Transcript Physics 2102 Lecture: 06 THU 11 FEB

Physics 2102
Jonathan Dowling
Physics 2102
Lecture: 08 THU 11 FEB
Capacitance I
25.1–4
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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
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!)
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 the plates: (Gauss’ Law)

Q
E 
0 0 A
Relate E to potential difference V:
Area of each
plate = A
Separation = d
charge/area = 
= Q/A
-Q
  d Q
Qd
V   E  dx  
dx 
 A
0 A
0
0 0
d
What is the capacitance C ?
Q 0 A
C 
V
d
 C2 m2  C2  CC C 
Units:  2
       
Nm m  Nm  J  V
+Q
Capacitance and Your iPhone!
Q 0 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!
 C2 m2  C2  CC C 
Units:  2
       F  Farad
Nm m  Nm  J  V
Isolated 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 is fixed!
• C decreases (=0A/d)
• V=Q/C; V increases.
+Q
–Q
Parallel Plate Capacitor & Battery
• 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.
Does the Electric Field Inside:
(a) Increase?
(b) Remain the Same?
• V is fixed by battery!
• C decreases (=0A/d)
(c) Decrease?
• Q=CV; Q decreases
• E = Q/0A decreases
+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
4 0 r
2
Relate E to potential difference
between the plates:
Concentric spherical shells:
Charge +Q on inner shell,
–Q on outer shell
b
  b kQ
 kQ 
V   E  dr   2 dr  

r
r

a
a
a
b
1 1
 kQ   
a b
Spherical Capacitor
What is the capacitance?
C = Q/V =
Q

Q 1 1
 

4 0  a b 
4 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  4 0 a
Cylindrical Capacitor
Quic kTime™ a nd a
d eco mp res so r
ar e n eed ed to see thi s p ictu re.
What is the electric field in between
the plates? Gauss’ Law!
E
Radius of outer
plate = b
Radius of inner
plate = a
Q
2 0 rL
Length of capacitor = L
+Q on inner rod, –Q on outer shell
Relate E to potential difference
between the plates:
20 L
b
C  Q /V 
 
V   E  dr
a
b
b 
ln 
a 
b
b 
 Q ln r 
Q

ln 

dr  


2

0 rL
 2 0 L  a 20 L a 
a
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e0 ab/(b-a)
Cylindrical:
C = 20 L/ln(b/a)]
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
• 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
PAR-V (Parallel V the Same)
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
Capacitors 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 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
C1=10 F
C2=20 F
C3=30 F
120V
Cpar  C1  C2  C3  10 20 30F  60F
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 (10F) (20F) (30F)
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