Exam 2 covers Ch. 27-33, Lecture, Discussion, HW, Lab
description
Transcript of Exam 2 covers Ch. 27-33, Lecture, Discussion, HW, Lab
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Exam 2 covers Ch. 27-33,Lecture, Discussion, HW, Lab
Chapter 27: The Electric Field Chapter 29: Electric potential & work Chapter 30: Electric potential & field
(exclude 30.7) Chapter 31: Current & Resistance Chapter 32: Fundamentals of Circuits
(exclude 32.8) Chapter 33: The Magnetic Field
(exclude 33.5-33.6, 33.9-10, & Hall effect)
Exam 2 is Tue. Oct. 27, 5:30-7 pm, 145 Birge
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Electric field lines
• Local electric field tangent to field line
• Density of lines proportional to electric field strength
• Fields lines can only start on + charge
• Can only end on - charge.
• Electric field lines can never cross
Oct. 22, 2009
Physics 208 Lecture 15
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QuestionHere is a picture of electric field lines. Which choice most accurately ranks the magnitude of the electric field at the different points?
A) E1=E3>E2=E4
B) E1=E2>E3>E4
C) E4=E3>E1=E2
D) E4=E2>E1>E3
E) E4<E3<E1<E2
1
2
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Oct. 22, 2009
Physics 208 Lecture 15
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Oct. 22, 2009 Physics 208 Lecture 15 4
Charge Densities Volume charge density: when a charge is distributed
evenly throughout a volume = Q / V dq = dV
Surface charge density: when a charge is distributed evenly over a surface area = Q / A dq = dA
Linear charge density: when a charge is distributed along a line = Q / dq = d
Electric fields and potentials from these charge elements superimpose
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Oct. 22, 2009 Physics 208 Lecture 15 5
+ + + + + + + + + + + + + + + + + + + +
Infinite line of charge, charge density λ
r
€
rE = 2ke
λ
r=
1
2πεo
λ
r
Infinite sheet of charge, charge density η
r
€
rE = 2π keη =
η
2εo
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Oct. 22, 2009 Physics 208 Lecture 15 6
Ring of uniform positive charge
y
x
z
€
E z = kzQ
z2 + R2( )
3 / 2
€
E z
Which is the graph of on the z-axis?
A)
B)
C)
D)
E)
z
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Properties of conductors
everywhere inside a conductor
Charge in conductor is only on the surface
surface of conductor
€
rE = 0
€
rE ⊥
----
-
-
+ +++++
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Electric potential: general
Electric field usually created by some charge distribution. V(r) is electric potential of that charge distribution
V has units of Joules / Coulomb = Volts
€
ΔU =r F Coulomb • d
r s ∫ = q
r E • d
r s ∫ = q
r E • d
r s ∫
Electric potential energy difference ΔU
€
ΔU /q ≡ ΔV = Electric potential difference
Depends only on charges that create E-fields
€
= r
E • dr s ∫
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Electric Potential
Q source of the electric potential, q ‘experiences’ it
Electric potential energy per unit chargeunits of Joules/Coulomb = Volts
Example: charge q interacting with charge Q
Electric potential energy
Electric potential of charge Q
€
=UQq = ke
r
€
=VQ r( ) =UQq
q= ke
Q
r
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Example: Electric PotentialCalculate the electric potential at B
Calculate the work YOU must do to move a Q=+5 mC charge from A to B.
Calculate the electric potential at A
x
+-
B
A
d1=3 m 3 m
d2=4 m
3 m
y
-12 μC +12 μC
d
€
VB = kq
d−
q
d
⎛
⎝ ⎜
⎞
⎠ ⎟= 0
€
VA = kq
d1
−q
3d1
⎛
⎝ ⎜
⎞
⎠ ⎟= k
2q
3d1
€
WYou = ΔU = UB −UA = Q(VB −VA ) = −k2qQ
3d1
€
=WE− field = −ΔUWork done by electric fields
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Potential from electric field
Electric field can be used to find changes in potential
Potential changes largest in direction of E-field.
Smallest (zero) perpendicular to E-field
€
dV = −r E • d
r l
€
dV = −r E • d
r l
€
dr l
€
rE
V=Vo
€
V = Vo −r E d
r l
€
V = Vo +r E d
r l
€
dr l
€
dr l
€
V = Vo
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Electric Potential and Field
Uniform electric field of
What is the electric potential difference VA-VB?
A) -12V
B) +12V
C) -24V
D) +24V
€
rE = 4 ˆ y N /C
A
x
y
2m
5m
2m
5m
B
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Capacitors
Energy stored in a capacitor:
€
U =Q2
2C=
1
2CΔV 2 =
1
2QΔV
C = capacitance: depends on geometry of conductor(s)
Conductor: electric potential proportional to charge:
€
V = Q /C
Example: parallel plate capacitor
€
ΔV = E insided =Qd
εoA= Q /C ⇒
€
C =εoA
d
+Q -Q
d
Area A
€
ΔV
13
€
rE inside =
1
2εo
Q
A−
1
2εo
−Q
A=
Q
εoA
€
rE outside = 0
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Question
What is the voltage across capacitor 1 after the two are connected?
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C2=3µF
V1=1V
C1=1µF
V2=0V
A. 1V
B. 2V
C. 0V
D. 0.25V
E. 4V
€
Q1before = Q1
after + Q2after
C1V1before = C1V
after + C2Vafter
V after =C1
C1 + C2
V1before =
1μF
1μF + 3μF1V = 0.25V
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Isolated charged capacitorPlate separation increased The stored energy 1) Increases2) Decreases3) Does not change
A)B)C)
Stored energy
q unchanged because C isolated
€
U =q2
2C
Cini =ε0A
d→ C fin =
ε0A
D⇒ C fin < Cini ⇒ U fin > U ini
q is the sameE is the same = q/(Aε0)ΔV increases = EdC decreasesU increases
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Conductors, charges, electric fields Electrostatic equilibrium
No charges moving No electric fields inside conductor. Electric potential is constant everywhere Charges on surface of conductors.
Not equilibrium Charges moving (electric current) Electric fields inside conductors -> forces on charges. Electric potential decreases around ‘circuit’
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Resistance and resistivity Ohm’s Law: ΔV = R I (J = σ E or E = ρ J ΔV = EL and E = J /A = ΔV/L R = ρL/A Resistance in ohms (Ω)
Question
5cm
2cm1cm
I
A block is made from a material with resistivity of 10-4Ω-m. It has 10 A of current flowing through it. What is the voltage across the block?
€
R = 10−4 Ω ⋅m( )0.05m
0.02m( ) 0.01m( )
= 0.025Ω
A. 0.1V
B. 0.25V
C. 0.5V
D. 1.0V
E. 5.0V
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Current conservation
Iin
Iout
Iout = Iin
I1
I2
I3I1=I2+I3
I2
I3
I1
I1+I2=I3
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Resistors in Series and parallel Series I1 = I2 = I Req = R1+R2
R1
R2
=R1+R2
2 resistors in series:R LLike summing lengths
R1R2
€
R = ρL
A€
1
R1
+1
R2
⎛
⎝ ⎜
⎞
⎠ ⎟
−1
=
I
I
II1 I2
I1+I2
Parallel V1 = V2 = V Req = (R1
-1+R2-1)-1
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Quick Quiz
What happens to the brightness of bulb A when the switch is closed?
A. Gets dimmer
B. Gets brighter
C. Stays same
D. Something else
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Quick QuizWhat is the current
through resistor R1?
R1=200Ω
R2=200Ω
R3=100Ω
R4=100Ω9V
Req=100Ω
Req=50Ω
9V
A. 5 mA
B. 10 mA
C. 20 mA
D. 30 mA
E. 60 mA
6V
3V
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Oct. 22, 2009 Physics 208 Lecture 15 22
Power dissipation (Joule heating)
Charge loses energy from c to d.
Ohm’s law:
Energy dissipated in resistor as Heat (& light) in bulb
Power dissipated in resistor =
€
E lost = −ΔE = − ΔKE + ΔU( ) = 0 − q Vd −Vc( )
€
Vc −Vd( ) = IR
€
dE lost
dt=
dq
dtIR = I2R Joules / s = Watts
€
E lost = qIR
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Capacitors as circuit elements
Voltage difference depends on charge Q=CV Current in circuit
Q on capacitor changes with time Voltage across cap changes with time
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Capacitors in parallel and series
ΔV1 = ΔV2 = ΔV Qtotal = Q1 + Q2
Ceq = C1 + C2
Q1=Q2 =Q ΔV = ΔV1+ΔV2 1/Ceq = 1/C1 + 1/C2
Parallel Series
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Example: Equivalent Capacitance
C1 = 30 μFC2 = 15 μFC3 = 15 μFC4 = 30 μF
C2V C3
C1
C4
Parallel combinationCeq=C1||C2
€
C23 = C2 + C3 =15μF +15μF = 30μF
€
C1, C23, C4 in series
€
⇒1
Ceq
=1
C1
+1
C23
+1
C4
€
1
Ceq
=1
30μF+
1
30μF+
1
30μF⇒ Ceq =10μF
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RC CircuitsR
Cε
€
q(t) = Cε(1− e−t / RC )
I(t) =ε
Re−t / RC
R
C
€
Vcap t( ) = ε 1− e−t / RC( )
€
Vcap t( ) = qo /C( )e−t / RC
€
q t( ) = qoe−t / RC
I t( ) =qo /C
Re−t / RC
Start w/uncharged CClose switch at t=0
Start w/charged CClose switch at t=0
Time constant
€
τ =RC
Charge Discharge
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Question
What is the current through R1 Immediately after the switch is closed?
10V
R1=100Ω
R2=100ΩC=1µF
A. 10A
B. 1 A
C. 0.1A
D. 0.05A
E. 0.01A
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Question
What is the current through R1 a long time after the switch is closed?
10V
R1=100Ω
R2=100ΩC=1µF
A. 10A
B. 1 A
C. 0.1A
D. 0.05A
E. 0.01A
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Question
What is the charge on the capacitor a long time after the switch is closed?
A. 0.05µC
B. 0.1µC
C. 1µC
D. 5µC
E. 10µC
10V
R1=100Ω
R2=100ΩC=1µF
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RC Circuits
What is the value of the time constant of this circuit?
A) 6 msB) 12 msC) 25 msD) 30 ms
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FB on a Charge Moving in a Magnetic Field, Formula
FB = q v x B FB is the magnetic force q is the charge v is the velocity of the
moving charge B is the magnetic field
SI unit of magnetic field: tesla (T)
CGS unit: gauss (G): 1 T = 104 G (Earth surface 0.5 G)
€
T =N
C ⋅m /s=
N
A ⋅m
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Magnetic Force on a Current
S
N
I
Current
Magnetic field
Magnetic force
€
rF = IBL
€
qr v ×
r B Force on each charge
Force on length of wire
€
dr s
€
Idr s ×
r B
Force on straight section of wire, length L
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Oct. 22, 2009 Physics 208 Lecture 15 33
Law of Biot-Savart
Each short length of current produces contribution to magnetic field.r
I in plane of pageds
€
dr B =
μo
4π
Idr s × ˆ r
r2
B out of page
ds
dB
r
€
μo = 4π ×10−7 N / A2= permeability of free space
r = distance from current element
Field from very short section of current
€
dr s
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Magnetic field from long straight wire:Direction What direction
is the magnetic field from an infinitely-long straight wire? I
x
y
€
rB =
μoI
2π r
€
μo = 4π ×10−7 N / A2= permeability of free space
r = distance from wire
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Oct. 22, 2009 Physics 208 Lecture 15 35
Magnetic field from loopWhich of these graphs
best represents the magnetic field on the axis of the loop?
Bz
Bz
Bz
Bz
A.
B.
C.
D.
z
z
z
z
z
x
y
I