Transcript Electric Potential Energy and Electric Potential

Electric Potential Energy
and Electric Potential
Chapter 16
Potential Energy
 Potential Energy is stored energy do to
an object’s position in a force field.
 Work = F·d = ΔPE = ΔU
 When work is done to move an object
into a force field, Potential Energy
Increases
Gravitational P.E. Ug
 To bring a mass in from
infinity to near the earth, we
would encounter the
Earth’s gravitational field.
 Work = F·d = ΔUg
 ΔUg = Uf – Ui = Ur – Uinf
 ΔUg= (GMem/r2)·r
 Ug = -GMm/r
Uniform Fields
 Near Earth’s surface, gravitational field is
uniform. g = 9.8 m/s2.
 We calculate Work = ΔU = mgh
 This is a simplification of Ug = -GMm/r
 Work is essentially F·d
Electric Potential Energy
 To bring a charge, q, in
from infinity to near +Q,
we would encounter the
charges electric field.
 Work = F·d = ΔUe
 ΔUe = Uf – Ui = Ur – Uinf
 ΔUe= (kQq/r2)·r
 Ue = kQq/r
Uniform Electric Field
 Parallel Plates with
equal charge provide
a uniform electric
field.
 Recall E = Fe/q
 Fe = Eq
Potential Energy – Uniform
Field (plate separation d)
 ΔU = Uf – Ui = UB – UA
 ΔU = Work = F·d
 ΔU = (Eq)d = Eqd
 E is electric field
between plates, d is
plate separation.
Uniform Gravitational and
Electric Field Comparison
 ΔUg = mgh for
constant g field
 ΔUe = qEd for
constant E field
 U is measured in
Joules
Non-Uniform Fields / PE
 Fg = GMm/r2
 Fe = kQq/r2
Ug = -GMm/r
Ue = kQq/r
Electric Potential, V
 Electric Potential is defined as potential
energy per unit charge.
 Think about adding charge, one charge at
a time, to a conducting sphere.
Electric Potential, V
 ΔV = ΔUe/q measured in Joules/Coulomb
 Joule/Coulomb = Volt!
 Uniform Electric Field: ΔV = qEd/q = Ed
 ΔV = Ed gives the voltage difference between two
parallel plates
 Non-Uniform E Field: ΔV = (kQq/r)/q = kQ/r
Summarize New Formulas
 Fe = kq1q2/r2
 Ue = F· (distance)
= kq1q2/r electric potential energy between two
charges (non uniform fields)
 Ue = qEd electric potential energy for parallel plates with
electric field E
 V = Ue/q
= kq/r electric potential due to charge q (non uniform
field)
 V = Ed
electric potential for parallel plates with electric
field E
Electric Potential Energy
for Several Charges
 To find the potential
energy of a system
of charges, add the
potential energy
between each pair of
charges.
 Ue = U12 + U13 + U23
Examples
 Read Examples 16.1, 16.2, 16.3, 16.4
 Copy these into your notes if you feel that
is helpful!
 Try # 11 – 14, 17, 18, 20 – 22, 24, 25, 28
page 562 { prepare to turn this
assignment in}
Equi-potential Surfaces15
 Surfaces where potential energy is
constant are known as equi-potential
surfaces.
 For the electric field, we are concerned
with electric potential, V, in addition to
electric potential energy, U. Equipotential surfaces are surfaces with the
same potential energy and the same
electric potential.
Conservative Fields
 A conservative force is a force
in which work done does NOT
depend on the path taken.
 Gravitational Force is a
conservative force.
 Electric Force is a conservative
force
 Friction is non-conservative.
 In moving an object along an
equi-potential surface, no work
is done.
Equi-potential Surfaces
Uniform Electric Field
 Recall E = F/q
F = Eq
 ΔU = Fd = Eqd
 ΔV = U/q
 For a uniform field,
ΔV = Ed = EΔx
Units
 E [Joule/Coulomb] or [Volt/meter]
 These are equivalent since
E = F/q = ΔV/d
Example:
 Normally the Earth is electrically charged. This
creates a constant electric field pointing down
near the Earth’s surface of 150 V/m.
A) What are the shapes of the equipotential
surfaces?
B) How far apart are two equi-potential
surfaces with 1000V difference between them?
Electron Volt measure for
Energy…
 An Electron-Volt is a common unit for
energy…
It is the amount of Kinetic energy
acquired by an electron if it is accelerated
through a voltage of 1 Volt.
 1eV = ΔKE = -ΔU = qV = 1.6 X 10 –19 J
Capacitance
 A capacitor is a device that stores charge.
 A good capacitor has the ability to store charge
without appreciably increasing the electric
potential.
 Work is done by a battery to move charge from
one parallel plate to another. Separation of
charge creates an electric field.
 Capacitance is defined as the amount of stored
charge per unit of potential difference.
Capacitance
 Capacitance = Charge stored / Voltage
 C = Q/V or Q = CV
 [Coulombs/Volt] = [Farad]
 1 Farad is a huge amount of capacitance.
It is most commonly measured in
microfarads = 10-6 F
Capacitance
 Capacitance of a parallel plate arrangement
depends on the area of and the distance
between the plates, as well as the material
between them. (dielectrics increase
capacitance)
 If the material between the plates is air, then
C = ξ0A/d
Where ξ0 is the permittivity of free space
And ξ0 = 8.85 X 10 –12 C2/Nm2
Example
 What would be the plate area of an air filled
1.0F parallel plate capacitor if the plate
separation were 1.0 mm?
Energy!
 Potential increases as
charge is added.
 Slope = ΔV/ ΔQ = 1/C
 Potential energy stored in a
capacitor = Work Done =
QVav = Q(V/2)
 Uc = ½ QV = ½ CV2 = Q2/2C
Example
 During a heart attack the heart beats
erratically. One way to get it back to
normal is to shock it with electrical
energy. About 300J of energy is required
to produce the desired effect. A
defibrillator stores this energy in a
capacitor charged by 5000V. What
capacitance is required? What is the
charge on the plates?
Homework!
 Read sections 16.4 and 16.5 pages 552
– 559.
 Do # 57, 59, 60, 61, 64, 65 – 67, 69- 72
Dielectric Materials
 Dielectric materials
placed between parallel
plates have several
purposes:
 Keep plates from coming
in contact
 Allow for flexible plates
 Increase capacitance of
the capacitor
 Dielectric constant, κ>1
 κ = C/C0
Two dielectric situations:
 Either the voltage difference is removed
once the plates are charged, then the
dielectric material inserted between
plates
 Or the dielectric material is inserted
between the plates while the voltage is
maintained.
 These are different situations, but both
result in increased capacitance.
Remove voltage then insert dielectric
 Voltage applied, V0,
separates charge, Q0 and
sets up electric field E0
 Dielectric inserted and
becomes polarized. Electric
field does work to polarize
dielectric molecules, which
set up a smaller electric field
in the opposite direction.
 Net electric field is reduced.
Therefore voltage is reduced.
Remove voltage then insert dielectric
 Charge, Q, doesn’t change
once dielectric is inserted.
 Induced electric field in dielectric
reduces original electric field to
E and original voltage difference
to V.
 κ = E0 / E = V0 /V
 Then C = Q/V = Q0/(V0/ κ)
= κ(Q0/V0) = κC0
 Uc = Q2/2C = Q2/2(κC0) = U0/ κ

Remove voltage then insert dielectric
 Capacitance increases by factor of κ
Since C = Q/V and voltage decreases,
this makes sense!
 Stored energy decreases by factor of κ
Some stored energy goes into
polarizing the molecules in the
dielectric, so this makes sense!
Insert dielectric and maintain constant voltage
 When capacitor is maintained at
constant voltage, the battery
continues to supply charge to
compensate for the induced
electric field.
 C = Q/V = κQ0/V0 = κC0
 Charge and capacitance
increase by a factor of κ.
 Uc = ½ CV2 = κC0V02
 Stored energy increased by κ
Maintain constant voltage across
the capacitor with dielectric
 Charge (and therefore capacitance)
increase by a factor of κ
 Stored energy ( in the capacitor)
increases by a factor of κ at the expense
of the battery, which does more work.
Capacitor Jewelry
Capacitance
 With dielectric, C = κC0 where C0 is
capacitance without dielectric.
 C = κ(ε0)A/d
Example
 Consider a capacitor with dielectric underneath
a computer key. The capacitor is connected to
12V and has a normal (uncompressed) plate
separation of 3.0 mm and plate area of 0.75
cm2.
a) What is the required dielectric constant if the
capacitance is 1.10 pF?
b) How much charge is stored on the plates
under normal conditions?
c) How much charge flows onto the plates if
they are compressed to a separation of 2.0
mm?
Circuitry – Capacitors in
Series and Parallel
 Capacitors in series are
connected one after
another.
 The voltage from the
battery is shared
between the series
capacitors.
 Want to find ‘equivalent
capacitance’.
Series and Parallel
Capacitors
 Capacitors in parallel are
connected in branches
parallel to one another.
 Each capacitor in
parallel receives the
same voltage from the
battery.
 Want to find the
‘equivalent capacitance’.
Series
 To find equivalent capacitance,
consider what is constant. For
series capacitors, the charge on
each capacitor must be constant.
(why?)
 Voltages across each capacitor
add to the total voltage supplied.
 V1 + V2 + V3 = Vtot
 Q/C1 + Q/C2 + Q/C3 =Q/Ceq
 1/Ceq = 1/C1 + 1/C2 + 1/C3
Parallel
 In parallel, each capacitor gets
the same voltage. With different
capacitances, the charge on
each is different.
 Charge adds to total charge
separated by battery
 Qtot = Q1 + Q2 + Q3
 CeffV = C1V + C2V + C3V
 Ceff = C1 + C2 + C3
Example
 Given two capacitors, one with a
capacitance of 2.5 μF and the other of
5.0 μF, what are the charge on each and
the total charge stored if they are
connected to a 12 V battery in
a) series
b) parallel
Example
 Three capacitors are connected as
shown on page 560. Find the voltage
across each capacitor.