capacitance and dielectrics chapter 241 any two conductors separated by an insulator forms a...
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Capacitance and Dielectrics
Chapter 24 1
Any two conductors separated by an insulator forms a Capacitor.Definition:
1F =1 farad = 1 C/V = 1 coulomb /volt
ab
QC
V
Chapter 24 2
Capacitors and Capacitance (Chapter 24, Sec 1)
Figure 24-2
A
QE
00
A
Q
QA
dd
A
QEdVab
00
abab CVVd
AQ 0
Coul/m2
Farads
1 F = 1 x 10-6 Farads
1 pF = 1 x 10-12 Farads
1 nF = 1 x 10-9 Farads
Chapter 24 3
12 2 2
2
120
0
1 1 .
1 /
1 /
1 V = 1J/C
/ .
1 F 1 C/V
1 F 1 C /J
8.85 10 /
8.85 10
J N m
E N C
E V m
C N m
x F m
x
Useful Definitions and Relationships
Chapter 24 4
Practical Capacitors
Practical Values100µF1µF0.01µF100pF1pF
Some examples of flat, cylindrical, and spherical capacitors
– See just how large a 1 F capacitor would be. Refer to Example 24.1.
– Refer to Example 24.2 to calculate properties of a parallel-plate capacitor.
– Follow Example 24.3 and Figure 24.5 to consider a spherical capacitor.
– Follow Example 24.3 and Figure 24.5 to consider a cylindrical capacitor.
Chapter 24 6
Capacitors in Series
11 C
QV
22 C
QV
212121
11
CCQ
C
Q
C
QVVV
21
11
CCQ
V
V
QCeq
Q
V
Ceq
1
21
111
CCCeq
All capacitors in series have the same charge Q
Chapter 24 7
Capacitors in ParallelAll Capacitors in Parallel have the same voltage V
Figure 24-7
VCQ 11 VCQ 22
VCCVCVCQQQ 212121
21 CCV
Q
21 CCV
QCeq
Chapter 24 8
Capacitors in Series and Parallel - Example 24.6
Chapter 24 9
Energy Storage in Capacitors
Let q equal the changing charge increasing from 0 to Q as the changing voltage v is increasingfrom 0 to Vab. We will determine the energy stored in the capacitor when the charge reaches Qand the voltage reaches Vab. (q and v are the intermediate charge and charging voltage
q
wv vqw v
qC
C
qv vdqdw
C
Cqdq
CdqC
qvdqW
QQQQ
22
11 2
0
2
000
joules
Chapter 24 10
Example 24-7, Page 827 Text Transferring charge and energy between capacitors
Calculate the initial chargeCalculate the initial stored energyConnect the capacitorsCalculate the resulting voltageCalculate the charge distributionCalculate the energy change
Chapter 24 11
Capacitor Dielectrics Solve Three Problems
1. Provides mechanical spacing between two large plates2. Increases the maximum possible potential between
plates.3. For a given plate area the dielectric increases the
capacitance.
Dielectrics change the potential difference
• The potential between to parallel plates of a capacitor changes when the material between the plates changes. It does not matter if the plates are rolled into a tube as they are in Figure 24.13 or if they are flat as shown in Figure 24.14.
Chapter 24 13
What Happens with a Dielectric
Figure 24-12
V V0 (Q unchanged)
AA
C0
C
00 V
QC
V
QC
Therefore: C C0
0C
CK (Definition of Dielectric Constant) (24-12)
V
V
C
CK 0
0
K
VV 0 (24-13)
Dielectric Constants
Field lines as dielectrics change• Moving from part (a) to
part (b) of Figure 24.15 shows the change induced by the dielectric.
Chapter 24 16
Induced Charge and Polarization
Inserting the dielectric increases permittivity by K, decreases E by 1/K and decreases energy density by 1/KThe E field does work on the dielectric as it is inserted. Removing the dielectric the energy is returned to the field.
Chapter 24 17
Dielectrics (Chapter 24, Sec 4)Induced Charge and Polarization
E
E
Ed
dE
V
V
C
CK 000
0
Therefore: E E0
0
0
KK
EEKEE 0
0 iE
00
E (24-15)
(24-18)
0
Chapter 24 18
Dielectric Breakdown
d
V = Ed
Vmax = Emax d
For dry air: Emax = 3 x 106 V/m
For Mylar, K=3.1, Emax = 9.3 x 106 V/m
where Emax is the dielectric strength of the dielectricin volts/meter.
Dielectric breakdown• A very strong electrical field can exceed the strength of the
dielectric to contain it. Table 24.2 at the bottom of the page lists some limits.
Electric Field Effect on Molecules
Polarization and electric field lines
Chapter 24 22
Chapter 24 23
Chapter 24 24