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Capacitors
February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 1
Physics for Scientists & Engineers 2 2
Review: Capacitance ! The definition of capacitance is
• Capacitance is a property of the capacitor
! 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 • d is the distance between the plates
qCV
=
0ACdε=
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
=
! The equivalent capacitance for n capacitors in series is =
! For two capacitors in series:
2/4/14 Physics for Scientists & Engineers 2, Chapter 23 5
Ceq = Ci
i=1
n
∑
1Ceq
= 1Cii=1
n
∑
Ceq =C1C2
C1 +C2
February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 8
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 9
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 10
System of Capacitors STEP 2
! We can see that C4 and C123 are in series ! We find for the equivalent capacitance:
! … and make a new drawing
1C1234
= 1C123
+ 1C4
⇒ C1234 =C123C4
C123 +C4
February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 11
System of Capacitors STEP 3
! We can see that C5 and C1234 are in parallel ! We find for the equivalent capacitance
! … and make a new drawing C12345 =C1234 +C5 =
C123C4
C123 +C4+C5 =
C1 +C2 +C3( )C4
C1 +C2 +C3 +C4+C5
February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 12
System of Capacitors STEP 4: INSERT THE NUMBERS
! So the equivalent capacitance of our system of capacitors
5+5+5( )55+5+5+5
+5⎛⎝⎜
⎞⎠⎟
nF = 8.75 nF
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:
February 5, 2014 Chapter 24 13
q5 = C5ΔV = 5 nF( ) 12 V( )= 60 nC
ΔV =ΔV123 +ΔV4 =
q4
C123
+q4
C4
= q41
C123
+1
C4
⎛
⎝⎜⎜⎜⎜
⎞
⎠⎟⎟⎟⎟⎟
System of Capacitors ! The charge on C4 is then:
! 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:
February 5, 2014 Chapter 24 14
q4 =ΔVC123C4
C123 + C4
=ΔVC1 + C2 + C3( )C4
C1 + C2 + C3 + C4
= 12 V( )15 nF( ) 5 nF( )
20 nFq4 = 45 nC
q1 = q2 = q3 =
45 nC3
=15 nC
February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 16
! 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
! The total work required to bring the capacitor to its full
charge q is
Energy Stored in Capacitors
dW = ΔV 'dq '= q '
Cdq '
Wt = dW∫ = q '
Cdq '
0
q
∫ = 12
q2
C
February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 17
! This work is stored as electric potential energy
! We define the electric energy density, u, as the electric potential energy per unit volume
! Taking the case of an ideal parallel plate capacitor
Energy Density in Capacitors
U = 1
2q2
C= 1
2C ΔV( )2 = 1
2q ΔV( )
u = U
volume
u = U
Ad=
12
C ΔV( )2
Ad=
C ΔV( )2
2Ad
February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 18
! Inserting our formula for the capacitance of a parallel plate capacitor we find
! 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
! 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
Energy Density in Capacitors
u =
ε0Ad
⎛⎝⎜
⎞⎠⎟ ΔV( )2
2Ad= 1
2ε0
ΔVd
⎛⎝⎜
⎞⎠⎟
2
u = 1
2ε0E2
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:
February 5, 2014 Chapter 24 19
C =ε0 Ad
=8.85⋅10−12 F/m( ) 2000 m( ) 3000 m( )
500 m= 0.11 µF
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.
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:
! So field strength is too low, except for trees, towers, etc. February 5, 2014 Chapter 24 20
ΔV =
qC
=80 C
0.11 µF= 7.3⋅108 V = 730 MV
E =ΔV
d=
730 MV500 m
=1.5 MV/m
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:
! This is enough energy to charge 500 Chevy Volt batteries.
February 5, 2014 Chapter 24 21
U =
12
q ΔV( )=12
80 C( ) 7.3⋅108 V( )= 2.9 ⋅1010 J
February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 22
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 23
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 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 24
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 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 25
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 26
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 27
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 28
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 29
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 30
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 31
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 32
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 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
February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 36
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
February 4, 2014 Physics for Scientists & Engineers 2, Chapter 24 37
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