Physics 122B
Electricity and Magnetism
Lecture 16
Review and Extension
May 2, 2007
Martin Savage
Lecture 16 Announcements
 Lecture HW Assignment #5 is due by
10pm this evening.
Lecture HW Assignment #6 is posted on
Tycho and is due next Wednesday at
10pm.
The second midterm exam is this Friday.
It will cover everything up to the end of
Wednesdays lecture, emphasizing the
most recent material, but assumes
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Physics
- Lecture 16
2
understanding of
all122Bmaterial
inclusive.
Current and Drift Velocity
If the electrons have an average drift
speed vd, then on the average in a time
interval Dt they would travel a distance Dx in the wire,
where Dx = vd Dt. If the wire has cross sectional area
A and there are n electrons per unit volume in the wire,
then the number of electrons moving through the cross
sectional area in time Dt is Ne = n A Dx = n A vd Dt = i Dt .
Therefore,
This table gives n
for various metals.
i  nAvd
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A Puzzle
We discharge a capacitor that
has been given a charge of Q = 16 nC,
using a copper wire that is 2 mm in
diameter and has a length of L = 20
cm. Assume that the electron drift
speed is vd = 10-4 m/s.
How long does it take to discharge
the capacitor? (Note that L/vd =
0.2m/10-4 m/s = 2000 s = 33.3 min.)
Points to consider:
1. The wire is already full of electrons.
2. The wire contains about 5x1022
conduction electrons.
3. Q = 16 nC requires about 1011 electrons.
4. A length L’ of wire that holds 16 nC
of conduction electrons is 4x10-13 m.
5. L’/vd = 4x10-9 s = 4 ns. That is roughly the discharge time.
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Establishing the
Electric Field in a Wire (2)
The figure shows the region of the wire near the neutral midpoint. The
surface charge rings become more positive to the left and more negative
to the right.
In Chapter 26, we found that a ring of charge makes an on-axis E field that:
1. Points away from a positive ring and toward a negative ring;
2. Is proportional to the net charge of the ring;
3. Decreases with distance from the ring.
The non-uniform surface charge distribution creates an E field
inside the wire. This pushes the electron current through the
wire
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A Model of Conduction (1)
Suppose E  0 : K  12 mv 2  32 kT
gives v  105 m/s. However, v  0.
Now turn on an E field. The straight-line
trajectories become parabolic, and because of the
curvature, the electrons begin to drift in the
direction opposite E, i.e., “downhill”.
ax=F/m=eE/m so
vx=vix+ axDt = vix+ Dt eE/m
This acceleration increases an electrons
kinetic energy until the next collision, a “friction”
that heats the wire….energy is imparted to the
atoms of the lattice.
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The Current Density in a Wire
I  ei  nevd A
I
J  current density   nevd
A
Example: A current of 1.0 A passes through a 1.0 mm diameter
aluminum wire. What is the drift speed of the electrons in the wire?
I
I
(1.0 A)
6
2
J  2 

1.3

10
A/m
A r
 (5.0 104 m)2
J
(1.3 106 A/m2 )
4
vd 


1.3

10
m/s
28
-3
19
ne (6.0 10 m )(1.60 10 C)
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Kirchhoff’s Junction Law
 I  I
in
I
i
out
 0; summed over all the currents to any "junction".
i
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Conductivity and Resistivity
The current density J = nevd is directly proportional to the electron
drift speed vd. Our microscopic conduction model gives vd = eE/m, where
 is the mean time between collisions. Therefore:
 e E  n e 
J  nevd  n e 
E

m
 m 
2
The quantity ne2/m depends only on the properties of the conducting
material, and is independent of how much current density J is flowing.
This suggests a definition:
2
ne 
conductivity:  
so J  J
 E=
m
E
This result is fundamental and tells us three things:
(1) Current is caused by an E-field exerting forces on charge carriers;
(2) Current density J and current I=JA depends linearly on E;
(3) Current density J also depends linearly on . Different materials
have different  values because n and  vary with material type.
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Resistivity and
Conducting Materials
For many applications, it is more convenient to use inverse of conductivity,
which is called the resistivity, denoted by the symbol r:
1
m
resistivity: r   2
 ne 
r

Thus, the current density
is J = E = E/r. Here are the
conducting properties of
common materials:
Units of resistivity are
Wm
Units:
ohms = W = Nm2/CA = Nm2s/C2
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Resistors and Resistance
Conducting material that carries current
along its length can form a resistor,
a circuit element characterized by an
electrical resistance R:
R ≡ rL/A
where L is the length of the conductor and A is
its cross sectional area.
R has units of ohms ( W ).
Multiple resistors may be combined in
series, where resistances add, or in parallel,
where inverse resistances add.
I
Rnet
Rnet
For identical resistors can
Series
Connection
[L]:
simply
add
the lengths
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Rnet  R1  R2  R3
For identical resistors can
simply Connection
add the areas
Parallel
[(1/A)]:
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1
1
1
1
 

Rnet R1 R2 R3
11
The Potential Energy of
Like and Unlike Charge Pairs
U elec 
Kq1q2
1 q1q2

r
4 0 r
This approach can be applied to pairs of electrically charged particles,
whether they have the same or opposite charges. However, for like-sign
particles (a) the system energy is positive and decreases with separation,
while for opposite-sign particles (b) the system is typically “bound”, so that
the net energy is negative and increases (closer to zero) with increasing
separation.
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The Electric Force as
a Conservative Force
The electrical force is a “conservative
force”, in that the amount of energy involved in
moving from point i to point f is independent of
the path taken.
This can be demonstrated in the field of a
single point charge by observing that tangential
paths involve no change in energy (because r is
constant). Therefore, an arbitrary path can be
approximated by a succession of radial and
tangential segments, and the tangential
segments eliminated.
What remains is a straight line path from
the initial to the final position of the moving
charge, indicating a net work that will be the
same for all possible paths.
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Multiple Point Charges
We have established that both energy and electrical forces
obey the principle of superposition, i.e., they can be added linearly
without “cross terms”. Therefore, for multiple point charges:
N
Kqi q j
i<j
ri j
U elec  
Here, “i<j” means that for summing over N particles, the sum
over i runs from 1 to N, and the sum over j runs from i+1 to N for
each value of i. This it a mathematical trick to avoid counting pairs
of point charges twice or having i=j terms, which would give a zero
in the denominator.
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The Electric Potential
In Chapter 25 we introduced
the concept of an electric field E,
which can be though of as a
normalized force, i.e., E = F/q,
the field E that would produce a
force F on some test charge q.
We can similarly define the
electric potential V as a chargenormalized potential energy, i.e.,
V=Uelec/q, the electric potential V
that would give a test charge q an
electric potential energy Uelec
because it is in the field of some
other source charges.
We define the unit of electric potential as the volt: 1 volt = 1 V =
1 J/C = 1 Nm/C. Other units are: kV=103 V, mV=10-3 V, and mV=10-6 V.
Example: A D-cell battery has a potential of 1.5 V between its terminals.
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The Electric Potential Inside
a Parallel Plate Capacitor
Consider a parallel-plate capacitor with
d  3.0 mm,  =4.42 10-9 C/m2


E   ,  to -    500 N/C, to right 
 0

U elec  U q+sources
 qEs for a charge q,
located at s from the - plate
(with U0=0)
DVC  V  V  Ed  1.5 V
DV
d
E  C ; DVC  Ed 
d
0
V  U elect / q  Es (V inside a parallel-plate
capaitor with V=0 @ - plate)
Note that 1 N/C = 1 V/m.
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Graphical Representations
of Electric Potential
Distance from + plate
DVC
x

V  Es 
(d  x)  DVC 1  
d
 d
This linear relation can be represented as a graph, a set of
equipotential surfaces, a contour plot, or a 3-D elevation graph.
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Field Lines and Contour Lines
Field lines and equipotential
contour lines are the most
widely used representations to
simultaneously show the E field
and the electric potential. The
figure shows the field lines and
equipotential contours for a
parallel plate capacitor.
Remember that for both the
field lines and contours , their
spacing, etc, is a matter of
choice.
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Rules for Equipotentials
1. Equipotentials never intersect
other equipotentials. (Why?)
2. The surface of any static
conductor is an equipotential
surface. The conductor volume
is all at the same potential.
3. Field line cross equipotential
surfaces at right angles. (Why?)
4. Dense equipotentials indicate a
strong electric field. The potential V decreases in the direction
in which the electric field E points, i.e., energetically “downhill”
for a + charge
5. For any system with a net charge, the equipotential surfaces
become spheres at large distances.
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Visualizing the Potential
of a Point Charge
The potential of a point charge can be
represented as a graph, a set of equipotential
surfaces, a contour map, or a 3-D elevation
graph.
Usually it is represented by a graph or a
contour map, possibly with field lines.
+
Spherical Shells
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The Electric Potential
of Many Charges
The principle of superposition allows us to
calculate the potentials created by many point
charges and then add the up. Since the potential V
is a scalar quantity, the superposition of potentials
is simpler than the superposition of fields.
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qi
V 
i 4 0 ri
1
21
Example: The Potential
of Two Charges
What is the potential at point p?
p
1 q1
1 q2
1  q1 q2 


  
4 0 r1 4 0 r2 4 0  r1 r2 
 (2.0 10-9 C) (1.0 10-9 C) 
9
 (9.0 10 Vm/C) 


(0.050
m)
(0.040
m)


Vp 
 135 V
Note that:
1/40 = 9.0 x 109 Nm2/C2
= 9.0 x 109 Vm/C,
which, for problems like this,
are more convenient units.
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Potential of a Disk of Charge
Von z axis 
Q
2 0 R 2

R2  z 2  z

 Q  R2  z 2  z


2

R
R
0


Von z axis  V0
R2  z 2  z
R
1   R / z  1
2
 V0
R / z
 1  1  R / z 2   1 
2

   1V R  1 Q
 lim V0 
 2 0
z 
R
/
z
z 4 0 z
 




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Example: The Potential of a Dime
A dime (diameter 17.5 mm) is given a charge of Q=+5.0 nC.
+ +
+ + + +
+ + + +
+ +
V0 
(a) What is the potential of the dime at its surface?
(b) What is the potential energy Ue of an electron 1.0 cm
above the dime (on axis)?
Q
2 0 R
Ve  V0
 4(9.0  109 Vm/C)(5.0 109 C) /(.0175 m)  10,300 V
R2  z 2  z
 3,870 V
R
U e  eVe  (1.6 1019 C)(3.87 103 V)  6.2 1016 J
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Finding E from V
DV 
DU q+sources
q
 Es  
Ex  
W

  Es Ds
q
DV
 dV 
 lim  

Ds Ds 0  ds 
dV
dV
dV
; Ey  
; Ez  
dx
dy
dz
 dV ˆ dV ˆ dV ˆ  In other words, the E field components are
E  
i
j
k  determined by how much the potential V
dy
dz  changes in the three coordinate directions.
 dx
dV
d  1 q
1 q
For a point charge: E  Er  
 

dr
dr  4 0 r  4 0 r 2
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Example:
Finding E from the Slope of V
An electric potential V in a particular
region of space where E is parallel to the
x axis is shown in the figure to the right.
Draw Ex vs x.
DV
Ex  
Dx
DV/Dx  (20 V)/(0.020 m)  1000 V/m
0<x<2 cm 
 Ex  1000 V/m
DV/Dx  (0 V)/(0.020 m)  0 V/m
2<x<4 cm 
 Ex  0 V/m
DV/Dx  (  20 V)/(0.040 m)  500 V/m
4<x<8 cm 
 Ex  500 V/m
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Example: Finding the E-Field
from Equipotential Surfaces
The figure shows a contour map of a
potential.
Estimate the strength and direction
of the electric field at points 1, 2, and 3.
E1  DV1/Dx  (100 V)/(0.020 m)  5000 V/m
"Downhill" direction  up
E2  DV2 /Dx  (100 V)/(0.040 m)  2500 V/m
"Downhill" direction  left
E3  DV3 /Dx  (100 V)/(0.025 m)  4000 V/m
"Downhill" direction  ~ 30 left of vertical
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Kirchhoff’s Loop Law
Since the electric field is
conservative, any path between
points 1 and 2 finds the same potential
difference. Any path can be approximated
by segments parallel and perpendicular to
equipotential surfaces, and the
perpendicular segments must cross the
same equipotentials.
Since a closed loop starts and ends at
the same point, the potential around the
loop must be zero. This is Kirchhoff’s Loop
Law, which we will use later.
DVloop    DV i  0
i
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A Conductor in
Electrostatic Equilibrium
A conductor is in electrostatic
equilibrium if all charges are at rest and
no currents are flowing. In that case,
Einside=0. Therefore, all of it is at a
single potential: Vinside=constant.
Rules for conductor.
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Forming a Capacitor
Any two conductors can form a
capacitor, regardless of their shape.
Q
C
DVC
The capacitance depends only on the geometry of the conductors,
not on their present charge or potential difference.
(In fact, one of the conductors can be moved to infinity, so the
capacitance of a single conductor is a meaningful concept.)
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Combining Capacitors
Parallel: Same DV, but different Qs.
Cparallel 
Q  Q2  Q3 
Q
 1
DVC
DVC
 C1  C2  C3 
Series: Same Q, but different DVs.
Cseries 
Q
Q

DVC DV1  DV2  DV3 

1
 DV1 / Q    DV2 / Q    DV3 / Q  

1
1/ C1  1/ C2  1/ C3 
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 C1 || C2 || C3 ||
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Energy Stored in a Capacitor
1
DU  dqDV  qdq
C
Q
2
1
1 Q
U C   qdq  2
C0
C
2
Q
U C  12
 12 C DVC 2
C
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Energy in the Electric Field
Volume of E-field
U C  C DV 
1
2
2
1
2
0 A
d
 Ed 
2

0
2
 Ad  E 2
energy stored U C  0 2
uE 

 E
storage volume Ad 2
Example: d=1.0 mm, DVC=500 V
E
DVC
500 V
5


5.0

10
V/m
d
1.0 10-3 m
uE 
0
2
E 
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2
1
2
 5.0 10
5
V/m  /  4  9.0 109 Vm/C   1.1 J/m 3
2
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Dielectric Materials*
There is a class of polarizable dielectric
materials that have an important application in
the construction of capacitors. In an electric
field their dipoles line up, reducing the E field
and potential difference and therefore
increasing the capacitance:
E off
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 0 A
Q
C

DVC
d
E on
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Electric Fields and Dielectrics
In an external field EO, neutral molecules can polarize. The induced
electric field E’ produced by the dipoles will be in the opposite direction
from the external field EO. Therefore, in the interior of the slab the
resulting field is E = EO-E’.
The polarization of the material has the net effect of producing a sheet
of positive charge on the right surface and a sheet of negative charge on
the left surface, with E’ being the field made by these sheets of charge.
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Capacitors and Dielectrics*
If a capacitor is connected to a
battery, so that it has a charge q,
and then a dielectric material of
dielectric constant e is placed in
the gap, the potential is unchanged
but the charge becomes eq.
If a capacitor is given a charge
q, and then a dielectric material of
dielectric constant e is placed in
the gap, the charge q is unchanged,
but the potential drops to V/e.
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Resistors and Ohm’s Law
J E 
I  JA 
R
rL
A
E
r
E
r
A
DV / L
r
I
A
DV
;
R
DV
 r L / A
DV  IR
Ohm’s Law
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Ohmic and Non-ohmic Materials
Despite its name, Ohm’s Law is not a law of
Nature (in the sense of Newton’s Laws). It is
a rule about the approximately linear
potential-current behavior of some materials
under some circumstances.
Important non-ohmic devices:
1. Batteries, where DV=E is determined by
chemical reactions independent of I;
2. Semiconductors, where I vs. DV can be very
nonlinear;
3. Light bulbs, where heating changes R;
4. Capacitors, where the relation between I and
DV differs from that of a resistor.
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The Ideal Wire Model
In considering electric circuits, we will make
the following assumptions:
1. Wires have very small resistance, so that
we can take Rwire=0 and DVwire=0 in circuits.
Any wire connections are ideal.
2. Resistors are poor conductors with constant
resistance values from 10 to 108 W.
3. Insulators are ideal non-conductors, with
R=∞ and I=0 through the insulator.
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Circuit Elements & Diagrams
These are some of the symbols we will use to represent objects in
circuit diagrams.
Other symbols: inductance, transformer, diode, transistor, etc.
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Applying Kirchhoff’s
Loop Law to Many Loops*
1. Define a minimum set of current
loops. Label all elements.
R1=
2. Write a loop equation for each
loop. (Battery or 0 = DV).
3. Solve equations for currents
4. Calculate other variables of
interest.
i1 – i3
R2=
i1
Vbat=
R4=
R3=
Loop equations:
R5=
i2
Loop 1: Vbat  R1i1  R2 (i1  i3 )  R3 (i1  i2 )
Loop 2: 0  R3 (i2  i1 )  R5 (i2  i3 )
Loop 3: 0  R2 (i3  i1 )  R4i3  R5 (i3  i2 )
i3
i1 – i2
Loop L:
 (i
L
 ii ,other ) Ri  0
i
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Applying Kirchhoff’s Junction
Law to Many Junctions*
1. Define a minimum set of junction
potentials. You can select one
ground point. Label all elements.
2. Write a junction equation for
each unknown junction .
J1 R5=
R1=
Vbat
V1
R3=
R2=
3. Solve these equations for the
unknown junction potentials.
4. Calculate the other variables of
interest.
R4=
Junction equations:
V  V 0  V1 V2  V1 V2  V1
J1: 0  bat 1 


R1
R2
R3
R5
J2: 0 
0  V2 V1  V2 V1  V2


R4
R3
R5
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J2 V2
V=0
Junction J:

i
Physics 122B - Lecture 16
Vi ,away  VJ
Ri
0
42
Energy and Power (1)
P  IE = I 2 R  E 2 / R
Example: A 90 W load
resistance is connected
across a 120 V battery.
How much power is delivered
by the battery?
P
dU
 power  rate of energy transfer
dt
DU  qDVbat  qE
Pbat 
E (120 V)

 1.33 A
R (90 W)
P  IE  (1.33 A)(120 V)  160 W
dU bat d
dq
 qDVbat 
DVbat  I DVbat
dt
dt
dt
7/19/2016
I
P  E 2 / R  (120 V)2 /(90 W)  160 W
Physics 122B - Lecture 16
43
Real Batteries (2)
DVbat  E  I rint  E
I
E
E

Req R  rint
DVR  IR 
DVbat
R
E
R  rint
DVbat  E  I rint  E 
rint
E
R  rint


r
R
= E 1  int  =
E
 R  rint  R  rint
7/19/2016
Question: How can you measure rint?
Answer: One (rather brutal) way is to vary
an external load resistance R until the
potential drop across R is ½E. Then R=rint
because each drops ½E.
Physics 122B - Lecture 16
44
Voltmeters vs. Ammeters
An ideal voltmeter has infinite
internal resistance. It must be
connected between circuit elements
to measure the potential difference
between two points in the circuit.
An ideal ammeter has zero
internal resistance. It must be
inserted by breaking a circuit
connection to measure the current
flowing through that connection in
the circuit.
7/19/2016
Physics 122B - Lecture 16
V
A
I
X
I
45
Example:
Analyzing a Complex Circuit (2)
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Physics 122B - Lecture 16
46
Lecture 16 Announcements
 Lecture HW Assignment #5 is due by
10pm this evening.
Lecture HW Assignment #6 is posted on
Tycho and is due next Wednesday at
10pm.
The second midterm exam is this Friday.
It will cover everything up to the end of
Wednesdays lecture, emphasizing the
most recent material, but assumes
7/19/2016
Physics
- Lecture 16
understanding of
all122Bmaterial
inclusive. 47