2. Electric Fields and Gauss’ Law

8/9/2019 2. Electric Fields and Gauss Law

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e 2

q = k

rE r

r

When calculating the electric field due to a

continuous charge distribution, we approach

the problem by dividing up the distribution

into tiny elements of charge q. Each of

these elements can be treated as a point

charge, which will yield an electric field of

The total field can then be approximated as

the sum off all such elements:

ie 2

i

qk

ri} iE r

r


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Sinc w hav a continuous dist ibution, w can g t a mo xact

valu by l tting th siz of th cha g l m nts app oach z o:

i

i

0i

dlim

ip

! ! iE r rr

wh re the integralisoverthe entire charge distribution. A

commonconditionofthese integralsisthatthe charge

distributionisuniform (constantcharge density). Inthiscase we

canma e the followingsubstitutions:

Ifthe charge Q isdistributedoveravolume V,we candefine the

volume charge densityas

Q/V


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Ifthe charge Q isdistributedoverasurface area A,we can

define the surface charge densityas

Q/A

Ifthe charge Q isdistributedoveraline oflengthl,we can

define the linearcharge densityas

Q/l

The amountsofcharge inadifferential elementofcharge d fora

volume,surface,orline thenbecome

d = dV, d = dA, d = dl


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Example: A rodoflengthl hasacharge Q uniformlydistributed

throughoutitslength. Findthe electricfieldalongthe atapoint

alongthe axisofthe rodadistance a awayfromone endoftherod.

Treating each elementasapointcharge,the electricfieldfor each

elementwillpointinthe xdirection. Since anintegralisbasically

asumofelementswe canassume thatallthe elementswillhave

fieldsthatwilladdupinthe same directionandthusthe total

electricfieldwillpointinthe xdirection (thiswillallowusto

avoidintegratingoverthe unitvectors).


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Each elementwillhave an electricfieldmagnitude of

e e2 2

d dxdE = k = k r x

e

e

2 2dx dx -1E = k k k e e ex x x

k Q1 1= k -

( )

x l a

l+a l+a

a ax a

a l + a a l a

!

!

! !

!

-

-

Soourtotal electricfield,atdistancesfrom P varyingbetweenx = aandx = l+ a willbe


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A thin ringofradiusacarries

acharge Q distributed

uniformlyalongitslength.

Findthe electricfieldatadistance xalongitscentral

axis.

e 2

dd k

r

The electricfielddue to each elementd willbe

The electricfieldwillhave componentsperpendiculartothe axis

ofthe ringandparalleltothe axisofthe ring (the xaxis,inour

case).We mentionthisbecause allcomponentsotherthanthose in

the xdirectionwillcancel.


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For everycharge element

we have acorresponding

elementonthe opposite side

ofthe ring. Bothare thesame distance awayfrom

the fieldpointandthuswill

have the same magnitude of

electricfield.

The directionofthe fieldissuchthatallcomponentswillcancelother

thanthe xcomponents.x e 2

e 2 2 2 2

e 2 2 3/ 2

ddE = dE = k (cos)

r

d x= k

x +a x +a

dk

(x +a )

!


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Normallywe needtosetd = dl,butsince every elementofcharge

is the same distance awayfromthe fieldpoint (inotherwords,x

and a are constant),

e e e2 2 3/ 2 2 2 3/ 2 2 2 3/ 2

d 1 QE k = k d = k

(x +a ) (x +a ) (x +a )!


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A convenientwaytovisualize an electricfieldisby representingthe

fielddirectionusinglinesthatareparalleltothe electricfieldvector

ataspecificpointinspace.These electricfieldlineswere first

introducedby Michael Faradayandare extremelyusefulinusing

symmetryto reduce aproblem.The twosimple rulesofthumbwhendrawing electricfieldlinesare:

1) The electricfieldvectoristangenttothe electricfieldline atany

point.2) The densityofthe fieldlinesisproportionaltothe electricfield

strength


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Some more usefulguidelinesare:

1. Electricfieldlinesoriginate onpositive chargesand

terminate onnegative charges.

2. The numberoflinesisproportionaltothe amountofcharge3. Notwofieldlinescancross.


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Electricfluxisa quantitythatmeasuresthe numberoffieldlines

penetratingasurface ofinterest.While notseeminglike an

importantconcept,thiswillleadustoan extremelyvaluable tool

forcalculatingthe electricfieldofhighlysymmetriccharge

distributions.

Imagine thatwe have asurface whichisinthepresence ofan

electricfield.We willfirstlookatthe fluxthroughsmall

elementsofthe surface andthen evaluate the fluxovertheentire surface area.


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Suppose an elementofarea, Ai,isbeing

penetratedbyand electricfieldEi.We can

representthis elementofareawithavector

havingthe magnitude ofthe areaitselfanda

directionnormal (perpendicular)tothe area

element.The fluxthroughthis elementis

definedas

We can evaluate the totalfluxthroughthe surfacebysummingover

allsuch elementsandmakingthe elementsize approachzero:

cosAEAEluxElectric iiiiE !|!TT

!! p AdEAElim iii0AE iTTTT


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Foraclosedsurface the

situationbecomesslightly

more complicated. Firstofall,

the normalvectorstoalltheelementsbyconventionpoint

outoftheboundedvolume.

Secondly,some ofthe dot

productswillyieldnegative

flux. A negative fluxsimplyrepresentsflowoffieldintoa

volume,positive flux

representsflowoutofa

volume. Ifwe have aclosed

surface,the fluxis representedslightlydifferently:

! AdE iETT


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A constant electricfield

flowsthroughasquare ofsidesl.The fieldpointsin

the xdirection. Findthe net

electricfluxthroughthe

entire cube.

Onlyfaces1and 2 willhave aflux,since E anddA are

perpendicularthroughoutall elementsonthe otherfacesofthe cube

(the dotproductinvolvesthe cosine ofthe anglebetweenE anddA).

2

1 d EdA cos180 = -E dA = -EA = -E! ! E Arr

g l

2

2 d E d cos0 E d -E E! ! E A

rrg l


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Thus,total = i = -El2 +El2 = 0

This resultmakessense,asthe source ofelectricfieldwas

outside ofourvolume. Inthatcase,we expectasmanyfield

linesflowingin (negative flux)aswe doflowingout (positive

flux).


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Gauss Lawgivesusa relationshipbetweenthe fluxthroughaclosedsurface andthe charge containedwithinthe surface. Imagine

thatwe are evaluatingthe electricfluxthroughasphericalsurface

withapointcharge atitscenter.

Animaginarysurface for evaluatingelectricfluxiscalledaGaussian surface.

Firstwe note thatthe electricfieldpoints

radiallyawayfromthepointcharge.We

alsonotice thatallthe areavectorsdAi onthe surface ofthe sphere alsopoint radially

awayfromthe charge. Finally,we knowthat

the electricfieldshouldbe constantonthe

surface ofthe sphere.


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Since E = keq/r2 forapointcharge,andthe surface areaofasphere

is 4r2,

2

E 2

0 0

q q = EA = 4r

4 r

! -

-

We usedthe substitution ke = 1/(40),where 0 isthepermittivity

offree space (8.85x10-12 C2/(Nm2))

We chose asphere tointegrate overbutthe shape couldbe a

completelyarbitraryone,aswe cansee inthe nextslide.

EAdAEcos0dAEAdEE !!!! TT


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Allthe shapeshave the same electricflux

since theyallcontainthe same numberof

fieldlines.Thus,foranyshape

surroundingourpointcharge q,

E

0

q =

Ifwe hadmanypointcharges,we couldsurround eachwitha

sphere whichwouldgive the same resultasthe one above.We

couldthensurroundallofthe individualGaussiansphereswithan

arbitraryvolume,whichwouldhave tocontainthe same amountoffluxasallthe individualspherescombined:

1 2E 1 2

0

q + q +..... = + +..... =

*


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We couldmake the same argumentforacontinuouscharge

distribution,aswe cansplititupintomanyinfinitesimal elements

andthenfindthe flux,yieldingthe same resultasbefore.

Gauss Law

The netfluxthroughanyclosedsurface is

where qin isthe netcharge containedwithinthe surface.

!!0

inE

qAdE

TT


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Imagine thatwe didnt knowthe fielddue

toapointcharge.We coulduse Gauss Law

tofindthe fieldatadistance rfromthe

charge. First,we constructasphere aroundthe chargebecause 1)allpointsonthe

spheressurface are the same distance away

fromthe charge,thusthe fieldshouldbe

constant,and 2)the elementsofareadA

point radiallyawayfromthe charge,inthesame directionas E.

ine2 2 2

0 0

q q qE= = = k

4 r 4 r r

0

in2

E

q)r4E(dAEcos0dAEAdEI

T !!!!! TT

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2. Electric Fields and Gauss’ Law