Gauss Law Lecture Notes

8/11/2019 Gauss Law Lecture Notes

1/13

!"#$$% '"(

)*+,-./ 0-"1.23 45/ 67 8 9

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4

:;?2$/

0"@=#@".2 .-2 2@2=.3+= A#B .-3C#,- " $#3D"=2E

F$2 !"#$$% '"( .C ="@=#@".2 .-2 2@2=.3+= G2@H

CD $IJJ2.3+= =-"3,2 H+$.3+;#>C*$E

F*H23$."*H -C( .C 32=C,*+$2 "*H #$2$IJJ2.3I +* !"#$$% '"( ="@=#@">C*$E

F$2 !"#$$% '"( .C #*H23$."*H .-2 13C123>2$

CD =C*H#=.C3$ +* "* 2@2=.3+= G2@HK

Electric Flux and Gauss' Law:28.3

Gauss' law ! method for calculatingE-field for even

quite complex charge distributions, provided they havereasonable degree of symmetry. Gauss' law relies onconcept of electric flux.

Define Electric Flux "

through surface S:

is vector normal to surface withmagnitude equal to area of A

= # of field lines passingthrough areaA

!

A

!

E

"

"

!

A

" =

!

E .

!

A =EAcos#


2/13

Electric Flux and Gauss' Law:28.3

!

A

!

E

"=0

" =

!

E .

!

A =EAcos#

" = EA cos#= EA

!

A

!

E

"=90!

" = EA cos#=0

For an arbitrary surface:

For a closed surface whichdoes not enclose anycharges:

# of field lines in is equal to# of field lines out

so net flux "= 0.

(surface normal points outwards everywhere)

" =!

E .d!

A#

d!

A

!

E

d!

A1

d!

A2

!

E

!

E!ve fluxcomponents

+ve fluxcomponents

For a closed surface thatdoes not enclose anycharges:

Negative flux contributionsfrom field lines entering

surface must exactlybalance positive fluxcontributions from fieldlines exiting surface.

d!

A1

d!

A2

!

E

!

E

-ve "

+ve "

0C*=21. L2$./

M ;#N23AI *2. +$ +* "

#*+DC3J 2@2=.3+= G2@HE !E

" $ $ - C ( * + * . - 2

"==CJ1"*I+*, G,#32K

L-2 3+JE " =+3=@2 CD 3"H+#$

aE + $ " @ + , * 2 H

12312*H+=#@"3 .C .-2

2@2=.3+= G2@HK O+*H "*

2B132$$ +C* DC3 .-2

2@2=.3+= A#BE !E "#$%&'#

(&)" "#* +*,+'-


3/13

Gauss' Law: 28.4

Electric flux, ", through arbitrary closed mathematical

surface with charge qwithin is:

For charge qenclosed withinsurface, # of field lines emerging thrusurface is q/ #

0

Charges outside volumecontribute no net flux

(integral over closed surface)

(NB. use signed value for qso that"can be +ve or -ve)

" =!

E .d!

A# =q

$0

q

+

+

q

Applying Gauss' Law:28.5

E-field for -ve point charge:

-

Choose spherical surface (Gaussiansurface) of radius r with point charge atcentre:

(- sign sinceEand dAopposite in direction)

NB. Problems usually simplified by careful choice of Gaussian surface, i.e.choose one for which field lines parallel or perpendicular to surface normal(s).

" =!

E .d!

A# = $E d!

A =# $q

%0

"q

"#E4$r2=

#q

%0

" E =1

4$%0

q

r2

d!

A!

E

0C*=21. L2$./

M 1C+*. =-"3,2 +$ 1@"=2H ".

C*2 =C3*23 CD " =#;2 CD $+H2

@2*,.- E "$ $-C(* +* .-2

"==CJ1"*I+*, H+",3"JK F$2

!"#$$% @"( "*H $IJJ2.3I

"3,#J2*.$ .C G*H .-2 "%"./

*/*0"$10 2&3 "#$%&'# "#*

0&4*K

a

+q

a

+q

4 cubes behind and 4 in front willcompletely surround charge q

E-field for line of charge:

linear charge density $C/m

d!

Aend2

d!

Aside

d!

A

end1

+

+

+

+

+

+

+

+

!

E

Eperpendicular to dAfor ends ofGaussian surface so they do notcontribution to flux.

Choose cylindrical Gaussiansurface to match symmetry ofcharge distribution (radial field).

r

h

" =!

E .d!

A# =q

$0

" E2#rh =$h

%0

" E =1

2#%0

$

r

Side area of cylinder

Charge in cylinder


4/13

E-field for large flat insulating sheet of charge:

Choose cylinder with axis perpendicular to plane of charge:

Fig.28.27 Knight, Physics for Scientists and

Engineers, 2ndEd.

" =!

E .d!

A# =q

$0

Eperpendicular todAfor side ofGaussian surfaceso only endscontribution to flux.

" EA + EA =#A

$0

" E =#

2$0

Areal charge density %

Gauss' Law and Symmetry: 28.1

Gauss' Law can be directly applied in situations thathave sufficient symmetry:

! Cylindrical (e.g. Wire)

! Spherical! Planar (e.g. Sheet of charge)

Fig.28.6 Knight, Physics for Scientists and Engineers, 2 ndEd.

Gauss' Law and Symmetry:

NB. It is desirable if possible to find a Gaussiansurface in which the surface normal is eitherperpendicular to or parallel to the electric field vectorat each point on the surface:

!

E

"=0

!

A

!

E

"=90!

!

A

Gauss' Law and Symmetry:

Gauss' Law gives the electric field at a Gaussiansurface arising from the charge, q, enclosed withinthe surface.

To get the resultant E-field at some point P in space,

need to consider influence of all charge distributions,both inside and outside the Gaussian surfacechosen, e.g. "


5/13

E=

20

sheet A

20

sheet B

= 0

For asymmetric distributions: Gauss' Law + Superposition

can sometimes be applied.

+

+

+

+

+

Sheet A

+

+

+

+

+

Sheet B

P

Gauss AGauss B

+

+

+

+

+

Sheet A

+

+

+

+

+

Sheet B

P

Symmetric Case with Two Charged Sheets:

" =!

E.d!

A# =q

$0

%EA +EA =&A +&A

$0

%E=&

$0

E.g. solid sphere with uniform charge density &C/m3

throughout volume (total charge Q)sphere

Spherical Gaussian surface

r " R : q =Q

4

3#R

3

4

3#r3 =Q

r 3

R3

"=

!

E .d

!

A#=

q

$0

"E4#r2=

Q

$0

r3

R3"E =

Q

4#$0

r

R3

(linear with r)

R

r

E.g. solid sphere with uniform charge density &C/m3

throughout volume (total charge Q)sphere

Gaussian surface

r" R :q =Q

"=

!

E .d

!

A#=

q

$0

" E4#r2=

Q

$0

" E =1

4#$0

Q

r2

(same as for field vs rfor point charge Q)

i.e. inverse square dependence as for pointcharge.

R

r


6/13

For concentric spherical cavityin sphere,E= 0within empty interior volume. Why?

" r # R : E=Q

4$%0

r

R3=

1

4$%0

q(r)

r2

r " R : q =Qr 3

R3# q r( ) =Q

r 3

R3

qas function of r

Same form as expression for r > R now.

Eat ronly depends on qinside radius rand if qis zero then so is E.

NB. Need to invoke symmetry arguments to show that the spherical shellcannot contribute to an electric field anywhere inside the cavity.

Exercise: Applying Superposition:

E.g. +ve charge Qdistributed uniformly over volume of solid sphere of radius R.Then spherical cavity, radius R/2, cut from sphere and associated material/chargediscarded.

Use superposition arguments to show thatE-field at Pis given by:

E=Q

4"#0

1

r2$

1

8 r $R /2( )2

%

&''

(

)** R

R/2

Superposition! field at Pgiven by:

R

R/2

Conductors in Electric Fields: 28.6

In conductor, large # of free electrons available!when

conductor placed inE-field, free electrons move inopposite direction to field leaving one surface +velycharged and other -vely charged:

conductor - - - - -

!

Eext


7/13

- - - - -

+ + + + +

Charge separation generates internalE-fieldEint

opposite

in direction to the external field.

Separation of chargecontinues until:

- - - - -+ + + + ++

- At equilibriumEnet= 0everywhere insideconductor

!

E int

!

Eext

!

Eint=

!

Eext

!

Enet=0

Similarly, excess charge on conductor resides at thesurface:

+ +

+

+ +++

+ + + ++

(same equilibrium condition

applies)

This is also what we would expect based on chargerepulsion and minimisation of potential energy (seelater).

!

Enet=0

+ + ++ + +

Also, at equilibrium, field lines atsurface of conductor areperpendicular to the surface atevery point

(otherwise there would be net lateral force oncharges and they would move)

E-field Outside Large Flat Conducting Plate:

++

+ ++ +

Charge density % C.m-2

E = 0inside conductor!only non-zero flux is thatthrough external end ofcylinder:

E-field is double that of insulating sheet with samecharge density. (Same # of charges per unit area !

same # of field lines BUT all field lines directedoutwards for conductor) OR consider that field outsideconductor has contributions from charges on top andbottom surfaces.

" EA + 0 =#A

$0

" E =#

$0

" =!

E .d!

A# =q

$0

!

E

= 0

!

E


8/13

E-field Outside Large Flat Conducting Plate:

Or, if you want to apply Gauss law in a symmetricalmanner:

" EA + EA =2#A

$0

" E =#

$0

" =!

E .d!

A# =q

$0

+ + + + + + + +

+ + + + + + + +

!

E

!

E

"

"

Charge on top and bottom surfaces

Hollow Conductor :

+

+

++

+

+

cavity

Since excess charge resideson surface andE=0everywhere inside conductor,must also be true for cavityanywhere inside conductor.

no net charge inside Gaussian surface

! metal shell or cage "shields" interior from electric field!

Faraday Cage

NB. The electrostatic potential will give us a better way of arguing that therecan be no electric field lines within an empty cavity in a conductor.

Faraday Cage:

Conductors shield their interior from E-field:

( Fig 25-19, Halliday, Resnick and Walker, Fundamentals Of Physics, Wiley2001)



9/13


E=0 inside metal cage

Charge Transfer in van der Graaf generator

++

+

+

+

++

+

!

!

!

!

Introduce charge into cavity ofconducting shell.

+


+

+

++

+

!

!

!

!

Connect inner conductor to shell.

Transfer charge to shell.

+ + +


+

++

+

To add more charge start the wholeprocess over again.

+


10/13



11/13

Three Minute Quiz:

R1

R2

R3

+Q

A solid conducting sphere of radiusR1is

enclosed within a concentric sphericalconducting shell of inner radius R

2and outer

radius R3, as shown. If charge +Qis

deposited on the central sphere:

(a) Sketch the equilibrium chargedistribution on the conductors and theelectric-field line distribution for all regions ofspace.(b) Use Gauss' law to find expressions forthe electric-field in each region and henceplot a graph of Evsradial distance r.

Lightning:

WhenE-field exceeds 3 x 106N/C (V/m) molecules inair can become ionised (neutral molecules broken into+vely and -vely charged ions):

+

-

these ions collide with neutral molecules! more ions

created! collision cascade ! spark.

E

E-Field of Earth:

Dry air E-Field of Earth ~ 100-200 V/m (N/C) pointingdownwards:

++

++ + + +

+

-- -

- - - --

-

Earth = Giant Spherical Capacitor

Lightning = discharge through volume of capacitor

Sources of Mobile Charges:

Upper Atmosphere: Cosmic Rays (very swift particlesfrom space, mostly protons) enter atmosphere andproduce dense collision cascades of charged particles(+ve and -ve).

Terrestrial: Ions produced by natural radioactive decayand other ionisation processes including induced

charge on droplets of water spray:

E

+ +++

-

+

--


12/13

Thunder Clouds

(From: http://littleswitzerlandweather.info/lightinfo/abtlightning.html)

Current understanding is that:Small ice particles are beingcharged positively and rapidlytransported upward.

~ +40 C

~ 40 C

Possible Charging Mechanisms

Ice particle polarised in E-field+ charge exchange with waterdroplet or another ice particle.

E -+

e-

Or -ve ions attracted towardsice particle and captured.

E -+

-

ground strike

Lightning Strike:

( Figs 24-13, 24-14, Halliday, Resnick and Walker, Fundamentals Of Physics,Wiley 2001)

Once column bridges gap, e- rapidly

transported:"current"e- collisions with air molecules"ionisation"

further currente- + air molecules

atom excitationspontaneous decayphotonsflash

- - - -

+ + + + +

Lightning:

Just prior to flash, electron avalanche descends to groundCharge density in column $~ -1 x 10-3C.m-1


13/13

and for line of charge

but visible region has radius ~ 0.5 m

"=1#10$3C/m

E=1

2"#0

$

r

"r =1

2#$0

%

E=

1

2#$0

1&10'3

3&106=6m

Discharge occurs first wherever E-field greatest:

+++

+++

+++ so hit the deck if this happensto you!

( Fig 25-24, Halliday, Resnick and Walker, Fundamentals Of Physics,Wiley 2001)

Concept Test

A cylindrical piece of insulating material is placed in an external electricfield, as shown. The net electric flux passing through the surfaceof the cylinder is

1. Positive2. Negative3. Zero

P#JJ"3I/

Q@2=.3+= O@#BK

!"#$$% '"(K

PIJJ2.3I "*H $#1231C$+>C*K

0C*H#=.C3 +* 2@2=.3+= G2@HK '+,-.*+*, "*H $."I+*, "@+?2K

R2"H )*+,-./ 0-"1.23 45

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Gauss Law Lecture Notes