Physics 2102 Physics 2102 Lecture: 08 THU 11 FEB Lecture: 08 THU 11 FEB
Capacitance I Capacitance I
Physics 2102
Jonathan Dowling
25.1–4
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are needed to see this picture.
Capacitors and Capacitors and CapacitanceCapacitance
Capacitor: any two conductors, one with charge +Q, other with charge –Q
+Q
–Q
Uses: storing and releasing electric charge/energy. Most electronic capacitors:micro-Farads (F),pico-Farads (pF) — 10–12 FNew technology: compact 1 F capacitors
Uses: storing and releasing electric charge/energy. Most electronic capacitors:micro-Farads (F),pico-Farads (pF) — 10–12 FNew technology: compact 1 F capacitors
Potential DIFFERENCE between conductors = V
Q = CCV where C = capacitance
Units of capacitance: Farad (F) = Coulomb/Volt
CapacitanceCapacitance
• Capacitance depends only on GEOMETRICAL factors and on the MATERIAL that separates the two conductors
• e.g. Area of conductors, separation, whether the space in between is filled with air, plastic, etc.
+Q–Q
(We first focus on capacitorswhere gap is filled by AIR!)
Electrolytic (1940-70)Electrolytic (new)
Paper (1940-70)
Tantalum (1980 on) Ceramic (1930 on)Mica (1930-50)
Variableair, mica
CapacitorsCapacitors
Parallel Plate CapacitorParallel Plate Capacitor
-Q
E field between the plates: (Gauss’ Law)
Relate E to potential difference V:
What is the capacitance C ?
Area of each plate = ASeparation = dcharge/area = = Q/A
∫ ⋅=d
xdEV0
rr
A
Qddx
A
Qd
00 0 εε∫ ==
d
A
V
QC 0ε==
d
A
V
QC 0ε==
A
QE
00 εε
==A
QE
00 εε
==
We want capacitance: C = Q/V
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Units :C2
Nm2
m2
m
⎡
⎣ ⎢
⎤
⎦ ⎥=
C2
Nm
⎡
⎣ ⎢
⎤
⎦ ⎥=
CC
J
⎡ ⎣ ⎢
⎤ ⎦ ⎥=
C
V
⎡ ⎣ ⎢
⎤ ⎦ ⎥ √
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Units :C2
Nm2
m2
m
⎡
⎣ ⎢
⎤
⎦ ⎥=
C2
Nm
⎡
⎣ ⎢
⎤
⎦ ⎥=
CC
J
⎡ ⎣ ⎢
⎤ ⎦ ⎥=
C
V
⎡ ⎣ ⎢
⎤ ⎦ ⎥ √
+Q
Capacitance and Your iPhone!Capacitance and Your iPhone!
d
A
V
QC 0ε==
Parallel Plate Capacitor — Parallel Plate Capacitor — ExampleExample
• A huge parallel plate capacitor consists of two square metal plates of side 50 cm, separated by an air gap of 1 mm
• What is the capacitance?
Lesson: difficult to get large valuesof capacitance without specialtricks!
Lesson: difficult to get large valuesof capacitance without specialtricks!
C = ε0A/d
= (8.85 x 10–12 F/m)(0.25 m2)/(0.001
m)
= 2.21 x 10–9 F
(Very Small!!)
€
Units :C2
Nm2
m2
m
⎡
⎣ ⎢
⎤
⎦ ⎥=
C2
Nm
⎡
⎣ ⎢
⎤
⎦ ⎥=
CC
J
⎡ ⎣ ⎢
⎤ ⎦ ⎥=
C
V
⎡ ⎣ ⎢
⎤ ⎦ ⎥≡ F[ ] = Farad
€
Units :C2
Nm2
m2
m
⎡
⎣ ⎢
⎤
⎦ ⎥=
C2
Nm
⎡
⎣ ⎢
⎤
⎦ ⎥=
CC
J
⎡ ⎣ ⎢
⎤ ⎦ ⎥=
C
V
⎡ ⎣ ⎢
⎤ ⎦ ⎥≡ F[ ] = Farad
IsolatedIsolated Parallel Plate Capacitor Parallel Plate Capacitor
• A parallel plate capacitor of capacitance C is charged using a battery.
• Charge = Q, potential difference = V.• Battery is then disconnected. • If the plate separation is INCREASED,
does Potential Difference V:
(a) Increase?(b) Remain the same?(c) Decrease?
+Q –Q
• Q is fixed!• C decreases (=ε0A/d)• V=Q/C; V increases.
Parallel Plate Capacitor & Parallel Plate Capacitor & BatteryBattery
• A parallel plate capacitor of capacitance C is charged using a battery.
• Charge = Q, potential difference = V.
• Plate separation is INCREASED while battery remains connected.
+Q –Q
• V is fixed by battery!• C decreases (=ε0A/d)• Q=CV; Q decreases• E = Q/ε0A decreases
Does the Electric Field Inside:(a) Increase?(b) Remain the Same?(c) Decrease?
Spherical CapacitorSpherical Capacitor
What is the electric field insidethe capacitor? (Gauss’ Law)
Radius of outer plate = bRadius of inner plate = a
Concentric spherical shells:Charge +Q on inner shell,–Q on outer shellRelate E to potential difference
between the plates:
204 r
QE
πε=
∫ ⋅=b
a
rdEVrr b
a
b
ar
kQdr
r
kQ⎥⎦⎤
⎢⎣⎡−==∫ 2 ⎥⎦
⎤⎢⎣⎡ −=
bakQ
11
Spherical CapacitorSpherical Capacitor
What is the capacitance?C = Q/V =
Radius of outer plate = bRadius of inner plate = a
Concentric spherical shells:Charge +Q on inner shell,–Q on outer shell
Isolated sphere: let b >> a,
⎥⎦⎤
⎢⎣⎡ −
=
baQ
Q11
4 0πε
)(
4 0
ab
ab
−=
πε
aC 04πε=
Cylindrical CapacitorCylindrical Capacitor
What is the electric field in between the plates? Gauss’ Law!
Relate E to potential differencebetween the plates:
Radius of outer plate = bRadius of inner plate = a
cylindrical Gaussiansurface of radius r
Length of capacitor = L+Q on inner rod, –Q on outer shell
rL
QE
02πε=
∫ ⋅=b
a
rdEVrr
b
a
b
a L
rQdr
rL
Q⎥⎦
⎤⎢⎣
⎡==∫
00 2ln
2 πεπε
€
=Q
2πε0Llnb
a
⎛
⎝ ⎜
⎞
⎠ ⎟
QuickTime™ and a decompressorare needed to see this picture.
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C =Q /V =2πε0L
lnb
a
⎛
⎝ ⎜
⎞
⎠ ⎟
€
C =Q /V =2πε0L
lnb
a
⎛
⎝ ⎜
⎞
⎠ ⎟
SummarySummary• Any two charged conductors form a capacitor.
•Capacitance : C= Q/V
•Simple Capacitors:
Parallel plates: C = ε0 A/d
Spherical: C = 4π e0 ab/(b-a)
Cylindrical: C = 2π 0 L/ln(b/a)]
Capacitors in Parallel: Capacitors in Parallel: V=ConstantV=Constant
• An ISOLATED wire is an equipotential surface: V=Constant
• Capacitors in parallel have SAME potential difference but NOT ALWAYS same charge!
• VAB = VCD = V
• Qtotal = Q1 + Q2
• CeqV = C1V + C2V
• Ceq = C1 + C2
• Equivalent parallel capacitance = sum of capacitances
A B
C D
C1
C2
Q1
Q2
CeqQtotal
V = VAB = VA –VB
V = VCD = VC –VD
VA VB
VC VD
V=V
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Cparallel =C1 +C2
PAR-V (Parallel V the Same)
Capacitors in Series: Capacitors in Series: Q=ConstantQ=Constant
• Q1 = Q2 = Q = Constant
• VAC = VAB + VBC
A B C
C1 C2
Q1 Q2
Ceq
Q = Q1 = Q2
21 C
Q
C
Q
C
Q
eq
+=
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1
Cseries
=1
C1
+1
C2
SERIES: • Q is same for all capacitors• Total potential difference = sum of V
Isolated Wire:Q=Q1=Q2=Constant
SERI-Q (Series Q the Same)
Capacitors in parallel and in Capacitors in parallel and in seriesseries
• In series : 1/Cser = 1/C1 + 1/C2
Vser = V1 + V2
Qser= Q1 = Q2
C1 C2
Q1 Q2
C1
C2
Q1
Q2
• In parallel : Cpar = C1 + C2
Vpar = V1 = V2
Qpar = Q1 + Q2 Ceq
Qeq
Example: Parallel or Example: Parallel or Series?Series?
What is the charge on each capacitor?
C1=10 F
C3=30 F
C2=20 F
120V
• Qi = CiV • V = 120V = Constant• Q1 = (10 F)(120V) = 1200 C • Q2 = (20 F)(120V) = 2400 C• Q3 = (30 F)(120V) = 3600 C
Note that:• Total charge (7200 C) is
shared between the 3 capacitors in the ratio C1:C2:C3
— i.e. 1:2:3
Parallel: Circuit Splits Cleanly in Two (Constant V)
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Cpar =C1 +C2 +C3 = 10 + 20 + 30( )μF = 60μF
Example: Parallel or Example: Parallel or SeriesSeries
What is the potential difference across each capacitor?
C1=10F C3=30FC2=20F
120V
• Q = CserV• Q is same for all capacitors• Combined Cser is given by:
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1
Cser
=1
(10μF)+
1
(20μF)+
1
(30μF)
• Ceq = 5.46 F (solve above
equation)
• Q = CeqV = (5.46 F)(120V) = 655
C
• V1= Q/C1 = (655 C)/(10 F) = 65.5
V
• V2= Q/C2 = (655 C)/(20 F) = 32.75
V
• V3= Q/C3 = (655 C)/(30 F) = 21.8
V
Note: 120V is shared in the ratio of INVERSE capacitances i.e. (1):(1/2):(1/3)
(largest C gets smallest V)
Series: Isolated Islands (Constant Q)
Example: Series or Example: Series or Parallel?Parallel?
In the circuit shown, what is the charge on the 10F capacitor?
In the circuit shown, what is the charge on the 10F capacitor?
10 F
10F 10V
10F
5F5F 10V
• The two 5F capacitors are in parallel
• Replace by 10F • Then, we have two 10F
capacitors in series• So, there is 5V across the 10 F capacitor of interest
• Hence, Q = (10F )(5V) = 50C
Neither: Circuit Compilation Needed!