Download PHY 184 lecture 15

PHY 184
Spring 2007
Lecture 15
Title: Direct Current
2/1/07
184 Lecture 15
1
Announcements
 Homework Set 4 is active and is due next
Tuesday morning at 8:00 am
 Today: Quick review of the material of the past 4
weeks and we will start with “Direct Currents”
 Midterm 1 will take place in class next Thursday
• Bring a calculator
• Bring a no. 2 pencil
• Bring your ID
2/1/07
184 Lecture 15
2
Outline
 Review of Chapters 16 – 19
 Introduction to Chapter 20 = Direct Current
2/1/07
184 Lecture 15
3
Review: Electrostatics (1)
 Electric charge can be either positive or negative; like
charges repel and unlike charges attract each other. An
object with equal amounts of positive and negative charge is
electrically neutral. The total charge of an isolated system
is always conserved.
 The electric force F between two charges, q1 and q2,
separated by a distance r is given by Coulomb’s Law:
q1q2
Opposite charges: F is attractive (-)
Fk 2
Like charges: F is repulsive (+)
r
 The constant k is called Coulomb’s constant and is given by
2
N

m
k  8.99 10 9
C2
2/1/07
k
1
4 0
184 Lecture 15
 0  8.85 10
12
C2
N  m2
4
Review: Electrostatics (2)
 We may define the unit of charge in terms of the
charge of one electron
• An electron is an elementary particle with charge q = -e
where: e = 1.60210-19 C
• A proton has the charge q = +e
2/1/07
184 Lecture 15
5
Clicker: Electrostatics
 Calculate the magnitude of the force (in N) between a
gold nucleus and an electron on an orbit with radius
4.88×10-12 m around the nucleus. The gold nucleus
has a charge of +79e.
A) 7.7 10-4 N
B) -1.56 10-3 N
C) 8.9 10-5 N
2
N

m
k  8.99 10 9
C2
e = 1.60210-19 C
2/1/07
184 Lecture 15
6
Clicker: Electrostatics
 Calculate the magnitude of the force (in N) between a gold
nucleus and an electron on an orbit with radius 4.88×10-12 m
around the nucleus. The gold nucleus has a charge of +79e.
A) 7.7 10-4 N
kq1q2
14
F  2  7.7  10 N
r
2/1/07
184 Lecture 15
7
Review: Electric Field (1)
 The electric force on a charge q due to an
electric field E is given by


F qE
 The electric field at any point is equivalent to the sum of
all the sources of electric field at that point:
 Electric field from a point charge:
 The electric field points radially away from a positive
charge and radially toward negative charges.
 A system of two oppositely charged point particles is


called an electric dipole.
•
•
•
•
2/1/07
p
p is the magnitude of the dipole moment
q is the magnitude of one of the opposite charges
d is the distance between the charges
p points from the negative to the positive charge
184 Lecture 15
qd
8
Review: Electric Field (2)
 The electric flux through a surface A is defined as
 Gauss’ Law:
0  q
Gauss’ Law says that the electric flux through a closed
surface is proportional to the net charge enclosed by this
 
surface.
 0 E  dA  q


2k 
E

2 0 r
r
conducting wire

E
2 0
Infinite non-conducting
charged sheet

E
0
Infinite conducting plane
The electric field inside a closed conductor is 0
2/1/07
184 Lecture 15
9
Review: Electric Field (3)
 The electric field inside a spherical shell of charge q is zero
 The electric field outside a spherical shell of charge q is the
same as the field from a point charge q
1
q
E
2
4 0 r
 Electric field from charge distributed uniformly throughout a sphere
of radius r
• r2 > r
• r1 < r
2/1/07
qt
qt
E r2   k 2 or just E  k 2
r2
r
qt r1
kqt r1
E r1  
 3
3
4 0 r
r
184 Lecture 15
10
Review - Potential Energy
 When an electrostatic force acts on charged particles,
assign an electric potential energy, U
 If the system is changed from initial state i to the final
state f, the electrostatic force does work, W
 The change in electric potential energy is U is equal to the
charge q times the change in electric potential V, U=qV
 Equipotential surfaces (or lines) represent adjacent points
in space that have the same potential.
 Calculate the change in the electric potential from the
electric field by integrating the field in a particular
direction,
2/1/07
184 Lecture 15
11
Review - Electric Potential
 Taking the convention that the electric potential is zero at infinity we
can express the electric potential in terms of the electric field as
 Calculate the electric field from gradients of the electric potential in
each component direction
V
V
V
Ex  
x
; Ey  
y
; Ez  
z
 The electric potential from a point charge q at a distance r is given by:
V
kq
r
 The electric potential can be expressed as an algebraic sum of all
sources of electric potential
n
n
kqi
V   Vi  
i 1
i 1 ri
n
In particular for a system of point charges:
2/1/07
184 Lecture 15
n
kqi
V   Vi  
i 1
i 1 ri
12
Clicker - Electric Field
Use:=qt/(Volume of sphere)
Volume=4/3R3
q1q2
Fk 2
r
Like charges: F is repulsive (+)
r1<R
A
B
C
2/1/07
184 Lecture 15
13
Clicker - Electric Field
Use:=qt/(Volume of sphere)
Volume=4/3R3
q1q2
Fk 2
r
Like charges: F is repulsive (+)
r1<R
A
B
CX
2/1/07
184 Lecture 15
14
Review: Capacitance (1)
 The definition of capacitance is
q
C
V
 The capacitance of a parallel plate capacitor
is given by
• A is the area of each plate
• d is the distance between the plates
 The capacitance of a spherical capacitor is
C
0 A
d
r1r2
C  4 0
r2  r1
• r1 is the radius of the inner sphere
• r2 is the radius of the outer sphere
2/1/07
184 Lecture 15
15
Review: Capacitance (2)
 The capacitance of an isolated spherical conductor is
C  4 0 R
 The capacitance of a cylindrical capacitor is
q
C 
V
L
2 0 L


ln r2 / r1  ln r2 / r1 
2 0
 Placing a dielectric between the plates of a capacitor
increase the capacitance by 
C  C
air
 The electric potential energy stored in a capacitor is
given by
1
2
U
2/1/07
2
CV
184 Lecture 15
16
Review: Capacitance (3)
 The equivalent capacitance for n capacitors in
parallel is
n
+

Ceq   Ci
i 1
+

+

 The equivalent capacitance for n capacitors in
series is
n
1
1

Ceq i 1 Ci
2/1/07
184 Lecture 15
+

+

+

17
Electric Current
Nature is simple – we can understand it and in some ways
even control it.
That is the origin of technology.
The human race has developed a remarkable technology of
electric current..
2/1/07
184 Lecture 15
18
Direct Current
 We will study charges in motion.
 Electric charge moving coherently from one region
to another is called electric current.
 Current is flowing through light bulbs, iPods, and
lightning strikes.
 Current usually consists of mobile electrons
traveling in conducting materials.
 Direct current is defined as a current that flows
only in one direction in the conductor.
Most of our electric technology is based on
Alternating Current – Chapter 24
2/1/07
184 Lecture 15
19
Electric Current
 We define the electric current i as the net charge
passing a given point in a given time.
 Random motion of electrons in conductors, or the
flows of electrically neutral atoms, are not electric
currents in spite of the fact that large amounts of
charge are moving past a given point.
 If net charge dq passes a point in time dt we
define the current i to be
dq
i
dt
2/1/07
184 Lecture 15
20
Electric Current (2)
 The amount of charge q passing a given point in
time t is the integral of the current with respect
to time given by
t
q   dq   idt
0
 We will use charge conservation, implying that
charge flowing in a conductor is never lost.
 Therefore the same amount of charge must flow
through one end of the conductor as the charge
that exits from the other end of the conductor.
2/1/07
184 Lecture 15
21
The Ampere
 The unit of current is coulombs per second, which
has been given the unit ampere, named after
French physicist André-Marie Ampère, (17751836)
 The ampere is abbreviated as A and is defined by
 Some typical currents are
•
•
•
•
2/1/07
1C
1A
1s
Flashlight - 1 A
The starter motor in a car - 200 A
iPod - 50 mA
In a lightning strike (for a short time) - 100000 A
184 Lecture 15
22
Batteries
 We use of batteries as devices that provide direct currents
in circuits.
 If you examine a battery, you will find its voltage written on
it.
 This voltage is the potential difference it can provide to a
circuit.
 You will also find their ratings in units of mAh.
 This rating provides information on the total charge that
they can deliver when fully charged.
 The quantity mAh is another unit of charge:
3
1 mAh  (10 A)(3600 s)  3.6 As  3.6 C
2/1/07
184 Lecture 15
23
2/1/07
184 Lecture 15
24
The half-reactions are:
At the cathode…
2 MnO2 + H2O + 2 e- —>Mn2O3 + 2 OHAt the anode…
Zn + 2 OH- —> ZnO + H2O + 2 e The overall reaction is:
Zn + 2MnO2 —> ZnO + Mn2O3 + [E=1.5 V]
Anode (negative terminal): Zinc powder
Cathode (positive terminal): Manganese dioxide (MnO2) powder
Electrolyte: Potassium hydroxide (KOH)
The flow of electrons is always from anode—to--cathode outside of the cell (i.e., in the
circuit) and from cathode—to--anode inside the cell. Inside a chemical cell, ions are
carrying the electrons from cathode—to--anode inside the cell.
2/1/07
184 Lecture 15
25
Alkaline battery
2/1/07
184 Lecture 15
Al Kaline batter
26
Current
 Current is a scalar.
 Current has a sign but not a direction.
 We will represent the direction of the current flowing in a
conductor using an arrow.
 This arrow represents whether the net current is positive
or negative in a conductor at a given point but does not
represent a direction in three dimensions.
 Physically, the charge carriers in a conductor are electrons
that are negatively charged.
 However, as is conventionally done, we define positive
current as the net flow of positive charge carriers past a
given point per unit time.
2/1/07
184 Lecture 15
27
Circuits
In this circuit, electrons flow around
the circuit counterclockwise. (The
conventionally defined current is
clockwise; remember, electrons are
negative charges.) The electrons
can’t disappear so the current
requires a whole loop!
+
Ohm’s Law
V=IR
Chemical action pumps electrons from the positive
terminal (+) to the negative terminal () in the battery.
The emf (electromotive force, or electric field) pushes
electrons around the wire from () to (+).
2/1/07
184 Lecture 15
28
Current Density
J
 Let’s consider current flowing in a conductor.
 Taking a plane through the conductor, we can define
the current per unit area flowing through the
conductor at that point as the current density J
 We take the direction of J as the direction of the
velocity of the charges crossing the plane.
 If the cross sectional area is small, the magnitude of
J will be large.
 If the cross section area is large, the magnitude of J
will be small.
2/1/07
184 Lecture 15
29
Current Density (2)
 The current flowing through the surface is
i
 
J  dA
 … where dA is the differential area element
perpendicular to the surface.
 If the current is constant and
perpendicular to the surface, then
and we can write an expression for
the magnitude of the current density
i
J
A
2/1/07
184 Lecture 15
30