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Chapter 24
Capacitance and
Dielectrics
PowerPoint® Lectures for
University Physics, Thirteenth Edition
– Hugh D. Young and Roger A. Freedman
Lectures by Wayne Anderson
Copyright © 2012 Pearson Education Inc.modified Scott Hildreth Chabot College 2016
Goals for Chapter 24
• To understand capacitors &
calculate capacitance
Goals for Chapter 24
• To understand capacitors &
calculate capacitance
• Unit: FARADs (Coulombs/Volt)
• Symbol:
Introduction
• How is a ROUND
capacitor like a
FLAT pair of
plates?
• UNROLL it! 
Goals for Chapter 24
• To analyze networks of capacitors
• To calculate energy stored in a
capacitor
• To examine dielectrics & how they
affect capacitance
Introduction
• How do camera flash
units store energy?
• Capacitors are
devices that store
electric potential
energy.
• Energy of capacitor is
stored in E field.
Introduction
Capacitors and capacitance
• Any two conductors separated by an “insulator”
form a capacitor.
• Insulator will allow E field between the conductors
• Insulator will not allow charge to flow from one
conductor through itself to the other.
Capacitors and capacitance
• Capacitors are like parking lots for
charge…
• They are built with physical dimensions in
mind, but how many cars are in them
depends upon the “pressure” to park there!
Capacitors and capacitance
• The more charge you can hold, the larger the
capacitor! “capacity”
• Charge a capacitor by pushing it there with a
potential voltage “pressure”
Capacitors and capacitance
• The definition of capacitance is C = Q/Vab
– Q = “charge stored” (in Coulombs)
+Q
– Q is held symmetrically on each plate
(same +Q as –Q)
-Q
a
b
– A voltage difference Vab is the electrical “pressure”
that pushes & keeps charge there
Capacitors and capacitance
• C = Q/Vab
so
Q = CVab
• Q increases as C is made larger, for a given pressure
– more capacity means more charge stored!
• Q increases as Vab increases
– For any particular capacitor, more pressure Vab
allows it to hold more charge!
• But no voltage applied means no charge stored…
Capacitors and capacitance
• The definition of capacitance is C = Q/Vab.
– Units of Capacitance: [Coulombs/Volt] = FARADs
– How much charge can you store per joule of work
to keep them there?
– But [Volt] = [Joules/Coulomb]
– Farads = Coulombs2/Joule
Parallel-plate capacitor
• TWO parallel conducting plates
• Separated by distance that is small compared to their
dimensions.
Batteries vs. Capacitors
• Batteries store MORE energy, for longer times, but
take more time to deliver it (or be recharged)
“Permanent Charge Storage”
• Capacitors store less energy, but deliver it instantly,
and recharge quickly. “Temporary Charge Storage”
Evolution in Capacitors
• Film dielectrics
• “Wet” electrolytic capacitors
• Ceramic Capacitors
• Polymer Capacitors
• Super Capacitors!
• Graphene!
Parallel-plate capacitor
• The capacitance of a parallel-plate capacitor is
C = 0A/d
Parallel-plate capacitor
• The capacitance of a parallel-plate capacitor is C = 0A/d.
• Note! C is engineered! You control Area & distance by
design!
• C increases with Area
• C decreases with separation
Parallel-plate capacitor
• The capacitance of a parallel-plate capacitor is C = 0A/d.
– So….0 = C x d/A = [Farads x meters/meters2]
– 0 has units of [Farads/meter]
• Recall “0” from Coulomb’s Law: Force = (1/4p 0) q1q2/r2
– Units of 0 were Coulomb2/Newton-meter2
• So [Farads/meter] = [Coulomb2/Newton-meter2]
•
[Farads] = [Coulomb2/Newton-meter]
•
[Farads] = [Coulomb2/Joule]
Parallel-plate capacitor
Example 24.2:
Plates 2.00 m2 in area; 5.00 mm apart;
10 kV applied Potential Difference.
•
C=?
•
Q on each plate = ?
•
E field between plates = ?
Parallel-plate capacitor
Example 24.2:
Plates 2.00 m2 in area; 5.00 mm apart;
10 kV applied Potential Difference.
•
C= 0 A/d = 8.85 x 10-12 F/m *2.00 m2/.005 m = 3.54 x 10-9 F
•
Q on each plate from C = Q/V so Q = CV = 3.54 x 10-5 C
•
E field between plates from V = Ed so
E = V/d = 2.00 x 10-6 V/m
•
Or… E = s/0 = (Q/Area)/0 = 3.54 x 10-5 C/2.00 m2 / 0
A spherical capacitor
• Example 24.3 – Two concentric spherical shells separated by
vacuum.. What is C?
A spherical capacitor
• Example 24.3 – Two concentric spherical shells separated by
vacuum.. What is C?
• C = Q/V so we need V!
• Get V from E field!
• V = kq/r in general for spherical charge distribution
• Vab = Va – Vb = “Voltage of a relative to b”
A spherical capacitor
• Example 24.3 – Two concentric spherical shells separated by
vacuum.. What is C?
• C = Q/V so we need V!
• Get V from E field!
• V = kq/r in general for spherical charge distribution
• Vab = Va – Vb = kQ/ra – kQ/rb = kQ (1/ra – 1/rb)
• Vab = Q/4p0 (1/ra – 1/rb)
• C = Q/V = 4p0/(1/ra – 1/rb) = 4p0 (ra rb)/ (rb – ra)
A spherical capacitor
• Example 24.3 – Two concentric spherical shells separated by
vacuum. What is C?
• C = Q/V = 4p0/(1/ra – 1/rb) = 4p0 (ra rb)/ (rb – ra)
•
Check!
– C increases as Area increases?
• YES!
– C decreases as separation increases?
• YES!!
A cylindrical capacitor
• Example 24.4 – Linear charge density -l on outer cylinder of
radius rb, + l on inner cylinder of radius ra. What is C?
A cylindrical capacitor
• Example 24.4 – Linear charge density - l on outer cylinder
of radius rb, +l on inner cylinder of radius ra. What is C?
• C = Q/Vab; find Q and find Vab!
• Q=lL
• V = [l/2p0 ]
ln (r0/r)
– r0 = distance where V was defined to be zero!
– Say r0 = rb here, so Vab = (l/2p0) ln (rb/ra)
• Does this make sense?? YES!
A cylindrical capacitor
• Example 24.4 – Linear charge density - l on outer cylinder
of radius rb, +l on inner cylinder of radius ra. What is C?
• C = Q/Vab = l L/[(l/2p0) ln (rb/ra)]
• C = 2p0 L/
ln (rb/ra)
• Check:
– Units = Farads/Meter x Meters = Farads
– C increases as L increases,
– and C increaes as rb closer to ra!
Capacitors in series
• Capacitors are in series if connected one after the other
Capacitors in series
• Capacitors are in series if connected one after the other
• NOTE:
• Charges are the same on all plates in the series (even with
different capacitances!)
Capacitors in series
• Capacitors are in series if connected one after the other
• Equivalent capacitance of a series combination:
1/Ceq = 1/C1 + 1/C2 + 1/C3 + …
Capacitors in series
• Capacitors are in series if connected one after the other
• Equivalent capacitance of a series combination:
1/Ceq = 1/C1 + 1/C2 + 1/C3 + …
• NOTE
• Ceq is always LESS than the smallest capacitor in series!
• Say C1 = 10 mF, C2 = 20 mF, C3 = 30 mF
• 1/Ceq = 1/10 + 1/20 + 1/30 = 6/60 + 3/60 + 2/60 = 11/60
• Ceq = 60/11 = 5.45 mF (< C1 !!)
• WHY?
Capacitors in parallel
• Capacitors are connected in parallel between a and b if
potential difference Vab is the same for all the capacitors.
Capacitors in parallel
• Capacitors are connected in parallel between a and b if
potential difference Vab is the same for all the capacitors.
• The equivalent capacitance of a parallel combination is the
sum of the individual capacitances: Ceq = C1 + C2 + C3 + … .
Capacitors in parallel
• Capacitors are connected in parallel between a and b if
potential difference Vab is the same for all the capacitors.
• The equivalent capacitance of a parallel combination is the
sum of the individual capacitances: Ceq = C1 + C2 + C3 + … .
• Note!
• Ceq is always MORE than largest capacitor in parallel!
• Say C1 = 10 mF, C2 = 20 mF, C3 = 30 mF
• Ceq = 10 + 20 + 30 = 60 mF > C3
• WHY??
Calculations of capacitance
• Example 24.6, a capacitor network:
• Find C eq?
Calculations of capacitance
• Example 24.6, a capacitor network:
• Find C eq?
Calculations of capacitance
• Example 24.6, a capacitor network:
• Find C eq?
Energy stored in a capacitor
• The potential energy stored in a capacitor is
• U = Q2/2C
(in JOULES)
• U increases with more stored charge;
• U is less for fixed charge on a larger capacitor
• U = Q2/2C but C = Q/V…. so
• U = 1/2 QV.
• U = Q2/2C and Q = CV … so
• U = 1/2 CV2
Energy stored in a capacitor
• Potential energy stored in capacitor:
U = Q2/2C = 1/2 CV2 = 1/2 QV
• Capacitor energy stored in electric field between the plates.
Energy stored in a capacitor
• Capacitor energy stored in electric field between the plates.
• Energy Density is u = 1/2 0E2
[Joules/m3]
How can we visualize energy in the field?
What happens when you pull plates apart?
+Q
+V
-
d
-Q
Pull capacitors apart when connected to a battery?
Start with a capacitor C = 0A/d, connected to
voltage source V, storing charge Q
+Q
+V
-
d
-Q
Pull capacitors apart when connected to a battery?
Pull plates apart - this takes work against field.
+V
+Q
-
-Q
Pull capacitors apart when connected to a battery?
One way to see this…..Capacitance decreases
C = 0A/d so if d
C
+q*
+V
-
d*
-q*
Pull capacitors apart when connected to a battery?
One way to see this…..Capacitance decreases
C = Q/V but V is fixed by battery, so C
+q*
+V
-
d*
-q*
=Q
Pull capacitors apart when connected to a battery?
+q*
+q
+V
-
d*
-q
-q*
Pull capacitors apart when connected to a battery?
• Suppose q stayed the same?
• Then E would be constant!
• So DV =
increases because dl increases!!
• So DVcap > DV battery!!
• And charge HAS to flow back into the battery!!!!
+q*
+q
+V
-
*
d
-q
-q*
Pull capacitors apart when connected to a battery?
d increases, A constant => C decreases
V fixed, so C = Q/V => Q = CV decreases
Q decreases so s decreases and E decreases
Energy = ½ QV decreases
+
q*
E
+V
-
d*
-q
q*
Pull capacitors apart when connected to a battery?
+q*
+V
E
-
d*
-q
-q*
Pull capacitors apart when NOT connected??
Start with a capacitor C = 0A/d, charged by
voltage source V, storing charge Q
+Q
+V
-
d
-Q
Pull capacitors apart when NOT connected??
Start with a capacitor C = 0A/d, charged by
voltage source V, storing charge Q,
then disconnect from battery!
+Q
d
-Q
Pull capacitors apart when connected to a battery?
Pull plates apart - this takes work against field.
But charge can’t go anywhere!
+Q
-Q
Pull capacitors apart when connected to a battery?
One way to see this…..Capacitance decreases
C = Q/V but Q stays the same so V increase
+Q
d*
-Q
Pull capacitors apart when connected to a battery?
Another way to see this…..Q same, so s same,
so E the same!
+q*
E
d*
-q*
Pull capacitors apart when connected to a battery?
Another way to see this…..Q same, so s same,
so E the same!
+q*
E constant
V*
d*
-q*
Energy density increases! (more volume in the capacitor)
Energy stored in a capacitor
• The potential energy stored in a capacitor is
U = Q2/2C = 1/2 CV2 = 1/2 QV.
• The capacitor energy is stored in the electric field between the
plates. The energy density is u = 1/2 0E2.
• The Z machine shown below can produce up to 2.9  1014 W
using capacitors in parallel!
Example 24.7
• C1 = 8.0 mF, charged to 120 V. Switch is open.
• What is Q on C1? Energy stored?
Example 24.7
• What is Q on C1?
• C = Q/V so Q = CV = 960 8.0 mC
•
What is energy stored?
• U = ½ CV2 or ½ QV = 0.58 J
Example 24.7
• Close the switch; after steady state, what is V across each
capacitor? What is charge on each?
• What is final energy of the system?
Example 24.7
• Close the switch; after steady state, what is V across each
capacitor? What is charge on each?
• Becomes a PARALLEL network; pressure will be unequal
initially, and charge flows from C1 to C2 until the
PRESSURE (voltages) are the same!
• Q0 spreads out over both plates = Q1 + Q2
Q0 decreases
to Q1
Q0 spreads
here to Q2
Example 24.7
• Q0 = Q1 + Q2 (and Q = CV!)
• Q0 = C1V1 + C2V2
• But in parallel, V1 = V2
• Q0 = (C1 + C2)V1 so V1 = V2 = 960 mC/(C1+C2) = 80V
• Q1 = C1V1 = 640 mC and Q2 = C2V2 = 320 mC
Q0 = 960 mC
decreases to
Q1 = 640 mC
Q0 spreads out
to Q2 = 320 mC
Dielectrics
• Dielectric is nonconducting material.
Most capacitors have dielectric between
plates.
• Dielectric increases the capacitance and
the energy density by a factor K.
• Dielectric constant of material is
K = C/C0 > 1.
• Dielectric reduces E field between the
plates, so for fixed voltage, more Q
• Dielectric reduces VOLTAGE between
plates w/ fixed Q on each (Q = CV)
Table 24.1—Some dielectric constants
Dielectrics increase energy
density
Dielectric breakdown
• If the electric field is strong enough, dielectric breakdown occurs
and the dielectric becomes a conductor.
• The dielectric strength is the maximum electric field the material
can withstand before breakdown occurs.
• Table 24.2 shows the dielectric strength of some insulators.
Molecular model of induced charge - I
•
Figures 24.17 (right)
and 24.18 (below)
show the effect of an
applied electric field
on polar and nonpolar
molecules.
Molecular model of induced charge - II
• Polarization of dielectric shows how induced charges reduce
magnitude of resultant electric field.