA EdA EA E

E 0 E zero

R

R

R R

Φ

π

π

Φ π π

= ⋅ = ⋅ + ⋅ + ⋅

⋅ = →− →− →− =−

⋅ = =

⋅ = → → =

∴ =− + + =

∫ ∫ ∫ ∫

∫ ∫ ∫ ∫

∫ ∫

∫ ∫ ∫

Example: with non uniform Example: with non uniform fieldfield

A non uniform E given by E = 3xi + 4j, A non uniform E given by E = 3xi + 4j, pierces the Gaussian cube shown in the pierces the Gaussian cube shown in the figure. What is the figure. What is the φφ through the right through the right face, the left face and the top face ?face, the left face and the top face ?

m

m

m

CN

dAdAjjxidAECNA

xdAdAijxidAE

CNAdAdAx

xdAdAijxidAE

ttop

l llleft

rr

r rrright

2

2

2

16

4)()43(

12)4)(3()1)(3(

3)()43(

36)4)(9(9)3)(3(3

3)()43(

⋅=

=⋅+=⋅=

⋅−=−=−=

−=−⋅+=⋅=

⋅=====

=⋅+=⋅=

∫ ∫∫Φ

∫ ∫∫Φ

∫∫

∫ ∫∫Φ

Charge on a conductor:Charge on a conductor:--

Charge will be on the surface of the Charge will be on the surface of the conductor .conductor .E=0 inside the conductorE=0 inside the conductor§§ E.dA = qE.dA = qenc enc

єєoo

Since E=0 Since E=0 ØØEE =0=0But q is not zero But q is not zero ……....Thus q must be on the out side surface of the conductor not Thus q must be on the out side surface of the conductor not

inside S. inside S.

conductor ++

+

+++

+

++

E=0

S

No charges on the inside surface of cavity.No charges on the inside surface of cavity.

sincesince E = 0 E = 0 ØØE E = 0 q must be on the out side = 0 q must be on the out side of the conductor.of the conductor.

+++

+ ++++

E=0

cavityS

Example :Example :--

Find E in a distance R from a uniform line Find E in a distance R from a uniform line charge distribution of infinite length the charge distribution of infinite length the charge per unit length is charge per unit length is λλ ..

using Gausses law. using Gausses law. ωω

E =?R

+

+

++

+

+

-ω

The answer:The answer:--§§E.dA = qE.dA = qenc enc EL2EL2ππR = R = λλ LL

єєo o єєoo

λλ == q E q E = = λλL L 22ππRRєєo o

§§E.dA = E.dA = §§E.dA + E.dA + §§E.dA + E.dA + §§E.dA E.dA

§§E.dA = E.dA = §§E.dA = 0E.dA = 0

§§E.dA = E.dA = §§E cos0 dA = E cos0 dA = §§EdA EdA

E E §§dA = EL2dA = EL2ππRR

a b c

a cCos 90 = 0

b b bsymmetry

b

+

+

++

-ω

ω

S

R R

E

A

E

A

A

Ea

c

b

For infinite line charge

Example:Example:--

Find E due to a non conducting Find E due to a non conducting infinite infinite

Plane with uniform charge per unit Plane with uniform charge per unit area area σσ ..

Using Gausses law.Using Gausses law.

Answer:Answer:--§§E.dA = qE.dA = qenc enc

2EA2EA = = σσAA

a b c

Cos 0=1Cos90=0Cos 0=1

a c

+

++

++

+

dA

E

dA

E

dA

E

c b

a

+

+

+

Conductors :Spherical

1- Spherical Shella) A shell of uniform charged attracts or repels a charged particle that is outside the shell as if all the shell’s charged were concentrating at the center of the shell.

Using Gauss’s law :

0

2

2

0

2

4

4

4 o

rr

r

π

π

πε

ε

ε

=∫=

= = =∫ ∫ ∫=

=

Q encE .dEQ enc qE .dE E dA cosφ E dA E

qEqE

q

S

E=?R++

+

++

+++

+

EdA

b) A shell of uniform charged exerts no electrostatic force on a charged particle that is located inside the shell.

Using Gauss’s law :

0ε=∫==∫

⇒ = =

QencE.dAS2Qenc zeroEdA zeroA#zero E 0,F 0

q

S1

R

EdA

S2E=?

2- Spheres :

a) A conducting sphere : 2o4 rπε

= qE

b) Non_conducting sphere :

* If r > a (outside the sphere):

* If r = a (on the sphere) :

* If r < a (inside the sphere) :

2o3 rε

=3ρ aE

3ε 0

= ρaE

2o4 rπε

=

3rQ aE or0ε= ρ rE 3