Transcript PowerPoint

Physics 212
Lecture 29
Course Review
• The Topics For Today
–
–
–
–
Electric Fields/Gauss’ Law/Potential
Faraday’s Law
RC/RL Circuits
AC Circuits
Physics 212 Lecture 29, Slide 1
Music
Who is the Artist?
A)
B)
C)
D)
E)
Ray Charles
Solomon Burke
Henry Butler
Johnny Adams
Otis Redding
“Rediscovered” Soul singer from 60’s
Absolutely Beautiful Voice
Why?
Highly Recommended DVD:
“One Night History of the Blues”
Great singer to end the course
Electric Field
Two curved rods each have charge +Q uniformly distributed over their length
Which statement best describes the electric field due to these two rods at the midpoint
between the two rods (marked by an X)
A. E points up
B. E points down
C. E = 0
48%
E from top arc points down
Horizontal components cancel
E from bottom arc points up
Top arc produces smaller horizontal components
Etotal points down
Calculation:
Etop  

dq
40 r
Q
2rq top
Etop 
2
q
cos q
Etop 
Q
sinq
q
1
Q
40 r 2rq top
2
sin q top
40 r 2 q top
q to p
2  rdq cos q
0
q
Physics 212 Lecture 29, Slide 3
Conductors, Potential Energy
A thin non-conducting spherical shell carries a positive charge Q spread uniformly
over the surface of the shell. A positive test charge q is located inside the shell.
Compare UA, the potential energy of the test charge in position A (near the shell)
with UB, the potential energy of the test charge in position B (center of the shell)
A. UA > UB
B. UA = UB
C. UA < UB
53%
Potential Energy is a measure of work done by E field
Spherical symmetry & Gauss’ law
E = 0 inside shell
E = 0 inside shell
No work done to move q
No change in potential energy !
Physics 212 Lecture 29, Slide 4
Electric Potential, Gauss’ Law
A thin non-conducting spherical shell of radius a carries a
uniformly distributed net surface charge 2Q. A second thin nonconducting shell of radius b carries a uniformly distributed net
surface charge 3Q. The two shells are concentric. Calculate the
potential difference V = V(a) – V(b) between the inner and outer
shells.
37%
(A)
1  2Q 3Q 



40  a
b 
(B)
1  3Q 2Q 



40  b
a 
(C)
(D)
0
(E)
ALWAYS START
FROM DEFINITION
OF POTENTIAL
a 
V    E  dr
1  2Q 2Q 



40  a
b 
1  2Q 2Q 



40  b
a 
Spherical symmetry &
Gauss’ law determines E
1 2Q
a < r < b: E 
40 r 2
b
2Q
V  
40
a dr
 2
br
2Q  1 1 
V 
  
40  a b 
Physics 212 Lecture 29, Slide 5
Electric Potential, Gauss’ Law
A thin non-conducting spherical shell of radius a carries a
uniformly distributed net surface charge 2Q. A second thin nonconducting shell of radius b carries a uniformly distributed net
surface charge 3Q. The two shells are concentric.
Calculate the potential difference V = V(a) – V(0) between the
inner shell and the origin.
45%
(A)
1  2Q 3Q 



40  a
b 
(B)
1  3Q 2Q 



40  b
a 
(C)
0
(E)
Spherical symmetry &
Gauss’ law determines E
r < a:
(D)
E0
1  2Q 2Q 



40  a
b 
1  2Q 2Q 



40  b
a 
 
V    E  dr  0
a
0
Physics 212 Lecture 29, Slide 6
Gauss’ Law, Conductors
A solid conducting sphere of radius a is centered on the
origin, and carries a total charge Q1. Concentric with this
sphere is a conducting spherical shell of inner radius b and
outer radius c, which carries a total charge Q2. What is the
surface charge density, b, on the inner surface of the outer
spherical shell (r = b)?
59%
 Q1
(D)  Q1
(A)
Q2  Q1
(C)
4b 2
4a 2
2
2
4 (b  a )
 Q1
 Q1
(B)
(E)
2
4a
4b 2
Charge must be
induced to insure
E = 0 within
conducting shell
Spherical symmetry &
Gauss’ law determines E
  Qenclosed
 E  dA 
0
E  4r 2  0
Qenclosed  Qa  Qb  Q1  ( Q1 )

 Q1
4b 2
Physics 212 Lecture 29, Slide 7
Gauss’ Law
Q1 = - 3 mC
A solid conducting sphere of radius a is centered on the
origin, and carries a total charge Q1. Concentric with this Q2 = 5 mC
sphere is a conducting spherical shell of inner radius b and
outer radius c, which carries a total charge Q2.
Which of the following graphs best represents the magnitude
of the electric field at points along the positive x axis?
70%
a < r < b: E = kQ1/r2
r > c: E = k(Q1+Q2)/r2
Q1 = -3mC
r < a: E = 0
Eliminate (a) and (c)
b < r < c: E = 0
Eliminate (e)
Q1 + Q2 = +2mC
For r > c,
E must be less than the
continuation of E from a to b
Physics 212 Lecture 29, Slide 8
Electric Potential
Q1 = - 3 mC
A solid conducting sphere of radius a is centered on the
origin, and carries a total charge Q1. Concentric with this Q2 = 5 mC
sphere is a conducting spherical shell of inner radius b and
outer radius c, which carries a total charge Q2.
If the inner conducting sphere were replaced with an
insulating sphere having the same charge, Q1, distributed
uniformly throughout its volume, the magnitude of the
potential difference |Vb – V0| would
53%
A. increase
B. decrease
ALWAYS START
FROM DEFINITION
OF POTENTIAL
b 
V    E  dr
0
C. stay the same
Break integral into two pieces
a  b 
V    E  dr   E  dr
0
conductor: = 0
insulator:
0
a
same for
conductor
& insulator
Physics 212 Lecture 29, Slide 9
RL Circuits
1
In the circuit below, V = 6 Volts, R = 10 Ohms, L = 100 mH. The
switch has been open for a long time. Then, at t = 0, the switch is
closed. What is the time constant for the current through the
inductor?
30%
A. R/L
B. R/2L
C. L/R
D. 2L/R
E. L/2R
2
I – I1
I
L
L
dI1
 ( I  I1 ) R  0
dt
1
I1
2
L
dI1
 IR  V  0
dt
IR  V  L
dI1
dt
dI1
dI
 V  L 1  I1R  0
dt
dt
2L
dI1
 I1R  V
dt
Strategy: Back to First Principles
• The time constant is determined from
a differential equation for the current
through the inductor.
• Equation for current through inductor
obtained from Kirchhoff’s Rules

" L" 2 L

" R" R
Physics 212 Lecture 29, Slide 10
Faraday’s Law
A rectangular wire loop travels to the right with constant velocity, starting in a region of no magnetic
field, moving into a region with a constant field pointing into the page (shaded rectangle below), and
continuing into a region of no magnetic field. Which plot below best represents the induced current
in the loop (Iloop) as it travels from the left through these three regions? Note: a positive Iloop
corresponds to a counter-clockwise current.
Current induced only when flux is changing.
Flux is changing because loop is moving.
As loop enters field, current will be induced
to reduce the increase in flux (Lenz’ law):
counterclockwise current generates B field
pointing out of the page.
When loop is completely inside field, flux is
constant, therefore, current is zero
As loop leaves field current will be induced
to produce B field pointing into the page.
Physics 212 Lecture 29, Slide 11
Faraday’s Law
Two fixed conductors are connected by a resistor R = 20 Ohms. The two fixed conductors are
separated by L = 2.5 m and lie horizontally. A moving conductor of mass m slides on them at a
constant speed, v, producing a current of 3.75 A. A magnetic field with magnitude 5 T points out of
the page. In which direction does the current flow through the moving conductor when the bar is
sliding in the direction shown?
A. to the right
B. to the left
R = 20 Ohms
H
v = 6 m/s
Flux is changing because area of the loop is
changing.
As the moving conductor slides downward
the area and therefore the flux increases.
Therefore a current must flow in the
clockwise direction to generate a magnetic
field pointing into the page
L
Physics 212 Lecture 29, Slide 12
Faraday’s Law
Two fixed conductors are connected by a resistor R = 20 Ohms. The two fixed conductors are
separated by L = 2.5 m and lie horizontally. A moving conductor of mass m slides on them at a
constant speed, v, producing a current of 3.75 A. A magnetic field with magnitude 5 T points out of
the page.
At what speed is the conductor moving?
A. 1 m/s
B. 3 m/s
R = 20 Ohms
C. 5 m/s
Faraday’s Law:
D. 6 m/s
E d  
E. 9 m/s
d
d
   IR 
dt
dt
As the moving conductor slides downward
the area of the loop, and therefore the flux,
increases.
v = 6 m/s
IR 
d
dA
dH
B
 BL
 BLv
dt
dt
dt
v
IR 3.75 A  20 

 6 m/s
BL
5 T  2.5 m
Physics 212 Lecture 29, Slide 13
Phasor diagram at t = 0
a
Phasers
What is VC at t = /(2w)
(A)
 VC max sin a
(C)
 VC max cos a
(B)
 VC max sin a
(D)
 VC max cos a
Phasor diagram at t = /(2w)
VR
a
VC
VL
Voltage is equal to
projection of phasor
along vertical axis
Physics 212 Lecture 29, Slide 14
.
Ampere’s Law Integrals
Two infinitely long wires carrying current run into the page as indicated. Consider a closed
triangular path that runs from point 1 to point 2 to point 3 and back to point 1 as shown.
Which of the following plots best shows B·dl as a function of position along the closed path?
B
B
The magnetic field points in the azimuthal direction and is oriented
clockwise.
No current is contained within the loop 1-2-3-1, therefore NOT (A)
The magnetic field falls off like 1/r (r is the distance from the wire).
From 3 to 1, B·dl is < 0, largest magnitude near wires.
From 1 to 2, B·dl is > 0
Physics 212 Lecture 29, Slide 15
62%
 
Flux definition:    B  d A  BA  Bwh
Faraday’s law:   
dB
 2T / s
dt
d
dB
  wh
dt
dt
I
 9/5

 .012 A
R 150
Physics 212 Lecture 29, Slide 16
65%
Current is determined by time rate of change of the flux
d/dt is determined by dB/dt
dB/dt (6s) = dB/dt (5 s) = -2 T/s
The induced currents at t = 5s and t = 6s are equal
Physics 212 Lecture 29, Slide 17
Faraday’s Law
76%
Current induced because flux is changing
Flux is changing beause B is changing
At t = 5 seconds, B is positive, but decreasing
Lenz’ law: emf induced to oppose change that brought it into being
Induced current must produce positive B field
Positive B field produced by counterclockwise current
Physics 212 Lecture 29, Slide 18