General Physics (PHY 2140)
Lightning Review
Last lecture:
Lecture 4
1. Equipotential surfaces.
¾ Electrostatics and electrodynamics
9
9
Capacitance and capacitors
9 capacitors with dielectrics
Electric current
9 current and drift speed
9 resistance and Ohm’s law
9 resistivity
They are defined as a surface in space on which the potential is the same
for every point (surfaces of constant voltage)
The electric field at every point of an equipotential surface is perpendicular
to the surface
http://www.physics.wayne.edu/~alan/2140Website/Main.htm
Chapter 16-17
5/16/2007
1
2
16.10 Capacitors with dielectrics
Lightning Review
A dielectric is an insulating material (rubber, glass, etc.)
Consider an insolated, charged capacitor
Last lecture:
1. Capacitance and capacitors
5/16/2007
C=
9 Combinations of capacitors
¾ Parallel
¾ Series
9 ParallelParallel-plate capacitor
9 Energy stored in a capacitor
Q
ΔV
1
1
1
= +
+ ...
Ceq C1 C2
−Q
+Q
Ceq = C1 + C2 + ...
C = ε0
A
d
1
Q2 1
U = QV =
= CV 2
2
2C 2
Review Problem: Consider an isolated simple parallel-plate capacitor whose plates
are given equal and opposite charges and are separated by a distance d.
Suppose the plates are pulled apart until they are separated by a distance D>d.
The electrostatic energy stored in a capacitor is
a. greater than
b. the same as
c. smaller than
5/16/2007
3
before the plates were pulled apart.
V0
+Q
−Q
Insert a dielectric
V
Notice that the potential difference decreases (κ = V0/V)
Since charge stayed the same (Q=Q0) → capacitance increases
C=
„
Q0
Q
κQ
= 0 = 0 = κ C0
V V0 κ V0
dielectric constant: κ = C/C0
Dielectric constant is a material property
5/16/2007
4
1
Capacitors with dielectrics - notes
Dielectric constants and dielectric
strengths of various materials at room
temperature
Capacitance is multiplied by a factor k when the
dielectric fills the region between the plates completely
E.g., for a parallelparallel-plate capacitor
A
C = κε 0
d
Material
Vacuum
Air
The capacitance is limited from above by the electric
discharge that can occur through the dielectric material
separating the plates
In other words, there exists a maximum of the electric
field, sometimes called dielectric strength,
strength, that can be
produced in the dielectric before it breaks down
5/16/2007
Water
Fused quartz
Dielectric
strength (V/m)
1.00
--
1.00059
3 ×106
80
1.5 ×107 *
3.78
9 ×106
For a more complete list, see Table 16.1
* For short duration (μsec) pulses
5
Example
Take a parallel plate capacitor whose plates have an area of 2.0 m2
and are separated by a distance of 1mm. The capacitor is charged
to an initial voltage of 3 kV and then disconnected from the
charging source. An insulating material is placed between the
plates, completely filling the space, resulting in a decrease in the
capacitors voltage to 1 kV. Determine the original and new
capacitance, the charge on the capacitor, and the dielectric constant
of the material.
5/16/2007
Given:
ΔV1=3,000 V
ΔV2=1,000 V
A = 2.00 m2
d = 0.01 m
C =?
C0=?
Q =?
κ =?
7
6
Take a parallel plate capacitor whose plates have an area of 2 m2 and are separated by a
distance of 1mm. The capacitor is charged to an initial voltage of 3 kV and then
disconnected from the charging source. An insulating material is placed between the
plates, completely filling the space, resulting in a decrease in the capacitors voltage to
1 kV. Determine the original and new capacitance, the charge on the capacitor, and the
dielectric constant of the material.
Find:
5/16/2007
Dielectric
constant, κ
5/16/2007
Since we are dealing with the parallel-plate capacitor,
the original capacitance can be found as
C0 = ε 0
A
2.00 m2
= 8.85 ×10−12 C 2 N ⋅ m2
= 18 nF
d
1.00 ×10−3 m
(
)
The dielectric constant and the new capacitance are
C = κ C0 =
ΔV1
C0 = 3 ⋅18nF = 54nF
ΔV2
The charge on the capacitor can be found to be
(
)
Q = C0 ΔV = 18 ×10−9 F ( 3000V ) = 5.4 ×10−5 C
8
2
+
−+
−
− + −+
−+
−
−+
−+
−+
−+
−+
−+
−+
−+
−+
+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
−+
How does an insulating dielectric material reduce electric fields
by producing effective surface charge densities?
Reorientation of polar molecules
−+
−+
−+
−+
17.1 Electric Current
Whenever charges of like signs move in a given
direction, a current is said to exist.
Consider charges are moving perpendicularly to a
surface of area A.
Definition: the current is the rate at which charge
flows through this surface.
surface.
Induced polarization of non-polar molecules
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
− +
+
+
Dielectric Breakdown: breaking of molecular bonds/ionization of
molecules.
5/16/2007
9
17.1 Electric Current - Definition
+
A
I
5/16/2007
10
17.1 Electric Current - Units
Given an amount of charge, ΔQ, passing through the area A in a
time interval Δt, the current is the ratio of the charge to the time
interval.
I=
+
+
The SI units of current is the ampere (A).
(A).
„
ΔQ
Δt
„
1 A = 1 C/s
1 A of current is equivalent to 1 C of charge passing through the
the
area in a time interval of 1 s.
+
+
+
+
+
A
I
5/16/2007
11
5/16/2007
12
3
17.1 Electric Current – Remark 1
17.1 Electric Current – Remark 2
Currents may be carried by the motion of positive or
negative charges.
charges.
It is conventional to give the current the same
direction as the flow of positive charge.
In a metal conductor such as copper, the current is due
to the motion of the electrons (negatively charged).
„
The direction of the current in copper is thus opposite the
direction of the electrons.
electrons.
-
-
-
-
v
I
5/16/2007
13
17.1 Electric Current – Remark 3
14
17.1 Electric Current – Remark 4
In a beam of protons at a particle
accelerator (such as RHIC at
Brookhaven national laboratory), the
current is the same direction as the
motion of the protons.
In gases (plasmas) and electrolytes
(e.g. Car batteries), the current is the
flow of both positive and negative
charges.
5/16/2007
5/16/2007
It is common to refer to a moving charge as a mobile
charge carrier.
carrier.
In a metal the charge carriers are electrons.
In other conditions or materials, they may be positive or
negative ions.
15
5/16/2007
16
4
17.1 Electric Current – Example
Current in a light bulb
The amount of charge that passes through the filament
of a certain light bulb in 2.00 s is 1.67 c. Find.
(A) the current in the light bulb.
The amount of charge that passes
through the filament of a certain
light bulb in 2.00 s is 1.67 c. Find.
(A) the current in the light bulb.
(B) the number of electrons that pass
through the filament in 1 second.
5/16/2007
I=
17
The amount of charge that passes through the filament of a certain
certain
light bulb in 2.00 s is 1.67 c. Find.
5/16/2007
18
The amount of charge that passes through the filament of a certain
certain
light bulb in 2.00 s is 1.67 c. Find.
(b) the number of electrons that pass through the filament in 1
second.
(b) the number of electrons that pass through the filament in 1
second.
Solution:
Reasoning:
In 1 s, 0.835 C of charge passes the crosscross-sectional
area of the filament.
This total charge per second is equal to the number
of electrons, N, times the charge on a single electron.
5/16/2007
ΔQ 1.67C
=
= 0.835 A
Δt
2.00 s
(
)
N q = N 1.60 ×10−19 C / electron = 0.835C
0.835C
1.60 ×10−19 C / electron
N = 5.22 ×1018 electrons
N=
19
5/16/2007
20
5
17.2 Current and Drift Speed
17.2 Current and Drift Speed (2)
Consider the current on a conductor of crosscross-sectional
area A.
A
q
Volume of an element of length Δx is : ΔV = A Δx.
Let n be the number of carriers per unit of volume.
The total number of carriers in ΔV is: n A Δx.
The charge in this volume is: ΔQ = (n A Δx)q.
x)q.
Distance traveled at drift speed vd by carrier in time Δt:
Δx = vd Δt.
Hence: ΔQ = (n A vd Δt)q.
The current through the conductor:
vd
Δx = vdΔt
I = ΔQ/ Δt = n A vd q.
5/16/2007
21
22
17.2 Current and Drift Speed - Example
17.2 Current and Drift Speed (3)
• In an isolated conductor, charge carriers move randomly in
all directions.
• When an external potential is applied across the conductor, it
creates an electric field inside which produces a force on the
electron.
• Electrons however still have quite a random path.
• As they travel through the material, electrons collide with
other electrons, and nuclei, thereby losing or gaining energy.
• The work done by the field exceeds the loss by collisions.
• The electrons then tend to drift preferentially in one direction.
5/16/2007
5/16/2007
23
Question:
A copper wire of crosscross-sectional area 3.00x10-6 m2 carries a current of
10. A. Assuming that each copper atom contributes one free electron
electron to
the metal, find the drift speed of the electron in this wire. The
The density of
3
copper is 8.95 g/cm .
5/16/2007
24
6
Question:
A copper wire of cross-sectional area 3.00x10-6 m2 carries a current of 10 A.
Assuming that each copper atom contributes one free electron to the metal, find
the drift speed of the electron in this wire. The density of copper is 8.95 g/cm3.
Question:
A copper wire of cross-sectional area 3.00x10-6 m2 carries a
current of 10 A. Assuming that each copper atom contributes
one free electron to the metal, find the drift speed of the
electron in this wire. The density of copper is 8.95 g/cm3.
Ingredients:
A = 3.00x10-6 m2 ; I = 10 A.; ρ = 8.95 g/cm3.; q = 1.6 x 10-19 C.
Reasoning: We know:
• A = 3.00x10-6 m2
• I = 10 A.
• ρ = 8.95 g/cm3.
• q = 1.6 x 10-19 C.
• n = 6.02x1023 atom/mol x 8.95 g/cm3 x ( 63.5 g/mol)-1
• n = 8.48 x 1022 electrons/ cm3.
5/16/2007
.
I
10.0C / s
vd =
=
nqA 8.48 ×1028 electrons m3 1.6 ×10−19 C 3.00 ×10−6 m2
n = 8.48 x 1022 electrons/cm3 = 8.48 x 1028 electrons/m3
(
25
5/16/2007
26
Consider a wire has a long conical shape. How does the velocity
of the electrons vary along the wire?
Electrons traveling at 2.46x10-4 m/s would would take 68 min to
travel 1m.
So why does light turn on so quickly when one flips a
switch?
5/16/2007
)
MiniMini-quiz
Drift speeds are usually very small.
Drift speed much smaller than the average speed
between collisions.
„
)(
= 2.46 ×10 m / s
17.2 Current and Drift Speed - Comments
„
)(
−4
Every portion of the wire carries the same current: as the cross
sectional area decreases, the drift velocity must increase to
carry the same value of current. This is dues to the electrical
field lines being compressed into a smaller area, thereby
increasing the strength of the electric field.
The info travels at roughly 108 m/s…
m/s…
27
5/16/2007
28
7
17.4 Resistance and Ohm’
Ohm’s Law - Intro
17.4 Definition of Resistance
When a voltage (potential difference) is applied across
the ends of a metallic conductor, the current is found to
be proportional to the applied voltage.
In situations where the proportionality is exact, one can
write.
ΔV = IR
I ∝ ΔV
• The proportionality constant R is called resistance
of the conductor.
• The resistance is defined as the ratio.
ΔV
R=
ΔV
I
I
5/16/2007
29
17.4 Resistance - Units
5/16/2007
30
17.4 Ohm’
Ohm’s Law
In SI, resistance is expressed in volts per ampere.
A special name is given: ohms (Ω
(Ω).
Resistance in a conductor arises because of collisions
between electrons and fixed charges within the material.
In many materials, including most metals, the resistance is
constant over a wide range of applied voltages.
This is a statement of Ohm’
Ohm’s law.
Example: if a potential difference of 10 V applied across
a conductor produces a 0.2 A current, then one
concludes the conductors has a resistance of 10 V / 0.2
A = 50 Ω.
Georg Simon Ohm
(1787-1854)
5/16/2007
31
5/16/2007
32
8
Linear or Ohmic Material
Ohm’s Law
Non-Linear or
Non-Ohmic Material
I
ΔV = IR
I
R understood to be independent of ΔV.
ΔV
ΔV
Most metals, ceramics
Semiconductors
e.g. diodes
5/16/2007
33
5/16/2007
34
Example:
Resistance of a Steam Iron
Definition:
Resistor: a conductor that provides a specified resistance in
an electric circuit.
All household electric devices are required to have a
specified resistance (as well as many other
characteristics…
characteristics…). Consider that the plate of a certain
steam iron states the iron carries a current of 7.40 A when
connected to a 120 V source. What is the resistance of the
steam iron?
The symbol for a resistor in circuit diagrams.
V = IR
I
+ -
R=
ΔV 120V
=
= 16.2 Ω
7.40 A
I
E
5/16/2007
35
5/16/2007
36
9
17.5 Resistivity - Intro
17.5 Resistivity - Definition
Electrons moving inside a conductor subject to an
external potential constantly collide with atoms of the
conductor.
They lose energy and are repeated rere-accelerated by the
electric field produced by the external potential.
The collision process is equivalent to an internal friction.
This is the origin of a material’
material’s resistance.
resistance.
5/16/2007
37
17.5 Resistivity - Remarks
The resistance of an ohmic conductor is proportional to
the its length, l, and inversely proportional to the cross
section area, A, of the conductor.
l
A
• The constant of proportionality ρ is called the
resistivity of the material.
R=ρ
5/16/2007
17.5 Resistivity - Units
Every material has a characteristic resistivity that
depends on its electronic structure, and the temperature.
Good conductors have low resistivity.
resistivity.
Insulators have high resistivity.
resistivity.
R=ρ
39
l
A
ρ=
RA
l
Resistance expressed in Ohms,
Length in meter.
Area are m2,
Resistivity thus has units of Ωm.
Analogy to the flow of water through a pipe.
5/16/2007
38
5/16/2007
40
10
Resistivity of various materials
(also see table 17.1)
Material
Resistivity (10-8 Ωm)
Material
Resistivity (10-8 Ωm)
Silver
Copper
Gold
Aluminum
1.61
1.70
2.20
2.65
Bismuth
Plutonium
Graphite
Germanium
106.8
141.4
1375
4.6x107
Pure
Silicon
Calcium
3.5
Diamond
2.7x109
3.91
1.8x1013
Sodium
Tungsten
Brass
Uranium
Mercury
4.75
5.3
7.0
30.0
98.4
Deionized
water
Iodine
Phosphorus
Quartz
Alumina
Sulfur
41
17.5 Resistivity - Example
(
5/16/2007
(b) If a potential difference of 10.0 V is maintained across a
1.0-m length of the nichrome wire, what is the current?
)
2
A = πr2 = π 0.321×10−3 m = 3.24×10−7 m2
I=
Resistivity (Table): 1.5 x 10−6 Ωm.
Resistance/unit length:
5/16/2007
42
17.5 Resistivity - Example
(a) Calculate the resistance per unit length of a 22-gauge
nichrome wire of radius 0.321 m.
Cross section:
Why do old light bulbs give less light than when
new?
Answer:
• The filament of a light bulb, made of tungsten, is kept at high
temperature when the light bulb is on.
• It tends to evaporate, i.e. to become thinner, thus decreasing in radius,
and cross sectional area.
• Its resistance increases with time.
• The current going though the filament then decreases with time – and
so does its luminosity.
• Tungsten atoms evaporate off the filament and end up on the inner surface
of the bulb.
• Over time, the glass becomes less transparent and therefore less
luminous.
1.3x1015
1x1017
1x1021
1x1022
2x1023
5/16/2007
MiniMini-quiz
ΔV 10.0V
=
= 2.2 A
4.6Ω
R
R ρ 1.5 ×10−6 Ωm
= =
= 4.6 Ω m
l A 3.24 ×10−7 m2
43
5/16/2007
44
11
17.6 Temperature Variation of Resistance
- Intro . . . next lecture
• The resistivity of a metal depends on many
(environmental) factors.
• The most important factor is the temperature.
• For most metals, the resistivity increases with
increasing temperature.
• The increased resistivity arises because of larger
friction caused by the more violent motion of the
atoms of the metal.
5/16/2007
45
12