Transcript Document

General Physics (PHY 2140)
Lecture 7
 Electrostatics and electrodynamics
 Capacitance and capacitors
capacitors with dielectrics
 Electric current
 current and drift speed
 resistance and Ohm’s law
http://www.physics.wayne.edu/~apetrov/PHY2140/
Chapter 16-17
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Reminder (for those who don’t read syllabus)
Reading Quizzes (bonus 5%):
It is important for you to come to class prepared, i.e. be familiar with the
material to be presented. To test your preparedness, a simple five-minute
quiz, testing your qualitative familiarity with the material to be discussed in
class, will be given at the beginning of some of the classes. No make-up
reading quizzes will be given.
There could be one today…
… but then again…
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Lightning Review
Last lecture:
1. Capacitance and capacitors



C
Parallel-plate capacitor
Combinations of capacitors
 Parallel
 Series
Energy stored in a capacitor
Q
V
A
C  0
d
Ceq  C1  C2  ...
1
1
1
 
 ...
1
Q2 1
Ceq C1 C2
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 then
b. the same as
c. smaller then
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before the plates were pulled apart.
16.10 Capacitors with dielectrics
A dielectrics is an insulating material (rubber, glass, etc.)
Consider an insolated, charged capacitor
-Q
Q
V0
Q
-Q
Insert a dielectric
V
Notice that the potential difference decreases (k = V0/V)
Since charge stayed the same (Q=Q0) → capacitance increases
Q0
Q0
 Q0
C


  C0
V V0  V0

dielectric constant: k = C/C0
Dielectric constant is a material property
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Capacitors with dielectrics - notes
Capacitance is multiplied by a factor k when the
dielectric fills the region between the plates completely
E.g., for a parallel-plate capacitor
A
C   0
d
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, that can be
produced in the dielectric before it breaks down
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Dielectric constants and dielectric
strengths of various materials at room
temperature
Material
Vacuum
Air
Water
Fused quartz
Dielectric
constant, k
Dielectric
strength (V/m)
1.00
--
1.00059
3 106
80
--
3.78
9 106
For a more complete list, see Table 16.1
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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.
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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.
Given:
V1=3,000 V
V2=1,000 V
A = 2.00 m2
d = 0.01 m
Find:
C=?
C0=?
Q=?
k=?
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Since we are dealing with the parallel-plate capacitor,
the original capacitance can be found as
2
A
2.00
m
C0   0  8.85 10-12 C 2 N  m2
 18 nF
-3
d
1.00 10 m


The dielectric constant and the new capacitance are
V1
C   C0 
C0  3 18nF  54nF
V2
The charge on the capacitor can be found to be


Q  C0 V  18  10-9 F  3000V   5.4 10-5 C
8
How does an insulating dielectric material reduce electric fields
by producing effective surface charge densities?
Reorientation of polar molecules
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
Induced polarization of non-polar molecules
- 
- 
- 
- 
- 
- 
- 
- 
- 
- 
- 
- 
- 
- 
- 
- 
Dielectric Breakdown: breaking of molecular bonds/ionization of
molecules.
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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.
+
+
+
+
+
A
I
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17.1 Electric Current - Definition
Given an amount of charge, Q, passing through the area A in a
time interval t, the current is the ratio of the charge to the time
interval.
Q
I
t
+
+
+
+
+
A
I
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17.1 Electric Current - Units
The SI units of current is the ampere (A).


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1 A = 1 C/s
1 A of current is equivalent to 1 C of charge passing through the
area in a time interval of 1 s.
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17.1 Electric Current – Remark 1
Currents may be carried by the motion of positive or
negative charges.
It is conventional to give the current the same
direction as the flow of positive charge.
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17.1 Electric Current – Remark 2
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.
-
-
-
-
v
I
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17.1 Electric Current – Remark 3
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 and electrolytes (e.g. Car
batteries), the current is the flow of
both positive and negative charges.
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17.1 Electric Current – Remark 4
It is common to refer to a moving charge as a mobile
charge carrier.
In a metal the charge carriers are electrons.
In other conditions or materials, they may be positive or
negative ions.
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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.
(B) the number of electrons that pass
through the filament in 1 second.
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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.
Q 1.67C
I

 0.835 A
t
2.00 s
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The amount of charge that passes through the filament of a 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.
Reasoning:
In 1 s, 0.835 C of charge passes the cross-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.
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The amount of charge that passes through the filament of a 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.
Solution:

-19

N q  N 1.60 10 C / electron  0.835C
0.835C
N
-19
1.60 10 C / electron
18
N  5.22 10 electrons
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15.2 Current and Drift Speed
Consider the current on a conductor of cross-sectional
area A.
A
q
vd
vdt
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15.2 Current and Drift Speed (2)
Volume of an element of length x is : V = A x.
Let n be the number of carriers per unit of volume.
The total number of carriers in V is: n A x.
The charge in this volume is: Q = (n A x)q.
Distance traveled at drift speed vd by carrier in time t:
x = vd t.
Hence: Q = (n A vd t)q.
The current through the conductor:
I = Q/ t = n A vd q.
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15.2 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.
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15.2 Current and Drift Speed - Example
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.
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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.
Reasoning: We know:
• A = 3.00x10-6 m2
• I = 10 A.
• r = 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.
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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.; r = 8.95 g/cm3.; q = 1.6 x 10-19 C.
n = 8.48 x 1022 electrons/ cm3.
I
10.0C / s
vd 

nqA 8.48 1022 electrons m3 1.6 10-19 C 3.00 10-6 m2




 2.46 10-6 m / s
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15.2 Current and Drift Speed - Comments
Drift speeds are usually very small.
Drift speed much smaller than the average speed
between collisions.

Electrons traveling at 2.46x10-6 m/s would would take 68 min to
travel 1m.
So why does light turn on so quickly when one flips a
switch?

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The info travels at roughly 108 m/s…
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Mini-quiz
Consider a wire has a long conical shape. How does the velocity
of the electrons vary along the wire?
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.
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17.3 Resistance and Ohm’s Law - Intro
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.
I  V
V
I
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17.3 Definition of Resistance
In situations where the proportionality is exact, one can
write.
V  IR
• The proportionality constant R is called resistance
of the conductor.
• The resistance is defined as the ratio.
V
R
I
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17.3 Resistance - Units
In SI, resistance is expressed in volts per ampere.
A special name is given: ohms (W).
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 W.
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17.3 Ohm’s Law
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’s law.
Georg Simon Ohm
(1787-1854)
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Linear or Ohmic Material
I
Non-Linear or
Non-Ohmic Material
I
V
Most metals, ceramics
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V
Semiconductors
e.g. diodes
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Ohm’s Law
V  IR
R understood to be independent of V.
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Definition:
Resistor: a conductor that provides a specified resistance in
an electric circuit.
The symbol for a resistor in circuit diagrams.
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Example:
Resistance of a Steam Iron
All household electric devices are required to have a
specified resistance (as well as many other
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?
V 120V
R

 16.2W
I
7.40 A
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