Capacitors
February 5, 2014 Physics for Scientists & Engineers 2, Chapter 24 1
September 17, 2008 Physics for Scientists & Engineers 2, Lecture 14 2
Review ! The electric potential energy stored in a capacitor
is given by
! The field energy density stored in a parallel plate capacitor is given by
! The field energy density in general is
212
U CV=
20
12
u Eε=
2
012
Vud
ε ⎛ ⎞= ⎜ ⎟⎝ ⎠
February 5, 2014 Physics for Scientists & Engineers 2, Chapter 24 3
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
! 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
C =κCair
C =
qΔV
February 5, 2014 Physics for Scientists & Engineers 2, Chapter 24 4
Parallel Plate Capacitor with Dielectric ! Placing a dielectric between the plates of a parallel plate
capacitor modifies the electric field as
! ε0 is the electric permittivity of free space ! ε is the electric permittivity of the dielectric material
! 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
! The capacitance is then
E = Eair
κ= qκε0A
= qεA
ε =κε0
ΔV = Ed = qd
κε0A
C = q
ΔV= κε0A
d
February 5, 2014 Physics for Scientists & Engineers 2, Chapter 24 5
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 6
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 7
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 8
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 10
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 11
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 12
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
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
February 5, 2014 Physics for Scientists & Engineers 2, Chapter 24 15
C =κ 2πε0L
ln r2
r1
⎛⎝⎜
⎞⎠⎟
= 2.6( )2π 8.85 ⋅10−12 F/m( ) 20.0 m( )
ln 2.00 ⋅10−3 m2.50 ⋅10−4 m
⎛⎝⎜
⎞⎠⎟
C =1.39 ⋅10−9 F=1.39 nF
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
February 5, 2014 Chapter 24 18
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Capacitor Partially Filled with a Dielectric Research
! The capacitance of a parallel place capacitor is: ! If a dielectric is placed between the plates we have:
! The capacitance of two capacitors in parallel is:
Simplify
! Putting our equations together gives us:
February 5, 2014 Chapter 24 19
C1 =
ε0 Ad
C2 =κ
ε0 Ad
C12 = C1 + C2
C12 = C1 =
ε0 Ad
+κε0 Ad
= κ+1( )ε0 Ad
Capacitor Partially Filled with a Dielectric
! The area of the plates for each capacitor is:
! Putting our expressions together gives us:
! Calculate
! Putting in our numerical values:
February 5, 2014 Chapter 24 20
A = L L / 2( )= L2 / 2
C12 = κ+1( )
ε0 L2 / 2( )d
=κ+1( )ε0L2
2d
C12 =
15.0+1( ) 8.85⋅10−12 F/m( ) 0.100 m( )2
2 0.00250 m( )= 2.832 ⋅10−10 F
Capacitor Partially Filled with a Dielectric Round
! Double-check
! To double-check our answer, we calculate the capacitance of the capacitor without any dielectric:
! Calculate the capacitance of the capacitor with dielectric:
! 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
C12 = 2.83⋅10−10 F = 283 pF
C1 =
8.85⋅10−12 F/m( ) 0.100 m( )2
0.00250 m( )= 35.4 pF
C1 = 15( )35.4 pF = 531 pF
February 5, 2014 Physics for Scientists & Engineers 2, Chapter 24 22
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