Fourth Edition
Chapter 24
Capacitance,
Dielectrics,
Electric Energy Storage
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✦ Capacitors
✦ Determination of Capacitance
✦ Capacitors in Series and Parallel
✦ Storage of Electric Energy
✦ Dielectrics
✦ Molecular Description of Dielectrics
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Capacitors
✦ A capacitor is a device which has two separated conducting plates which
can be charged (e.g. parallel plates in previous E field discussions)
10x – 20x
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Capacitors
✦ A capacitor is a device which has two separated conducting plates which
can be charged (e.g. parallel plates in previous E field discussions)
✦ The capacitance, C, of a capacitor is a measure of how much charge it
holds for a given voltage.
C = Q/V
(and if we know V(Q), Q’s cancel!)
✦ It is measured in Farads, F, and depends on geometry. ( 1 F = 1 Q / V )
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Capacitors
✦ A main purpose of a capacitor is
to store electric charge (energy)
✦ By connecting a battery to the
capacitor plates, charge is pulled
from one plate to the other
Batteries supply a potential
difference, ΔVbat, that produces a
ΔVC
After a period of “charging up”,
ΔVC = ΔVbat
✦ After the battery is disconnected, the
capacitor stays charged (This is why the light
The wires, plates, and battery
terminals are equipotentials
(conductors)
on your computer’s AC adapter stays on for a bit
after you unplug it!)
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Capacitors: real life
✦ Simplest use of a capacitor with a battery
Wire (conducting metal with an
insulator around it)
Cartoon
version
Schematic
version
We will also look at capacitors again when we study circuits
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Calculating capacitance
✦ The capacitance depends primarily on the geometry (shape) of
the capacitor.
✦ For a parallel plate capacitor, it’s fairly easy to work out.
Referring back to earlier results for a parallel plate geometry
we found that V = Ed and E = σ/ε0 = (Q/ε0A)
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Clicker
A parallel plate capacitor with square plates has a capacitance C.
The area of the plates is doubled, and the new capacitance C’ is:
a) C’ = C/2
b) C’ = C
c) C’ = 2C
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Clicker
A parallel plate capacitor with square plates has a capacitance C.
The area of the plates is doubled, and the new capacitance C’ is:
a) C’ = C/2
b) C’ = C
c) C’ = 2C
Answer: c; C’ = ε0A’/d = ε0(2A)/d = 2ε0A/d = 2C
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Clicker
If (for the starting capacitor, C) the gap between the plates is
doubled, the new capacitance C’ is
a) C’ = C/2
b) C’ = C
c) C’ = 2C
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Clicker
If (for the starting capacitor, C) the gap between the plates is
doubled, the new capacitance C’ is
a) C’ = C/2
b) C’ = C
c) C’ = 2C
Answer, a; C’ = ε0 A/d’ = 2ε0A/(2d) = C/2
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Clicker?
An isolated capacitor has a charge Q. The plates are then pulled
apart so that the distance between them is larger.
After the plates are pulled apart,
A. The charge increases and the electric field decreases.
B. The charge decreases and the electric field increases.
C. Both the charge and the field increase.
D. Both the charge and the field decrease.
E. The charge and the field remain constant.
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Clicker?
An isolated capacitor has a charge Q. The plates are then pulled
apart so that the distance between them is larger.
After the plates are pulled apart,
A. The charge increases and the electric field decreases.
B. The charge decreases and the electric field increases.
C. Both the charge and the field increase.
D. Both the charge and the field decrease.
E. The charge and the field remain constant.
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Let’s put some numbers
(a) Calculate the capacitance of a parallel-plate capacitor whose
plates are 20 cm × 3.0 cm and are separated by a 1.0-mm air gap.
(b) What is the charge on each plate if a 12-V battery is connected
across the two plates?
(c) What is the electric field between the plates?
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Let’s put some numbers
(a) Calculate the capacitance of a parallel-plate capacitor whose
plates are 20 cm × 3.0 cm and are separated by a 1.0-mm air gap.
(b) What is the charge on each plate if a 12-V battery is connected
across the two plates?
(c) What is the electric field between the plates?
a) C = ε0A/d = (8.85 x 10-12)(0.20 x 0.03)/0.0010 = 53 pF (p = 10-12)
b) Q = CV = 53 pF x 12V = 636 pC = 6.4 x 10-10 C
c) E = V/d = 12 V /0.001 m = 12,000 V/m
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More calculations
✦ Your (cable) TV is probably connected to the wall with
a coaxial cable. What is it’s capacitance?
C = q/V, so we need V(q) which, if we don’t
know it, we can get (e.g.) from
r<R0, where E=0 (inside conductor), No contribution
R0 < r < R1, from Gauss’s Law,
Total length, L >> R1
-q
+q R
0
R1
And then plug V into the def’n of C
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More calculations
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𝝀
More calculations
using Gaus’ law
=Q/L
-
L
L
C=Q/V(Q)
Capacitance
per meter:
C/L
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Energy in a capacitor
✦ Because charge can be pushed on to a capacitor (by a battery,
expending chemical energy), or drawn off to turn on a light bulb
(creating thermal energy) a capacitor must store (electrical) energy.
Energy is equal to the work done storing the charge
✦ The energy U stored in the electric field in a capacitor is given by
U = ½ QV = ½ CV2= ½ Q2/C
✦ And we can also calculate the energy density, u = U/Volume and find
that
u = ½ ε0E2
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Energy in a capacitor
A 2.5 µF capacitor is charged to 5V. How much energy is
stored in the capacitor?
E = ½ CV2 = ½ (2.5 x 10-6) 52 = 31 μJ
next time we will see how to increase the energy stored
(by increasing the capacitance)
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Energy in a capacitor
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break
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Capacitors in circuits
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Capacitors in parallel
✦ Capacitors in parallel have the same
voltage across each one.
✦ The equivalent capacitor is one that
stores the same (total) charge when
connected to the same battery:
Q = Q1 + Q2 + Q3 = C1V + C2V + C3V
= (C1 + C2 + C3)V = CeqV
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Capacitors in series
✦ Capacitors in series have the same
charge.
✦ The equivalent capacitor has the
same charge across the full voltage
drop
V0 = V1 + V2 + V3 = Q/C1 + Q/C2 + Q/C3
= Q(1/C1 + 1/C2 + 1/C3)
= Q/Ceq
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Example
+
V0
C1 = 10 pF
C2 = 20 pF
C3 = 20 pF
C4 = 10 pF
The figure shows four capacitors in a circuit.
What is the equivalent capacitance?
A) 1/3 pF
B) 3 pF
C) 30 pF
D) 60 pF
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Example
+
V0
C1 = 10 pF
C2 = 20 pF
C3 = 20 pF
C4 = 10 pF
The figure shows four capacitors in a circuit.
What is the equivalent capacitance?
A) 1/3 pF
B) 3 pF
C) 30 pF
D) 60 pF
Answer, D: They are in parallel, and the general formula for capacitors in
parallel is that the equivalent capacitance C’ = C1 + C2 + … so for this case
C’ = 10 pF + 20 pF + 20 pF + 10 pF = 60 pF
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Clicker
The picture to the right shows
three capacitors hooked
together. The capacitances
are shown. What is the
effective capacitance of this
group?
C1 = 12 nF
C2 = 9 nF
C3 = 7 nF
C1
C2
C3
A) 28 nF
B) 1/28 nF
C) 0.34 nF
D) 3 nF
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Clicker
The picture to the right shows
three capacitors hooked
together. The capacitances
are shown. What is the
effective capacitance of this
group?
A) 28 nF
B) 1/28 nF
C) 0.34 nF
D) 3 nF
C1 = 12 nF
C2 = 9 nF
C3 = 7 nF
C1
C2
C3
Answer, D) The capacitors are in series (why?) and the
general form in this case is 1/C’ = 1/C1 + 1/C2 + …, so
1/C’ = 1/12 + 1/9 + 1/7 = 0.34 nF
And then
C’ = 3 nF
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Example
+
V0
C1 = 10 pF
C3 = 20 pF
C2 = 20 pF
C4 = 10 pF
What is the effective
capacitance of the circuit
shown?
Answer:
C1 and C2 are in series, so the equivalent is 1/C12 = 1/C1 + 1/C2, so C12 = 6.7 pF
C3 and C4 are in series, so the equivalent is 1/C34 = 1/C2 + 1/C4, so C34 = 6.7 pF
But then C12 and C34 are in parallel so the final equivalent capacitance is
C’ = C12 + C34 = 13.4 pF
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Infinite room for complication
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Dielectric
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Dielectrics
✦ We can increase the capacitance
of a capacitor by placing material
between the plates
❖ The atoms in the material
become polarized
❖ This produces an electric field
that opposes the field produced
by the plates
✦ Lower field means needs more charge to reach the same voltage
==> capacitance increases
✦ The material MUST be an non conductor to keep the plates insulated
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Dielectrics
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Dielectrics
✦ The change is characterized by the
dielectric constant, κ (always κ>1)
Cwith dielectric = κCno dielectric
More generally when adding a
dielectric, replace
by κ
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Clicker
A parallel plate capacitor has some stored charge, Q. If a dielectric
is placed between the plates, the electric potential between the
plates will
A. Increase
B. Decrease
C. Stay the same
D. Not enough information to tell
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Clicker
A parallel plate capacitor has some stored charge, Q. If a dielectric
is placed between the plates, the electric potential between the
plates will
A. Increase
B. Decrease
C. Stay the same
D. Not enough information to tell
Answer, B: V = Q/C, Q doesn’t change and C gets bigger,
so V gets smaller
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Partly filled capacitor
d
L = Left
R = Right
How much charge is in each side of the capacitor if the total
charge is
?
The plates are like wires, so
the voltage must be the
same, so they are in parallel
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Partly filled capacitor
d
L = Left
R = Right
How much charge is in each side of the capacitor if the total
charge is
?
The plates are like wires, so
the voltage must be the
same, so they are in parallel
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Partly filled capacitor
d
L = Left
R = Right
How much charge is in each side of the capacitor if the total
charge is
?
The plates are like wires, so
the voltage must be the
same, so they are in parallel
This is like 2 capacitors
in parallel, and is a
“charge divider”!
What if no dielectric, K=1?
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Clicker: energy consideration
Before
+Q
After
d
d
-Q
How much work is done adding the dielectric to a charged capacitor?
Before
After
If you are the person inserting the
dielectric, do you have
a) push on it
b) pull back during insertion
to keep it from flying through?
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Energy consideration
Before
+Q
After
d
d
-Q
How much work is done adding the dielectric to a charged capacitor?
Before
After
If you are the person inserting the
dielectric, do you have push on it or
pull back during insertion to keep it
from flying through?
The cap has lost stored energy
(like a ball rolling down a hill).
So you have to pull back on it!
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Food for thoughts
Partially filling with an insulator
when the battery is still
connected.
How much does this change if I
insert a conductor? [ careful not
to touch both sides at the same
time!!! ]
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Summary
C = Q/V
= Kε0A/d Parallel plates
ΔU
ΔV =
q
kq1q2
Ue =
r
kQ
ΔV =
r
Point charge
Q
ΔV = Ed
Parallel charged
plates (capacitor)
U = ½ CV2
= ½ Q2/C
Potential from a set of point charges is
just the sum of the potential from the
each charge
And also conservation of energy…
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