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Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Unit 2: Electric Forces and Fields
Electric Forces
Other forces can act on objects beside the gravitational force
– e.g.
 Magnetic force – causing a nail to accelerate towards a
magnet

Electric force – a rubbed comb with sweater attracts
small pieces of paper
There are two kinds of charges.
1. Positive Charge
An object is positively charged if it has more protons than
electrons.
2.
Negative Charge
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
An object is negatively charged if it has more electrons than
protons.
Neutral objects
An object is neutral when the number of electrons is equal
to the number of protons.
 Protons do not leave their positions and are not free to
move. But electrons can be added or removed from
atoms.
 Static electricity (electricity at rest) - Electrostatic
charges are collected and held in the object.
Law of Electrostatics
1. Like charges repel.
2. Unlike charges attract.
3. Charged objects attract neutral objects
http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l1c.html
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Methods of Charging
1. Friction – electrons move from one object to another when
it is rubbed. Electrons go to the object that has stronger
attraction to electrons (protons do not leave their positions).
Ex: Ebonite rubbed with fur becomes negatively charged
(Electrons move from the fur to ebonite.) Glass rod
rubbed with silk becomes positively charged (Glass loses
electrons to silk)
Law of Conservation of Charge - In an isolated system the
total amount of charge remains constant. Charges only
separate or electrons move from one object to another.
http://www.regentsprep.org/Regents/physics/phys03/atribo/default.htm
http://www.regentsprep.org/Regents/physics/phys03/aeleclab/chargstp.htm
http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l2a.html
2. Conduction or Direct Contact – An object gets the same
charge as the charging object by contact. Protons do not
move from atom to atom.
http://www.regentsprep.org/Regents/physics/phys03/aeleclab/escope.htm
http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l2c.html
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Electroscope – device to identify and measure the amount and
type of charge.
- Pith ball electroscope
http://www.fiu.edu/~phydemo/j1-04.htm
- Metal leaf electroscope
http://www.engr.uky.edu/~gedney/courses/ee468/expmnt/escope.html
http://rene.balderacchi.club.fr/illustrations/Electroscope%2002.jpg
Charging a metal leaf electroscope by contact
http://regentsprep.org/Regents/physics/phys03/aeleclab/induct.htm
http://cougar.slvhs.slv.k12.ca.us/~pboomer/labsphys/physlabook/lab36.htm
l
http://www.toppermost.biz/electroscope.gif
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
3. Induction – There is no contact between the charging
object and the charged object. This is a rearrangement of
charges. By induction, an object receives the opposite
charge.
http://www.physicsclassroom.com/mmedia/estatics/isop.html
http://www.physicsclassroom.com/mmedia/estatics/itsn.html
http://www.physicsclassroom.com/mmedia/estatics/epn.html
http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l2b.html
http://encarta.msn.com/media_701504655_761566543_1_1/Charging_by_Induction.html
http://www.shep.net/resources/curricular/physics/P30/Unit2/electroscope.html
Grounding charged object by providing a path for the charges
to move to/from Earth.
http://www.physicsclassroom.com/mmedia/estatics/gep.html
http://www.physicsclassroom.com/mmedia/estatics/gen.html
http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l2b.html
Why does a charged object attract a neutral object?
The electrons in the neutral object rearrange themselves
so that unlike charges are close to each other.
http://www.ux1.eiu.edu/~cfadd/1160/Ch17/Friction.html
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Lightning – is a discharge (transfer of electrons) between 2
clouds or a cloud and the earth.
http://www.regentsprep.org/Regents/physics/phys03/alightnin/default.htm
p. 539 – key terms (define)
Movie:
http://www.bchs.calgary.ab.ca/site.php?p=/curriculum/science/physics30/
Charging electroscope and pithball
http://departments.weber.edu/physics/amiri/director/dcrfiles/electricity/pith
BallS.dcr
http://departments.weber.edu/physics/amiri/director/dcrfiles/electricity/elect
roscopeS.dcr
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Coulomb’s Law
Coulomb’s law states that the electric force between two
charges is directly proportional to the product of the charges
and inversely proportional to the square of the distance
between them.
Fe q1q2
Fe 1/R2
Fe = k
q1q 2
R2
Fe = Electric Force in N
q1 and q2 are charges in Coulombs (C)
R = distance between the charges in m
k = Coulomb’s Constant = 8.99 x 109 N.m2/C2
1 C = 6.25 x 1018 elementary charges (e- or e+)
Thus, the charge on an electron and proton is:
1 e- = 1.60 x 10-19 C
1 e+ = 1.60 x 10-19 C
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1

Double the distance, force drops by ¼.

Double both charges, force increases by a factor of 4.
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
http://www.usm.maine.edu/~newton/Chy251_253/Lectures/CoulombsLaw/Coulo
mb.html
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Example
Find the force between a 50 C charge and a -30 C
charge separated by a distance of 20 cm.
Fe = kq1q2/R2
= (8.99 x 109)(5.0 x 10-5)(3.0x10-5) / (2.0 x 10-1)2
= 337.1 N
F = 3.4 x 102 N (attraction)
An Elementary Charge is
The charge on an electron is
The charge on a proton is
The charge on an alpha particle is
1.60 x 10-19C
1.60 x 10-19 C
1.60 x 10-19 C
3.20 x 10-19 C
Find the force between a proton and another
proton separated by a distance of 5.0 x 10-11 m.
Ex:
Fe = kq1q2/R2
= (8.99 x 109)(1.60 x 10-19 )(1.60x10-19 ) / (5.0 x 10-11 )2
F = 9.2 x 10-8 N (repulsion)
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Ex: Find the electric force between a nucleus (+20e)
and an electron separated by a distance of 5.0 x 10-11
m.
Fe = kq1q2/R2
= (8.99 x 109 )(20)(1.60 x 10-19 )(1.60 x 10-19 ) /
(5.0x10-11 )2
F = 1.8 x 10-6 N (attraction)
Newton’s Law of Universal Gravitation
The gravitational force between two masses is directly
proportional to the product of the masses and inversely
proportional to the square of the distance between them
Fg 1/R2
Fg  m1m2
Fg Gm1m2/R2
Fg = Gravitational Force (N)
m1, m2 = Mass 1, 2 (kg)
R = Distance between the centers of masses (m)
G = Gravitational Constant = 6.67 x 10-11 N.m2/kg2
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Example Find
the gravitational force between a 20 kg
object and a 50 kg object if the distance between them
is 30 cm.
Fg= 7.4 x 10-7 N
Differences between Gravitational Force and Electric Force
Gravitational Force
Electric Force
Weak Force
Strong Force
Only attractive
Both attractive repulsive
Inversely proportional to the square of Inversely proportional to the square of
distance between the masses.
distance between the charges.
Proportional to the product of the
Proportional to the product of the
masses
charges
p. 530 #1, p.531 #1, p. 533 #1, 2 p. 538 #5 - 7
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
The redistribution of charges on spheres after they touch
Example
1. Sphere 1 of charge 10 C is touched to sphere 2 of
charge –20 C and then separated. Then sphere 1 is
touched to sphere 3 of 25 C and separated again. Find the
charge on each sphere.
10
-20
-5
-5
(1)
(2)
(1)
(2)
Thus,
Sphere 1
Sphere 2
Sphere 3
=
=
=
10 C
-5.0 C
10 C
-5
25
10
10
(1)
(3)
(1)
(3)
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
2. A conducting sphere X that has an initial charge of 2.0x10-8 C and an identical
conducting sphere Y has an initial charge of –3.0x10-8 C are touched together.
(a) After they are separated, what is the charge on sphere X?
(b) If the distance between the charges is 2.0 cm after they are separated, what is
the electric force between the charges?
(a) -5.0x10-9 C
(b) 5.6x10-4 N
Vector addition of electric forces
Example: Three charges are collinear as shown.
FBC
FAB
3.0C
A
4.0C
1.0 m
B
-5.0C
2.0 m
Find the net electric force on B.
C
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
2
FAB=kq1q2/R
= (8.99x109)(3.0x10-6 )(4.0x10-6 )/12 = 0.10788 N (repulsion)
FBC = (8.99x109 )(4.0x10-6 )(5.0x10-6 )/22 = 0.04495 N (attraction)
Net force on B = 0.10788+0.04498
FNet = 0.15 N (right)
B
FAB
FBC
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Example
Three charges of 30 C, 40C, and 50 C are placed on the
vertices of a right triangle as shown. Find the net force on the
charge at the right angle.
30 C
450
50 C
1.5 m
θ
FR
F1
F2
40C
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Electric Fields
 The region around a charge in which there is an
electric force experienced by a neutral object or a
charged object is the electric field.
 Electric field is a vector quantity.
(Gravitational field produced by a mass is also a vector
quantity)
 Examples for scalar fields are heat and sound
fields.
 The direction of an electric field, produced by a
charged object, is the direction of the electric force
on a positive test charge.
This direction representation is a convention used by
all scientists.
+
+
Producing
charge
+
Electric fields are drawn
away from positive charges
Producing
charge
-
Test charge
Electric fields are drawn
toward negative charges.
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
E
E
++
++
http://www.mta.ca/faculty/science/physics/suren/FieldLines/FieldLines.html
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
To calculate the magnitude of the electric field intensity:
ee
Electric force per unit charge
E = Electric Field Intensity (N/C)
Fe = Electric Force (N -Newton)
q = Test Charge (C - Coulombs)
Rem: Gravitational field intensity is gravitational force per unit mass, g = Fg/m
charge of 2.0 C is placed in an electric field and it
experiences a force of 20 N. Find the electric field intensity at
that point.
Ex1: A
E = Fe/q
= 20 N / 2.0 x 10-6 C
= 1.0 x 107 N/C
charge of 5.0 C is placed in an electric field (3.0 x 104
N/C). Find the electric force.
Ex2: A
E = Fe/q
3.0 x 104 = Fe/ 5.0
Fe = 1.5 x 105 N
F = Eq
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Electric Field intensity due to a point charge
(the charge producing the field)
r
Q
E = kQ/r2
E = Electric field intensity (N/C)
Q = Charge producing the electric field (C - Coulombs)
R = Distance from the charge to the point (m)
Ex3: Find the electric field intensity at a point 20 cm from a
charge of 5.0 C.
E = kQ/r2
= 8.99 x 109 (5.0) / (0.2)2
E = 1.1 x 1012 N/C
Ex4: Find the electric field intensity 10 cm from an alpha
particle.
E= kQ/r2=8.99x109x3.20x10-19/0.102
E = 2.9x10-7N/C
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
p. 548 #1, 2 p. 549 #1, 2 p. 550 #1, 2 p. 553 #1
E
-
-
F
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Electric Field Intensity due to two Charged Plates
Two parallel plates, one charged positively and the other
negatively, creates a uniform electric field.
+++++++++++
d
E (uniform)
---------------
E = V/d
E = Electric Field Intensity (N/C or V/m)
V = Potential Difference between the plates (V = Volts)
d = distance between the plates (m)
For charged plates use only these two formulas
involving the electric field:
E=V
d
E=F
q
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Potential difference or Voltage is the amount of work
done to move a unit charge from one point to another
against the field. (If the charge is moved perpendicularly
through the field no work is done – same as gravity).
V=W
q
V = Potential Difference (V - Volts)
W = Work done (J - Joules)
q = Amount of charge (C-Coulombs)
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Work is also equal to the change in energy (energy used or released). Thus,
V = ΔE
q
1 V = 1 J/C
Find the electric field intensity between two plates with a
potential difference of 50 V and a distance of 5.0 cm between
them.
Ex:
E = V/d
= 50 V /5.0 x 10-2 m
= 1000 V/m
= 1.0 x 103 V/m
Note:
The electric field intensity between two parallel plates is
uniform. The electric field intensity inside a conductor is
zero. Charges usually reside on the outer surface of a metal
sphere.
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
1 eV = 1.60 x 10-19 J (eV = electron volt)
p. 540 #11, 12 p. 548 #1, 2 p. 549 #1, 2
p. 550 #1, 2 p. 553 #1
p. 565 #1, 2 p. 566 #1, 2
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
p. 434 #5-8
p. 444 #10-13
p. 703 #1-5,8-12
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Electric field Intensity Practice
Example
What is the magnitude and direction of the electric field intensity at a point 3.0 m
to the right of a positive point charge of 5.4x10-4 C?
E=kQ/R2=8.99x109x5.4x10-4/3.02
=5.4x105 N/C, right
Example
50C
-10C
X
X
0.45m Y 0.30 m Z
What is the magnitude and direction of the electric field at a point Z in the
diagram above, due to the charged spheres at points X and Y?
E due to X=kQ/R2=
8.99x109 x50x10-6/0.752
=799111.1N/C, right(direction taken by a positive test charge)
E due to the Y=kQ/R2=
8.99x109x10x10-6/0.302=
998888.888 N/C, left(direction taken
by a positive test charge)
Net electric field intensity=998888.88-799111.11
=2.0x105N/C, left
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Example
Determine the magnitude and direction of the electric field at point
Z in the diagram below, due to the
charges at points X and Y.
Z
4.0 cm
X
3.0 cm
3.0cm Y
Charge on X=2.0x10-8 C
Charge on Y=2.0x10-8 C
XZ=YZ=5.0 cm
E due to X=kQ/R2=
8.99x109x2.0x10-8/.052=
71920 N/C
E due to Y is the same.
0=53.1301
Ex(x)=71920 cos 53.1301=43152 N/C
Ex(y)= 71920 sin 53.1301= 57535.998 N/C
Ey(x)= -71920 cos 53.1301= -43152 N/C
Ey(y)= 71920 sin 53.1301= 57535.998 N/C
The x-components cancel each other.
The sum of the y-components= 115071.996 N/C
1.2x105 N/C (up)
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
#14. What is the magnitude and direction of the electric field at Z due to X and
Y?
X
0.60m
Y
0.30 m Z
-5
-6
X=2.0x10 C Y=8.0x10 C
Ex= 221975.30864 N/C to the right
Ey= 799111.11111N/C, right
E resultant= 1021086.419 N/C to the right =1.0 x106 N/C
Unit 2: Electric Forces and Fields
Chapter 2: Electric Field
v 1.1
Physics 30 – Electric field intensity practice problems:
1. What is the electric field intensity 0.50m away from a small sphere with a charge of 1.6 x 10-8 C?
2.
Calculate the electric field intensity midway between two negative charges of 3.2 x 10-9 C and 6.4 x
10-9 C that are 30 cm apart.
3. What is the magnitude and direction of the electric field at point Z due to charges at points X and Y.
-5μC
+1.5μC
X
Y
Z
4.5 m
3.0 m
4. Calculate the electric field intensity at point Z. If an electron is placed at point Z, calculate the
electron’s acceleration.
A= 2.0 μC
B= 1.2 μC
C= 3.5 μC
B
30 cm
A
70 cm
Z
30 cm
C
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
Electric Power and Electrical
Energy
Ex: The wire in a house circuit is rated at
15.0 A that has a resistance of 0.150 
a) What is its power rating?
b) How much heat does the wire give off in
10.0 mins?
a)
P = I2R
P = (15.0)2(0.150)
= 33.8 W
b)
E = Pt
E = (33.8)(600)
E = 2.03 x 104 J
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
Ex: A heating coil has a resistance of 4.0  and
operates on 120 V:
a) What is the current in the coil while it is
operating?
b) What energy is supplied to the coil in 5.0
min?
c) If the coil is immersed in an insulated
container holding 20.0 kg of water, what
will be the increase in temperature of the
water in 5.0 minutes?
“c” (specific heat) of water = 4190 J/(kg.oC)
a) I = V/R
= 120 V / 4.0 
= 30 A
b) E = VIt
= (120 V)(30 A)(300 s)
= 1.1 x 106 J
c) Electrical Energy = Heat Energy
E = mcT
1.1 x 106 = 20.0(4190)Δt
Δt = 13 °C
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
*E
(Heat gained or lost)= mct
This is Science 10 formula
Ex: An electric heater is rated at 500 W
a) How much energy is delivered to the
heater in half an hour?
b) The heater is being used to heat a room
containing 50.0 kg of air. If the specific
heat of air is 1100 J/kg°C, what is the
change in air temperature after half an hour?
Only 50% of the energy heats the air.
a) E = Pt
E = 500 W(1800 s)
= 9.00 x 105 J
b) E = mct
9.00 x 105x(50%) = 50.0 kg(1100)Δt
= 8.18 °C
Cost of Electricity
Calculated in terms of kW. h (or MJ).
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
Ex: Find the cost of operating a 60 W bulb for 30 days at 8
hours a day if the cost of electricity is 9.0 cents/kW.h
E = Pt
Cost = kW.h x Price
= (60 /1000 kW)(240 h) = 14.4 kW.h
Cost= 14.4 kWh x 9.0 cents/kWh
= $ 1.30
Ammeters and Voltmeters
An ammeter is a device that measures currents.
It is always connected in series.
Voltmeter is a device, which measures potential
difference. It is always connected in parallel.
Safety devices in electric circuits
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
1. Grounding Wire – Appliances are
connected to the ground, so that any extra
electric current goes to the ground.
2. Fuses – Fuse is a metal that melts when a
certain amount of current passes through it.
If excess current passes through the metal,
the fuse melts and the circuit becomes open.
3. Circuit Breakers – A breaker can carry a
certain amount of current. If the current
exceeds the limit, then the circuit breaker
goes off.
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
Millikan’s Oil Drop Experiment
Millikan wanted to find the elementary charge. He believed that all charges
are multiples of this elementary charge.
http://wps.prenhall.com/wps/media/objects/602/616516/Media_Assets/Chapte
r02/Text_Images/FG02_04.JPG
http://www.britannica.com/nobel/cap/omillik001a4.html
http://www68.pair.com/willisb/millikan/experiment.html
http://www.juliantrubin.com/bigten/millikanoildrop.html
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
Millikan’s Oil Drop Experiment
Ex: An oil drop whose mass is found to be 4.95 x 10-15 kg is balanced between two large, horizontal
parallel plates 1.0 cm apart, by a potential difference of 510 V, with the upper plate being positive.
What is the charge on the drop, both in coulombs and electrons? Is it an excess or deficit of electrons?
E = V/d
Eq = mg
= 510 V/(1.0 x 10-2 m)
= 51000 V/m
1. Oil Drop Stationary
Fe = Fg (Oil drop is stationary)
Eq = mg
Eq=mg
51000q=4.95x10-15x9.81
51000q=4.86x10-14
q=9.5x10-19 C=6 e(Excess)
2. Oil Drop Accelerating Upward
Fe > Fg
Fe – Fg = ma
Eq – mg = ma
3. Oil Drop Accelerating Downward
Fe < Fg
Fg – Fe = ma
mg – Eq = ma
4. Oil Drop Moving at a Constant Speed
Fe – Fg = ma
Fe – Fg = 0
Fe = Fg
Eq = mg
Ex: An oil drop whose mass is 5.70 x 10-16 kg accelerates upward at a rate of 2.90 m/s2 when placed
between two horizontal parallel plates that are 3.50 cm apart. If the potential difference between the
plates is 792 V, what is the magnitude of the
charge on the oil drop?
E = V/d
= 792 V / (3.50 x 10-2 m)
= 2.26 x 104 N/C
Eq – mg = ma
(2.26 x 104 J)q – (5.70 x 10-16 kg)(9.81 m/s2) = (5.70 x 10-16 kg)(2.90 m/s2)
q = 3.20 x 10-19 C
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
Potential Difference
Potential Difference (PD) is defined as energy
per unit charge – commonly known as Voltage.
Formula: V = ∆E
q
Units:
∆E = change in energy in J
q = charge in C
J/C or Volt (V)
1 V = 1 J/C
The first battery was produced by Volta
in 1800.
Current Electricity
Electrical energy can be transferred over long distances
and then can be changed to other forms of energy.
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
When two conducting spheres touch, charges flow from
one sphere (high potential) to the other sphere (low
potential) until the potentials are equal.
Current – Electrons flow continuously from the
negative to the positive in a closed circuit. This is
known as the electron flow.
The flow of positive charges from the positive to the
negative was known as the conventional current.
http://www.windows.ucar.edu/tour/link=/spaceweather/images/water_pump2_
lg_jpg_image.html
Electric Circuits
Parts:
1. Sources of Electricity - provides electric potential
difference. The change in potential energy of the
charge is converted to other energies.
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
a) Electrochemical Cells – Chemical potential
energy is converted into electrical energy.
i) Cells in Series
The total potential difference of the
battery is the sum of the potential
differences of the cells.
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
ii)
Cells in Parallel
The potential difference of the battery is the same as the potential
difference of one cell.
b)
Generators – Mechanical energy is
converted into electrical energy
c)
Photoelectric Cells – solar (light) energy is
converted into electricity
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
2. Resistors – resist the flow of electrons. Resistors
reduce potential energy of charges flowing through
it. Unit of resistance is the Ohm (Ω)
Factors that determine the amount of resistance in a
wire:
1. Thickness – As the thickness increases, resistance
decreases.
2. Type of wire – Copper – low resistance.
Nichrome – High resistance
3. Length of Wire – Longer the wire, greater the
resistance.
4. Temperature – As the temperature increases,
resistance increases.
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
Resistors in series
Resistors in
parallel
Current is the amount of charge transferred in a second.
I=q
t
I = Current (A-Ampere)
q = charge (C)
t = time (s)
1 A is the electric current when 1 C of charge moves
past a point in a conductor in 1 s.
Ex: Find the number of electrons transferred in a circuit if a current of 2.0 A
passes through a point in 2 min.
I = q/t
2.0 A = q/120 s
q = 240 C
240 C / 1.60 x 10-19 C/Electrons
= 1.5 x 1021 Electrons
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
Ohm’s Law states that the ratio between the potential
difference and current is a constant. This constant is
called the resistance.
V/I =k=R
R=V
I
V = Potential difference (V = Volts)
I = Current (A = Amperes)
R = Resistance (Ω = Ohms)
(Ohm’s law)
Electric power = VI measured in watts (W)
P = VI
P = power in W
I = current in A
V = voltage in V
Proof: P = W
t
P = Vq
t
But W = ∆E
P =VI
and ∆E = Vq
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
Ex: A circuit has a resistance of 20 Ω when the PD is 120 V. What is the
current through the circuit?
R = 20 Ω
I=V
V = 120 V
R
I=?
= 120/20
I=6A
p. 451 #1-4
p. 454 #5-9
p. 465 #1-6 (problems)
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
Electrical Power
P = VI
From Ohm’s law V = IR and I = V
R
Thus,
P = (IR) I
P=V V
R

P = I2R
P = V2
R
kw·h measures energy.
1 kw·h = 3,600,000 J
Not on formula
sheet.
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
e.g.
What is the heat produced in a 7 Ω resistor in 1.5 min if
there is a current of 10 A?
R=7Ω
E = Pt
t = 90 s
I = 10 A
P = I2R = (10)27 = 700 W
E=?
E = (700)(90)
E = 63,000 J = 63 kJ
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
Kirchoff’s Voltage Law
Around any complete path through an electric circuit, the sum of the
increases in electric potential is equal to the sum of the decreases in
electric potential.
Kirchoff’s Current law
At any junction point in an electric circuit, the total electric current
into the junction is equal to the total electric current out.
Kirchoff’s Laws
Series
Parallel
Req = R1 +R2+R3+......
V = V1 +V2+V3+.......
I = I1 =I2=I3= ... (I is equal throughout)
1/Req = 1/R1 +1/R2+1/R3+......
V = V1 = V2=V3= .....(V is equal throughout)
I = I1 +I2+I3+.....
Simple circuits
Example
Find the equivalent resistance when a 4.0 bulb and an8.0 bulb are
connected:
(a)in series
(b)in parallel.
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
(a)R =R1+R2=4.0+8.0=12
(b)1/R=1/R1+1/R2=1/4+1/8
R=2.7
Example
Two 20- resistors are connected in parallel with a 50- resistor in series.
Find the total resistance.
1/R=1/R1+1/R2=1/20+1/20
R=10
R=10+50=60
Example
Find: R=
I=
I1=
I2=
V1=
V2=
V 3=
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
I3=
Example 2
Find: V=
V1=
V2=
V3=
I=
R=
Example
Find: V1=
I1=
R=
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
V2=
I2=
R2=
Example
Find:
R=
R3=
V1=
V3=
I1=
I 3=
V2=
Example
V=
V1=
I=
I2=
I2=
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
R=
R2=
Example
R1=
Find:
V=
V2=
I=
I2=
Example
V 1=
V3=
I1=
R=
Find: R=
I=
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
V1=
V3=
I2=
V2=
I1=
I3=
Example
Find: R=
V1=
V3=
V5=
I2=
I4=
Example
I=
V2=
V4=
I1=
I3=
I5=
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
Find: R=
V1=
V3=
I2=
Example
I=
V2=
I1=
I3=
Unit 2: Electric Forces and Fields
Chapter 4: Current Electricity
v 1.1
Find: R=
V1=
V3=
I2=
I=
V2=
I1=
I3 =