1 Exam 2 covers Ch. 27-32, Lecture, Discussion, HW, Lab Chapter 27: Electric flux & Gauss’ law...
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Transcript of 1 Exam 2 covers Ch. 27-32, Lecture, Discussion, HW, Lab Chapter 27: Electric flux & Gauss’ law...
1
Exam 2 covers Ch. 27-32,Lecture, Discussion, HW, Lab
Chapter 27: Electric flux & Gauss’ law Chapter 29: Electric potential & work Chapter 30: Electric potential & field Chapter 28: Current & Conductivity Chapter 31: Circuits Chapter 32: Magnetic fields & forces
(exclude 32.6,32.8,32.10)
Exam 2 is Wed. Mar. 26, 5:30-7 pm, 2103 Ch: Adam(301,310), Eli(302,311), Stephen(303,306), 180 Science Hall: Amanda(305,307), Mike(304,309), Ye(308)
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
Field outside uniformly-charged sphere
Field direction: radially out from charge Gaussian surface:
Sphere of radius r
Surface area where
Value of on this area:
Flux thru Gaussian surface:
Charge enclosed:
€
rE • d
r A ≠ 0 :
€
4π r2
€
rE • d
r A
€
E
€
Q
€
E 4π r2
Gauss’ law:
€
E 4π r2 = Q /εo ⇒ E =1
4πεo
Q
r2
5
Electric potential energy
€
Whand = ΔU since they repel! potential energy increases
Work is Force x distance (taking into account cosθ between 2 vectors!)
>0
If opposte charges they attract => W <0 and potential energy decreases
6
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
7
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
8
A. W = +19.8 mJB. W = -19.8 mJC. = 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
9
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
10
Electric Potential and Field• Uniform electric field of E = 4i+3j N/C
• Points A at 2m and B at 5m on the x axis.
• What is the potential difference VA - VB?
B(5,0)
E = 4i N/C x(m)
A(2,0)0
A) -12VB) +12VC) -24VD) +24V
€
E = −∇V ⇒ Ex = −dV
dx⇒ dV
A
B
∫ = −Ex dxA
B
∫ ⇒
VA −VB = 4 × 3 =12V
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
12
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
13
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
15
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)
16
Current conservation
Iin
Iout
Iout = Iin
I1
I2
I3I1=I2+I3
I2
I3
I1
I1+I2=I3
17
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
18
Quick Quiz
How does brightness of bulb B compare to that of A?
A. B brighter than A
B. B dimmer than A
C.Both the same
Battery maintain constant potential difference
Extra bulb makes extra resistance -> less current
19
Quick QuizWhat happens to the brightness of bulb B
when the switch is closed?
A. Gets dimmer
B. Gets brighter
C. Stays same
D. Something else
Battery is constant voltage,not constant current
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
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
22
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
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
24
Calculate the equivalent Capacitance C1 = 10 μF
C2 = 20 μFC3 = 30 μFC4 = 40 μFV = 50 Volts
C2V C3
C1
C4
€
1
Ceq
=1
C1
+1
C2 + C3
+1
C4
⇒ Ceq = 6.9μF
€
V = V1 + V23 + V4 V23 = V2 = V3
Q = Q1 = Q23 = Q4 Q23 = Q2 + Q3
V =Q
Ceq
=Q
C1
+Q
C2 + C3
+Q
C4 parallel
C1, C23, C4 in series
25
RC Circuits
What is the value of the time constant of this circuit?
A) 6 msB) 12 msC) 25 msD) 30 ms
26
Magnetic fields and forces
€
FB = qvBsinθ
€
F = qv × B
I
B
€
FB = Ids × B Magnetic force
on current-carrying wire
Magnetic torque on current loop
Magnetic force on moving charged particle
I
B
27
Effect of uniform magnetic field
Effect of uniform B-field on charged particle If charged particle
is not moving - no effect If particle is moving:
force exerted perpendicularto both field and velocity
€
F = qv × B
vector ‘cross product’
12/09/2002 U. Wisconsin, Physics 208, Fall 2006 28
Lorentz force
Electron moves in plane of screen the page.
B- field is in the plane of screen to the right.
Direction of instantaneous magnetic force on electron is
A) toward the top of the page
B) into the page
C) toward the right edge of the page
D) out of the pagev
B
F
electron
29
Trajectory in Constant B Field
FFv
x x x x x x
x x x x x x
x x x x x xv B
q
x x x x x x
x x x x x x
x x x x x x
• Charge enters B field with velocity shown. (vB)
• Force is always to velocity and to B.
Path is a circle.
Radius determined by velocity:
€
R =mv
qB
30
Current loops & magnetic dipoles Current loop produces magnetic dipole field. Magnetic dipole moment:
€
rμ
€
rμ =IA
currentArea of loop
€
rμ
magnitude direction
Effect of uniform magnetic field
Magnetic field exerts torqueTorque rotates loop to align with
€
rτ =
rμ ×
rB ,
€
rμ
€
rB
€
rτ =
rμ
rB sinθ
12/09/2002 U. Wisconsin, Physics 208, Fall 2006 31
Which of these loop orientations has the largest magnitude torque?
(A) a (B) b (C) c
Question on torque
Answer: (c). all loops enclose same area and carry same current ⇒ magnitude of μ is the same for all.
(c) μ upwards, μ ⊥ B and τ = μB. (a), τ = 0 (b) τ = μBsin
a bc
μμ ττ