Download Review MTE 2

Exam 2 covers Ch. 27-33,
Lecture, Discussion, HW, Lab
Exam 2 is Tue. Oct. 27, 5:30-7 pm, 145 Birge



Chapter 27: The Electric Field
Chapter 29: Electric potential & work
Chapter 30: Electric potential & field



Chapter 31: Current & Resistance
Chapter 32: Fundamentals of Circuits


(exclude 30.7)
(exclude 32.8)
Chapter 33: The Magnetic Field

(exclude 33.5-33.6, 33.9-10, & Hall effect)
1
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
2
Question
Here 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
4
C) E4=E3>E1=E2
3
1
D) E4=E2>E1>E3
2
E) E4<E3<E1<E2
Oct. 22,
2009
Physics 208 Lecture 15
3
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
Oct. 22, 2009
Physics 208 Lecture 15
4
r

1 
E  2ke 
r 2o r
Infinite line of charge,
charge density λ
r
+ + + + + + + ++ + + + + + + + + + + +
r

E  2 ke 
2o
r
Infinite sheet of charge,
charge density η

Oct. 22, 2009
Physics 208 Lecture 15
5
A)
Ring of uniform positive charge
z
B)
y
Which is the graph of
E z on the z-axis?
x
C)

D)
z
Ez  k
E)
Oct. 22, 2009
Physics 208 Lecture 15
zQ
z
2
R

2 3/2
6
Properties of conductors



E  0 everywhere inside a conductor
Charge in conductor is only on the surface
E  surface of conductor
---
++
+
+
++
7
Electric potential: general
Electric potential energy difference U
U 

F
Coulomb
 ds 
 qE  ds  q  E  ds
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 = Volts
8
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
 ke
Electric potential of charge Q  VQ r 
q
r
Q source of the electric
it
potential, q ‘experiences’

9
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.

2qQ
WYou  U  UB UA  Q(VB VA )  k
3d1
Work done by electric fields  W E  field  U
10
Potential from electric field
dV  E  d



Electric field can be
used to find changes
in potential
V  Vo
Potential changes
largest in direction of

E-field.

Smallest (zero)
perpendicular to
E-field
d
d
E

d
V  Vo  E d

V=Vo
 V  Vo  E d

11
Electric Potential and Field
Uniform electric field of E  4 yˆ N /C
What is the electric potential difference VA-VB?
y

A) -12V
B) +12V
C) -24V
D) +24V
5m
A
B
2m
2m
5m
x
12
Capacitors
V  Q /C
Conductor: electric potential proportional to charge:
C = capacitance: depends on geometry of conductor(s)
Example: parallel plate capacitor  +Q
r
1 Q 1 Q Q
E inside 


2o A 2o A o A
r
E outside  0
V  E insided 
Qd
A
 Q/C  C  o
o A
d
-Q
Area A
d
V
Q2 1
1
2
Energy stored in a capacitor: U    CV  QV
2C 2
2

13
Question
What is the voltage
across capacitor 1
after the two are
connected?
C1=1µF
C2=3µF
V1=1V
V2=0V
A. 1V
B. 2V
Q1before  Q1after  Q2after
C. 0V
C1V1before  C1V after  C2V after
D. 0.25V
E. 4V
V
after
C1
1F
before

V1

1V  0.25V
C1  C2
1F  3F
14
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
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’
16
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:
Question
A. 0.1V
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?
B. 0.25V
C. 0.5V
D. 1.0V
E. 5.0V
5cm
R  104  m
0.05m
0.02m0.01m
 0.025
1cm
2cm
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
R2
I2
=
1
2 resistors in series:
RL
Like summing lengths
 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
R2=200Ω
R3=100Ω
B. 10 mA
C. 20 mA
6V
D. 30 mA
E. 60 mA
Req=100Ω
3V
9V
Req=50Ω
21


Power dissipation (Joule heating)

Charge loses energy from c to d.
E lost  E  KE  U   0  qVd  Vc 

Ohm’s law:


Vc  Vd   IR
Elost  qIR

Energy dissipated in resistor as
 Heat (& light) in bulb

Power dissipated in resistor =
dE lost dq

IR  I 2 R
dt
dt
Oct. 22, 2009
Joules / s = Watts
Physics 208 Lecture 15
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
Parallel
Q1=Q2 =Q
ΔV = ΔV1+ΔV2
1/Ceq = 1/C1 + 1/C2
Series
24
Example: Equivalent Capacitance
C1 = 30  F
C2 = 15  F
C3 = 15  F
C4 = 30  F
C23  C2  C3 15F 15F  30F
C1, C23, C4 in series
1
1
1
1

 

Ceq C1 C23 C4

1
1
1
1



 Ceq  10F
Ceq 30F 30F 30F
C1
C2
V
C3
C4
Parallel
combination
Ceq=C1||C2
25
Charge
Discharge
R
RC Circuits
Time constant
C

  RC
 C
Start w/uncharged
Close switch at t=0
q(t)  C(1 e
I(t) 

e
t / RC
)
t / RC
R
Vcap t   1 et / RC 
R
C
Start w/charged C
Close switch at t=0
qt   qoet / RC
qo /C t / RC
It  
e
R
Vcap t   qo /Cet / RC
26
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
27
Question
What is the current through R1 a long time
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
28
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
29
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
30
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)
31
Magnetic Force on a Current
I

Force on each charge qv  B

Force on length ds of wire
Ids  B
N

 Force on straight section

of wire, length L

F  IBL
Current
Magnetic force
S
Magnetic 
field
32
Law of Biot-Savart
B out of page
r


ds
Each short length of current
produces contribution to
magnetic field.
I in
plane of page
dB
r
ds
r o Idrs  rˆ
dB 
2
4 r
Field from very short
r
section of current ds
r = distance from current element
o  4 107 N / A 2 = permeability of free space
Oct. 22, 2009

Physics 208 Lecture 15
33
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 107 N / A 2 = permeability of free space
34
Magnetic field from loop
Bz
A.
Which of these graphs
best represents the
magnetic field on the
axis of the loop?
z
Bz
B.
y
x
z
Bz
I
z
C.
z
Bz
D.
z
Oct. 22, 2009
Physics 208 Lecture 15
35