Transcript Physics 2102 Lecture: 06 THU 11 FEB

Physics 2113
Jonathan Dowling
Physics 2113
Lecture: 16 MON 23 FEB
Capacitance I
Capacitors and Capacitance
Capacitor: any two conductors, one with
charge +Q, other with charge –Q
–Q
Potential DIFFERENCE between
conductors = V
Q = CV where C = capacitance
For fixed Q the larger the capacitance
the larger the stored voltage.
Units of capacitance:
Farad (F) = Coulomb/Volt
+Q
Uses: storing and releasing
electric charge/energy.
Most electronic capacitors:
micro-Farads ( F),
pico-Farads (pF) — 10–12 F
New technology:
compact 1 F capacitors
Capacitance
• 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 capacitors
where gap is filled by AIR!)
Capacitance depends only on
GEOMETRICAL factors and on
the MATERIAL that separates
the two conductors. It does not
depend on the voltage across it
or the charge on it at all.
(a) Same
(b) Same
Electrolytic (1940-70)
Electrolytic (new)
Paper (1940-70)
Capacitors
Variable
air, mica
Tantalum (1980 on)
Ceramic (1930 on)
Mica (1930-50)
Parallel Plate Capacitor
We want capacitance: C = Q/V
E field between conducting plates:
Area of each
plate = A
Separation = d
charge/area = 
= Q/A
s
Q
E= =
e0 e0 A
Relate E to potential difference V:
+Q
s
Qd
= ò dx =
e0 0
e0A
d
What is the capacitance C ?
Q e0 A
C= =
V
d
σ
d
é C2 m2 ù é C2 ù éCCù é C ù
Units : ê
ú=ê
ú=ê ú=ê ú Ö
2
ëNm m û ëNmû ë J û ëV û
–Q
Capacitance and Your iPhone!
Q e0 A
C= =
V
d
Parallel Plate Capacitor — Example
• 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?
C = 0A/d
= (8.85 x 10–12 F/m)(0.25 m2)/(0.001 m)
= 2.21 x 10–9 F
(Very Small!!)
Lesson: difficult to get large values
of capacitance without special
tricks!
é C2 m2 ù é C2 ù éCCù é C ù
Units : ê
ú=ê
ú = ê ú = ê ú º [F] = Farad
2
ëNm m û ëNmû ë J û ëV û
C=
•
•
•
Isolated Parallel Plate Capacitor: ICPP
Q e0 A
V
=
d
A parallel plate capacitor of capacitance C is
charged using a battery.
Charge = Q, potential voltage difference = V.
Battery is then disconnected.
If the plate separation is INCREASED, does the
capacitance C:
• Q is fixed!
(a) Increase?
• d increases!
(b) Remain the same?
• C decreases (= 0A/d)
(c) Decrease?
• V=Q/C; V increases.
If the plate separation is INCREASED, does the
capacitance C:
(a) Increase?
(b) Remain the same?
(c) Decrease?
+Q
–Q
Parallel Plate Capacitor & Battery: ICPP
• 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.
• V is fixed constant by battery!
Does the Electric Field Inside: • C decreases (= A/d)
0
(a) Increase?
• Q=CV; Q decreases
• E = σ/0 = Q/0A decreases
(b) Remain the Same?
(c) Decrease?
Battery does work on
capacitor to maintain
constant V!
Q e0 A
C= =
V
d
s
Q
E= =
e0 e0 A
+Q
–Q
Spherical Capacitor
What is the electric field inside
the capacitor? (Gauss’ Law)
E=
Radius of outer
plate = b
Radius of inner
plate = a
Q
4pe 0 r
2
Relate E to potential difference
between the plates:
b
Concentric spherical shells:
Charge +Q on inner shell,
–Q on outer shell
b
kQ
é kQ ù
= ò 2 dr = ê- ú
r
r
ë
ûa
a
é1 1ù
= kQ ê - ú
ëa bû
Spherical Capacitor
What is the capacitance?
C = Q/V =
Q
=
Q é1 1ù
- ú
ê
4pe 0 ë a b û
4pe 0 ab
=
(b - a )
Radius of outer
plate = b
Radius of inner
plate = a
Concentric spherical shells:
Charge +Q on inner shell,
–Q on outer shell
Isolated sphere: let b >> a,
C = 4pe 0 a
Cylindrical Capacitor
What is the electric field in between
the plates? Gauss’ Law!
E=
Radius of outer
plate = b
Radius of inner
plate = a
Q
2pe 0 rL
Length of capacitor = L
+Q on inner rod, –Q on outer shell
Relate E to potential difference
between the plates:
2pe0 L
C = Q /V =
æ bö
lnç ÷
è aø
b
æbö
é Q ln r ù
Q
=ò
dr = ê
ú = 2pe L lnçè a ÷ø
2pe 0 rL
0
ë 2pe 0 L û a
a
b
Q
cylindrical
Gaussian
surface of
radius r
Summary
• Any two charged conductors form a capacitor.
•Capacitance : C= Q/V
•Simple Capacitors:
Parallel plates:
C = 0 A/d
Spherical:
C = 40 ab/(b-a)
Cylindrical:
C = 20 L/ln(b/a)]
•Parallel plates:
Spherical:
•
Cylindrical:
C = 0 A/d
C=
1
1 é1 1ù
- ú
ê
4pe 0 ë a b û
Hooked to battery
V is constant.
Q=VC
C = 20 L/ln(b/a)]
(a) d increases -> C decreases -> Q decreases
(b) a inc -> separation d=b-a dec. -> C inc. -> Q increases
(c) b increases -> separation d=b-a inc.-> C dec. -> Q decreases
Capacitors in Parallel: V=Constant
• An ISOLATED wire is an equipotential
surface: V = Constant
• Capacitors in parallel have SAME
potential difference but NOT ALWAYS
same charge!
V = VAB = VA –VB
Q1
A
• VAB = VCD = V PAR-V (Parallel V the Same)
• Qtotal = Q1 + Q2
• CeqV = C1V + C2V
• Ceq = C1 + C2
C
C1
VA
Q2
VB
C2
VC
VD
B
D
V = VCD = VC –VD
• Equivalent parallel capacitance =
sum of capacitances
Cparallel = C1 + C2
PARALLEL:
• V is same for all capacitors
• Total charge = sum of Q
Qtotal
V=V
Ceq
Capacitors in Series: Q=Constant
•
Q1 = Q2 = Q = Constant
•
VAC = VAB + VBC
Isolated Wire:
Q=Q1=Q2=Constant
Q1
SERI-Q (Series Q the Same)
Q
Q Q
= +
Ceq C1 C2
1
Cseries
B
A
1
1
= +
C1 C2
SERIES:
• Q is same for all capacitors
• Total potential difference = sum of V
Q2
C1
C
C2
Q = Q 1 = Q2
Ceq
PARALLEL:
• V is same for all capacitors
• Total charge = sum of Q
SERIES:
• Q is same for all capacitors
• Total potential difference =
sum of V
(a) Parallel: Voltage is same V on each but charge is q/2 on
each since
q/2+q/2=q.
(b) Series: Charge is same q on each but voltage is V/2 on each
since V/2+V/2=V.
“Compiling” in parallel and in series
• In parallel :
Cpar = C1 + C2
Vpar = V1 = V2
Qpar = Q1 + Q2
Q1
C1
Qeq
Q2
C2
Ceq
• In series :
1/Cser = 1/C1 + 1/C2
Vser = V1 + V2
Qser= Q1 = Q2
Q1
Q2
C1
C2
Example: Parallel or Series?
Parallel: Circuit Splits Cleanly in Two (Constant V)
What is the charge on each capacitor?
• Qi = CiV
• V = 120V = Constant
• Q1 = (10 μF)(120V) = 1200 μC
• Q2 = (20 μF)(120V) = 2400 μC
• Q3 = (30 μF)(120V) = 3600 μC
Note that:
• Total charge (7200  C) is shared
between the 3 capacitors in the ratio
C1:C2:C3 — i.e. 1:2:3
C1=10  F
C2=20  F
C3=30  F
120V
Cpar = C1 + C2 + C3 = (10 + 20 + 30)mF = 60mF
Example: Parallel or Series
Series: Isolated Islands (Constant Q)
What is the potential difference across each capacitor?
• Q = CserV
• Q is same for all capacitors
• Combined Cser is given by:
1
1
1
1
=
+
+
Cser (10mF) (20mF) (30mF)
C1=10 F
C2=20 F
C3=30 F
120V
• 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)
Example: Series or Parallel?
Neither: Circuit Compilation Needed!
In the circuit shown, what is
the charge on the 10 F
capacitor?
• The two 5 F capacitors are in
parallel
• Replace by 10 F
• Then, we have two 10 F
capacitors in series
• So, there is 5V across the 10  F
capacitor of interest
• Hence, Q = (10 F )(5V) = 50 C
5 F
10 F
5 F
10V
10  F
10 F
10V