electricity magnetism lecture 2: electric fields lecture 02... · the electric field e at a point...
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Electricity & Magnetism Lecture 2: Electric Fields
Today’sConcepts: A)TheElectricField
B)Con9nuousChargeDistribu9ons
Electricity&Magne/smLecture2,Slide1
★ Couldyoupleasereschedulethefinalexamtooneweeklater?Asit'sawfullyscheduledbetweenmyotherexams,anditsalsothesameweekschoolends,whichmakesit's9mingevenmorehec9c.
★ Andalsothesecondmidterm,couldyoupleasemakeitinsteadofJuly18thhaveitbeeitherWednesdayorFridayofthatveryweekthatthesecondmidtermison?(thelaterthebeNer:D)
★ Thanks!
There’sawriNenhomeworkassignmentforUnit19now.
Suddenly,terriblehaiku:Posi9vetestchargeRepelledbyop9mistsCuphalfelectric
I had problems understanding the concept and calculation of charge on and infinite line
aliNlebitconcentra9onitrequires.
defineElectricfieldandelectricforceandchargeinwordsplease
Posi9vesandnega9ves,whatdotheyreallymean?Itseemslikewe,ashumans,justdecidedtonamethemthis.
??Infinitelinesofchargee???itwouldhelpalottogooverthose,everythingelseissimmplle
Your Comments
Electricity&Magne/smLecture2,Slide2
I almost forgot how exciting smartphysics was, yay
Irony?
Inthediagramsbelow,themagnitudeanddirec9onoftheelectricfieldisrepresentedbythelengthanddirec9onofthebluearrows.Whichofthediagramsbestrepresentstheelectricfieldfromanega%vecharge?
3equal–chargesareatthecornersofanequilateraltriangle
WhereistheEfield0?Theblacklineis1/2waybetweenthebaseandthetopcharge.A.Abovetheblackline
B. BelowtheblacklineC. ontheblackline
D.nowhereexceptinfinity
E. somewhereelse
!
Electric Field PhET
Electric Field Line Applet
TheelectricfieldEatapointinspaceissimplytheforceperunitchargeatthatpoint.
Electricfieldduetoapointchargedpar/cle
Superposi/on
E2
E3
E4
EFieldpointstowardnega/veandAwayfromposi/vecharges.
“Whatexactlydoestheelectricfieldthatwecalculatemean/represent?““Whatistheessenceofanelectricfield?“
Electric Field
q4
q2
q3
Electricity&Magne/smLecture2,Slide4
Twoequal,butoppositechargesareplacedonthexaxis.Theposi/vechargeisplacedtotheleLoftheoriginandthenega/vechargeisplacedtotheright,asshowninthefigureabove.
Whatisthedirec/onoftheelectricfieldatpointA?
oUpoDownoLeLoRightoZero
CheckPoint: Electric Fields1
Electricity&Magne/smLecture2,Slide5
A
Bx+Q −Q
ABCDE
Twoequal,butoppositechargesareplacedonthexaxis.Theposi/vechargeisplacedtotheleLoftheoriginandthenega/vechargeisplacedtotheright,asshowninthefigureabove.
Whatisthedirec/onoftheelectricfieldatpointA?
oUpoDownoLeLoRightoZero
CheckPoint Results: Electric Fields1
Electricity&Magne/smLecture2,Slide6
A
Bx+Q −Q
ABCDE
Twoequal,butoppositechargesareplacedonthexaxis.Theposi/vechargeisplacedtotheleLoftheoriginandthenega/vechargeisplacedtotheright,asshowninthefigureabove.
Whatisthedirec/onoftheelectricfieldatpointB?
oUpoDownoLeLoRightoZero
CheckPoint: Electric Fields2
Electricity&Magne/smLecture2,Slide7
A
Bx+Q −Q
ABCDE
Twoequal,butoppositechargesareplacedonthexaxis.Theposi/vechargeisplacedtotheleLoftheoriginandthenega/vechargeisplacedtotheright,asshowninthefigureabove.
Whatisthedirec/onoftheelectricfieldatpointB?
oUpoDownoLeLoRightoZero
CheckPoint Results: Electric Fields2
Electricity&Magne/smLecture2,Slide8
A
Bx+Q −Q
Polarization Demo
ABCDE
CheckPoint: Magnitude of Field (2 Charges)
InwhichofthetwocasesshownbelowisthemagnitudeoftheelectricfieldatthepointlabeledAthelargest?
oCase1oCase2oEqual
+Q
+Q
+Q
−Q
AA
Case1 Case2
Electricity&Magne/smLecture2,Slide9
ABC
InwhichofthetwocasesshownbelowisthemagnitudeoftheelectricfieldatthepointlabeledAthelargest?
oCase1oCase2oEqual
“Theupperle^+Qonlyaffectsthexdirec9oninbothandthelowerright(+/-)Qonlyaffectstheydirec9onsoinboth,nothingcancelsout,sothey'llhavethesamemagnitude.”
CheckPoint Results: Magnitude of Field (2 Chrg)
E
E
Electricity&Magne/smLecture2,Slide10
+Q
+Q
+Q
−Q
AA
Case1 Case2
ABC
CheckPoint: Magnitude of Field (2 Charges)
InwhichofthetwocasesshownbelowisthemagnitudeoftheelectricfieldatthepointlabeledBthelargest?
oCase1oCase2oEqual
+Q
+Q
+Q
−Q
AA
Case1 Case2
Electricity&Magne/smLecture2,Slide9
B B
CheckPoint: Magnitude of Field (2 Charges)
InwhichofthetwocasesshownbelowisthemagnitudeoftheelectricfieldatthepointlabeledBthelargest?
oCase1oCase2oEqual
+Q
+Q
+Q
−Q
AA
Case1 Case2
Electricity&Magne/smLecture2,Slide9
B B
CheckPoint: Magnitude of Field (2 Charges)
InwhichofthetwocasesshownbelowisthemagnitudeoftheelectricfieldatthepointlabeledBthelargest?
oCase1oCase2oEqual
+Q
+Q
+Q
−Q
AA
Case1 Case2
Electricity&Magne/smLecture2,Slide9
B B
Twochargesq1andq2arefixedatpoints(-a,0) and(a,0) asshown.Togethertheyproduceanelectricfieldatpoint(0,d) whichisdirectedalongthenega/vey-axis.
x
y
q1 q2(−a,0) (a,0)
(0,d)
Whichofthefollowingstatementsistrue:
A)Bothchargesarenega/veB)Bothchargesareposi/veC)ThechargesareoppositeD)Thereisnotenoughinforma/ontotellhowthechargesare
related
Clicker Question: Two Charges
Electricity&Magne/smLecture2,Slide11
_ _+ +
_+
Electricity&Magne/smLecture2,Slide12
CheckPoint Results: Motion of Test Charge
Electricity&Magne/smLecture2,Slide13
Aposi/vetestchargeqisreleasedfromrestatdistancerawayfromachargeof+Qandadistance2rawayfromachargeof+2Q.HowwillthetestchargemoveimmediatelyaLerbeingreleased?
oTotheleLoTotherightoStays/lloOther
“Theforceispropor9onaltothechargedividedbythesquareofthedistance.Therefore,theforceofthe2Qchargeis1/2asmuchastheforceoftheQcharge.“
“Eventhoughthechargeontherightislarger,itistwiceasfaraway,whichmakestheforceitexhertsonthetestchargehalfthatasthechargeonthele^,causingthechargetomovetotheright.”
Thera9obetweentheRandQonbothsidesis1:1meaningtheywillresultinthesamemagnitudeofelectricfieldac9nginoppositedirec9ons,causingqtoremains9ll.
Ex
= k
q
d2� q�p
2d�2 cos
⇡
4
!
Ey = k
0BBBBBBBB@
qd2 �
q⇣p
2d⌘2 sin ⇡4
1CCCCCCCCA
Electric Field Example
CalculateEatpointP.
d
P
d
A) B) C) D)Needtoknowd
Needtoknowd&qE)
Whatisthedirec/onoftheelectricfieldatpointP,theunoccupiedcornerofthesquare? −q +q
+q
Electricity&Magne/smLecture2,Slide14
About r
~r12θ1
2
r12
y
x
j
i ~r12
=r12
cos ✓ ˆi + r12
sin ✓ ˆj
r12
=r
12
cos ✓ ˆi + r12
sin ✓ ˆjr
12
r12
= cos ✓ ˆi + sin ✓ ˆj
For example
~r12(1,2)
(5,5)
r12
θ
~r12
=4
ˆi + 3
ˆj
r12
=p
4
2 + 3
2 = 5
cos ✓ = 4
5
sin ✓ = 3
5
r12
=r
12
cos ✓ ˆi + r12
sin ✓ ˆjr
12
r12
= 4
5
ˆi + 3
5
ˆj
λ = Q/L
Summa/onbecomesanintegral(becarefulwithvectornature)
“Idon'tunderstandthewholedqthingandlambda.”
WHATDOESTHISMEAN?
Integrateoverallcharges(dq)
risvectorfromdqtothepointatwhichEisdefined
r
dE
Continuous Charge Distributions
LinearExample:
charges
ptforE
dq = λ dx
Electricity&Magne/smLecture2,Slide15
Clicker Question: Charge Density
Linear(λ =Q/L)Coulombs/meter
Surface(σ = Q/A)Coulombs/meter2
Volume(ρ=Q/V) Coulombs/meter3
Whathasmorenetcharge?.A)Aspherew/radius2metersandvolumechargedensityρ=2C/m3
B)Aspherew/radius2metersandsurfacechargedensityσ=2C/m2
C)BothA)andB)havethesamenetcharge.
“Iwouldliketoknowmoreaboutthechargedensity.”
SomeGeometry
Electricity&Magne/smLecture2,Slide16
Prelecture Question
A)
B)
C)
D)
E)
What is the electric field at point a?
“Howistheintegra/onofdEoverLworkedout,stepbystep?”
Clicker Question: Calculation
A) B) C) D) E)
Whatis?
Chargeisuniformlydistributedalongthex-axisfromtheorigintox=a.ThechargedensityisλC/m.Whatisthex-componentoftheelectricfieldatpointP:(x,y)=(a,h)?
x
y
a
h
P
x
r
dq = λ dx
Electricity&Magne/smLecture2,Slide19
Weknow:
Clicker Question: Calculation
Weknow:
Whatis?
A) B) C) D)
Chargeisuniformlydistributedalongthex-axisfromtheorigintox=a.ThechargedensityisλC/m.Whatisthex-componentoftheelectricfieldatpointP:(x,y)=(a,h)?
xa
P
x
r
θ1 θ2
θ2
dq = λ dx
h
y
Electricity&Magne/smLecture2,Slide20
k� cos ✓2
Z 1
�1
dx
(a � x)
2 + h
2
k� cos ✓2
Za
0
dx
(a � x)
2 + h
2
Weknow:
cosθ2 DEPENDSONx!
Clicker Question: Calculation
Whatis?
A) B)
C) noneoftheabove
Chargeisuniformlydistributedalongthex-axisfromtheorigintox=a.ThechargedensityisλC/m.Whatisthex-componentoftheelectricfieldatpointP:(x,y)=(a,h)?
xa
P
x
r
θ1 θ2
θ2
dq = λ dx
h
y
Electricity&Magne/smLecture2,Slide21
Weknow:
Clicker Question: CalculationChargeisuniformlydistributedalongthex-axisfromtheorigintox=a.ThechargedensityisλC/m.Whatisthex-componentoftheelectricfieldatpointP:(x,y)=(a,h)?
xa
P
x
r
θ1 θ2
θ2
dq = λ dx
h
y
Electricity&Magne/smLecture2,Slide22
Whatis?
A) B) C) D)
E
x
(P) = k�
1 � hp
a
2 + h
2
!
Weknow:
Calculation
Whatis?
xa
P
x
r
θ1 θ2
θ2
dq = λ dx
h
y
Electricity&Magne/smLecture2,Slide23
Chargeisuniformlydistributedalongthex-axisfromtheorigintox=a.ThechargedensityisλC/m.Whatisthex-componentoftheelectricfieldatpointP:(x,y)=(a,h)?
E
x
(P) = k�
Za
0dx
a � x
p(a � x)2 + h
2
E
x
(P) =k�
h
1 � hp
h
2 + a
2
!E
x
(P) =k�
h
(1 � sin ✓1)
E
x
(P) =k⇥
h
Z ⇤/2
�1
d� cos �
Exerciseforstudent:
Changevariables:writexintermsofθ
Result:obtainsimpleintegralinθ
Observation
NotethatourresultcanberewriNenmoresimplyintermsofθ1.
Chargeisuniformlydistributedalongthex-axisfromtheorigintox=a.ThechargedensityisλC/m.Whatisthex-componentoftheelectricfieldatpointP:(x,y)=(a,h)?
xa
P
x
r
θ1 θ2
θ2
dq = λ dx
h
y
Electricity&Magne/smLecture2,Slide24