Physics Around the Home

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Physics Around the Home

• Electrostatics.
• Electricity.
• Heat and Temperature.

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What is Electric Charge?

• The unit of electric charge is the coulomb.
• Ordinary matter is made up of atoms which have
  positively charged nuclei and negatively charged
  electrons surrounding them.
• Charge is quantized as a multiple of the electron
  or proton charge
• Charge on a Proton p 1.602 x 10-19 coulombs
• Charge on an Electron e - 1.602 x 10-19
  coulombs

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The Key to Electrostatics

• Like Charges Repel, Unlike Charges Attract.
• 1 Coulomb of charge contains 6.25 x 1018
  elementary charges.

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Coulombs Force Law

• The force between two charges is proportional to
  the size of those charges and inversely
  proportional to the square of the distance
  between them. This is known as an inverse square
  law.

Q1
Q2
d - separation
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Problem using Coulombs Law

• What is the magnitude of the electric force on
  the electron (Q1 1.6 x 10-19C) of a hydrogen
  atom exerted by the single proton (Q2
  1.6x10-19 C) that is its nucleus, when the
  electron orbits the proton at its average
  distance of 0.53 x 10-10 m.

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Problem II Coulombs Law involving 3 charges

• Calculate the net electric force on particle 3
  (the 4.0 mC on the right) due to the other two
  charges.

0.30m
0.20m
Q1 -3.0µC
Q2 5.0µC
Q3 -4.0µC
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Solution to 3 charge problem

• Logic The net force on the 4.0 mC particle will
  be the vector sum of the force exerted by the
  3.0 mC particle and the force exerted by 5.0 mC
  particle.

-4.0mC
Due to the 5mC charge
Net force of 4.1N
Due to the -3mC charge
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Electric Fields

• The Electric Field
• Energy and Electric Potential
• Millikans Oil Drop Experiment

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Electric Field Strength

• Field around a positive charge

• The Electric Field Strength at any point in space
  is defined as the force F exerted on a tiny
  positive test charge at that point divided by the
  magnitude of the charge.
• E F/q

q

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Electric Field Strength

• Field around a negative charge

• Electric Field Strength has units of Newtons per
  coulomb (N/C)
• It is a vector quantity

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Electric Field Strength

• The Electric Field Strength E due to a single
  point charge Q at a distance d is given by
• .

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Properties of Field Lines

• 1. The field lines indicate the direction of the
  electric field the field points in the direction
  tangent to the field lines at any point.
• 2. The lines are drawn so that the magnitude of
  the electric field, E, is proportional to the
  number of lines crossing unit area perpendicular
  to the lines. The closer the lines, the stronger
  the field.
• 3. Electric Field lines start on positive chares
  and end on negative charges and the number
  starting or ending is proportional tot he
  magnitude of the charge.

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Fields between charges

• Field between two point charges

• Field between two parallel plates.

positive
negative
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Electric Potential

• Consider two parallel plates.

High P.E.

- - - - - - - - - - - - - -
Low P.E
Where is a small test charge
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Relationship between Electric Potential and
Electric Fields.

• The work done W by the electric field to move a
  positive charge q from b to a is
• W q Vba (1)
• We can also write the work done W as the force
  F times distance d
• W Fd (2)
• The Force F on a small positive test charge q
  in an electric field E is
• F q E (3)

• Consider two parallel plates.

High PE

b
a
- - - - - - - - - - - -
Low PE
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Relationship between Electric Potential and
Electric Fields.

• Substituting (3) into (2) we get
• W q E d (4)
• where d is the distance between points a and
  b.
• We now equate equation (4) to equation (1), we
  get
• qVba q Ed

? V Ed
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The Electron Volt - A Unit of Energy

• The joule is a very large unit when dealing with
  energies of electrons, atoms or molecules.
• One electron volt is defined as the energy
  acquired by an electron as a result of moving
  through a potential difference of 1 V.
• 1 eV 1.6 x 10-19 J.

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Electric Potential Due to a Single Point Charge.

• The electric Potential at a distance r from a
  single point charge Q is given by

r
Q
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Millikans Oil-drop Experiment.

• Consider two charged parallel plates.

• Millikan measured the charge of an electron.
• A drop is suspended if the electric and
  gravitational forces are balanced.
• Fw mg
• Fe Eq
• ? Eq mg

Fe
Fw
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Electricity

• Electric Current.
• Voltage
• Resistance
• Series and Parallel Ciruits.
• Kirchhoffs Laws

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Electric Current.

• The flow of charge in a wire is called current.
• It is expressed in terms of the number of
  coulombs per second going past a given point on a
  wire.
• One coulomb/sec equals 1 ampere (symbol A) and
  corresponds to a flow of about 6.24 1018
  electrons per second past any point of the
  circuit
• The unit of electric current is named after the
  French physicist André Marie Ampère.

André Marie Ampère (1775-1836)
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Voltage.

• When 1 coulomb of charge travels across a
  potential difference of 1 volt, the work done
  equals 1 joule.
• The unit named after the English physicist James
  Prescott Joule (1818-1889).

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Resistance and Ohms Law

• The unit used for expressing the quantity of
  resistance is the ohm. The symbol for the ohm is
  the Greek letter ?, omega.
• It is defined as the amount of resistance that
  will limit the flow of current to 1 ampere in a
  circuit with a potential difference of 1 volt.
• OHMS LAW V I R, in which V is the votage (in
  volts), I is the current in amperes, and R is the
  resistance in ohms.

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Resistors

• Measuring the value of the Resistance.
• Colour Code
• Series and Parallel Circuits

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Voltmeter-Ammeter Method

• The ratio of V/I gives the resistance.
• A graph of V vs I will produce a straight line
  with the slope being the resistance, in ?

A
R1
V
V
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Wheatstone Bridge

• This is a slide-wire potentiometer.
• When the bridge is balanced, no current flows
  through the Galvanometer.

L1
Rx
G
V
R1
L2
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Colour Code used on Resistors.

• Black 0 Brown 1 Red 2 Orange 3 Yellow
  4 Green 5 Blue 6 Violet 7 Grey 8
  White 9
• Associated error gold 5 silver 10 none 20

tolerance
multiplier
Second digit
First digit
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Series Circuits

• Rt R1 R2 R3 . .
• Vt V1 V2 V3 . .
• It I1 I2 I3 . .

R1
R2
V
R3
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Parallel Circuits

• Vt V1 V2 V3 . .
• It I1 I2 I3 . . .

R1
R2
V
R3
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Kirchhoffs JUNCTION Rule for Circuits

• G.R.Kirchhoff (1824-1887)
• It is based on the conservation of charge
• At any junction point, the sum of all currents
  entering the junction must equal the sum of all
  the currents leaving the junction.

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Kirchhoffs Loop Rule for Circuits

• It is based on the conservation of energy.
• The algebraic sum of the changes in potential
  around any closed path of the circuit must be
  zero.

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Power in Electric Circuits

• The electric power in watts associated with a
  complete electric circuit or a circuit component
  represents the rate at which energy is converted
  from the electrical energy of the moving charges
  to some other form, e.g., heat, mechanical
  energy, or energy stored in electric fields or
  magnetic fields.

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Heat and Temperature.

• Heat and Thermal Energy
• Law of heat exchange
• Specific Heat and Latent Heat.

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Heat as Energy Transfer

• Heat refers to energy that is transferred from
  one body to another because of a difference in
  temperature.
• It is measured in JOULES.
• A common unit for heat was the CALORIE. It was
  defined as the amount of heat necessary to raise
  the temperature of 1 gram of water by 1 degree
  Celsius.
• 1 Calorie 4.189 Joules.

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Specific Heat Calorimetry

• The amount of heat Q required to change the
  temperature of a system is found to be
  proportional too the mass m of the system and to
  the temperature change ?T, and can be expressed
  in the equation
• Q m c ?T
• where c is a quantity characteristic of the
  material called specific heat.

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Specific Heat Problem
Problem - How much heat is required to raise the
temperature of 20 kg of iron form 10oC to
90oC? Solution - Data - specific Heat for Iron c
450 J/kgoC Temperature Change ?T
(90-10) 80oC. Thus Q m c ?T 20 x 450 x 80
7.2 x 105 J. Therefore the amount of heat
required was 720 kJ.
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Law of Heat Exchange

• The law of heat exchange has its origins in the
  law of conservation of energy.
• If the system is completely isolated, no energy
  can flow into or out of it the heat lost by one
  part of the system is equal to the heat gained by
  the other part of the system.
• Heat lost heat gained.

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Heat Exchange Problem

• Problem - If 200 cm3 of tea at 95oC is poured
  into a 150 g glass initially at 25oC, what will
  be the final temperature of the mixture when
  equilibrium is reached, assuming that no heat
  flows to the surroundings?

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Solution Since tea is mainly water, its
specific heat is 4180 J/kg oC and its mass is its
density times its volume m ? V 1x103 x 200 x
10-6 0.20 kg Using the law of heat exchange
(and let T be the final Temperature of the
mixture) heat lost heat gained mtea ctea ?T
mcup ccup ?T 0.20 x 4180 x (95-T) 0.15 x 840 x
(T-25) 79 420 - 836T 126 T - 3150 T 85.8 oC
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Heat Exchange Problem

• We wish to determine the specific heat of a new
  alloy. A 0.150 kg sample of the alloy is heated
  to 540oC. It is then quickly placed in 400g of
  water at 10.0oC which is contained in a 200 g
  aluminum calorimeter cup. The final temperature
  of the mixture is 30.5oC. Calculate the specific
  heat of the alloy.

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Data Required for this problem to find Calloy

• Mass of alloy malloy 0.150 kg
• Temperature Change for alloy
• ?Talloy (540-30.5) 509.5oC
• Mass of water mwater 0.40kg
• Specific Heat of water cwater 4180 J/kg oC
• Temperature Change for water
• ?Twater (30.5 - 10) 20.5oC
• Mass of the calorimeter cup mcup 0.20kg
• Specific Heat of the cup ccup 900 J/kg oC
• Temperature Change for the cup
• ?Tcup (30.5 - 10.0)20.5oC

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Solution to the problem

• Heat Lost Heat Gained
• Heat lost by alloy heat gained by water and cup
• maca?Ta mwcw?Tw mccc?Tc
• 0.15 x ca x 509.5 0.4 x 4180 x 20.5 0.2 x 900
  x 20.5
• 76.4 ca (34 300 3700)
• ca 500 J/kg oC.
• Therefore the specific heat of the alloy was 500
  J/kg oC.

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Latent Heat

• The heat required to change the phase of a
  substance (eg from solid to a liquid) is known as
  Latent Heat.
• When the substance is changing phase, there is no
  change in the substances temperature.
• The formula is simply
• Q m L
• where Q is the heat (in Joules), m is the mass
  (in kg) and L is the Latent Heat (in J/kg)

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Latent Heat of Fusion Lv

• The Latent heat of fusion is the amount of heat
  required to change 1.0kg of a substance from the
  solid to the liquid state.
• For example, when 5.00kg of water freezes at 0oC,
  the water must release
• Q m L 5.00 x 3.33 x 105
• ? Q 1.67 x 106 J of energy

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Latent Heat of Vaporisation Lv

• The Latent heat of vaporisation is the amount of
  heat required to change 1.0kg of a substance from
  the liquid to the gaseous state.
• For example, when 5.00kg of water boils at 100oC,
  the water must absorb
• Q m L 5.00 x 22.6 x 105
• ? Q 1.13 x 107 J of energy

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Problem involving Latent Heats.

• Problem - How much energy does a refrigerator
  have to remove from 1.5kg of water at 20oC to
  make ice at -12oC?
• Data -
• mass of water mw 1.5 kg
• specific heat of water cw 4180 J/kgoC
• Temperature change of water ?Twater 20oC
• latent heat of water/ice Lf 3.33 x 105 J/kg
• mass of ice mice 1.5 kg
• specific heat of ice ci 2100 J/kgoC
• Temperature change of the ice ?Tice 12oC

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Solution
Q mwater cwater ?Twater m Lf mice cice
?Tice Q 1.5 x 4180 x 20 1.5 x 3.33 x 105
1.5 x 2100 x 12 Q 6.6 x 105 J Therefore 660 kJ
must be removed by the refrigerator in order to
freeze ice, initially at 20oC down to -12oC.
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Amount of Heat Flow

• How much energy does it take to boil 1.0 litre of
  water in a 1500W electric jug if the water is
  originally at 20.0oC?
• Data
• C 4180 J/kgoC
• m 1.0 kg
• ?T 100 20 80.0oC
• P 1500 W
• Find time, t

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Linear Thermal Expansion

• The change in length of a 1m length of a
  substance due to a temperature change of 1oC is
  called the coefficient of linear expansion (?).
• The formula to help find the change in length of
  a substance due to thermal expansion of solids is
  
• ?L Li ? ?T

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Linear Expansion Problem

• An electric company strung an aluminum wire
  between two piers 200m apart on a day when the
  temperature was 25oC. They strung it tight so
  that it would not sag. Find the length of the
  wire when the temperature fell to -25oC on a cold
  winters night? What might happen? What should
  have been done to prevent this from occurring?

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Volumetric Thermal Expansion

• The coefficient of volume expansion (?) is
  normally associated with liquids. Liquids take
  the shape of the container hence we are mainly
  interested in the volume changes of liquids with
  temperature.
• As with solids, the change in volume of liquids
  can be found using the formula
• ?V Vi ? ?T

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Volume Expansion Problem

• What would be the increase in the volume of 0.20L
  of acetone if it was heated form 10oC to 40oC?
• Solution
• ?V Vi ? ?T
• 14.87 x 10-4 x 0.20 x (40-10)
• 89.2 x 10-4 L
• 8.9 x 10-3 L

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References

• All photographs, including those of the
  Scientists are from the following sources
• Microsoft Encarta.