Capacitors
February 5, 2014
Physics for Scientists & Engineers 2, Chapter 24
1
Review
!  The electric potential energy stored in a capacitor
is given by
1
U = CV 2
2
!  The field energy density stored in a parallel plate capacitor is
given by
1 ⎛V ⎞
u = ε0 ⎜ ⎟
2 ⎝d⎠
2
!  The field energy density in general is
1
u = ε0E2
2
September 17, 2008
Physics for Scientists & Engineers 2, Lecture 14
2
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 5, 2014
Physics for Scientists & Engineers 2, Chapter 24
3
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 5, 2014
q κε 0 A
C=
=
ΔV
d
Physics for Scientists & Engineers 2, Chapter 24
4
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 5, 2014
Physics for Scientists & Engineers 2, Chapter 24
5
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 5, 2014
Physics for Scientists & Engineers 2, Chapter 24
6
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
!  A nonpolar dielectric material
is composed of atoms or
molecules that have no inherent
electric dipole moment
Dipole moment is induced by
external electric field
February 5, 2014
Physics for Scientists & Engineers 2, Chapter 24
7
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 5, 2014
Physics for Scientists & Engineers 2, Chapter 24
8
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 5, 2014
Physics for Scientists&Engineers 2
10
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 5, 2014
Physics for Scientists&Engineers 2
11
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 5, 2014
Physics for Scientists&Engineers 2
12
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 5, 2014
Physics for Scientists & Engineers 2, Chapter 24
14
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 5, 2014
Physics for Scientists & Engineers 2, Chapter 24
15
Capacitor Partially Filled with a
Dielectric
PROBLEM:
!  A parallel plate capacitor is constructed of two square
conducting plates with side length L = 10.0 cm.
!  The distance between the plates is d = 0.250 cm.
!  A dielectric with dielectric constant κ =15.0 and thickness
0.250 cm is inserted between the plates.
!  The dielectric is L = 10.0 cm wide and L/2 = 5.00 cm long.
!  What is the capacitance of this capacitor?
February 5, 2014
Chapter 24
17
Capacitor Partially Filled with a
Dielectric
SOLUTION:
!  We have a capacitor partially filled with a dielectric.
!  We can treat this capacitor as two capacitors in parallel.
!  One capacitor is a parallel plate capacitor with plate area
A = L(L/2) and air between the plates.
!  The second capacitor is a parallel plate capacitor with plate
area A = L(L/2) and a dielectric between the plates.
!  Sketch
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Permission required for reproduction or display.
February 5, 2014
Chapter 24
18
Capacitor Partially Filled with a
Dielectric
Research
!  The capacitance of a parallel place capacitor is:
ε0 A
C1 =
d
!  If a dielectric is placed between the plates we have:
ε0 A
C2 = κ
d
!  The capacitance of two capacitors in parallel is:
C12 = C1 + C2
Simplify
!  Putting our equations together gives us:
ε0 A
ε0 A
ε0 A
C12 = C1 =
+κ
= (κ +1)
d
d
d
February 5, 2014
Chapter 24
19
Capacitor Partially Filled with a
Dielectric
!  The area of the plates for each capacitor is:
A = L( L / 2) = L2 / 2
!  Putting our expressions together gives us:
C12 = (κ +1)
ε0 ( L2 / 2)
d
=
(κ +1)ε0 L2
2d
!  Calculate
!  Putting in our numerical values:
(15.0 +1)(8.85⋅10 F/m)(0.100 m)
C12 =
2(0.00250 m)
2
−12
February 5, 2014
Chapter 24
= 2.832⋅10−10 F
20
Capacitor Partially Filled with a
Dielectric
Round
C12 = 2.83⋅10−10 F = 283 pF
!  Double-check
!  To double-check our answer, we calculate the capacitance of
the capacitor without any dielectric:
8.85⋅10
(
C =
−12
1
F/m)(0.100 m)
(0.00250 m)
2
= 35.4 pF
!  Calculate the capacitance of the capacitor with dielectric:
C1 = (15)35.4 pF = 531 pF
!  Our result for the partially filled capacitor is half of the sum
of these two results, so it seems reasonable.
February 5, 2014
Chapter 24
21
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 5, 2014
Physics for Scientists & Engineers 2, Chapter 24
22