1. Chapter 8

2.  Electric charge is a fundamental quantity – we
don’t really know what it is
◦ But we can describe it, use it, control it
◦ Electricity runs motors, lights, heaters, A/C, stereos,
TV’s, computers, etc.
 Electric Forces – at the microscopic level they
hold atoms and molecules together
◦ Electric Forces hold matter together
◦ Gravitational Forces hold the universe together
 Magnetism is also closely associated with
electricity
Intro

3.  Experimental evidence leads us to conclude that
there are two types charges
◦ Positive (+)
◦ Negative (-)
 All matter is composed of atoms, which in turn are
composed of subatomic particles
◦ Electrons (-)
◦ Protons (+)
◦ Neutron (neutral)
Section 8.1

4.  Note that the Proton and Neutron each have
about 1000x more mass than the Electron
 If the atom has the same number of protons and
electrons it is electrically neutral
Section 8.1

5.  Named after a French scientist – Charles
Coulomb (1736-1806)
 Note that the charge on a single electron (-) or
proton (+) is 1.60 x 10-19
C (very small!)
 q usually designates electric charge
◦ Excess of positive charges +q
◦ Excess of negative charges -q
Section 8.1

6.  An electric force exists between any two charged
particles – either attractive or repulsive
Section 8.1

7.  Law of Charges – like charges repel, and unlike
charges attract
◦ Two positives repel each other
◦ Two negatives repel each other
◦ Positive and negative attract each other
 The force between two charged bodies is directly
proportional to the product of the two charges &
inversely proportional to the square of their
distance apart.
Section 8.1

8.  Force =
Section 8.1
constant x charge 1 x charge 2
(distance between charges)2
λ
In equation form F = kq1q2 / r2
λ
k is a constant = 9.0 x 109
Nm2
/C2

9.  An object with an excess of electrons is said to be
negatively charged
 An object with a deficiently of electrons is said to
be positively charged
 Photocopier (xerography) – practical and
widespread use of electrostatics
Section 8.1

10. Two negative charges
repel
Two positive
charges repel
One negative and one
positive attract
Repulse Repulse Attract
Section 8.1

11.  Force of attraction/repulsion between two charged
bodies is directly proportional to the product of the
two charges and inversely proportional to the
square of the distance between them
 F = (kq1q2) / r2
◦ F = force of attraction or repulsion
◦ q1 = magnitude of the first charge
◦ q2 = magnitude of the second charge
◦ r = distance between charges
◦ k = constant = 9.0 x 109
N-m2
/C2
Section 8.1

12.  Equations look similar
◦ F = kq1q2 / r2
& F = Gm1m2/ r2
 Both depend on r2
 Coulomb’s law can describe either an attractive or
repulsive force – gravity is always positive
 Electrical charges are much stronger than
gravitational forces
Section 8.1

13.  I = charge/time = q/t
◦ I = electric current (amperes)
◦ q = electric charge flowing past a point (coulombs)
◦ t = time for the charge to pass point (seconds)
 1 ampere (A) = flow of 1 Coulomb per second
 Rearrange equation above:
◦ q = It or 1 coulomb = 1 ampere x 1 second
 Therefore, 1 coulomb is the amount of charge that
flows past a given point in 1 second when the
current is 1 ampere
Section 8.2

14.  Electrical conductor – materials in which an
electric charge flows readily (most metals, due to
the outer, loosely bound electrons)
 Electrical insulator – materials that do not conduct
electricity very well due to very tight electron
bonding (wood, plastic, glass)
 Semiconductor – not good as a conductor or
insulator (graphite)
Section 8.2

15.  A wire carries a current of 0.50 A for 2 minutes.
a) how much (net) charge goes past a point in the wire in
this time?
b) how many electrons make up this amount of charge?
 GIVEN: I = 0.50 A, t = 2 minutes (120 seconds)
 WANTED: q (charge) & n (number of electrons)
 (a) q = It = (0.50 A)(120 s) = 60 C (coulombs)
Section 8.2

16.  A wire carries a current of 0.50 A for 2 minutes.
a) how much (net) charge goes past a point in the wire
in this time?
b) how many electrons make up this amount of charge?
 To solve for (b), we know that each electron has
a charge of 1.6 x 10-19
C and we know the total
charge from part (a) = 60 Coulombs
 n = 60/(1.6 x 10-19
C / electron) = 3.8 x 1020
electrons
Section 8.1

17.  When work is done to separate positive and
negative charges, we have electric potential
energy
Section 8.2

18.  Instead of measuring electric potential energy, we
measure the potential difference, or voltage
 Voltage – the amount of work it would take to
move a charge between two points, divided by the
value of the charge
 Voltage = work / charge = V = W/q
 Measured in volts (V) = 1 joule/Coulomb
 When we have electric potential energy, this may
be used to set up an electrical current
Section 8.2

19.  Whenever there is an electrical current, there is
resistance (R) within the conducting material
◦ R is due to atomic/subatomic collisions
 Georg Ohm (1787-1854) – formulated a simple
relationship between voltage, current, and
resistance
 Ohm’s Law  V = IR
◦ V = voltage in volts, I = current in amperes, and R =
resistance in ohms
 1 ohm = 1 volt/1 ampere (R=V/I)
Section 8.2

20.  Electrons flow from negative terminal to positive
terminal (provided by the chemical energy of the
battery) -- negative to positive
 Open switch – not a complete circuit and no flow
of current (electrons)
 Closed switch – a complete circuit and flow of
current (electrons) exists
 Closed Circuit Required – to have a sustained
electrical current
Section 8.2

21. The light bulb offers resistance. The kinetic energy of the
electric energy is converted to heat and radiant energy.
Section 8.2

22. Section 8.2

23.  Recall: P = W/t (Power = work/time) – Ch. 4
 V = W/q (Voltage = work/charge) – Ch. 8
 Rearrange V=W/q  W = qV
 Substitute W = qV into P = W/t equation
 Result  P = qV/t
 Recall that I = q/t (Current = charge/time)
 ∴ P = IV (Electric power = current x voltage)
 also P = I2
R (substitute in V = IR)
 P in watts, I in amperes, R in ohms, V in volts
Section 8.2

24.  Find the current and resistance of a 60-W, 120-V
light bulb in operation.
 Given: P = 60W, V = 120 V
 Find: I (current in amperes), R (resistance in
ohms)
 Since P = IV  I = P/V = 60 W/120 V = 0.50 A
 Since V = IR  R = V/I = 120 V/0.50A = 240 Ω
 Since P = I2
R  R = P/I2
= 60 W/(0.50 A)2 = 240
Ω (same answer as above)
Section 8.2

25.  A coffeemaker draws 10 A of current operating at
120 V. How much electrical energy does the
coffeemaker use each second.
 Given: I = 10 A, V = 120 V
 Find: P (electrical energy in watts)
 P = IV
 P = (10 A) x (120 V) = 1200 W or 1200 J/s
Section 8.2

26.  Current: I = q/t
◦ I = current (amperes),
◦ q = electric charge (coulombs),
◦ t = time (seconds)
 Coulomb’s Law: F = kq1q2/r2
 Voltage: V = W/q [Work is in joules (J)]
 Ohm’s Law: V= IR
 Electrical Power: P = IV = I2
R
Section 8.2

27.  Direct Current (dc) – the electron flow is always in
one direction, from (-) to (+)
◦ Used in batteries and automobiles
 Alternating Current (ac) – constantly changing the
voltage from positive to negative and back
◦ Used in homes.
◦ 60 Hz (cycles/sec) and Voltage of 110-120 V
Section 8.3

28.  Plugging appliances into a wall outlet places them
in a circuit – either in series or in parallel
 In a series circuit, the same current (I) passes
through all the resistances.
◦ The total resistance is simply the sum of the individual
resistances.
Section 8.3

29.  Rs = R1+R2+R3+…
 If one bulb were to burn out the circuit would be
broken, and all the lights would go out
Section 8.3
Same current
all the way

30.  In a parallel circuit, the voltage (V) across each
resistance is the same, but the current (I) through
each may vary.
Section 8.3

31. Current Divides, Voltage is the same
Section 8.3
I = I1+I2+I3+… for R in parallel: 1/Rp = 1/R1+1/R2+…
for 2 R in parallel: Rp = (R1R2)/(R1+ R2)

32.  Three resistors have values of R1=6.0Ω, R2=6.0Ω,
and R3=3.0Ω. What is their total resistance when
connected in parallel, and how much current will
be drawn from a 12-V battery if it is connected to
the circuit?
 1/Rp = 1/R1+1/R2+1/R3 = 1/6 + 1/6 + 1/3
 1/Rp = 1/6 + 1/6 + 2/6 = 4/6 Ω
 Rp = (6 Ω)/4 = 1.5 Ω = Total Resistance
 I = V/Rp = (12 V)/(1.5 Ω) = 8.0 A = current
drawn
Section 8.3

33.  Wired in parallel independent branches any
particular circuit element can operate when others
in the same circuit do not.
Section 8.3

34.  Wires can become hot as more and more current
is used on numerous appliances.
 Fuses are placed in the circuit to prevent wires
from becoming too hot and catching fire.
 The fuse filament is designed to melt (and thereby
break the electrical circuit) when the current gets
too high.
 Two types of fuses: Edison and S-type
 Circuit Breakers are generally now used.
Section 8.3

35.  A dangerous shock can
occur if an internal ‘hot’
wire comes in contact
with the metal casing of
a tool.
 This danger can be
minimized by grounding
the case with a
dedicated wire through
the third wire on the
plug.
Section 8.3

36. Section 8.3

37. Section 8.3

38.  Closely associated with electricity is magnetism.
 In fact electromagnetic waves consists of both
vibrating electric and magnetic fields. These
phenomena are basically inseparable.
 A bar magnet has two regions of magnetic
strength, called the poles.
 One pole is designated “north,” one “south.”
Section 8.4

39.  The N pole of a magnet is “north-seeking” – it
points north.
 The S pole of a magnet is “south-seeking” – it
points south.
 Magnets also have repulsive forces, specific to
their poles, called …
 Law of Poles – Like poles repel and unlike poles
attract
◦ N-S attract
◦ S-S & N-N repel
Section 8.4

40. All magnets have
two poles – they are
dipoles
Section 8.4

41.  Magnetic field - a set of imaginary lines that
indicates the direction in which a small compass
needle would point if it were placed near a magnet
 These lines are indications of the magnetic force
field.
 Magnetic fields are vector quantities.
Section 8.4

42.  The arrows indicate the direction in which the
north pole of a compass would point.
Section 8.4

43.  The source of magnetism is moving and spinning
electrons.
 Hans Oersted, a Danish physicist, first discovered
that a compass needle was deflected by a current-
carrying wire.
◦ Current open  deflection of compass needle
◦ Current closed  no deflection of compass needle
 A current-carrying wire produces a magnetic field:
stronger current  stronger field
 Electromagnet – can be switched on & off
Section 8.4

44.  Most materials have many electrons going in
many directions, therefore their magnetic effect
cancels each other out  non-magnetic
 A few materials are ferromagnetic (iron, nickel,
cobalt) – in which many atoms combine to create
magnetic domains (local regions of magnetic
alignment within a single piece of iron)
 A piece of iron with randomly oriented magnetic
domains is not magnetic.
Section 8.4

45.  The magnetic
domains are
generally random,
but when the iron
is placed in a
magnetic field the
domains line up
(usually
temporarily).
Section 8.4

46.  A simple electromagnet
consists of an insulated
coil of wire wrapped
around a piece of iron.
 Stronger current 
stronger magnet
 Electromagnets are
made of a type of iron
that is quickly
magnetized and
unmagnetized – termed
“soft.”
Section 8.4

47.  Materials cease to be ferromagnetic at very high
temperatures – the “Curie temperature” (770o
C for
iron)
 Permanent magnets are made by permanently
aligning the many magnetic domains within a
piece of material.
 One way to create a permanent magnet is to heat
the ferromagnetic material above the Curie
temperature and then cool the material under the
influence of a strong magnetic field.
Section 8.4

48.  This planet’s magnetic field exists within the earth
and extends many hundreds of miles into space.
 The aurora borealis (northern lights) and aurora
australis (southern lights) are associated with the
earth’s magnetic field.
 Although this field is weak compared to magnets
used in the laboratory, it is thought that certain
animals use it for navigation.
Section 8.4

49.  Similar to the
pattern from a
giant bar magnet
being present
within the earth
(but one is not
present!)
Section 8.4

50.  The origin of the earth’s magnetic field is not
completely understood.
◦ Probably related to internal currents of electrically
charge particles in the liquid outer core of the earth, in
association with the earth’s rotation
 The temperatures within the earth are much hotter
than the Curie temperature, so materials cannot
be ferromagnetic.
 The positions of the magnetic poles are constantly
changing, suggesting “currents.”
Section 8.4

51.  Magnetic declination – the angle between
geographic (true) north and magnetic north
 The magnetic declination varies depending upon
one’s location on Earth.
◦ In the northern hemisphere the magnetic North Pole is
about 1500 km from the geographic North Pole.
 Therefore, a compass does not point to true north,
but rather magnetic north.
◦ An adjustment (magnetic declination) must be made to
determine true north from a compass.
Section 8.4

52. Section 8.4

53.  Resistance in Series: Rs = R1+R2+R3+…
 Resistance in Parallel: 1/Rp = 1/R1+1/R2+…
 For 2 R in parallel: Rp = (R1R2)/(R1+R2)
Section 8.4

54.  Electromagnetism – the interaction of electrical
and magnetic effects
 Two basic principles:
1) Moving electric charges (current) give rise to magnetic
fields (basis for an electromagnet).
2) A magnetic field will deflect a moving electric charge
(basis for electric motors and generators).
Section 8.5

55.  Telephone transmission involves the conversion
of sound waves into varying electric current
 Sound waves at the transmitting end vibrate a
diaphragm that puts pressure on carbon grains
 The varying pressure on the carbon, results in
varying amounts of resistance in the circuit
 By Ohm’s law (V=IR), varying resistance results in
varying current
Section 8.5

56.  The varying current gives an electromagnet at the
receiving end varying magnetic strengths
 The electromagnet is drawn toward/away a
permanent magnet as a function of the electric
current in the line
 The electromagnet is attached to a diaphragm that
vibrates, creating sound waves that closely
resemble the original sound waves
Section 8.5

57. Section 8.5

58.  When a moving charge (a current) enters a
magnetic field, the moving charge will be deflected
by the magnetic field
 The magnetic force (Fmag) is perpendicular to the
plane formed by the velocity vector (v) of the
moving charge and the magnetic field (B)
Section 8.5

59.  Electrons entering a
magnetic field
experience a force
Fmag that deflects
them “out of the
page”
Section 8.5

60.  A beam of electrons is
deflected downward due
to the introduction of the
magnetic field of a bar
magnet
Section 8.5

61.  Since a magnetic field deflects a moving charge, a
current-carrying wire loop will experience a force
Section 8.5

62. Section 8.5

63. Section 8.5

64.  Electric motor – a device that converts electrical
energy into mechanical work
 When a loop of coil is carrying a current within a
magnetic field, the coil experiences a torque and
rotates
Section 8.5

65.  The inertia of the spinning loop carries it through the
positions where unstable conditions exist
 The split-ring commutator
reverses the loop current
every half-cycle, enabling
the loop to rotate
continuously
Section 8.5

66.  When a wire is moved
perpendicular to a
magnetic field, the
magnetic force causes
the electrons in the wire
the move – therefore a
current is set up in the
wire
Section 8.5

67.  Generator – a device that converts mechanical
energy into electrical energy
 Generators operate on the principle of
electromagnetic induction
 Electromagnetic induction was discovered by the
English scientist, Michael Faraday in 1831
◦ When a magnet is moved toward a loop or coil of wire,
a current is induced in the wire
◦ The same effect is obtained if the magnetic field is
stationary and the loop is rotated within it
Section 8.5

68.  A current is induced in the circuit as the magnet is
moved toward and into the coil
Section 8.5

69.  When the loop is rotated
within the magnetic field
a current is induced in
the loop
 The current varies in
direction every half-cycle
and is termed alternating
current (ac)
Section 8.5

70.  Transformers serve to step-up or step-down the
voltage of electric transmissions
 A transformers consists of two insulated coils of
wire wrapped around an iron core
◦ The iron core serves to concentrate the magnetic field
when there is a current in the input coil
 The magnetic field resulting from the primary coil
goes through the secondary coil and induces a
voltage and current
Section 8.5

71.  Since the secondary
coil has more
windings, the induced
ac voltage is greater
than the input
voltage.
 The factor of voltage
step-up depends on
the ratio of the
windings on the two
coils.
Section 8.5

72.  The main reason is to reduce the amount of
current going through the lines
◦ As the current is reduced, the amount of resistance is
also reduced
◦ If the amount of resistance is reduced, the amount of
heat (called ‘joule heat’ or ‘I2
R’ losses)
 For example, if the voltage is stepped-up by a
factor of 2, the resulting I2
R losses would be
reduced four-fold
Section 8.5

73.  The voltage change for a transformer is given by
the following:
 V2 = (N2/N1)V1
 V1 = primary voltage
 V2 = secondary voltage
 N1 = Number of windings in primary coil
 N2 = Number of windings in secondary coil
Section 8.5

74.  A transformer has 500 windings in its primary coil
and 25 in its secondary coil. If the primary voltage
is 4400 V, find the secondary voltage.
 Use equation V2 = (N2/N1)V1
 Given: N1 = 500, N2 = 25, V1 = 4400 V
 V2 = (25/500)(4400 V) = 220 V
 This is typical of a step-down transformer on a
utility pole near homes
Section 8.5

75. Voltage is dramatically stepped-up at the generating plant to minimize
joule heat loss during long-distance transmission. The voltage must
then be stepped-down for household use.
Section 8.5

76.  F = kq1q2/r2
Coulomb’s Law
 I = q/t Current
 V= W/q Voltage
 V= IR Ohm’s Law
 P= IV = I2
R Electric Power
Review

77.  Rs = R1 + R2 + R3… Resistance in Series
 1/Rp = 1/R1 + 1/R2 + 1/R3 …. Resistance in Parallel
 Rp = (R1R2)/(R1 + R2) Two Resistances in Parallel
 V2 = (N2/N1)V1 Transformer (Voltages and Turns)
Review