Transcript Exam Review I - UW-Madison Department of Physics

Physics 208
Exam Review I
03/10/2006
1
Chapters 23-28
•
•
•
•
•
•
•
•
•
•
Coulomb force between charges
Work done to move a charge
Electric potential as U/q
Electric field as F/q
Motion of charged particle in electric field
Relation between electric field and electric potential
Electric field lines and equipotentials
Electric fields in and near conductors
Electric flux and Gauss' law
Electric field and potential of a point charge, line charge, sheet charge,
spherical shell
• Capacitance of an isolated conductor, and of a pair of conductors
• Calculating capacitance, parallel plate, cylinder, sphere
• Energy in a capacitor, energy density, energy in E-field
• Dielectrics
• Current and resistance, Ohms law
• Power dissipation in a resistor
•03/10/2006
Simple circuits (parallel and series) and RC circuits
2
Charging processes
•Triboelectric:
-glass rod rubbed with silk  positive
-rubber rod rubbed with fur  negative
•Induction
•Conduction
+- +
+++ + ++ + +++++ - -+ +- +
electron flow
+ +
+ + ++ + + +
+
+ +
Less positively charged rod
03/10/2006
Positively
charged
metal
3
Quiz on Charge Conservation
If you rub an inflated balloon against your hair, the two
materials attract each other, as shown in this figure. Fill in
the blank: the amount of charge present in the system of the
balloon and your hair after rubbing is _____ the amount of
charge present before rubbing.
(a) less than
(b) the same as
(c) more than
03/10/2006
4
Quiz on charge sign
3 pithballs suspended from thin threads are charged by
touching them with a charged object. It is found that
pithballs 1 and 2 repel each other and pithballs 2 and 3
repel each other.
Which of the following statements are correct?
1) 1 and 3 carry opposite charge
2) All of them carry charges of the same sign
3) We need to do more experiments to determine the sign
of the charges
1
+
03/10/2006
2
3
+
+
1
-
2
3
-
-
1
+
2
3
+
5
Quiz on charge sign cont.
3 objects are brought close to each other, 2 at a time. When
objects A and B are brought together, they attract. When
objects B and C are brought together, they repel. From this,
we conclude that:
(a) objects A and C possess charges of opposite sign.
(b) all three of the objects possess charges of the same
sign.
(c) we need to perform additional experiments to
determine information about the charges on the objects.
A
B
C
A
B
C
+
-
-
-
+
+
03/10/2006
6
Answer
Answer: (c). In the first experiment, objects A and B may
have charges with opposite signs, or one of the objects may
be neutral since they can attract one the other even due to
induction.
The second experiment shows that B and C have charges
with the same signs, so that B must be charged. But we still
do not know if A is charged or neutral.
03/10/2006
7
Insulators and conductors
One of these isolated charged spheres is copper and the other is rubber. Label
which is which and support your answer with an explanation.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
a) A is rubber and B is copper.
b) A is copper and B rubber
c) They are both copper
Charged rubber rods are placed near a neutral conducting sphere, causing a
redistribution of charge on the spheres. Which of the diagrams below depict the
proper distribution of charge on the spheres? List all that apply.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
a) A and D
b) C and B
c) E
03/10/2006
8
Coulomb’s Law
• Electrical force between two stationary charged particles
• The SI unit of charge is the coulomb (C ), µC = 10-6 C
• 1 C corresponds to 6.24 x 1018 electrons or protons
• ke = Coulomb constant ≈ 9 x 109 N.m2/C2 = 1/(4πeo)
 eo  permittivity of free space = 8.854 x 10-12 C2 / N.m2
03/10/2006
9
Electric Force
03/10/2006
q1q2
Fe  k 2
r
m1m2
Fg  G 2
r
10
Electric Force magnitude
The electric force between 2 like charged objects is F=0.02 N. If
the charge of one of the 2 objects is halved and the distance
separating the 2 objects is doubled, what is the new force?
Q1Q2
Q1Q2
F
3
F  k 2  F' k
  2.5 10 N
2
r
2(2r)
8
03/10/2006
11
Electric Force: Newton’s 3rd law
03/10/2006
12
Forces as vectors
2 small beads with positive charges +3q
and +q are fixed at the opposite ends of
an horizontal rod. A 3rd small charged
bead is free to slide on the rod.
a) Does the result depend on the
sign of the charge on the bead?
b) Find its equilibrium position.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
F2
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
+
F1
d2
d1
1
2
F1
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
03/10/2006
a) The result is independent on the
sign of the charge of the bead
b) Equations to solve the problem
(q’ charge of the bead):
-
F2

3qq'
qq'
F1  F2  k 2  k 2
3 3

d

d  0.63d

d1
d2
1
2

d  d1  d2

13
Electrostatic Potential of 2 like charges
03/10/2006
Electrostatic potential
Q
V (r)  k (point charge Q)
r
14
E field, potential difference and energy conservation
An electron initially travels horizontally in a uniform electric field pointing
upwards with 3 possible directions of the initial velocity of the same
magnitude. Which of the following relation is correct?
a) vb > v > va
b) va > v > vb
c) vb = va > v
d) vb = v = va
Answer: a) the greatest change
x
---------------------
L
in potential V occurs for the case
+
with the max vertical deflection
from the initial position.
This occurs when the electron is projected
at an angle below the horizontal.
K = -U=-qV
03/10/2006
b
+ + + + + + + + + + + + + + ++ a
dV
E 

dx
b
b
 dV  E  dx  V  V
a
a
 Vb  EL
a
15
Example E Fields and Potential
What is the electric field at point B?
q
qd1
cos


2k

d2
d3
12106
9
 1810  2
 3  5.18103 N /C
2 3/2
(3  4 )

E  2k
What is the electric field at
point A?
y
B
x
d2=4 m
d
+12 mC
-12 mC
-
d1=3 m
3m
+
A
3m
 1
1  3 q
E  kq 2 
 k 2
2 
d1 (2d1)  4 d1
Calculate the force on a Q=-3mC charge placed at point A:
F=QE
03/10/2006
16
Example E Potential
B
Calculate the electric potential at B
q q 
VB  k   0
d d 
Calculate the electric potential at A
q
q 
q
VA  k 
 k
d1 2d1  2d1
y
x
d
-12 mC
d2=4 m
-
+12 mC
+
d1=3 m 3 m
A
3m
Calculate the work YOU must do to move a Q=+5 mC charge
from A to B.

03/10/2006
qQ
W  U A  U B  QVA  VB   k
2d1
17
Gauss’ Law
03/10/2006
18
Electric Flux
03/10/2006
19
Plane charge density
The electric charge density for plate 1 is - and for plate 2
is +The magnitude of the electric field associated with
plate 1 is: /(2e0) and the filed lines are shown in the figure.
When they are placed parallel one to the other, the
Magnitude of the field is:
1. /e0 between and ± /(2e0) outside
2. /e0 between and 0 outside
3. None of the above
03/10/2006
20
Insulator and Conductor sphere
A solid nonconductive spherical ball of diameter d = 2R = 1.0 cm is uniformly charged
such that is has an electric field of 20 kV/m at its surface.
1.
How much charge is on the ball? Chose gauss surface with r = R
E 
 E  dA  e
Q
 E 4R 2 
0
3
ER2 2010  0.5 10
Q

k
9 109

Q
e0

R
2 2
 5.551011 C
2. What is the magnitude of the electric field halfway to the center of the ball?
kQ'
inside
E(r)  2
rR
r
Q
Q'
r3


 Q' Q 3
4 R 3 4 r 3
R
3
3
the insulating shell E(r)  k
uniform charge
distribution
r
Qr
R3
R
kQ E
 E' 2   10kV /m
2
2R
2
3. If the ball were made of a conductive material, how would the answers to 1) and 2)
change?
1. Same (all charge on surface)
2. E
= 0 (field inside a conductor is 0)
03/10/2006
21
Capacitors
• C  Q/V
• Parallel Plate C = e0A/d
• Adding
– Series
• Veq=V1+V2
• Qeq = Q1=Q2
1
1
1
 
Ceq C1 C2
– Parallel
• Veq=V1=V2
• Qeq = Q1+Q2
03/10/2006
Ceq  C1  C2
22
Energy stored in a parallel plate
capacitor
The parallel plates of a capacitor with
vacuum between them are given equal and
opposite charge. If the plates are pushed
closer from a distance d to D < d, does the
energy stored in the capacitor
1) Increase
2) Decrease
3) Remain equal
Q2
U
2C
e0 A
e0 A Dd
Ci 
 Cf 

C f  Ci  U f  Ui
d
D
03/10/2006
23
Metal slab in a capacitor
In the slab E = 0 and we have 2 equivalent capacitors in series
1
1 1
d
  2
Ceq C C
3e0 A
In the other case the capacitance between the 2 bottom plates is (for the other
3e0 A
2 plates at the same potential 1/C 0)
C 
03/10/2006

eq
d
24
Capacitor with slab of dielectric
A dielectric slab is introduced between the plates of a
capacitor.
The capacitor is charged and then the dielectric is
removed.
Which of the following statements are correct:
1) The stored energy increases
2) The capacitance increases
3) The potential difference decreases
C0 
03/10/2006
C

V0  V
0 = quantities in vacuum
>1
Q2
U
2C
25
Calculate the equivalent Capacitance
C1 = 10 mF
C2 = 20 mF
C3 = 30 mF
C4 = 40 mF
V = 50 Volts
1
1
1
1
 

 Ceq  6.9mF
Ceq C1 C2  C3 C4
C1
Compare V1 with V4
V  V1  V23  V4
Q  Q1  Q23  Q4
V
V23  V2  V3
Q23  Q2  Q3
Q
Q
Q
Q
 

Ceq C1 C2  C3 C4
C1<C4 V1>V4
03/10/2006
C2
V
C3
C4
parallel
C1, C23, C4 in series
26
Resistors
•
•
•
•
R = V/I
R= L/A
P = RI2
Adding
– Series
• Veq=V1+V2
• Ieq = I1=I2
– Parallel
• Veq=V1=V2
• Ieq = I1+I2
03/10/2006
Req  R1  R2
1
1 1
 
Req R1 R2
27
Resistors in Series
03/10/2006
28
Resistors in Parallel
1
1

Req
R
03/10/2006
29
Light bulbs in Series
Consider the
resistance of the
wire negligible
Since the wire has R =0 all current will flow through it and bulb B will fade.
Instead A will look brigther since the current in A increaseas from
I = V/(RA+RB) to I = V/RA
03/10/2006
30
Light Bulbs of Different Power in
Parallel
A 100 W
B 60 W
Which of the following statements are
correct?
1. A looks dimmer than B
2. A looks as bright as B
3. A looks brighter than B
RA < RB  more current flows in A: IA>IB
RAIA2 > RBIB2
What happens if A and B are connected in series?
03/10/2006
31
Resistors in series and parallel
03/10/2006
32
Kirchhoff’s Rules
I1 = I2 + I3
• Junction Rule:
SIin = SIout
• A statement of Conservation of Charge
• Loop Rule:
e
closed
loop
k

V
i
closed
loop
• A statement of Conservation of Energy

03/10/2006
-
33
Kirchhoff’s laws
R1 = 10 W
R2 = 20 W
R3 = 30 W
V1 = 50 Volts
V2 = 10 Volts
2 loops
Chose 2 currents and their verse
V1  R1I1  R2 (I1  I2 )  R3 I2


 V2  R3 I2  R2 (I2  I1 )
(R2  R3 )I2  V2
I1 
 2.45A
R2
V2
I1
+
-
V1
-+
R1
I2
R2
R3
R2V1  (R1  R2 )V2
I2 
 1.18A
R1(R2  R3 )  R2 R3
Calculate total power delivered by the batteries
Ptot  V1I1  V2I2 134.5W
PR,tot  R1I12  R2 (I1  I2 )2  R3I32 134.5W
03/10/2006
34
RC Circuits: CHARGE OF C
Calculate current immediately after
switch is closed.
R
C
e
I = e/R = 2 A
S1
Calculate current after switch has
been closed for 2 time constants
e
R=10W, C=30 mF
=RC = 0.3 s I(t  2 )  e  0.27A and e=20 Volts
R
dq q
Calculate current after switch has been closed
eR 
dt C
for a very long time
I(t  )  0
2
Calculate
 charge on capacitor after switch
has been closed for a long time
03/10/2006
q(t  )  Ce  0.6C

q(t)  Ce(1 e t / RC )

I(t) 
e
R
e t / RC
35
RC Circuits
You are given a 5-pack of R=2W resistors and a
5-pack of C=2mF capacitors. How do you
configure them to produce a circuit of 21 ms
time constant RC circuit?
Time constant,  = RC = 21 ms = 7 * 3 ms = 7 W * 3 mF
03/10/2006
R 7
3R   R  3W
2 2
C
3
C C
2
2
36