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Capacitance
and
Dielectrics
Chapter 26 HW: 7th/6th
4/6, 9/11, 11/13 (on 9 and 11 start from
Gauss’s Law!), 17/21, 19/23, 25/29,
28/33, 32/40,39/45,41/49,42/50,
44/53,45/73,49/58, 58/68, 59/67
Recall: The Infinite Charged Plane
σ
E=
2ε 0
Potential Difference in a Uniform
Field
r r
V − V = ΔV = − ∫ E ⋅ ds = −E ∫ ds = −Ed
B
„
„
„
A
B
B
A
A
Electric field lines always point in
the direction of decreasing
electric potential
When the electric field is directed
downward, point B is at a lower
potential than point A
When a positive test charge
moves from A to B, the chargefield system loses potential
energy
Parallel Plates: In the center we assume the electric
field is constant and uniform and that the potential
difference between any two points depends on the
distance d between those points!
σ
E=
ε0
ΔV = − Ed
Parallel Plate Assumptions
„
„
„
The assumption that the electric field is uniform is
valid in the central region, but not at the ends of the
plates
If the separation between the plates is small
compared with the length of the plates, the effect of
the non-uniform field can be ignored
The charge density on the plates is
σ = Q/A
„ A is the area of each plate, which are equal
„ Q is the charge on each plate, equal with opposite
signs
Makeup of a Capacitor
„
„
„
„
„
„
A capacitor consists of two conductors
„ These conductors are called plates
„ When the conductor is charged, the plates
carry charges of equal magnitude and
opposite directions
A potential difference exists between the plates
due to the charge
Capacitance will always be a positive quantity
The capacitance of a given capacitor is constant
The farad is a large unit, typically you will see
microfarads (μF) and picofarads (pF)
A capacitor stores electrical energy
Capacitors
„
„
Capacitors are devices that store
electric charge and energy
Examples of where capacitors are used
include:
„
„
„
radio receivers
filters in power supplies
energy-storing devices in electronic flashes
Some Uses of
Capacitors
„
Defibrillators
„
„
„
When fibrillation occurs, the
heart produces a rapid, irregular
pattern of beats
A fast discharge of electrical
energy through the heart can
return the organ to its normal
beat pattern
In general, capacitors act as
energy reservoirs that can be
slowly charged and then
discharged quickly to provide
large amounts of energy in a
short pulse
Charging a Parallel Plate Capacitor
„
„
Each plate is
connected to a
terminal of the
battery
If the capacitor is
initially uncharged,
the battery
establishes an
electric field in the
connecting wires
Active Figure 26.10
AF_2604.html
Capacitance
„
The capacitance, C, of a capacitor is defined as the ratio
of the magnitude of the charge on either conductor to the
potential difference between the conductors
Q
C=
ΔV
„
„
The capacitance is a measure of the capacitor’s ability to
store charge
The SI unit of capacitance is the farad (F)
Capacitance
„
The capacitance is proportional
to the area of its plates and
inversely proportional to the
distance between the plates
Q
C=
ΔV
εo A
Q
Q
Q
Q
C=
=
=
=
=
d
ΔV Ed σ d (Q / εo A ) d
εo
εo A
C=
d
Circular Plates
Q
C=
ΔV
εo A
C=
d
(plate radius = 10 cm) ε0= 8.85x10 –12 Farad/m
Capacitors in Series
„
„
„
When a battery is connected
to the circuit, electrons are
transferred from the left plate
of C1 to the right plate of C2
through the battery
As this negative charge
accumulates on the right
plate of C2, an equivalent
amount of negative charge is
removed from the left plate
of C2, leaving it with an
excess positive charge
All of the right plates gain
charges of –Q and all the left
plates have charges of +Q
thus:
Series: Q1 = Q2 = Q
Q
C=
ΔV
Capacitors in Series
„
The potential differences
add up to the battery
voltage
ΔV = V1 + V2 + …
1
1
1
=
+
+K
Ceq C1 C2
„
Q
ΔV =
Ceq
The equivalent capacitance of a
series combination is always less
than any individual capacitor in the
combination
Active Figure 26.08
Capacitors in Series
Capacitors in Parallel
„
„
The total charge is
equal to the sum of the
charges on the
capacitors
„ Qtotal = Q1 + Q2
The potential difference
across the capacitors is
the same
„ And each is equal to
the voltage of the
battery:
ΔV = ΔV1 = ΔV2
Capacitors in Parallel
The capacitors can be
replaced with one capacitor
with a capacitance of Ceq
Qtotal = Ceq ΔV
Qtotal = Q1 + Q2
Ceq ΔV = C1ΔV + C2 ΔV
Ceq = C1 + C2 + …
Essentially, the areas are
combined
Active Figure 26.07
Capacitors in Parallel
Capacitor Summary
Capacitors in Parallel:
ΔV = ΔV1 = ΔV2
Qtot = Q1 + Q2
Ceq = C1 + C2 + K
Capacitors in Series:
ΔV = ΔV1 + ΔV2 + ⋅ ⋅ ⋅
Q1 = Q2 = Q
1
1
1
=
+
+K
Ceq C1 C2
Equivalent Capacitance
Draw the reduced circuit it in
stages!
Three capacitors are connected to a battery as shown. Their capacitances are C1 = 3C, C2 = C, and C3 = 5C. (a) What is the equivalent capacitance of this set of capacitors? (b) State the ranking of the capacitors according to the charge they store, from largest to smallest. (c) Rank the capacitors according to the potential differences across them, from largest to smallest. Capacitors in Parallel:
ΔV = ΔV1 = ΔV2
Qtot = Q1 + Q2
Ceq = C1 + C2 + K
Capacitors in Series:
ΔV = ΔV1 + ΔV2 + ⋅ ⋅ ⋅
Q1 = Q2 = Q
1
1
1
=
+
+K
Ceq C1 C2
Fig P26-22, p.824
You Try:
Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in Figure P26.27. Take C1 = 5.00 μF, C2 = 10.0 μF, and C3 = 2.00 μF.
−1
1 ⎞
⎛ 1
+
= 3.33 μ F
Cs = ⎜
⎟
⎝ 5.00 10.0 ⎠
C p1 = 2 ( 3.33) + 2.00 = 8.66 μ F
C p 2 = 2 (10.0 ) = 20.0 μ F
Ceq
Fig P26-27
1 ⎞
⎛ 1
=⎜
+
⎟
8.66
20.0
⎝
⎠
−1
= 6.04 μ F
Series? Parallel? Neither Both?
P26.72 Assume a potential difference across a and b, and notice that the potential difference
across the 8.00 μ F capacitor must be zero by symmetry. Then the equivalent capacitance can be
determined from the following circuit:
⇒
⇒
FIG. P26.72
You Try:
Find the Equivalent Capacitance
P26.75 By symmetry, the potential difference across 3C is zero, so the circuit reduces to
1⎞
⎛ 1
+
C eq = ⎜
⎝ 2C 4C ⎟⎠
−1
=
8
4
C=
C .
6
3
Energy in a Capacitor
„
„
„
Consider the circuit to
be a system
Before the switch is
closed, the energy is
stored as chemical
energy in the battery
When the switch is
closed, the energy is
transformed from
chemical to electric
potential energy
Energy Stored in a Capacitor
„
„
Assume the capacitor is being charged and, at some
point, has a charge q on it
The work needed to transfer a charge from one plate
to the other is
q
dW = ΔVdq = dq
C
„
The total work required is
W =∫
Q
0
„
2
q
Q
dq =
C
2C
The work done in charging the capacitor appears as
electric potential energy U:
Q2 1
1
U=
= Q ΔV = C ( Δ V ) 2
2C 2
2
Energy Stored in a Capacitor
1
Q2 1
U=
= Q ΔV = C ( Δ V ) 2
2C 2
2
„
„
„
„
„
„
This applies to a capacitor of any geometry
The energy stored increases as the charge increases and
as the potential difference increases
The energy can be considered to be stored in the electric
field
The total work done by a battery charging a capacitor is
QV. Half the energy is either dissipated as heat or radiated
as electromagnetic waves.
For a parallel-plate capacitor, the energy can be expressed
in terms of the field as U = ½ (εoAd)E2
It can also be expressed in terms of the energy density
(energy per unit volume)
uE = ½ εoE2
Energy Problem
1
Q2 1
U=
= Q ΔV = C ( Δ V ) 2
2C 2
2
Determine the energy stored in C2 when C1 = 15 µF, C2 =
10 µF, C3 = 20 µF, and V0 = 18 V.
a.
b.
c.
d.
e.
0.72 mJ
0.32 mJ
0.50 mJ
0.18 mJ
1.60 mJ
1
Q2 1
U=
= Q ΔV = C ( Δ V ) 2
2C 2
2
You Try:
If VA – VB = 50 V, how much energy is stored in
the 36-µF capacitor?
a.
b.
c.
d.
e.
50 mJ
28 mJ
13 mJ
8.9 mJ
17 mJ
Example 26.4: Rewiring Two
Charged Capacitors
Fig. 26-12, p. 733
Energy Problem
1
Q2 1
U=
= Q ΔV = C ( Δ V ) 2
2C 2
2
A 3.0-µF capacitor charged to 40 V and a 5.0-µF
capacitor charged to 18 V are connected to each
other, with the positive plate of each connected to
the negative plate of the other. What is the final
charge on the 3.0-µF capacitor?
a.
11 µC
b.
15 µC
c.
19 µC
d.
26 µC
e.
79 µC
Problem
What is the maximum voltage that can be sustained
between 2 parallel plates separated by 2.5 cm of dry air?
Dry air supports max field strength of 3x 106 V/m .
V = Ed
6
= (3x10 V / m)(.025m)
= 7.5 x10 V
4
= 75kV
More than this and the air breaks down and
becomes a conductor. LIGHTENING!
The dielectric breakdown strength of dry air, at
Standard Temperature and Pressure (STP), between
spherical electrodes is approximately 33 kV/cm.
Dielectric Breakdown of Air
Dielectrics
A dieletric is an insulator that increases the capacitance.
C=
κε 0 A
d
Reduced E field prevents breakdown
& discharge between plates.
The dieletric constant is the ratio of the net electric field
magnitude without the dielectric,E0, and with the dielectric, E.
The dielectric reduces the net electric field and potential
difference across the plates.
κ Air ~ 1
κ Paper = 3.7
κ Water = 80
E0
κ=
E
E = E0 / κ
Reduced E field prevents breakdown
& discharge between plates.
Dielectrics
If the field becomes too great, the dielectric breaks down
and becomes a conductor and the plates discharge.
Reduced E field prevents breakdown
& discharge between plates.
Dielectrics – An Atomic View
„
„
The molecules that
make up the
dielectric are
modeled as dipoles
The molecules are
randomly oriented in
the absence of an
electric field
Dielectrics – An Atomic View
„
„
„
„
„
„
„
An external electric field is applied which
produces a torque on the molecules
The molecules partially align with the
electric field
The degree of alignment of the molecules
with the field depends inversely upon
temperature and directly with the
magnitude of the field
The degree of alignment of the molecules
with the field depends on the polarization
of the molecules.
Molecules are said to be polarized when a
separation exists between the average
position of the negative charges and the
average position of the positive charges
Polar molecules are those in which this
condition is always present
Molecules without a permanent
polarization are called nonpolar
molecules
E0 = σ / ε 0
Water Molecules
„
„
„
„
„
A water molecule is an
example of a polar
molecule
The center of the
negative charge is near
the center of the oxygen
atom
The x is the center of
the positive charge
distribution
The average positions of the positive and negative
charges act as point charges
Therefore, polar molecules can be modeled as
electric dipoles
Electric Dipole
„
„
„
„
An electric dipole consists
of two charges of equal
magnitude and opposite
signs
The charges are
separated by 2a
The electric dipole
moment (p) is directed
along the line joining the
charges from –q to +q
The electric dipole
moment has a
magnitude of p = 2aq
2aqk
Ey = 3
x
Electric Dipole in an External E Field
„
„
„
„
„
Each charge has a force of
F = Eq acting on it
The net force on the dipole is
zero
The forces produce a net torque
on the dipole that makes it align
with the external electric field.
The magnitude of the torque is:
τ = 2(Fa sin θ) = 2Eqa sin θ = pE sin θ
The torque can also be expressed as the cross product of
the moment and the field: τ = p x E
Induced Polarization
„
„
„
„
A symmetrical molecule has no
permanent polarization (a)
Polarization can be induced by
placing the molecule in an
electric field (b)
Induced polarization is the
effect that predominates in
most materials used as
dielectrics in capacitors
Induced polar molecules can
also be modeled as electric
dipoles
Dielectrics – An Atomic View
„
„
An external field can
polarize the dielectric
whether the molecules
are polar or nonpolar
The charged edges of the
dielectric act as a second
pair of plates producing
an induced electric field
in the direction opposite
the original electric field,
thus reducing the net
electric field in the
dielectric:
E = E0 − Eind
E0 = σ / ε 0
Eind = σ ind / ε 0
E = E0 / κ
Dielectrics – An Atomic View
E = E0 − Eind
E0 = σ / ε 0
σ
σ σ ind
= −
κε 0 ε 0 ε 0
σ ind
κ −1
=
σ
κ
κ > 1 → σ ind < σ
Eind = σ ind / ε 0
E = E0 / κ
Induced Charge and Field
E = E0 − Eind
„
„
„
The electric field due to the
plates is directed to the right
and it polarizes the dielectric
The net effect on the
dielectric is an induced
surface charge that results
in an induced electric field
If the dielectric were
replaced with a conductor,
the net field between the
plates would be zero
Problem
Determine (a) the
capacitance and (b) the
maximum potential
difference that can be
applied to a Teflon-filled
parallel-plate capacitor
having a plate area of
1.75 cm2 and plate
separation of 0.040 0
mm.
(a)
C=
(b)
κ ∈0 A
d
=
(
)(1.75 × 10
2.10 8.85 × 10−12 F m
−4
−5
4.00 × 10
(
m
ΔV m ax = Em axd = 60.0 × 106 V m
m
2
) = 8.13 × 10
−11
)( 4.00 × 10
−5
)
F = 81.3 pF
m = 2.40 kV
Dielectric Problem
Capacitance of a Cylindrical Capacitor
„
„
„
From Gauss’s Law,
the field between
the cylinders is
E = 2keλ / r
ΔV = -2keλ ln (b/a)
The capacitance
becomes
Q
l
=
C0 =
ΔV0 2ke ln ( b / a )
Q
≡l
=
C = ≡ C0 = ≡
ΔV 2ke ln ( b / a )
Coaxial Cable
Coaxial cables are often used as a transmission line for
radio frequency signals. In a hypothetical ideal coaxial cable
the electromagnetic field carrying the signal exists only in
the space between the inner and outer conductors. Practical
cables achieve this objective to a high degree. A coaxial
cable provides protection of signals from external
electromagnetic interference, and effectively guides signals
with low emission along the length of the cable.
Radio-grade flexible coaxial cable.
A: outer plastic sheath
B: copper screen
C: inner dielectric insulator
D: copper core
Types of Capacitors – Tubular
„
„
Metallic foil may be
interlaced with thin
sheets of paraffinimpregnated paper
or Mylar
The layers are rolled
into a cylinder to
form a small
package for the
capacitor
Types of Capacitors – Oil
Filled
„
„
Common for highvoltage capacitors
A number of
interwoven metallic
plates are immersed
in silicon oil
Types of Capacitors –
Electrolytic
„
„
Used to store large
amounts of charge
at relatively low
voltages
The electrolyte is a
solution that
conducts electricity
by virtue of motion
of ions contained in
the solution
Variable Capacitors
„
„
„
„
Variable capacitors
consist of two
interwoven sets of
metallic plates
One plate is fixed and
the other is movable
These capacitors
generally vary between
10 and 500 pF
Used in radio tuning
circuits
Next Week’s Lab
Circular Plates
C=
κε 0 A
d
C = κ C0
(plate radius = 10 cm) ε0= 8.85x10 –12 Farad/m
ΔV = ΔV0 / κ
Example 26.7: The Effect of a
Metallic Slab
Fig. 26-24, p. 742