Chapter 24Capacitance, dielectrics and
electric energy storage • Basic circuit devices
– Resistors
– Capacitors
– Inductors
– Power supply (Battery, Generator)
• Use our knowledge of electric fields, potentials, and energy to describe how capacitors work.
Parallel Plate Capacitor
Fig 26-4b, p.800
Active Figure 26.4
(SLIDESHOW MODE ONLY)
Potential difference and electric fields in a uniform electric field
+Q
-Qa
b
d
d d
bao o o0 0
ˆ ˆ ˆV j dxi dyj dy d
enclosedE
o
QE dA
o
AE A
o
E
Potential difference and electric fields in a uniform electric field
+Q
-Qa
b
d
bao o
QdV d
A
baQ V
baQ CV
The constant of proportionality is called “capacitance.”For a parallel plate capacitor, the capacitance is:
0AC
d
Factors affecting capacitance
• Size of the capacitor (A, d)• Geometric arrangement
– Plates– Cylinders
• Material between conductors– Air– Paper– Wax
Units of capacitance
Coulomb CFarad F
Volt V
ba
QC
V
A Farad is a lot of capacitance. Typical capacitors are“micro, nano, pico-Farad
Capacitance – Isolated Sphere
• Assume a spherical charged conductor
• Assume V = 0 at infinity
• Note, this is independent of the charge and the potential difference
4/ o
e e
Q Q RC πε R
V k Q R k
Cylindrical capacitor
b a
enclosedE
o
QE dA
+Q on center conducting cylinder
-Q on outer conducting cylinder
o
QE 2 rL
o
QE
2 rL
L
b b
bao o oa a
Q Q dr Q bˆ ˆV r drr ln
2 rL 2 L r 2 L a
Capacitance of a cylindrical capacitor
abQ CV
b a
L
o
ab
o
2 LQ QC
Q b bV ln ln2 L a a
L
Example
• How strong is the electric field between the plates of a 0.80 F air gap capacitor if they are 2.0 mm apart and each has a charge of 72 C?
Capacitors in Parallel
total 1 2 3Q Q Q Q
a b
abV V
1Q
3Q
2Qtotal eqQ C V
1C
2C
3C
1 1Q C V 2 2Q C V 3 3Q C V
eq 1 2 3C V C V C V C V
eq 1 2 3C C C C
1 2 3V V V V
Capacitors in Series
1 2 3V V V V
a b
abV V
1Q 3Q2Q
eqeq
QQ C V V
C 1C 3C2C
11
QV
C
eq 1 2 3
Q Q Q Q
C C C C
eq 1 2 3
1 1 1 1
C C C C
22
QV
C 3
3
QV
C
total 1 2 3Q Q Q Q
Capacitor circuit example
2 F
2 F3 F 4 F
What single capacitor can replace the four shown here?How much charge can the system hold?How much charge is on one of the 2 F capacitors?
3 V
Energy Storage in Capacitors
bb
U dUV V
q dq
dU Vdq
(Like problem 23-50)
q Q
q 0
U Vdq
bao
QdV
A
bao
Q aV ln
2 L b
Parallel Plates
Concentric Cylinders
Qq Q 2 2
q 0 0
q 1 q QU dq
C C 2 2C
QV
CIn General
Change in potential energywhile charging capacitor
Alternate Energy Expressions
2QU
2C Q
VC
2 22C V 1
U CV2C 2
21 Q 1U V QV
2 V 2
Energy Density
Energy per unit volume: Uu
Volume
Consider a Parallel Plate Capacitor: 21U CV
2
0AC
d
V Ed
2 2 200
A1 1U E d Ad E
2 d 2
20
U U 1u E
Volume Ad 2
Dielectrics•A dielectric is a nonconducting material that, when placed between the plates of a capacitor, increases the capacitance
•Materials with Dipoles that can align with an external electric Field. Dielectrics include rubber, plastic, and waxed paper
+ ++
--
-
+ ++
---
oE
indE
oDielectric o ind
EE E E
is the Dielectric Constant
Measure of the degree of dipole alignment in the material
Dielectrics
Example values of dielectric constant
“Dielectric strength” is themaximum field in the dielectric before breakdown.(a spark or flow of charge)
max maxE V / d
Effect of a dielectric on capacitance
oDielectric
EE
o
Dielectric
EE d d
oDielectric
VV
Potential difference with a dielectric is less than the potential difference across free space
oo
Q QC C
V V Results in a higher capacitance.
Allows more charge to be stored before breakdown voltage.
Effect of the dielectric constant
Parallel Plate Capacitoro o
o o
A AC C C
d d
o
AC
d
Material permittivity measures degree to which the material permits induced dipolesto align with an external field
2 2 2o 0 0
1 1 1u E u E E
2 2 2
Example modificationsusing permittivity
d
3
What is new capacitance?
Example – Parallel Plate Capacitor
+Q
-Q
d
A
oAC
d
DipolesThe combination of two equal charges of opposite sign, +Q and –Q, separated by a distance l
p Q2a p
2a
-Q +Q
1p
2p
1 2p p p
Dipoles in a Uniform Electric Field
p
2a
-Q
+Q F QE
F QE
E
F a sin F a sin QEa sin QEa sin Q2aEsin pEsin
x y z
ˆ ˆ ˆi j k
r F x y z
F F F
r F sin
p E
Work Rotating a Dipole in an Uniform Electric Field
p
2a
-Q
+Q F QE
F QE
E
dW d pEsin d dU
oLet: U( =90 ) 0 U pE cos
U p E
0U pE sin d pE cos U
0U 0
Example P26.9
When a potential difference of 150 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of 30.0 nC/cm2. What is the spacing between the plates?
0 AQ Vd
12 2 2
09 2 4 2 2
8.85 10 C N m 150 V4.42 m
30.0 10 C cm 1.00 10 cm m
Vd
Example P26.21
Four capacitors are connected as shown in Figure P26.21.
(a) Find the equivalent capacitance between points a and b.
(b) Calculate the charge on each capacitor if ΔVab = 15.0 V.
1
2.50 F
2.50 6.00 8.50 F
1 15.96 F
8.50 F 20.0 F
s
p
eq
C
C
C
1 1 115.0 3.00sC
5.96 F 15.0 V 89.5 CQ C V
89.5 C4.47 V
20.0 F15.0 4.47 10.53 V
6.00 F 10.53 V 63.2 C on 6.00 F
QV
C
Q C V
89.5 63.2 26.3 C
Example P26.27
Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in Figure P26.27. Take C1 = 5.00 μF, C2 = 10.0 μF, and C3 = 2.00 μF.
1
1
2
1
1 13.33 F
5.00 10.0
2 3.33 2.00 8.66 F
2 10.0 20.0 F
1 16.04 F
8.66 20.0
s
p
p
eq
C
C
C
C
Example P26.35
A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled?
2 12d d 2 112
C C stored energy doubles,
. Therefore, the
Example P26.43Determine (a) the capacitance and (b) the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.75 cm2 and plate separation of 0.040 0 mm.
12 4 2110
5
2.10 8.85 10 F m 1.75 10 m8.13 10 F 81.3 pF
4.00 10 m
AC
d
6 5max max 60.0 10 V m 4.00 10 m 2.40 kVV E d
Example P26.59A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is 3.00 and whose dielectric strength is 2.00 × 108 V/m. The desired capacitance is 0.250 μF, and the capacitor must withstand a maximum potential difference of 4 000 V. Find the minimum area of the capacitor plates.
8 maxmax 2.00 10 V m
VE
d
60 0.250 10 FA
Cd
62max
12 80 0 max
0.250 10 40000.188 m
3.00 8.85 10 2.00 10
C VCdA
E
3.00