Download Lecture 5 Capacitance

Friday July 11
8:30 AM 9:0 AM Catch up
Lecture 3
Slide 5 Electron projected in electric field problem
Chapter 23 Problem 29 Cylindrical shell problem surrounding wire
Show Faraday Ice Pail no chrage inside shell
Lecture 4
Slide 34 Equipotential surfaces,
Sharp points
Total energy of charge distribution of charge
9:00 AM Lord Kelvin Water drop Generator explanation Lecture 5 Capacitance
Reminder Put Microphone On
1
Example: An electron is projected perpendicularly to a downward electric
field of E= 2000 N/C with a horizontal velocity v=106 m/s. How much is the
electron deflected after traveling 1 cm.
V •e
d
E
E
2
Chapter 23 Problem 29
3
Kelvin Water Drop Generator
Electrons follow red path as
the rings and the cans become
increasingly polarized through
induction
4
The figure belows shows a thin plastic rod of length L and uniform positive charge
Q lying on an x axis. Find the electric field at point P1 on the axis, at distance d from
one end of the rod. Find the x and y components.
x0
L
V  k 
0
Ex  
dx 
Lx
 k  ln(
)
x  x
x
dV
L
1
 k
 kQ
dx
(x)(L  x)
(x)(L  x)
x0
L
V  k 
0
Ex  
x
0
dx 
Lx
 k  ln(
)
x  x
x
dV
L
1
 k
 kQ
dx
(x)(L  x)
(x)(L  x)
5
6
7
Lecture 5 Capacitance Ch. 25
•Cartoon - Capacitance definition and examples.
•Opening Demo - Discharge a capacitor
•Topics
•Capacitance - What is it?
•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
• Isolated sphere capacitor
• Circular parallel plate capacitor
•Cylindrical capacitor
•Dielectric Slab sliding into demo
• Elmo Problems
•Polling
8
What is Capacitance?
Capacitance is a characteristic of a single isolated conducting object
or a pair of conducting objects or even three objects. To keep it simple, suppose I
have sphere and I put some charge on it say an amount
q. Then it will have some voltage V. If now I double the q,
the voltage will double. The point is that the ratio q/V is constant
and it is called the capacitance. C= q/V. This will be true for any combination of
objects. Even if there is no charge on the object it still has a capacitance as you will
see.
To help you understand capacitance, consider the isolated sphere
again in more detail. If I put a chare q on the isolated sphere, the
potential at any point r where r>R, the radius of the sphere, is given by
the potential for a point charge.
9
q0
+
Isolated conducting sphere with total charge uniformly
distributed on its surface. Recall we said that the potential for r
> R is given by a point charge q at the center
q
At r > R
+
+
+
At r  R
V=
kq
r
V=
kq
R
R
+
+
+
+
W  q0V
W
V
q0
q R
C 
V k
+
10
Isolated conducting sphere Continued
q
R
R
C   4 0 R
k
1
 4 0
k
Units of Capacitance is 1F =1 Farad = 1 C/Volt
k = 8.99 X 109 Nm2/C2 and R is in
meters we have
C
R
10
12

10
R(m)

10
R(cm)
10
10
q
 4 R
V
 0  8.85  10 12
F
m
Earth: C = (6x108 cm)pF = 600 F
Conducting Marble: 1 pF
You: 30 pF
C  R(cm)pF
11
How does the capacitance of the charged conducting sphere
change when another neutral sphere is nearby.
q0
q
d
a
W  q0V
W
V
q0
+
a
d
+
+
+
+
+
+
Notice the redistribution of charge on both spheres. In effect
the potential V of the original sphere is lowered for the same
amount of charge. This means its effective capacitance has increased
in the presence of the neutral sphere. Try to demonstrate it.
12
What is the capacitance of two charged conducting spheres?
One of charge +q and one of charge - q.
q
d
a
a
d
C = q/V = 40a (1/(1-m))
d =20 cm
a =10 cm
m = a/d = 0.5
C=10-10(.1)(1/(1-m))
C= 0.2 x 10-10 F
C= 0.02 nF =20 pF
The effective capacitance increases
for the paired conductor case and that
is why we make capacitors the way
we do.
If d gets very large, then C= 10 pF
13
for the isolated sphere.
Two isolated conducting objects such as two
parallel plates
1
Qi  CijV j
 a11
C
 a21
-Q
a12 
a22 
2
+Q
Q1  a11V1  a12V2
Q2  a21V1  a22V2
 b b 
C
 b b 
Q1  bV
Q2  bV
V  V2  V1
14
Capacitance
The most widely used capacitor 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
.
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
15
Parallel Plate Capacitor
16
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
Area of plate =A
q
0 A
f
Vf  Vi    E  dr
i
Integrate from - charge to + charge so that
E  dr  Edr

V f  Vi    Edr  Ed

Coulomb/Volt = Farad
V  Ed
17
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
 0  8.85  10 12
11 C 2
Nm 2
.03m 2
)
.001m
F
m
Coulomb
Volt
Farad
C  3  10 10 F
C  300pF
p = pico = 10-12
18
Demo Continued
Connect Battery to aluminum plate to charge it up with plates close together.
Connect voltmeter to plates.
Disconnect battery and move plates apart and watch voltage increase
Recall
C
0A
d

q
V
1.
As d increases, C decreases and V increases because q is constant.
2.
If you leave battery connected, V is constant and q decreases.
Demonstrate that a piece of wire has capacitance by touching electroscope.
19
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.
20
Induced dipoles
Permanent dipoles
_ ++ _
E0 = the applied field
E’ = the field due to
induced dipoles
E = E0 - E’
21
Dielectrics
The amount that the field is reduced defines the dielectric constant  from the
E
formula
the dielectric.
E0

, where E is the new field and E0 is the old field without
Since the electric field is reduce, the voltage difference is reduced, V 
and the capacitance is increased.
C
V0

Q
Q

  C0
V V0 
where  is typically between 2 – 6 with water equal to 80. See pg 669 for table of
dielectric constants
Repeat demo with dielectric slab sliding in between plates. Watch how
capacitance and voltage change. Also show aluminum slab.
22
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
23
Gauss’s Law for
Dielectrics
Induced charge
r r
0 —
 E  dA  q  q'
r r
0 —
  E  dA  q
Induced charge factored into kappa
24
Model of coaxial cable for calculation of capacitance
Outer metal braid
Signal wire
- to +
25
Find the capacitance of a ordinary piece of coaxial cable (TV cable)
f
Vf  Vi    E  dr
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. dr   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

26
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)
27
Capacitance of two concentric spherical shells
a
a
b
b
V  Va  Vb    E  dr    Edr
dr
Integration path
-q
+q
b
as E. dr  Eds cos180  E(dr)
a
a
a
a
kq
dr
Va  Vb    Edr    2 dr  kq  2
r
r
b
b
b
1
1 1
ba
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
28
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 
29
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
30
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.
1
1
1
U  QV  ( AE)(Ed)   E 2 (Ad)
2
2
2
volume occupied by E
U
1
 E2  
Ad 2
1
2
  E 2
We are now
including dielectric
effects: 
Electrostatic energy density general result for all
geometries.
To get total energy you need to integrate over volume.
31
How much energy is stored in the Earth’s
atmospheric electric field?
(Order of magnitude estimate)
E  100 Vm  102
1
U   0 E 2  Volume
2
atmosphere
20 km
h
Volume  4R 2 h
Earth
R
Volume  4 (6 106 ) 2 (2 104 )  8.6 1018 m3
U
R = 6x106 m
1 11 C 2
(10 Nm2 )(10 2 Vm )2 (8.6  1018 m 3 )
2
U  4.3 1011 J
The total solar influx is 200 Watts/m2
Usun  200  3.14(6  106 )2  2  1016 J s  2  1021 J day
Only an infinitesimal fraction gets converted to electricity.
32
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.
+Q1
+Q2
Ceq
-Q1
Ceq 
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
 C 1  C 2  Ceq  C 1  C 2
V
33
Series Combination of Capacitors
2
1
+Q
-Q +Q
V1
eq
-Q
V2
Q
V
Q
V
Ceq
Ceq 
What is the equivalent capacitor Ceq?
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
Ceq
1
1
1
C 1C 2
So
Therefore,


Ceq 
Ceq C1 C 2
C1  C 2
V  V1  V 2 
34
Sample problem
C1 = 10 F
C2 = 5.0 F
C3 = 4.0 F
a) Find the equivalent capacitance of the three capacitors
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
35
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
36
Chapter 25 Problem 28
In Fig. 25-28, a potential difference V = 100 V is applied across a
capacitor arrangement with capacitances C1 = 10.0 µF, C2 = 5.00 µF,
5 uF
and C3 = 15.0 µF. Find the following values.
(a) the charge on capacitor 3
10 uF
100 V
15 uF
(b) the potential difference across capacitor 3
(c) the stored energy for capacitor 3
37
5 uF
(d) the charge on capacitor 1
(e) the potential difference across capacitor 1
(f) the stored energy for capacitor 1
10 uF
100 V
15 uF
(g) the charge on capacitor 2
(h) the potential difference across capacitor 2
(i) the stored energy for capacitor 2
38
Chapter 25 Problem 34
An air-filled parallel-plate capacitor has a capacitance of 1.4 pF.
The separation of the plates is doubled and wax is inserted
between them. The new capacitance is 2.3 pF. Find the dielectric
constant of the wax.
39
Chapter 25 Problem 40
You are asked to construct a capacitor having a capacitance near 1 nF
and a breakdown potential in excess of 10,000 V. You think of using the
sides of a tall Pyrex drinking glass as a dielectric, lining the inside and
outside curved surfaces with aluminum foil to act as the plates. The
glass is 17 cm tall with an inner radius of 3.1 cm and an outer radius of
3.5 cm. The dielectric strength is 14 kV/mm.
(a) What is the capacitance?
(b) What is the breakdown potential of this capacitor?
40