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Honors Physics
Electrical Potential Energy
 PE associated with a charge due to its position in an
electric field.
 Analogous to PEg
 PEg of an object results from its position in a
gravitational field (mgh)
 Is a component of mechanical energy
 ME = KE + PEgrav + PEelastic + PEelectric
Similarity of PEelectric and PEg
 PEg = mgh
 m is mass
 g is gravitational field (ag)
 h is distance above a reference point
 PEelect = -qEd
 q is charge
 E is electric field strength
 d is component of displacement in the direction of the
electric field from reference point
 Using dimensional analysis, what is the unit of PEelect?
Electric Work
 Whenever a force moves an object, work is done on the
object.
 When an electric force moves a charge, work is done
on that charge.
 It is the electric field, E, that exerts force on a charge
 Therefore, the electric field, E, does work on a charge.
 This results in a change in PEelect.
Electric PE in a Uniform Electric Field
 A uniform field is one that has the same magnitude
and direction at all points, such as between two
parallel plates
 Remember: electric field lines are always directed
from away from positive and toward negative
Electric Potential Energy
 Recall that ΔPE = -W
 When charge q is released at point a,
electric force will move the charge to
b, i.e.
 The electric field does work on the
charge q
Electric Potential Energy
 W = Fd
 Since F = qE (remember E = F/q)
 W = qEd
 PEb-PEa= -qEd
 ΔPE = -qEd
 Significance of the (-) sign: PEelect
 Increases if charge is (-)
 Decreases if charge is (+)
PE as a charge moves in a
uniform electric field
Movement of charge
+ charge
- charge
Along E
Loses PE
Gains PE
Opposite E
Gains PE
Loses PE
ΔPE = -qEd
Negative sign indicates that PE will
increase if the charge is negative and
decrease if the charge is positive
Potential Difference
 Electric potential (V) is the ratio of PEelect to charge q
PEelect
V
q
 Represents the work needed to move a charge against
electric forces from a reference point to some other point in
an electric field, divided by the charge
 The SI units of electric potential are what? Which is a …?
Potential difference
 The change in electric potential
 The difference in electrical potential between two
points
PEelect
V 
q
 Is the work that must be done against electric forces to
move a charge from one point to another divided by
the charge
 Is the change in energy per unit charge
Potential Difference
 Unit is the volt (V)
PEelect
V 
q
?
1 volt 
?
 Remember:
 PEelect is a quantity of energy
 Electrical potential is a measure of energy per unit
charge
 Potential difference describes change in energy per unit
charge
Potential Difference in a Uniform
Electric Field
 Varies in a uniform field with displacement from a
reference point
 Where d is displacement parallel to the field
 Use this equation to determine potential difference
between two points in a field
PE
PE  qEd and V 
q
So, V   Ed
Potential Difference at a Point Near a
Charge
 One point is near the
charge
 The other point is at
infinity
q
V  k C
r
Electric potential due to multiple
charges
 Electric potentials are scalar quantities (whew!)
 So….
 Total potential at some point in a field is the
simple sum of the potentials due to each charge
 Keep track of signs!
Sample Problem
 As a charge moves xa = 4.0 cm to xb = 6.0 cm in a
uniform field of 350 N/C, it loses 4.5 x 10-16 J of
potential energy.
 What is the magnitude of the charge?
 6.4 x 10-17C
PE
PE  qEd  q  
Ed
 What is the potential difference between the two
points a and b?
 -7.0V
PE
V 
or V   Ed
q
17.2 Capacitance
 Capacitors are devices that store electrical PE
 Often constructed of parallel metal plates
 When connected to a battery, the plates become
charged
 When fully charged, ∆Vcap = ∆Vbat
Schematic Representation of a
Capacitor and Battery
Intro to Capacitor
Construction of a Capacitor
 Parallel plates
 Parallel plates separated
by an insulator
(dielectric material)
rolled into a cylinder
saves space
Capacitance
 Ability of a conductor to store energy in the form of
separated charges
Q
C
V
 Unit of capacitance is the Farad, named for Michael
Faraday
 1F = 1C/V
 1 Farad is a large amount of capacitance so…
 Often use pF, nF, or µF
 Supplemental instruction on capacitance
 View on your own, ~ 17 min.
Capacitance of a Parallel-Plate
Capacitor in a Vacuum
 When no material exists
between the plates
 ε0 is the permittivity of the
medium between the plates
 A measure of ability to
develop an electrical field,
permitting transfer of charges
 ε0 in a vaccuum is 8.85 x 10-12
C2/Nm2
A
C  0
d
Dielectric Materials
 Materials placed
between the plates of a
capacitor can increase
capacitance.
 Typically these are
insulating materials
A
C  0
d
Dielectric Constants (K)
 Dielectric materials have
different values of
“dielectric constant” (K).
 Increase capacitance
A
C  K 0
d
Performance of Dielectric
Materials
 Molecules of the dielectric are
polarizable
 As charge builds on the capacitor
plates, dielectric molecules orient
to the electric field
 This effectively reduces the charge
on the plates….
 allowing more charge to be
carried by each plate
Capacitor Discharge
 The opposite of charging,
releasing stored charge
 Electrical potential of the
capacitor is used to do
electrical work such as …
 The flash of a camera
 Signaling the stroke of a
computer keyboard
Capacitance of a Sphere
 R is radius
 Because the earth has a large
radius, it has a very large
capacitance
 i.e., the earth can accept or supply
a very large amount of charge
without changing its electrical
potential
 This is why the earth is “ground,”
(reference point for measuring
potential differences)
Q
R
Csphere 

V kC
Energy and Capacitors
 Because work is done to
1
move charges to and
 W  PE  QV
2
from opposite plates…
Q
 A charged capacitor
 Q  CV
holds electrical potential Since C 
V
energy
1
 PE stored in a charged
PE  CV 2
2
capacitor is equal to the
(–) work done to charge
it
Breakdown voltage
 Voltage at which discharge begins, i.e. charges
move
Energy and Capacitors
PE Stored in a Charged Capacitor
1
PE  QV
2
1
2
PE  C V 
2
2
1Q
PE 
2 C
Current and Resistance
 Current is the rate of movement of charge
 Rate of movement of electrons through a cross-
sectional area
Q
I
t
coulomb
1ampere  1
second
Sample Problem
 If current flowing through a light bulb is 0.835 A, how
long does it take for 1.67 C of charge to pass through
the filament of the bulb?
 2.00 seconds
Conventional Direction of Current
 Depending upon the circumstances, either positive,
negative, or both can move.
 Particles that move are called charge carriers
 By convention, direction of current is defined as the
direction a positive charge moves or would move if it
could.
 In metals, only electrons can move.
 Good conductors permit charge carriers to move easily
 Electrons in metals
 Ions in solution (electrolytes)
Conventional Direction of Current
Drift Velocity
http://www.bbc.co.uk/staticarchive/4e6786539008e5012ff9c723c4255ae6fc6c1b9f.gif
 Recall the structure of metals
 Valence electrons move about randomly due to their
thermal energy
 Their net movement is zero
 But if an electric field is established in the wire, there
is a net movement of electrons against the electric
field (toward +)
 Drift velocity animation
Drift Velocity
Consider motion of an
electron through a wire
 It is the electric field that exerts force and thereby
sets charge carriers in motion
 E propagates very rapidly (near speed of light)
 Charge carriers move more slowly, in an erratic
path,
 Called drift velocity
 Slow: e.g. in a copper wire carrying a 10.0 A
current, vdrift = 0.246 mm/s
Resistance to Current
 Opposition to electric
current
 Unit of electrical
resistance is the ohm (Ω)
 More commonly known
as Ohm’s law
V
R
I
volt
1 ohm  1
amp
V  IR
Ohmic and Non-ohmic Materials
 Materials which follow ohm’s law
are ohmic materials
 Resistance is constant over a
wide range of potential
differences (linear)
 Non-ohmic materials have
variable resistance (non-linear)
 Diodes are constructed of nonohmic materials
Other Factors Affecting Resistance
Function of Resistance
 From Ohm’s Law, changing resistance can
change current
V  IR
 So, if current needs to be reduced in a
circuit, you can increase the resistance
 In many cases, ∆V is constant, so changing
resistance is the only option for reducing
current.
Electrical Resistance in the Body
 Electrical resistance is reduced as the body
becomes wet or sweats
 This is due to the greater availability of ions to
conduct current
 Practical applications:
 Your body is more susceptible increased current
when wet
 Lie detectors
 EKGs, etc
Potentiometers
 Devices that have variable resistance
 “Pots”
 Applications
 Control knobs on electronic devices
 Stereos, dimmer switches, joy sticks,
etc.
17.4 Electric Power
 A potential difference (∆V) is necessary to cause
current (I)
 Batteries supply chemical energy (PEchem) which can be
converted into electical PE
 Generators convert mechanical energy into electrical
PE
 E.g. hydroelectric power plants
 Coal or natural gas powr plants
 Nuclear power plants
Direct and Alternating Current
 DC current flows in one direction only
 Electrons move toward the (+) terminal
 Conventional current directed from (+) to (-)
 AC current
 Terminals of source of ΔV constantly switch
 Causing constant reversal of current, e.g. 60 Hz
 Rapid switching causes e-s to vibrate rather than
have a net motion.
DC and AC
 DC
 constant
 uni-directional
 AC
 not constant
 bi-directional
Energy Transfer
 In a DC circuit
 Electrons leave the
battery with high PE
 Lose PE as flow through
the circuit
 Regain PE when
returned to battery
 (battery supplies PE
through electrochemical
reactions)
Electric Power
 The rate of
W PE
conversion of
P

t
t
electrical energy
PE
 SI unit is the watt
V 
 PE  qV
(W)
q
qV
q
P
Since
I
t
t
P  IV
Other Formulas for Power
Beginning with P  IV
Using Ohm' s Law...
PI R
2

V 
P
2
R
Kilowatt-hours
 How utility companies measure energy consumed
 Is the energy delivered in one hour a constant rate of
one kW
 1kWh=3.6 x 106 J
 What is the cost to light a 100 W light bulb for 1 full
day if the electric utility rate is $0.0600 per kWh?
100 W  24 h  2400 Wh  2.4 kWh
$0.0600
2.4 kWh 
 $0.144
kWh
Transmission Lines
 Transit at high voltage
and low current to
minimize energy lost
during transmission
 P=I2R