Download Ch. 24 Capacitance

Ch. 24 Capacitance
& Dielectrics
AP Physics C
Capacitance
• Ability of a capacitor to store
charge
• Ratio of charge stored to potential
difference
Q
C
• 1 C/V = 1 Farad (F)
V
Capacitor
• A device that stores electric charge
• Used with resistors in timing circuits
because it takes time for a capacitor to
fill with charge
• Used to smooth varying DC supplies by
acting as a reservoir of charge
• Used in filter circuits because
capacitors easily pass AC (changing)
signals but they block DC (constant)
signals
Types of Capacitors
• Polarized
• Unpolarized
Polarized Capacitors
• Electrolytic
capacitors are
polarized and
they must be
connected the
correct way.
• Examples:
• Schematic
Symbol
Un-polarized
Capacitors
• Small value
capacitors are
un-polarized and
may be
connected either
way round.
• Examples:
• Schematic
Symbol
Charging/Discharging
• http://micro.magn
et.fsu.edu/electro
mag/java/capacit
or/index.html
Factors Affecting
Capacitance:
• The capacitance of a capacitor is
affected by three factors:
– The area of the plates
– The distance between the plates
– The dielectric constant of the
material between the plates
Area of the plates:
• Larger plates provide greater
capacity to store electric charge.
Therefore, as the area of the
plates increase, capacitance
increases.
Distance between
the plates:
• Capacitance is directly
proportional to the electrostatic
force field between the plates.
This field is stronger when the
plates are closer together.
Therefore, as the distance
between the plates decreases,
capacitance increases. As the
distance between the plates
increases, capacitance decreases.
The dielectric
between the plates:
• The ability of the dielectric to
support electrostatic forces is
directly proportional to the
dielectric constant. Therefore, as
the dielectric constant increases,
capacitance increases.
Capacitance:
• http://micro.magnet.fsu.edu/electro
mag/java/capacitance/index.html
C 
o A
d
Sample Problem 1
• An air-filled parallel-plate capacitor
has a plate area of 2.0 cm2 and a
plate separation of 1.00 mm.
– Find its capacitance.
– If a potential difference of 100 V is
applied across the plates, what is
• The charge on each plate and
• The electric field between the plates?
– What is its capacitance for a plate
separation of 3.00 mm?
Sample Problem 2
• A spherical conductor consists of
a spherical conducting shell of
radius b and charge – Q
concentric with a smaller
conducting sphere of radius a and
charge Q. Find the capacitance of
this device.
• What is the magnitude of the
electric field outside the spherical
capacitor?
Sample Problem 3
• A solid cylindrical conductor of
radius a and charge Q is coaxial
with a cylindrical shell of negligible
thickness, radius b > a, and
charge – Q. Find the capacitance
of this cylindrical capacitor if its
length is l.
• What is the magnitude of the
electric field outside the cylindrical
capacitor?
Energy Storage
• The work it takes to charge up a
capacitor is equal to the electrical
energy stored in a capacitor
• Alternate forms:
Capacitors in
Circuits:
• Series Circuit– Same charge
– Potential differences add
Capacitors in
Circuits:
• Parallel Circuit– Charges add
– Potential differences are the same
Sample Problem 4
• Two capacitors, 6 μF and 10 μF,
are connected in series across
a potential difference of 24 V.
– Determine:
• Equivalent capacitance,
• Charge stored on each capacitor,
• Potential difference across each
capacitor, and
• Energy stored on each capacitor.
Sample Problem 5
• Two capacitors, 6 μF and 10 μF,
are connected in parallel
across a potential difference of
24 V.
– Determine:
• Equivalent capacitance,
• Charge stored on each capacitor,
• Potential difference across each
capacitor, and
• Energy stored on each capacitor.
Sample Problem 6
• Given the following values:
– C1 = 10 pF
– C2 = 20 pF
– C3 = 60 pF
– ΔV = 120 V
• Find the equivalent capacitance, charge
stored on each capacitor, energy stored
in each capacitor and the potential
difference across each capacitor.
Sample Problem 7
• A capacitor with a capacitance of 8.0
μF is charged by connecting it to a
source of potential difference 120 V.
Once the capacitor is charged fully, it is
disconnected from the source.
– What is the charge stored on the 8.0 μF
capacitor?
– What is the energy stored in the capacitor?
Sample Problem 8
• The charged capacitor in the previous
problem is connected to an uncharged
4.0 μF capacitor.
– Explain how the charge is distributed once the
connection is made.
– How much charge is stored on each capacitor?
– What is the potential difference across each
capacitor?
– What is the total energy stored on this combination?
– Compare this to the energy stored before the
connection was made.
Conceptual Questions:
• A parallel-plate
capacitor is fully
charged and then
disconnected
from the source.
When the plates
are pulled apart,
do the following
quantities
increase,
decrease, or stay
the same?
• Capacitance?
• Charge stored?
• Energy stored?
• Potential difference?
• Electric field
strength?
Conceptual Questions:
• A parallel-plate
capacitor is fully
charged and
remains connected
from the source.
When the plates
are pulled apart, do
the following
quantities increase,
decrease, or stay
the same?
• Capacitance?
• Charge stored?
• Energy stored?
• Potential difference?
• Electric field
strength?
Dielectrics:
• What are they? • What are the three
functions of a
– Nonconducting
dielectric?
materials
– Increases capacitance
– Increases maximum
operating voltage
– Mechanical support
that prevents the plates
from touching; Allows
the plates to be even
closer together
Dielectric Strength:
• The Dielectric Strength is the
maximum Electric Field that the
material can withstand before the
material breaks down as an
insulator and permits current to
flow through the material.
Material
Dielectric Constant
k = e / eo
Dielectric Strength
(V/m)
Vacuum
1
Air (1 atm)
1.00054
Air (100 atm)
1.0548
Bakelite (typical)
4.9
24x106
Glass (Pyrex)
4.5-5.5
15x106
Polystyrene (typical)
2.6
25x106
Mylar (typical)
3.5
6-13 x10 6
Paper (typical)
3.5
14x106
Porcelain (typical)
7
4x106
Teflon (typical)
2
60x106
Mineral Oil (typical)
4.5
12x106
Water (20 oC)
80.4
Conductor
1-3 x10 6
Dielectric Constant:
• Depends on the permittivity-The
permittivity of a substance is a
characteristic which describes how it
affects any electric field set up in it. A
high permittivity tends to reduce any
electric field present. We can increase
the capacitance of a capacitor by
increasing the permittivity of the
dielectric material.

 
o
Voltage remains constant:
• A fully-charged parallel-plate capacitor
remains connected to a battery while a
dielectric is slide between the plates. Do the
following quantities increase, decrease, or
stay the same?
–
–
–
–
–
Capacitance
Charge stored
Electric field strength between the plates
Potential difference
Energy stored in the capacitor
Charge remains constant:
• A fully-charged parallel-plate capacitor is
disconnected from a battery and then a
dielectric is slid between the plates. Do the
following quantities increase, decrease, or
stay the same?
–
–
–
–
–
Capacitance
Charge stored
Electric field strength between the plates
Potential difference
Energy stored in the capacitor
Sample Problem 9:
• A parallel-plate capacitor has plates of
dimensions 2.0 cm by 3.0 cm
separated by a 1.0 mm thickness of
paper. The dielectric constant of paper
is 3.7.
– Find its capacitance.
– What is the maximum charge that can be
placed on the capacitor? The dielectric
strength of paper is 16 x 106 V/m.
– What is the maximum energy that can be
stored in the capacitor?
Sample Problem 10
• Suppose that the capacitance in the
absence of a dielectric is 8.50 pF and
that the capacitor is charged to a
potential difference of 12.0 V. If the
battery is disconnected and a slab of
polystyrene (dielectric constant of 2.56)
is inserted between the plates, what is
the Uo – U?
Polarization of a Dielectric:
• If a material contains polar molecules, they
will generally be in random orientations when
no electric field is applied. An applied electric
field will polarize the material by orienting the
dipole moments of polar molecules.
Parallel-Plate with Dielectric:
• The insertion of a dielectric will cause the
effective electric field to decrease and the
capacitance to increase.
Sample Problem 12
• A parallel-plate capacitor with a
plate separation of d has a
capacitance Co in the absence of
a dielectric. What is the
capacitance when a slab of
dielectric material of dielectric
constant K and thickness 1/3d is
inserted between the plates?
Gauss’s law in dielectrics
• The enclosed charge,
Qenclosed = (σ – σinduced)A
• Applying Gauss’s law, gives
EA = (σ – σinduced)A/εo
Gauss’s law in dielectrics
• Using: E = Eo/K
• Eo = σ/εo
• E = (σ – σinduced)/εo
• You get that: (σ – σinduced) = σ/K
• EA = (σ – σinduced)A/εo
• Becomes: EA = σA/Kεo
KEA = σA/εo
• Gauss’s law in dielectrics:
∫KE·dA = Qencl-free/ εo
Works Cited:
• http://www.kpsec.freeuk.com/capacit.htm
• http://micro.magnet.fsu.edu/electromag/electri
city/capacitance.html
• http://www.ac.wwu.edu/~vawter/PhysicsNet/T
opics/TopicsMainTemplate.html
• http://www.matter.org.uk/schools/SchoolsGlos
sary/permittivity.html
• http://hyperphysics.phyastr.gsu.edu/hbase/electric/dielec.html