Transcript Lecture 5 Capacitance

Lecture 5 Capacitance Chp. 26
•Cartoon - Capacitance definition and examples.
•Opening Demo - Discharge a capacitor
•Warm-up problem
•Physlet
•Topics
•Demos
• Circular parallel plate capacitior
•Cylindrical capacitor
•Concentric spherical capacitor
•Leyden jar capacitor
•Dielectric Slab sliding into demo
•Show how to calibrate electroscope
Capacitance
•
Definition of capacitance
A capacitor is a useful device in electrical circuits that allows us to
store charge and electrical energy in a controllable way. The simplest to
understand consists of two parallel conducting plates of area A
separated by a narrow air gap d. If charge +Q is placed on one plate,
and -Q on the other, the potential difference between them is V, and
then the capacitance is defined as C=Q/V. The SI unit is C/V, which is
called the Farad, named after the famous and creative scientist Michael
Faraday from the early 1800’s.
• Applications
–
–
–
–
–
Radio tuner circuit uses variable capacitor
Blocks DC voltages in ac circuits
Act as switches in computer circuits
Triggers the flash bulb in a camera
Converts AC to DC in a filter circuit
Parallel Plate Capacitor
Electric Field of Parallel Plate Capacitor
S
Area A
+++++++++++ + q
E
-------------------- - q
Gauss Law

E
0

q
A
qS
V  ES 
0 A


 A
q
q
C   qS  0
V
S
0 A

E
q
0 A
q  0 EA

Coulomb/Volt = Farad
Show Demo Model, calculate its capacitance , and show
how to charge it up with a battery.
Circular parallel plate capacitor
r
r = 10 cm
r
A = r2 = (.1)2
A = .03 m 2
s
S = 1 mm = .001 m
C
0A
S
C  (10 11 )
.03 Coulomb
.001 Volt
Farad
C  3 1010 F
C  300pF
p = pico = 10-12
Demo Continued
Demonstrate
1.
As S increases, voltage increases.
2.
As S increases, capacitance decreases.
3.
As S increases, E0 and q are constant.
Dielectrics
• A dielectric is any material that is not a conductor, but polarizes
well. Even though they don’t conduct they are electrically active.
– Examples. Stressed plastic or piezo-electric crystal will produce a
spark.
– When you put a dielectric in a uniform electric field (like in between
the plates of a capacitor), a dipole moment is induced on the
molecules throughout the volume. This produces a volume
polarization that is just the sum of the effects of all the dipole
moments. If we put it in between the plates of a capacitor, the
surface charge densities due to the dipoles act to reduce the electric
field in the capacitor.
Dielectrics
• The amount that the field is reduced defines the dielectric
constant  from the formula E = E0 / , where E is the new field
and E0 is the old field without he dielectric.
• Since the electric field is reduced and hence the voltage
difference is reduced (since E = Vd), the capacitance is
increased.
– C = Q / V = Q / (V0 / ) =  C0
–  is typically between 2 – 6 with water equal to 80
– Show demo dielectric slab sliding in between plates. Watch how
capacitance and voltage change. Also show aluminum slab.
Permanent
dipoles
Induced
dipoles
_ ++ _
E0 = the applied field
E’ = the field due to
induced dipoles
E = E0 - E’
Cq
V
V  E 0S
S

E0 
0
 qA
E0 
q
0A
E
E0
V 
E 0S
V


V0

V
C
qS
0A
0A
S
Cq
C
V
q
V0
C  C 0
Find the capacitance of a ordinary piece of coaxial cable (TV cable)
2k
Er 
r
For a long wire we found that
where r is radial to the wire.
a
• r
a
Va  Vb    E. ds  2k 
b
b

E. ds  Edscos180 Eds Edr
dr
 2k ln r
r
outer insulator
metal braid
with - q
•
a
radius b
b
ds = - dr because path of integration is radially inward
Va Vb  2k ln
or
b
V  2k ln
a

Q
b
V
ln
20L a
Va is higher than Vb

 QL
k
1

4 0  air

a = 0.5 mm
C 2 0
 b
L ln a
a
b

Q2

0
L
C  QV 
Qln ba
C
2 0 L
ln ba
signal
wire
radius a
with + q
Insulator
(dielectric )
b = 2.0 mm
2
C 6 1011 6 1011


L
ln 4
1.38
C
 43 PF m
L
C
 86 PF m
L
0 (air)
=2
Model of coaxial cable for calculation of capacitance
Outer metal braid
Signal wire
Spherical capacitor or sphere
Recall our favorite example for E and V is spherical symmetry
Q
R
The potential of a charged sphere is V = (kQ)/R with V = 0 at r = 
.
The capacitance is
C
Q
Q
R

  4 0 R
V kQ R k
Where is the other plate (conducting shell)?
It’s at infinity where it belongs, since that’s where the electric lines of flux terminate.
k = 1010 and R in meters we have
C
R
10
12
10
R(m)
10
R(cm)
10
10
C  R(cm) PF
Earth: C = (6x108 cm)PF = 600
F
Marble: 1 PF
Basketball: 15 PF
You: 30 PF

Demo: Leyden jar capacitor
Demo: Show how you measured capacitance of electroscope
Capacitance of two concentric spherical shells
-q
Integration path
E
+q
a
a
a
b
b
Va  Vb    E. ds    Edr
E. ds  Edscos180 Eds Edr
b
ds = - dr
a
a
a
b
b
b
2
 Va  Vb    Edr    kq/r dr  kq 
1
1 1
b  a)
V  kq  kq(  )  kq(
rb
a b
ab
a


ab
ab
C  q /V 
 4 0
k(b  a)
ba
dr
r2
Electric Potential Energy of Capacitor
•
As we begin charging a capacitor, there is initially no potential difference between
the plates. As we remove charge from one plate and put it on the other, there is
almost no energy cost. As it charges up, this changes.
+q
-q
At some point during the charging, we
have a charge q on the positive plate.
+ -
The potential difference between the plates is V = q/C. As
we transfer an amount dq of positive charge from the
negative plate to the positive one, its potential energy
increases by an amount dU.
dU  Vdq  q dq.
C
The total potential energy increase is
Q
q
q2
U   dq 
C
2C
0
Also
Q2

2C
1
1
1 Q2
2
U  QV  CV 
2
2
2 C
using C = Q/ V
Graphical interpretation of integration
dU  Vdq
Q
U   Vdq where V = q/C

0
V
V = q/c
q/c
q
dq
Q
1 N
U   qiqi = Area under the triangle
C i 1
Q
Area under the triangle is the value of the integral
Area of the triangle is also = 1/2 bh
1
1
Q 1 Q2
Area = (b)(h)  (Q)( ) 
2
2
C
2 C
Q

q
q2 Q Q2
0 C dq  2  2C
0
q
0 C dq
Where is the energy stored in a capacitor?
• Find energy density for parallel plate capacitor. When we charge a
capacitor we are creating an electric field. We can think of the work
done as the energy needed to create that electric field. For the
parallel plate capacitor the field is constant throughout, so we can
evaluate it in terms of electric field E easily.
Use U = (1/2)QV
E

 Q and V = ES
 
 0  A
We are now
including dielectric
effects: 
Solve for Q = AE, V = ES and substitute in
1
1
1
U  QV  (AE )( ES )  E 2 ( AS )
2
2
2
U
1
volume occupied by E
 E 2  
AS 2
Electrostatic energy density general result for all
1 2
geometries.
  E
2
To get total energy you need to integrate over volume.
How much energy is stored in the Earth’s atmospheric
electric field?
(Order of magnitude estimate)
atmosphere
20 km
h
Earth
R
R = 6x106 m
E  100Vm  102
1
U   0 E 2  Volume
2
Volume 4R 2 h
Volume 4 (6 106 )2 (2 104 )  8.6 1018 m3
1
U  (10 11 )(10 2 )(8.6 1018 )
2
U  4.3 1011 J
This energy is renewed daily by the sun. Is this a lot?
The total solar influx is 200 Watts/m2
Usun  200 3.14(6 106 )  2 1016 J s  2 1021 J day
U Usun  2 1010
Only an infinitesimal fraction gets converted to electricity.
World consumes
about 1018 J/day.
This is 1/2000 of
the solar flux.
Parallel Combination of Capacitors
Typical electric circuits have several capacitors in them. How do they
combine for simple arrangements? Let us consider two in parallel.
V
+
-
C1
Q1
C2
Q2
V
+
-
C1
C = Q/V
We wish to find one equivalent capacitor to replace C1 and C2.
Let’s call it C.
The important thing to note is that the voltage across each is the
same and equivalent to V. Also note what is the total charge
stored by the capacitors? Q.
Q  Q1  Q2  C1V  C 2V  (C1  C 2)V
Q
 C1  C 2  C  C1  C 2
V
Series Combination of Capacitors
Q
+
V
+
-
Q
-
+
C1
V1
Q
-
C2
V2
+
V
+
-
-
C
V
C
Q
V
V
Q
C
What is the equivalent capacitor C?
Voltage across each capacitor does not have to be the same.
The charges on each plate have to be equal and opposite in sign by
charge conservation.
The total voltage across each pair is:
V  V1 V 2 
So
Q Q
1
1
1

 Q(  )  Q( )
C1 C 2
C1 C 2
C
1
1
1 ; Therefore,
 
C1 C1 C 2
C
C 1C 2
C1  C 2
Sample problem
C1 = 10 F
C1
V
+
-
C3
C2
C2 = 5.0 F
C3 = 4.0 F
a) Find the equivalent capacitance of the entire combination.
C1 and C2 are in series.
1
1
1
C1C 2
 
 C12 
C12 C1 C 2
C1  C 2
C12 
10  5 50

 3.3F
10  5 15
C12 and C3 are in parallel.
Ceq  C12  C 3  3.3  4.0  7.3F
Sample problem (continued)
C1 = 10 F
C1
V
+
-
C3
C2
C2 = 5.0 F
C3 = 4.0 F
b) If V = 100 volts, what is the charge Q3 on C3?
C = Q/V
Q3  C3V  4.0 106 100
Q3  4.0 104 Coulombs
c) What is the total energy stored in the circuit?
1
1
U  CeqV 2  1.3 10 6 10 4  3.6 10  2 J
2
2
U  3.6 102 J
Warm up set 5
1. HRW6 26.TB.03. [119752] A capacitor C "has a charge Q". The actual charges on its plates are:
Q/2, Q/2
Q, -Q
Q/2, -Q/2
Q, 0
Q, Q
2. HRW6 26.TB.13. [119762] Pulling the plates of an isolated charged capacitor apart:
increases the potential difference
increases the capacitance
does not affect the capacitance
decreases the potential difference
does not affect the potential difference
3. HRW6 26.TB.25. [119774] Let Q denote charge, V denote potential difference and U denote stored energy.
Of these quantities, capacitors in series must have the same:
Q only
Q and U only
U only
V and U only
V only
What is the electric field in a sphere of uniform
distribution of positive charge. (nucleus of protons)

R
r

E




Q
4 3
R
3
EdA 
qenc
0
4 r 3

2
3
E 4 r 
0
r
Q
E

r
3
30 4 0 R