Download Lecture #23 04/26/05

Announcements
Today
Selected Review of older material
Even the full notes from today (online) are not
comprehensive
Some Basic definitions
Maxwell’s Equations
Circuits
Final
Comprehensive!!! Special emphasis last few
lectures
12 Multiple Choice
5 Short Answer
1 Problem (chosen from a few) hard!
Coulomb’s Law
•Like charges repel, unlike charges attract
•Force is directly along a line joining the two charges
q1
q2
r
ke q1q2
Fe 
2
r
ke = 8.988109 Nm2/C2
•An inverse square law, just like gravity
•Can be attractive or repulsive – unlike gravity
•Constant is enormous compared to gravity
Quiz
Two test charges are brought separately into the vicinity of a charge +Q.
First, test charge +q is brought to point A a distance r from charge +Q. Next, the +q
charge is removed and a test charge +2q is brought to point B a distance 2r from
charge +Q. Compared with the electric field of the charge at A, the electric field of
the charge at B is:
• T
+Q
+q
+Q
A
+2q
B
A) Greater
B) Smaller
C) The same.
Electric Field Lines
•Graphical Illustration of Electrical Fields
•Lines start on positive charges and end on negative
•Number of lines from/to a charge is proportional to that charge
•Density of lines tells strength of field.
+
-
+
-
Electric Field Lines
Consider the four field patterns below:
Assuming that there are no charges in the region of space depicted,
which field pattern(s) could represent electrostatic field(s)?
Currents make Magnetic Fields
Magnetic
Field B
Current I
•Magnetic Fields are created by electric currents
•They are perpendicular to both the direction of the current and the
separation between the wire and the point
P
Magnetic
•Falls off with distance from wire
Field B
r̂
Current I
•Direction of magnetic field is given by right-hand rule
The Biot-Savart Law
0 Ids  rˆ
dB 
2
4 r
0 I ds  rˆ
B
2

4
r
0  4 107 T  m/A
•Must add up, i.e. integrate, the contribution from the entire
current distribution
Magnetic
Field B
ds
Current I
Fields we know and love: Wires
•Infinite straight wire
0 I
B
2 a
•Inside the wire
B=0I(r)/(2R2)
•Finite Straight wire
0 Ikˆ
B
 cos1  cos 2 
4 a
Fields we know and love: Other
•A single circular loop (Current Coil), has an axial field
B( z ) 
0iR 2
2( R 2  z 2 )3 / 2
•A solenoid has a uniform field inside
B
 0 NI
L
•A plane of current has a uniform field
B
 0 NI
2L
Inductance and Inducers
N turns
Length l
•Any loop of wire (especially a coil) will
have a certain reluctance to change its
current
•This reluctance is called the inductance
•Denoted by the letter L
dI
V  L
dt
V s
1 H 1
A
•Unit is Vs/A also called a henry (H)
•Circuit diagram looks like this:
R
Current I
Capacitors
•Capacitors accumulate separated charge
•Capacitance is measure of how well a capacitor stores
charge.
•Farad (F) is C/V, measures capacitance
•Capacitors also store energy
•Typical capacitances measured in mF,F or pF
2
1
Q
2
U  C  V  
2
2C
Q  C V
The capacitance depends on the geometry of the capacitor and the
presence or absence of a dielectric medium .
Resistance
•Define resistance as the ratio of the voltage to the current
V  IR
•Resistance is measured in units of Ohms ()
1V
1 
1A
•Resistance is always positive
•Current always flows from positive to negative
•Note: This is not Ohm’s law! We can (in principle) always use this
Electric Field E
Current I
Length L
Area A
Kirchoff’s Rules
•The total current flowing into a point must equal the total
current flowing out of a point [conservation of charge]
•The total voltage change around a loop must total zero
I2
I3
V1
V2
–
+
I1
I3=I2+I1
V3
V1 + V2 + V3 = 0
Using Kirchoff’s Rules
•Draw a circuit diagram and label everything known or
unknown!
•To every series of components, assign a direction to the
current I (don’t worry if you get it wrong, the result will be
correct just negative)
•You must be consistent however after you assign a
direction!
•Write down conservation of charge at each vertex
•Write down one equation for each loop
•Solve all equations
You might end up with many equations, but I trust that
you can solve simultaneous equations.
Note on Kirchoff’s Rules
• For complex circuits one obtains differential equations
• Z.B. RC,LC, RL (RLC not discussed) circuits
• Brute force! Will always work (sans mistakes)
• You can make life easier by
1) using obvious parallel and series relations
2) Using thoughfully picked loops
A)Remember the Res-monster maze in class
(Ch28 Circuits p653)
Example
Find the equivalent capacitance of the combination in the
figure below.
Assume that C1 = 4 µF, C2 = 4 µF, and C3 = 2 µF.
What is the equivalent capacitance of the whole system?
A) 10 F
C) 4 F
B) 1 F
D) 2.5 F
Series Circuits
•Two or more resistors connected together can be treated as
one giant resistor
•For resistors in series, the current is the same through both of
them
R
R
1
2
V1 = R1 I
V2 = R2 I
V = V1+ V2 = R1 I + R2 I = (R1 + R2 )I
R = R1 + R2
In Series:
C1
C2
C3
1 1
1
1
 

C C1 C2 C3
Charges are the same on each capacitor
Parallel Circuits
•Two or more resistors connected together can be treated as
one giant resistor
•For resistors in parallel, the voltage is the same across both of
them
V = R2 I2
1 1
R2
R1
V = R1 I1
1
 
R R1 R2
1 1 
V V
 V   
I  I1  I 2 

R1
R2
 R1 R2 
In Parallel:
C1
C2
C3
Voltages are the same also
Same Voltages
across each capacitor
C  C1  C2  C3
RL - Circuits
I
•What happens when the
switch S is closed at t = 0?
•Let I be the current in
the circuit
R
+
E
L
–
•Use Kirchoffs rule for loops on the circuit
dI E R
dI
  I
0  E  RI  L
dt L L
dt
E
E
 Rt / L
t /
I  t   1  e
 1  e 

R
R
  L/ R
S
Does this look familiar?
RC–Circuits vs RL-Circuits
dq
q

R 0
dt
c
q  C 1  e

t
RC
   t RC 
I  e

R



dI
0  E  RI  L
dt
 
 tR 
I  1  e L 
R

RC–Circuits vs RL-Circuits
   t RC 
I  e

R

•At t=0, ordinary wire
•As t-> infinity, broken wire
 
 tR 
I  1  e L 
R

•At t=0, broken wire, little current
for small t
•As t-> infinity, ordinary wire
Quiz
The
Thediagrams
diagramsshow
showthree
threecircuits
circuitswith
withidentical
identicalbatteries,
batteries,identical
identicalinductors,
inductors,
and
after
the
switch
isisclosed
which
the
andidentical
identicalresistors.
resistors.Just
A
Just
very
after
long
the
time
switch
later,
which
closedhas
which
thehas
has
least
theleast
greatest
current
currentthrough
throughthe
thebattery?
battery?
•What if we had capacitors instead of
inductors? Think about this yourself.
LC - Circuits
+
E
C
–
•Switch S1 is closed, then
opened.
•At t = 0, switch S2 is
closed.
•What happens?
I
L
S2
S1
V  t  0  E
Q  C V
d  V 
d Q
dI
V  L   L 2  CL
dt
dt
dt 2
2
2
V  E cos t 
E
C
–
d 2  V 
V  CL
2
dt
+
LC - Circuits
L
S1
S2
  1/ LC
dQ
d
I
 C  V   C sin  t 
dt
dt
These equations
Electric Potential
Point A
•Path you choose does not matter.
(conservative)
B
U  q  E  ds
A
•Factor out the charge – then you
have electric potential V
Electric Field E
Point B
U
V 
   E  ds
q
A
B
Potential difference in constant field
B
V    E  ds
V  E  s
A
Equipotential surfaces
•Electric Fields are
perpendicular to
equipotential surfaces
+
•Positively charged particles
are attracted to low
potential areas (and vice
versa)
Electric
Field
-
High Potential
Low Potential
Binding Energy and mass
•The mass of a nuclei is less than the mass of its parts.
•There must exist a binding energy, i.e., an energy keeping the nuclei
together. Remember the energy is from electrostatic and nuclear
forces!
•The binding energy is so large that there is a measurable mass
difference. (Unlike binding due to other forces!)
•Eb(MeV)=(Zmp+Nmn-MA) x 931.494MeV/amu
•Binding energy per nucleon is the binding energy divided by the
mass number
Maxwell’s Equations
Integral Form
qin
 E  dA  
S
 B  dA  0
S
0
dE
 B  ds  0 I  0 0 dt
dB
 E  ds   dt
Differential Form

E 
0
 B  0
dE
  B  0 J  0 0
dt
dB
 E  
dt
Maxwell and Lorentz Force Law
Differential Form

E 
0
 B  0
dE
  B  0 J  0 0
dt
dB
 E  
dt
~
~ ~ ~
F  qE  qv  B
Gauss’s Law
•Flux out of an enclosed region
depends only on total charge inside
E 
qin
0
charge q
A positive charge q is set down outside a sphere. Qualitatively, what
is the total electric flux out of the sphere as a consequence?
A) Positive
B) Negative
C) Zero
D) It is impossible to tell from the given information
Gauss’s Law II
•Magnetic field lines are always loops, never start or end on
anything (no magnetic monopoles)
•Net flux in or out of a region is zero
 B  dA  0
•Gauss’s Law for electic fields
 E   E  dA  4 ke qin
Ampere’s Law Generalized
•When there is a net current flowing into a region, the charge
in the region must be changing, as must the electric field.
•By Gauss’s Law, the electric flux must be changing as well
•Change in electric flux creates magnetic fields, just like currents do
•Displacement current – proportional to time derivative of electric
flux
dE
Id  0
dt
dE
 B  ds  0 I  0 0 dt
Magnetic fields are created both
by currents carried by conductors
(conduction currents) and by
time-varying electric fields.
Quiz: Ampere’s Law
•Consider three wires with
current flowing in/out as shown
•Consider three different loops
surrounding the wires
X
Y
Which of the loops has the largest and
smallest integrals of the magnetic
field around the loops drawn?
A) X > Y > Z
C) Y > Z > X
B) X > Z > Y
D) Y > X > Z
2A
3A
1A
Z