Transcript Lecture - Galileo

Lecture 5 Capacitance Ch. 25
•Cartoon - Capacitance definition and examples.
•Opening Demo - Discharge a capacitor
•Warm-up problem
•Physlet
•Topics
•Capacitance
•Parallel Plate Capacitor
•Dielectrics and induced dipoles
•Coaxial cable, Concentric spheres, Isolated sphere
•Two side by side spheres
•Energy density
•Graphical integration
•Combination of capacitance
•Demos
•Super VDG
•Electrometer
•Voltmeter
• Circular parallel plate capacitor
•Cylindrical capacitor
•Concentric spherical capacitor
•Dielectric Slab sliding into demo
•Show how to calibrate electroscope
1
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
Q
between them is V, and then the capacitance is defined as C  .
V
Q
The SI unit is
, which is called the Farad, named after the famous and creative
V
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
2
Parallel Plate Capacitor
3
Electric Field of Parallel Plate Capacitor
Gaussian
surface
+q
+++++++++++
d
E
A
- - - - - - - - - - - -q
EA 
q
E
0
qd
V  Ed 
0 A
q
q
C   qd
V
0 A
C
0 A
d
q
0 A
Area of plate =A
q  0EA
f
Vf  Vi    E  dr̂
i
Integrate from - charge to + charge so that
E  dr̂  Edr

V f  Vi    Edr  Ed

Coulomb/Volt = Farad
V  Ed
4
Show Demo Model, calculate its capacitance, and
show how to charge it up with a battery.
Circular parallel plate capacitor
r = 10 cm = 0.1m
r
A = r2 = (.1m)2
r
A = .03 m 2
d
d = 1 mm = .001 m
C
0A
d
C  (10
11 C 2
Nm 2
.03m 2
)
.001m
Coulomb
Volt
Farad
C  3  10 10 F
C  300pF
p = pico = 10-12
5
Demo Continued
Demonstrate
1.
As d increases, voltage increases.
2.
As d increases, capacitance decreases.
3.
As d increases, E0 and q are constant.
6
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.
7
Induced dipoles
Permanent dipoles
_ ++ _
E0 = the applied field
E’ = the field due to
induced dipoles
E = E0 - E’
8
Dielectrics
The amount that the field is reduced defines the dielectric constant  from
the formula
E
E0 , where E is the new field and E is the old field
0

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
Q

 C0
V  V0 
 
 
where  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.
9
Cq
V
V  E0 d
d

E0 
0
 qA
E0 
q
0A
E
E0
V
E0
V
d
V0

C
0A
d
Cq


qd
V
0A
C
V
q
V0
C  C 0
10
Find the capacitance of a ordinary piece of coaxial cable (TV cable)
f
Vf  Vi    E  dr̂
i
E. dr̂  Edr cos180  Edr Er 
r̂
2k
r
Integrate from b to a or - to +
a
a
b
b
a
Va  Vb    E. dr̂   Edr  2k  
b
Va  Vb  2k  ln
a
Va is higher than Vb

b
b
dr
 2k  ln r
r
a
 QL
k
1
4 0
 air

11
capacitance of a coaxial cable cont.
So,
V
Q
b
ln
20L a
C
Q Q20L

V
Q ln ba
C
20L
ln ba
a = 0.5 mm
Now if a=0.5mm and b=2.0mm, then
C 20

L
ln ba
2
And if  = 2, then
C 6  10 11 6  10 11


L
ln 4
1.38
pF
C
 43
L
m
b = 2.0 mm
pF
C
 86
L
m
For  = 2
0 (for air)
12
Model of coaxial cable for calculation of capacitance
Outer metal braid
Signal wire
- to +
13
Capacitance of two concentric spherical shells
a
a
b
b
V  Va  Vb    E  dr̂    Edr
dr
Integration path
-q
+q
b
as
E. dr̂  Eds cos180  Edr
a
a
a
a
kq
dr
Va  Vb    Edr    2 dr  kq  2
r
r
b
b
b
b  a)
1
1 1
V  kq  kq(  )  kq(
rb
a b
ab
a
E
ab
C  q / V  4 0
ba
Let b get very large. Then
C  4 0 a
for an isolated sphere
14
Spherical capacitor or sphere
Recall our favorite example for E and V is spherical symmetry
Q
The potential of a charged sphere is V 
R
kQ
with V = 0 at r =  .
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
R
C  10 1010 R(m) 1012 R(cm)
10
C  R(cm )pF
Earth: C = (6x108 cm)pF = 600
F
Marble: 1 pF
Basketball: 15 pF

Demo: Show how you measured
capacitance of electroscope
You: 30 pF
15
Capacitance of one charged conducting sphere of radius a relative to another oppositely
charged sphere of radius a
d
a
a
d
C = 40a (1+m+m2+m3+m4+…..)
m= a/d
d >> a
If d gets very large, then C= 10 pF
d =20 cm
a =10 cm
m =0.5
C=10-10(.1) (1+.5 +.25 +.125….)
C=10-10(.1)(1/(1-m))
C= 0.2 x 10-10 F
C= 0.02 nF =20 pF
16
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.
At some point during the charging, we
have a charge q on the positive plate.
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.
The potential difference between the plates is V 
dU  Vdq 
q
dq.
C
The total potential energy increase is
Q
q
q2
U   dq 
C
2C
0
Also
Q2

2C
2
1
1
1
Q
U  QV  CV 2 
2
2
2 C
using C 
17
Q
V
Graphical interpretation of integration
dU  Vdq
V
V = q/c
q/c
dq
q
Q
U   Vdq where V  q
C
 0
Q
1 N
U   qiqi = Area under the triangle
C i 1
Q
q
q2
0 C dq  2
Q
0
Q2

2C
Q
Area under the triangle is the value of the integral
1
Area of the triangle is also = 2 b  h
q
0 C dq
1
1
Q 1 Q2
Area = (b)(h)  (Q)( ) 
2
2
C 2 C
18
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

 Q and
V  ES
E
 
 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
  E 2
geometries.
To get total energy you need to integrate over volume.
19
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  100 Vm  102
1
U   0 E 2  Volume
2
Volume  4R 2 h
Volume  4 (6 106 ) 2 (2 104 )  8.6 1018 m3
U
1
2 V
18
3
C2
(10 11 Nm
)(
8
.
6

10
m
)
2 )(10
m
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  2  1016 J s  2  1021 J day
World consumes
about 1018 J/day.
This is 1/2000 of
the solar flux.
U Usun  2 1010
Only an infinitesimal fraction gets converted to electricity.
20
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.
C
Q1
Q
V
Q2
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  Q 2  C1V  C 2V  (C1  C 2)V
Q
 C1  C 2  C  C1  C 2
V
21
Series Combination of Capacitors
Q
V1
Q
V2
Q
V
Q
V
C
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:
Q Q
1
1
1

 Q(  )  Q( )
C1 C 2
C1 C 2
C
1
1
1
C1C 2
Therefore,
 
C
C1 C1 C 2
C1  C 2
V  V1 V 2 
So
22
Sample problem
C1 = 10 F
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
10  5 50
C12 

 3.3F
10  5 15
C12 and C3 are in parallel.
Ceq  C12  C 3  3.3  4.0  7.3F
23
Sample problem (continued)
C1 = 10 F
C2 = 5.0 F
C3 = 4.0 F
b) If V = 100 volts, what is the charge Q3 on C3?
C = Q/V
Q3  C 3V  4.0 106 100
Q3  4.0 104 Coulombs
c) What is the total energy stored in the circuit?
U
1
1
CeqV 2   7.3  10 6 F  10 4V 2  3.6  10 2 J
2
2
U  3.6 10 2 J
24