Debating the Olympics: For or Against London 2012
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Electricity and
Magnetism
Unit III
I Electrostatics
0
The study of electric charges at rest and their electric
fields and potentials
0
Charges at rest if there is no net transfer of charge
A.
Microstructure of Matter
0
The smallest unit of matter is an atom
0
Atoms are made up of protons, neutrons and electrons
0
The electron is the fundamental negatively charged
particle of matter
0
The proton is the fundamental positively charged
particle of matter
0
The elementary charge (e) equals
0
The charge on an electron (-e)
0
The charge on a proton (+e)
0
Because all atoms are electrically neutral, atoms
contain equal number of protons and neutrons
0
Neutrons have no charge
B. Charged Objects
0
Protons and neutrons cannot be removed from an
atom
0
Electrically charged objects are formed when neutral
objects gain or lose electrons
0
When an atom becomes a charged particle, it is called
an ion
0
Excess electrons creates negative ions
0
Loss of electrons creates positive ions
0
Objects with the same sign of charge are repelled
0
Objects with opposite signs are attracted
C. Transfer of Charge
0
If a system only contains neutral objects, the total net
charge is zero
0
If the objects are rubbed together, electrons are
transferred
0
The system as a whole remains neutral
Example
3 spheres have initial charges
0
When R and S touch, excess electrons (-8e) move to
neutral (0e)
0
When separated each sphere “splits” the excess
0
(-8e + 0e) ÷ 2 spheres
0
Each sphere becomes -4e
0
When S and T touch, excess electrons (-4e) moves to
(+6e)
0
When separated each sphere “splits” the excess
0
(-4e + 6e) ÷ 2 spheres
0
Each sphere becomes +1e
0
TOTALS R = -4e S = +1e T = +1e
0e
R
-8e
S
+6e
T
D. Quantity of Charge
0
SI unit of charge is a coulomb (C)
0
1C = 6.25 x 10
18
elementary charges (electrons)
0
The charge on an electron (-e) = 1.6 x 10
-19
C
0
The net charge on a charged object is always a multiple
of the charge on an electron
E. Coulomb’s Law
0
The electrostatic force between 2 charges is
0
Directly related to the product of the charges
0
Inversely related to the square of the distance between
them
0
F…electrostatic force (newtons)
0
q…charge (coulombs)
0
r…distance between charges (meters)
0
k…electrostatic constant (8.99 x 10
9
N m
2
/C
2
)
F =
kq
1
q
2
r
2
II. Electric Fields
o
Region around a charged particle
o
Exerts a force on other charged particles
o
Represented by field lines
o
Direction a positive charge(+) would move
o
If lines are curved, charge would move tangent to
the point on the field line
o
Begin (+) and end (-)
o
Never cross
o
Energy varies inversely squared with
distance
o
Strength
Where
E = electric field strength (N/C)
F = force the charge experiences (N)
q = the charge (C)
E = F
q
What is the magnitude of the electric field at a
point in a field where an electron experiences a
1.0 N force?
E = 6.3 x 10
18
N/C
E = F
q
E = 1.0 N
1.60 x 10
-19
C
A. Potential Difference
o
Work done as charge is moved against a
field
o
Where W is work in joules
o
q is charge in coulombs
o
V is
potential difference
o
Joules/coulomb is the derived unit volt
o
1 J/C = 1 volt
o
If the charge is an elementary charge
(ex.
electron)
the
potential difference is an
electron volt (eV)
o
1 eV = 1.60 x 10
-19
J
V = W
q
Moving a point charge of 3.2 x 10
-19
coulombs
between two points in an electric field requires
4.8 x 10
-18
joules. What is the potential
difference between these two points?
V = 15 V
q
V = W
3.2 x 10
-19
C
V = 4.8 x 10
-18
J
III Electric Current
A. Electric current
o
The rate that a charge passes a point in a circuit
o
Represented using the symbol, I
o
Unit is an ampere (A)
o
Formula I = q
t
o
I is current in amperes
o
q = charge in coulombs
o
t = time in seconds
o
Current is measured using an ammeter
B. Electric circuit
o
The closed path that a charged particle moves along
o
Potential difference is also needed for an electric
current
o
Supplied by
o
Cell
o
Battery
o
Measured using a voltmeter
o
Electron flow determines the direction of current
o
A switch is used to make, break or change the
connections in a circuit
C. Resistance
o
Electrical resistance (R) is the opposition of
electron flow
o
Ratio of potential difference to current flow
o
R = V
I
o
Where
o
V = potential difference in volts
o
I = current in amperes
o
R = resistance in volts per ampere
•
Ohm (
Ω
)represents 1 volt per
ampere
o
Resistance is affected by temperature
A current of 0.10 amperes flow through a lamp
connected to a 12.0-volt source. What is the resistance
of the lamp?
I = 0.10 A
V = 12.0 V
R = V
I
R = 12.0 V
0.10 A
R = 120
Ω
Factors Affecting Resistance
Length of Wire
o
Increasing length, increases the collisions of electrons with
atoms in the wire
o
Directly related
Thickness of Wire
o
Increasing thickness creates more spaces for electrons to
travel through, decreasing the resistance
o
Inverse relationship with cross-sectional area
Resistivity
o
Characteristic of the material
o
Good conductors have low resistivities
o
As temperature increases, resistivity increases
o
Values found in reference tables
Resistance of a Wire
R =
ρ
L
A
Where
•
R is resistance in ohms (
Ω
)
•
ρ
is the resistivity in ohm-meters (
Ω
-m)
•
L is length in meters (m)
•
A is cross-sectional area in square meters (m
2
)
Determine the resistance of a 4.00-meter piece
of copper wire that has a diameter of 2.00 mm
at 20°C.
ρ
copper
= 1.72 x 10
-2
L = 4.00 m
Diameter = 0.002 m
Area =
π
r
2
Area =
π
(.001)
2
Area = 3.14 x 10
-6
m
2
R = (1.72 x 10
-2
x 4.00)
3.14 x 10
-6
R =2.19 x 10
-2
Ω
Resistor
•
Device to have a definite amount of resistance
•
Used to limit current flow
•
Provide a potential drop
•
Symbol
Electric Power
Power (P) = Work/time
Units
Work,,,joules
Time sec
Power watts
Electrical power is the product of VI
Since V=IR then
P = I2R
In January 2012, a writing task challenged students to take a stance on the upcoming Olympic Games in London. This content provides guidance on constructing a formal argument in a letter to a national newspaper. Explore viewpoints for and against the Olympics, plan arguments, format a formal letter, and include essential elements for a top-grade response.
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Presentation Transcript
I Electrostatics 0 The study of electric charges at rest and their electric fields and potentials 0 Charges at rest if there is no net transfer of charge A. Microstructure of Matter 0 The smallest unit of matter is an atom 0 Atoms are made up of protons, neutrons and electrons 0 The electron is the fundamental negatively charged particle of matter 0 The proton is the fundamental positively charged particle of matter
0 The elementary charge (e) equals 0 The charge on an electron (-e) 0 The charge on a proton (+e) 0 Because all atoms are electrically neutral, atoms contain equal number of protons and neutrons 0 Neutrons have no charge
B. Charged Objects 0 Protons and neutrons cannot be removed from an atom 0 Electrically charged objects are formed when neutral objects gain or lose electrons 0 When an atom becomes a charged particle, it is called an ion 0 Excess electrons creates negative ions 0 Loss of electrons creates positive ions 0 Objects with the same sign of charge are repelled 0 Objects with opposite signs are attracted
C. Transfer of Charge 0 If a system only contains neutral objects, the total net charge is zero 0 If the objects are rubbed together, electrons are transferred 0 The system as a whole remains neutral
Example 3 spheres have initial charges 0e R -8e S +6e T 0 When R and S touch, excess electrons (-8e) move to neutral (0e) 0 When separated each sphere splits the excess 0 (-8e + 0e) 2 spheres 0 Each sphere becomes -4e 0 When S and T touch, excess electrons (-4e) moves to (+6e) 0 When separated each sphere splits the excess 0 (-4e + 6e) 2 spheres 0 Each sphere becomes +1e 0 TOTALS R = -4e S = +1e T = +1e
D. Quantity of Charge 0 SI unit of charge is a coulomb (C) 0 1C = 6.25 x 1018 elementary charges (electrons) 0 The charge on an electron (-e) = 1.6 x 10-19C 0 The net charge on a charged object is always a multiple of the charge on an electron
E. Coulombs Law 0 The electrostatic force between 2 charges is 0 Directly related to the product of the charges 0 Inversely related to the square of the distance between them kq1q2 r2 F = 0 F electrostatic force (newtons) 0 q charge (coulombs) 0 r distance between charges (meters) 0 k electrostatic constant (8.99 x 109 N m2/C2)
II. Electric Fields oRegion around a charged particle oExerts a force on other charged particles oRepresented by field lines oDirection a positive charge(+) would move oIf lines are curved, charge would move tangent to the point on the field line oBegin (+) and end (-) oNever cross oEnergy varies inversely squared with distance
oStrength E = F q Where E = electric field strength (N/C) F = force the charge experiences (N) q = the charge (C)
What is the magnitude of the electric field at a point in a field where an electron experiences a 1.0 N force? E = F q E = 1.0 N 1.60 x 10-19C E = 6.3 x 1018 N/C
A. Potential Difference oWork done as charge is moved against a field V = W q oWhere W is work in joules oq is charge in coulombs oV is potential difference oJoules/coulomb is the derived unit volt o 1 J/C = 1 volt oIf the charge is an elementary charge (ex. electron) thepotential difference is an electron volt (eV) o1 eV = 1.60 x 10-19J
Moving a point charge of 3.2 x 10-19 coulombs between two points in an electric field requires 4.8 x 10-18 joules. What is the potential difference between these two points? V = W q V = 4.8 x 10-18 J 3.2 x 10-19C V = 15 V
III Electric Current A. Electric current oThe rate that a charge passes a point in a circuit oRepresented using the symbol, I oUnit is an ampere (A) oFormula I = q t oI is current in amperes oq = charge in coulombs ot = time in seconds oCurrent is measured using an ammeter
B. Electric circuit oThe closed path that a charged particle moves along oPotential difference is also needed for an electric current oSupplied by o Cell o Battery oMeasured using a voltmeter oElectron flow determines the direction of current oA switch is used to make, break or change the connections in a circuit
C. Resistance oElectrical resistance (R) is the opposition of electron flow oRatio of potential difference to current flow oR = V I oWhere oV = potential difference in volts oI = current in amperes oR = resistance in volts per ampere Ohm ( )represents 1 volt per ampere oResistance is affected by temperature
A current of 0.10 amperes flow through a lamp connected to a 12.0-volt source. What is the resistance of the lamp? I = 0.10 A V = 12.0 V R = V I R = 12.0 V 0.10 A R = 120
Factors Affecting Resistance Length of Wire o Increasing length, increases the collisions of electrons with atoms in the wire o Directly related Thickness of Wire o Increasing thickness creates more spaces for electrons to travel through, decreasing the resistance o Inverse relationship with cross-sectional area Resistivity o Characteristic of the material o Good conductors have low resistivities o As temperature increases, resistivity increases o Values found in reference tables
Resistance of a Wire R = L A Where R is resistance in ohms ( ) is the resistivity in ohm-meters ( -m) L is length in meters (m) A is cross-sectional area in square meters (m2)
Determine the resistance of a 4.00-meter piece of copper wire that has a diameter of 2.00 mm at 20 C. copper = 1.72 x 10-2 L = 4.00 m Diameter = 0.002 m Area = r2 Area = (.001)2 Area = 3.14 x 10-6 m2 R = (1.72 x 10-2 x 4.00) 3.14 x 10-6 R =2.19 x 10-2
Resistor Device to have a definite amount of resistance Used to limit current flow Provide a potential drop Symbol
Electric Power Power (P) = Work/time Units Work,,,joules Time sec Power watts Electrical power is the product of VI Since V=IR then P = I2R
