exam 2 covers ch. 28-33, lecture, discussion, hw, lab
Post on 29-Jan-2016
21 Views
Preview:
DESCRIPTION
TRANSCRIPT
1
Exam 2 covers Ch. 28-33,Lecture, Discussion, HW, Lab
Chapter 28: Electric flux & Gauss’ law Chapter 29: Electric potential & work Chapter 30: Electric potential & field
(exclude 30.7) Chapter 31: Current & Conductivity Chapter 32: Circuits
(exclude 32.8) Chapter 33: Magnetic fields & forces
(exclude 33.3, 33.6, 32.10, Hall effect)
Exam 2 is Tue. Oct. 28, 5:30-7 pm, 2103 Ch
2
Electric flux Suppose surface make angle surface normal
E = EA cos E =0 if E parallel A
E = EA (max) if E A
Flux SI units are N·m2/C
€
rE = E ||
ˆ s + E⊥ˆ n
€
ˆ n
€
ˆ s
€
rA = A ˆ n
Component || surface
Component surface
Only component‘goes through’ surface
€
E =r E •
r A
3
Gauss’ law
net electric flux through closed surface = charge enclosed /
€
E =r E • d
r A ∫ =
Qenclosed
εo
4
Properties of conductors
everywhere inside a conductor
Charge in conductor is only on the surface
surface of conductor
€
rE = 0
€
rE ⊥
----
-
-
+ +++++
5
Gauss’ law example: Charges on parallel-plate capacitor
Determine fields by superposition
Q -Q
Area A=Length X Width
€
E =Q
Aεo
€
E = 0
€
E = 0
Apply Gauss’ law: E=0 inside metal E=0 to left No charge on outer surface
€
⇒ E = 0, Qencl = 0
Apply Gauss’ law: E=0 inside metal E=Q/Ao in middle
€
E =Q
Aεo
Asurf , Qencl = η inner Asurf
⇒ η inner = Q / A
6
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 proportional to charge q that work is done on
€
ΔU /q ≡ ΔV = Electric potential difference
Depends only on charges that create E-fields
€
= r
E • dr s ∫
7
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
8
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
9
A. W = +19.8 mJB. W = -19.8 mJC. W= 0
Work and electrostatic potential energy
−μC
−μC
−μC m
m m
Question: How much work would it take YOU to assemble 3 negative charges?
Likes repel, so YOU will still do positive work!
q3
q2q1
€
W1 = 0
W2 = kq1q2
r12
= 9 ×109 −1×10−6 × −2 ×10−6
5= 3.6mJ
W3 = kq1q3
r13
+ kq2q3
r23
=16.2mJ
W tot = kq1q2
r12
+ kq1q3
r13
+ kq2q3
r23
= +19.8mJ
€
UE =19.8mJ electric potential energy of the system increases
10
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
11
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
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 = Q /C
€
C =εoA
d
+Q -Q
d
Area A
€
ΔV
13
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
14
Spherical capacitor
Charge Q moved from outer to inner sphere
Gauss’ law says E=kQ/r2 until second sphere
Potential difference
+ ++
++ +
++
+
€
ΔV = E • dsa
b
∫
Along path shown
€
ΔV =kQ
r2a
b
∫ = −kQ1
r a
b
= kQ1
a−
1
b
⎛
⎝ ⎜
⎞
⎠ ⎟
€
C =Q
ΔV= k
1
a−
1
b
⎛
⎝ ⎜
⎞
⎠ ⎟−1
€
C =Q
ΔV= k
1
a−
1
b
⎛
⎝ ⎜
⎞
⎠ ⎟−1
Gaussian surface to find E
Path to find ΔV
15
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’
Electric current
Current density J= I/A = nqvd
(direction of + charge carriers)
L
SI unit: ampere 1 A = 1 C / s
Average current:
Instantaneous value:
n = number of electrons/volumen x AL electrons travel distance L = vd Δt
Iav = ΔQ/ Δt = neAL vd /L
17
Resistance and resistivity
Ohm’s Law: ΔV = R I (J = σ E or E = ρ J ΔV = EL and E = ρ J ρ /A = ΔV/L R = ρL/A Resistance in ohms (Ω)
18
Current conservation
Iin
Iout
Iout = Iin
I1
I2
I3I1=I2+I3
I2
I3
I1
I1+I2=I3
19
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
20
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
21
Quick QuizWhat is the current
through resistor R1?
R1=200Ω
R1=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
22
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
23
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
24
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
25
RC CircuitsR
C
€
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
26
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
27
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
28
RC Circuits
What is the value of the time constant of this circuit?
A) 6 msB) 12 msC) 25 msD) 30 ms
29
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
30
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
31
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
32
Current loops & magnetic dipoles
• Current loop produces magnetic dipole field.
• Magnetic dipole moment:
€
rμ
€
rμ =IA
currentArea of loop
€
rμ
magnitude direction
In a uniform magnetic field
Magnetic field exerts torqueTorque rotates loop to align with
€
rτ =
rμ ×
rB ,
€
rμ
€
rB
€
rτ =
rμ
rB sinθ
top related