Capacitors
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
1
Review: Capacitance
!  The definition of capacitance is
•  Capacitance is a property of the capacitor
q
C=
V
!  The capacitance of a parallel plate capacitor
depends on the plate area and the plate separation and is given by
•  A is the area of each plate
ε0 A
•  d is the distance between the plates
C=
Physics for Scientists & Engineers 2
d
2
Circuits
!  An electric circuit consists of wires that connect circuit
elements.
!  These elements can be capacitors.
!  Other important elements include resistors, inductors,
ammeters, voltmeters, diodes, and transistors.
!  Circuits usually need a power source.
!  A battery can provide a fixed potential difference commonly
called voltage.
!  An AC power source provides a time-varying potential
difference.
February 5, 2014
Chapter 24
3
Circuit Symbols
!  Circuit elements are represented by commonly used
symbols.
February 5, 2014
Chapter 24
4
Capacitors in series and in parallel
!  The equivalent capacitance for n capacitors in parallel is
n
Ceq = ∑ Ci
=
i=1
!  The equivalent capacitance for n capacitors in series is
n
1
1
=∑
Ceq i=1 Ci
=
!  For two capacitors
in series:
C1C2
Ceq =
C1 + C2
2/4/14
Physics for Scientists & Engineers 2, Chapter 23
5
System of Capacitors
PROBLEM
!  Consider the circuit shown
!  What is the combined
capacitance of this set of
five capacitors?
!  If each capacitor has a
capacitance of 5 nF, what is the
equivalent capacitance of the
arrangement?
SOLUTION
!  This problem can be solved by sequential steps, using the
rules for equivalent capacitances of capacitors
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
8
System of Capacitors
STEP 1
!
!
!
!
We can see that C1 and C2 are in parallel
and that C3 is also in parallel with C1 and C2
We find C123 = C1 + C2 + C3
… and make a new drawing
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
9
System of Capacitors
STEP 2
!  We can see that C4 and C123 are in series
!  We find for the equivalent capacitance:
1
C1234
1
1
C123C4
=
+
⇒ C1234 =
C123 C4
C123 + C4
!  … and make a new drawing
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
10
System of Capacitors
STEP 3
!  We can see that C5 and C1234 are in parallel
!  We find for the equivalent capacitance
C1 + C2 + C3 )C4
(
C123C4
C12345 = C1234 + C5 =
+ C5 =
+ C5
C123 + C4
C1 + C2 + C3 + C4
!  … and make a new drawing
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
11
System of Capacitors
STEP 4: INSERT THE NUMBERS
!  So the equivalent capacitance of our system of capacitors
⎛ (5 + 5 + 5 ) 5 ⎞
⎜⎝ 5 + 5 + 5 + 5 + 5⎟⎠ nF = 8.75 nF
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
12
System of Capacitors
Step 5: Calculate the charges – Reconstructing the circuit in reverse
!  C1234 and C5 are in parallel so they have the same potential
difference across them, 12 V.
!  The charge on C5 is then:
!  C1234 is composed of C123 and C4 in series so C123 and C4
must have the same charge q4:
q5 = C5ΔV = (5 nF)(12 V) = 60 nC
⎛ 1
q4
q4
1 ⎞⎟
⎜
ΔV = ΔV123 + ΔV4 =
+ = q4 ⎜⎜
+ ⎟⎟
⎜⎝ C123 C4 ⎟⎠
C123 C4
February 5, 2014
Chapter 24
13
System of Capacitors
!  The charge on C4 is then:
(C1 + C2 + C3 )C4
(15 nF)(5 nF)
C123C4
q4 = ΔV
= ΔV
= (12 V)
C123 + C4
C1 + C2 + C3 + C4
20 nF
q4 = 45 nC
!  C123 is equivalent to three capacitors in parallel and it has
the same charge as C4, 45 nC.
!  The three capacitors C1, C2, and C3 have the same
capacitance and the same potential difference across them
and the sum of the charge on these three capacitors is 45 nC.
!  Therefore the charge on C1, C2, and C3 is:
45 nC
q1 = q2 = q3 =
= 15 nC
3
February 5, 2014
Chapter 24
14
Energy Stored in Capacitors
!  A battery must do work to charge a capacitor
!  We can think of this work as changing the electric potential
energy of the capacitor
!  The differential work dW done by a battery with a potential
difference ΔV’ to put a differential charge dq’ on a capacitor
with capacitance C is
q'
dW = ΔV 'dq' = dq'
C
!  The total work required to bring the capacitor to its full
charge q is
Wt = ∫ dW = ∫
q
0
February 4, 2014
q'
1 q2
dq' =
C
2C
Physics for Scientists & Engineers 2, Chapter 24
16
Energy Density in Capacitors
!  This work is stored as electric potential energy
1 q2 1
1
2
U=
= C ( ΔV ) = q ( ΔV )
2C 2
2
!  We define the electric energy density, u, as the electric
potential energy per unit volume
U
u=
volume
!  Taking the case of an ideal parallel plate capacitor
1
2
2
C
ΔV
(
)
C
ΔV
U
(
)
u=
=2
=
Ad
Ad
2Ad
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
17
Energy Density in Capacitors
!  Inserting our formula for the capacitance of a parallel plate
capacitor we find
2
⎛ ε0 A ⎞
ΔV
(
)
2
⎜⎝
⎟⎠
1
ΔV
d
⎛
⎞
u=
= ε0 ⎜
⎟⎠
⎝
2Ad
2
d
!  Recognizing that ΔV/d is the magnitude of the electric
field, E, we obtain an expression for the electric energy
density for parallel plate capacitor
1
u = ε0E 2
2
!  This result, which we derived for the parallel plate capacitor,
holds for the electric potential energy stored in any electric
field per unit volume occupied by that field
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
18
Thundercloud
!  Suppose a thundercloud with horizontal dimensions of
2.0 km by 3.0 km hovers at an altitude of 500 m over a flat
area.
!  The cloud carries a charge of 160 C.
!  The ground has no charge.
PROBLEM 1:
!  What is the potential difference between the cloud and
the ground?
!  SOLUTION 1:
!  We approximate the cloud-ground systems as a parallel
plate capacitor with capacitance:
ε0 A
C=
=
d
February 5, 2014
−12
8.85⋅10
F/m)(2000 m)(3000 m)
(
500 m
Chapter 24
= 0.11 µF
19
Thundercloud
!  A parallel plate capacitor has +q on one plate and –q on the
other plate.
!  We take +80 C on the clouds and -80 C on the ground so
that q = 80 C.
q
80 C
ΔV = =
= 7.3⋅108 V = 730 MV
C 0.11 µF
PROBLEM 2:
!  Lightning requires an electric field strength of 2.5 MV/m.
!  Are the conditions right for lightning?
SOLUTION 2:
!  The electric field is:
ΔV 730 MV
E=
=
= 1.5 MV/m
d
500 m
!  So field strength is too low, except for trees, towers, etc.
February 5, 2014
Chapter 24
20
Thundercloud
PROBLEM 3:
!  What is the total electric potential energy contained in the
field between the cloud and ground?
SOLUTION 3:
!  The total electric potential energy is:
1
1
U = q (ΔV ) = (80 C)(7.3⋅108 V) = 2.9⋅1010 J
2
2
!  This is enough energy to charge 500 Chevy Volt batteries.
February 5, 2014
Chapter 24
21
Capacitors with Dielectrics
!  So far, we have discussed capacitors with air or
vacuum between the plates
!  However, most real-life capacitors have an insulating
material, called a dielectric, between the two plates
!  The dielectric serves several purposes:
•  Provides a convenient way to maintain mechanical separation
between the plates
•  Provides electrical insulation between the plates
•  Allows the capacitor to hold a higher voltage
•  Increases the capacitance of the capacitor
•  Takes advantage of the molecular structure of the dielectric
material
!  Demo
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
22
Capacitors with Dielectrics
!  Placing a dielectric between the plates of a capacitor
increases the capacitance of the capacitor by a numerical
factor called the dielectric constant, κ
!  We can express the capacitance of a capacitor with a
dielectric with dielectric constant κ between the plates as
C = κ Cair
!  Where Cair is the capacitance of the capacitor without the
dielectric
!  Placing the dielectric between the plates of the capacitor has
the effect of lowering the electric field between the plates
and allowing more charge to be stored in the capacitor
q
C=
ΔV
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
23
Parallel Plate Capacitor with Dielectric
!  Placing a dielectric between the plates of a parallel plate
capacitor modifies the electric field as
E=
Eair
q
q
=
=
κ κε 0 A ε A
!  ε0 is the electric permittivity of free space
!  ε is the electric permittivity of the dielectric material
ε = κε 0
!  Note that the replacement of ε0 by ε is all that is needed to
generalize our expressions for the capacitance
!  The potential difference across a parallel plate capacitor is
ΔV = Ed =
qd
κε 0 A
!  The capacitance is then
February 4, 2014
q κε 0 A
C=
=
ΔV
d
Physics for Scientists & Engineers 2, Chapter 24
24
Dielectric Strength
!  The dielectric strength of a material measures the ability of
that material to withstand potential difference
!  If the electric field strength in the dielectric exceeds the
dielectric strength, the dielectric will break down and begin
to conduct charge between the plates via a spark, which
usually destroys the capacitor
!  A useful capacitor must contain a dielectric that not only
provides a given capacitance but also enables the device to
hold the required potential difference without breaking
down
!  Capacitors are usually specified in terms of their capacitance
and by the maximum potential difference that they are
designed to handle
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
25
Dielectric Constant
!  The dielectric constant of vacuum is defined to be 1
!  The dielectric constant of air is close to 1 and we will use
the dielectric constant of air as 1 in our problems
!  The dielectric constants of common materials are listed
below (more are listed in the book in Table 24.1)
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
26
Microscopic Perspective on Dielectrics
!  Let’s consider what happens at the atomic and molecular
level when a dielectric is placed in an electric field
!  A polar dielectric material is
composed of molecules that
have a permanent electric dipole
moment
!  Normally the directions of the
electric dipoles are random
!  When an electric field is applied
to these polar molecules, they
tend to align with the
electric field
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
27
Microscopic Perspective on Dielectrics
!  A nonpolar dielectric material
is composed of atoms or
molecules that have no inherent
electric dipole moment
!  These atoms are molecules can
be induced to have a dipole
moment by an external electric
field
!  The electric field acts on the
positive and negative charges in
the atom or molecule and
produce and induced dipole
moment
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
28
Microscopic Perspective on Dielectrics
!  In both polar and non-polar dielectrics, the fields resulting
from aligned electric dipole moments tend to partially
cancel the original electric field
+
!  The resulting electric field inside the capacitor then is the
original field minus the induced field
  
Er = E − Ed
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
29
Example I
!  A parallel plate capacitor whose capacitance C is 13.5pF is charged by
a battery to a potential difference of V =12.5V between its plates. The
battery is now disconnected and material with κ=6.5 is slipped
between the plates.
(a) What is the potential energy before the material is inserted?
(b) What is U after the material has been inserted?
(a)
(b) Key Idea: Because the battery is disconnected, the charge on the
capacitor cannot change!
February 4, 2014
Physics for Scientists&Engineers 2
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Example II
!  A parallel plate capacitor whose capacitance C is 13.5pF is charged by
a battery to a potential difference of V =12.5V between its plates. The
battery is now disconnected and material with κ=6.5 is slipped
between the plates.
(a) What is the potential energy before the material is inserted?
(b) What is U after the material has been inserted?
(b) Key Idea: Because the battery is disconnected, the charge on the
capacitor cannot change, but the capacitance does change (C--> κC)!
February 4, 2014
Physics for Scientists&Engineers 2
31
Example III
!  A parallel plate capacitor whose capacitance C is 13.5pF is charged by
a battery to a potential difference of V =12.5V between its plates. The
battery is now disconnected and material with κ=6.5 is slipped
between the plates.
(b) What is U after the material has been inserted?
The potential energy decreased by a factor κ. The “missing” energy, in
principle, would be apparent to the person inserting the material. The
capacitor would exert a tiny tug on the material and would do work
on it, in amount
February 4, 2014
Physics for Scientists&Engineers 2
32
Capacitance of a Coaxial Cable
!  Coaxial cables are used to transport signals between devices
with minimum interference
!  A 20.0 m long coaxial cable is composed of a conductor and
a coaxial conducting shield around the conductor
!  The space between the conductor and the shield is filled with
polystyrene
!  The radius of the conductor is
0.250 mm and the radius of the shield
is 2.00 mm
!  PROBLEM
!  What is the capacitance of the coaxial
cable?
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
35
Capacitance of a Coaxial Cable
SOLUTION
!  We can think of the coaxial cable as a cylinder
!  The dielectric constant of polystyrene is 2.6
!  We can treat the coaxial cable as a cylindrical capacitor with
r1 = 0.250 mm and r2 = 2.00 mm, filled with a dielectric with
κ = 2.6
!  The capacitance of the coaxial cable is
2π ( 8.85⋅10−12 F/m )( 20.0 m )
2πε 0 L
C =κ
= ( 2.6 )
⎛ r2 ⎞
⎛ 2.00⋅10−3 m ⎞
ln ⎜
ln ⎜ ⎟
⎝ 2.50⋅10−4 m ⎟⎠
⎝ r1 ⎠
C = 1.39⋅10−9 F=1.39 nF
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
36
Supercapacitor / Ultracapacitor
!  Supercapacitors (ultracapacitors) are made using a material
with a very large surface area between the capacitor plates
!  Two layers of activated charcoal are given opposite charge
and are separated by an insulating material
!  This produces a capacitor with capacitance millions of times
larger than ordinary capacitors
!  However, the potential difference can only be 2 to 3 V
February 4, 2014
Physics for Scientists & Engineers 2, Chapter 24
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