Review MTE 2 - UW-Madison Department of Physics
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Transcript Review MTE 2 - UW-Madison Department of Physics
Exam 2 covers Ch. 28-33,
Lecture, Discussion, HW, Lab
Exam 2 is Tue. Oct. 28, 5:30-7 pm, 2103 Ch
Chapter 28: Electric flux & Gauss’ law
Chapter 29: Electric potential & work
Chapter 30: Electric potential & field
Chapter 31: Current & Conductivity
Chapter 32: Circuits
(exclude 30.7)
(exclude 32.8)
Chapter 33: Magnetic fields & forces
(exclude 33.3, 33.6, 32.10, Hall effect)
1
Electric flux
Suppose surface make angle surface normal
E E|| sˆ E nˆ
A Anˆ
nˆ
Component || surface
Component surface
Only component
‘goes through’ surface
E = EA cos
E =0 if E parallel A
E = EA (max) if E A
sˆ
Flux SI units are N·m2/C
E E A
2
Gauss’ law
net electric flux through closed surface
= charge enclosed /
E
E dA
Qenclosed
o
3
Properties of conductors
E 0 everywhere inside a conductor
Charge in conductor is only on the surface
E surface of conductor
---
++
+
+
++
4
Gauss’ law example:
Charges on parallel-plate capacitor
Determine fields by
superposition
Apply Gauss’ law:
E=0 inside metal
E=0 to left
E 0, Qencl 0
-Q
Q
E
No charge on outer surface
Apply Gauss’ law:
E 0
Q
A o
E 0
E=0 inside metal
E=Q/Ao in middle
Q
E
Asurf , Qencl inner Asurf
Ao
inner Q / A
Area A=Length X Width
5
Electric potential: general
U
F
Coulomb
ds
qE ds q E ds
Electric potential energy difference U
proportional to charge q that work is done on
U /q V Electric potential difference
E ds
Depends only on charges that create E-fields
Electric field usually created
by some charge
distribution.
V(r) is electric potential of that charge
distribution
V has units of Joules / Coulomb = Volts6
Electric Potential
Electric potential energy per unit charge
units of Joules/Coulomb = Volts
Example: charge q interacting with charge Q
Qq
Electric potential energy UQq ke
r
UQq
Q
Electric potential of charge Q VQ r q ke r
Q source of the electric
it
potential, q ‘experiences’
7
Example: Electric Potential
Calculate the electric potential at B
q q
VB k 0
d d
Calculate the electric potential at A
q
q
2q
VA k
k
d1 3d1 3d1
B
y
x
d
-12 C
d2=4 m
-
+12 C
A
+
d1=3 m
3m
3m
Calculate the work YOU must do to move a Q=+5 mC charge
from A to B.
WYou
2qQ
U U B U A Q(VB VA ) k
3d1
Work done by electric fields WE field U
8
Work and electrostatic potential energy
Question: How much work would it take YOU to
assemble 3 negative charges?
A. W = +19.8 mJ
B. W = -19.8 mJ
C. W= 0
Likes repel, so YOU will
still do positive work!
q3
W1 0
W2 k
6
C
6
q1q2
110 2 10
9 109
3.6m J
r12
5
qq
qq
W 3 k 1 3 k 2 3 16.2m J
r13
r23
W tot k
q1q2
qq
qq
k 1 3 k 2 3 19.8m J
r12
r13
r23
U E 19.8mJ
5m
q1
C
5m
q2
5m
C
electric potential energy of the system increases
9
Potential from electric field
dV E d
Electric field can be
used to find changes V V
o
in potential
Potential changes
largest in direction of
E-field.
Smallest (zero)
perpendicular to
E-field
d
d
E
V Vo E d
d
V=Vo
V Vo E d
10
Electric Potential and Field
Uniform electric field of E 4 yˆ N /C
What is the electric potential difference VA-VB?
A) -12V
B) +12V
C) -24V
D) +24V
y
5m
A
B
2m
2m
5m
x
11
Capacitors
V Q /C
Conductor: electric potential proportional to charge:
C = capacitance: depends on geometry of conductor(s)
+Q
Example: parallel plate capacitor
V Q /C
C
-Q
o A
Area A
d
d
Energy
stored in a capacitor:
Q2 1
1
2
U
CV QV
2C 2
2
V
Stored energy
Isolated charged capacitor
Plate separation increased
The stored energy
A)
1) Increases
B)
2) Decreases
C)
3) Does not change
q2
U
2C
Cini
0 A
d
C fin
0 A
D
q unchanged because C isolated
q is the same
E is the same = q/(Aε0)
ΔV increases = Ed
C decreases
U increases
C fin Cini U fin U ini
13
Spherical capacitor
Charge Q moved from outer to inner sphere
Gauss’ law says E=kQ/r2
until second sphere
Potential difference V
b
E ds
a
Along path shown
Gaussian surface
to find E
1 1
kQ
1
V 2 kQ kQ
a b
ra
a r
b
b
1 1 1
Q
C
k
a b
V
Path to find V
+ + +
+
+
+
+ + +
14
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’
15
L
Electric current
Average current:
Instantaneous value:
SI unit: ampere 1 A = 1 C / s
n = number of electrons/volume
n x AL electrons travel distance L = vd Δt
Iav = Q/ t = neAL vd /L
Current density J= I/A = nqvd
(direction of + charge carriers)
Resistance and resistivity
V = R I (J = E or E = ρ J
V = EL and E = J /A = V/L
R = ρL/A Resistance in ohms ()
Ohm’s Law:
17
I2
Current conservation
Iin
I1
I3
I1=I2+I3
I1
I3
Iout
Iout = Iin
I2
I1+I2=I3
18
Resistors in Series and parallel
Series
I1 = I2 = I
Req = R1+R2
Parallel
V1 = V2 = V
Req = (R1-1+R2-1)-1
I1+I2
I
R1
R1+R2
R2
=
I
I1
R1
I
2 resistors in series:
RL
Like summing lengths
R2
I2
=
1 1 1
R1 R2
L
R
A
19
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
20
Quick Quiz
What is the current
through resistor R1?
9V
A. 5 mA
R1=200Ω
R4=100Ω
R1=200Ω
R3=100Ω
B. 10 mA
C. 20 mA
6V
D. 30 mA
E. 60 mA
Req=100Ω
3V
9V
Req=50Ω
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
Capacitors in parallel and series
ΔV1 = ΔV2 = ΔV
Qtotal = Q1 + Q2
Ceq = C1 + C2
Parallel
Q1=Q2 =Q
ΔV = ΔV1+ΔV2
1/Ceq = 1/C1 + 1/C2
Series
23
Example: Equivalent Capacitance
C1 = 30 F
C2 = 15 F
C3 = 15 F
C4 = 30 F
C23 C2 C3 15F 15F 30F
C1, C23 , C4 in series
1
1
1
1
Ceq C1 C23 C4
1
1
1
1
Ceq 10F
Ceq 30F 30F 30F
C1
C2
V
C3
C4
Parallel
combination
Ceq=C1||C2
24
Charge
Discharge
R
RC Circuits
Time constant
C
RC
Start w/uncharged
C
Close switch at t=0
q(t) C(1 e
I(t)
e
t / RC
)
t / RC
R
Vcap t 1 et / RC
R
C
Start w/charged C
Close switch at t=0
qt qoet / RC
qo /C t / RC
It
e
R
Vcap t qo /Cet / RC
25
Question
What is the current through R1 Immediately
after the switch is closed?
A. 10A
R1=100Ω
B. 1 A
10V
C. 0.1A
D. 0.05A
C=1µF
R2=100Ω
E. 0.01A
26
Question
What is the charge on the capacitor a long
time after the switch is closed?
A. 0.05µC
R1=100Ω
B. 0.1µC
10V
C. 1µC
D. 5µC
C=1µF
R2=100Ω
E. 10µC
27
RC Circuits
What is the value of the time constant of this circuit?
A) 6 ms
B) 12 ms
C) 25 ms
D) 30 ms
28
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)
N
N
T
C m /s A m
CGS unit: gauss (G): 1 T = 104 G (Earth surface 0.5 G)
29
Magnetic Force on a Current
I
Force on each charge qv B
Force on length ds of wire
Ids B
N
Current
Force on straight section
of wire, length L
F IBL
Magnetic force
S
Magnetic
field
30
Magnetic field from
long straight wire:
Direction
What direction
is the magnetic
field from an
infinitely-long
straight wire?
o I
B
2 r
y
x
I
r = distance from wire
o 4 107 N / A2 = permeability of free space
31
Current loops & magnetic dipoles
• Current loop produces magnetic dipole field.
• Magnetic dipole moment:
IA
current
Area of
loop
magnitude
direction
In a uniform magnetic field
Magnetic field exerts torque B, B sin
Torque rotates loop to align with B
32