1 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
Transcript
1
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
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CapacitanceDefinition 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 .
The SI unit is , 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
VQC =
VQ
3
Parallel Plate Capacitor
4
Electric Field of Parallel Plate Capacitor
EA =qε0
EAq 0ε=Aq
E0ε
=
V =Ed=qdε0A
Area of plate =A
+ + + + + + + + + + +
- - - - - - - - - - -
E
+ q
- q
C =qV
=q
qdε0A
Coulomb/Volt = Farad
dA
Gaussian surface
C =ε0Ad
Vf −Vi =−
rE ⋅dr̂
i
f
∫Integrate from - charge to + charge so that
rE ⋅dr̂ =−Edr
Vf −Vi =+ Edr−
+
∫ =Ed
V =Ed
5
Circular parallel plate capacitor
r
r
d
r = 10 cm = 0.1m
A = r2 = (.1m)2
A = .03 m 2
d = 1 mm = .001 m
C =ε 0Ad
F103C 10−×=
pF300C = p = pico = 10-12
Show Demo Model, calculate its capacitance, and show how to charge it up with a battery.
mmC
NmC
001.03.)10(
211
2
2−= }Farad Volt
Coulomb
6
Demo Continued
Demonstrate
1. As d increases, voltage increases.
2. As d increases, capacitance decreases.
3. As d increases, E0 and q are constant.
7
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.
8
Permanent dipoles Induced dipoles
++_ _
E0 = the applied field
E’ = the field due to
induced dipoles
E = E0 - E’
9
Dielectrics
The amount that the field is reduced defines the dielectric constant from
the formula , 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 ), the capacitance is increased.
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.
0E
E =
VdE =
00
CVQ
VQC
=⎟⎠
⎞⎜⎝
⎛==
10
VqC =
V =E0d
00
εσ
=E
Aq=σ
A
qE
00
ε=
V =qdε 0A
C =ε 0Ad
0E
E =
V =E0
d
0V
V =
VqC =
0V
qC
=
0CC =
d
11
Find the capacitance of a ordinary piece of coaxial cable (TV cable)
€
Er =2kλ
r
Va −Vb =−
rE. dr̂
b
a
∫ = Edrb
a
∫ =2kλdrrb
a
∫ =+2kλ lnrb
a
Va is higher than Vb
LQ=λ
€
k =1
4πε 0 air
Va −Vb =2kλ lnba
rE. dr̂ =Edrcos180 =−Edr
r̂ Vf −Vi =−
rE ⋅dr̂
i
f
∫
Integrate from b to a or - to +
12
capacitance of a coaxial cable cont.
abln
L2QV ,So
0ε=
ab
0
lnQL2Q
VQC ε==
ab
0
lnL2C ε=
ab
0
ln2
LC ε=
38.1106
4ln106
LC 1111 −− ×=×=
mpF
LC 43=
mpF
LC 86=
ε0 (for air)
For = 2
a = 0.5 mm
b = 2.0 mm
2
Now if a=0.5mm and b=2.0mm, then
And if = 2, then
13
Model of coaxial cable for calculation of capacitance
Signal wire
Outer metal braid
- to +
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Capacitance of two concentric spherical shells
Integration path
Va−Vb =+ Edrb
a
∫ =+kqr2
b
a
∫ dr =+kqdrr2
b
a
∫
C =4ε0a
rE. dr̂ =Edscos180 =−Edr
ab)ab
(kq)b1
a1(kq
r1kqV
a
b
−=−==
dr
b
a
+q
-q
E
V =Va−Vb =−
rE ⋅dr̂
b
a
∫ =+ Edrb
a
∫
as
Let b get very large. Then for an isolated sphere
C =q /V =4ε0
abb−a
15
Spherical capacitor or sphere
Recall our favorite example for E and V is spherical symmetry
The potential of a charged sphere is with V = 0 at r = .
The capacitance is
Rk
R
RkQ
Q
V
QC 04ε====
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
1010=10−10R(m) =10−12R(cm)
pFcmRC )(=
Earth: C = (6x108 cm)pF = 600 F
Marble: 1 pF
Basketball: 15 pF
You: 30 pFDemo: Show how you measured capacitance of electroscope
RkQV =
Q
R
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Capacitance of one charged conducting sphere of radius a relative to another oppositely charged sphere of radius a
d
C = 4ε0a (1+m+m2+m3+m4+…..)
m= a/d
d >>a
a a
d
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 pFIf d gets very large, then C= 10 pF
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Electric Potential Energy of CapacitorAs 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.
The potential difference between the plates is
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.
.dqCq
VdqdU ==
The total potential energy increase is
C
Q
C
qdqC
qU
Q
22
22
0
===∫
Also C
QCVQVU
22
2
1
2
1
2
1=== using
VQC =
Cq
V =
18
Graphical interpretation of integration
∫=Q
VdqU0
where
Qdq
q/c
q
V
V = q/c
∑=
Δ=N
i
ii qqC
U1
1= Area under the triangle
Area under the triangle is the value of the integral dqC
qQ
∫0
Area of the triangle is also =
€
dU =Vdq
€
Area =1
2(b)(h) =
1
2(Q)(
Q
C) =
1
2
Q2
C
Cq
V =
hb⋅21
q
Cdq =
q2
2 0
Q
0
Q
∫ =Q2
2C
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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
A
QE
εεσ
εσ
===0
and We are now including dielectric effects: ε
Solve for Q = εAE, V = ES and substitute in
)(2
1))((
2
1
2
1 2 ASEESAEQVU εε ===
volume occupied by Eηε == 2
2
1E
AS
U
2
2
1Eεη =
Electrostatic energy density general result for all geometries.
To get total energy you need to integrate over volume.
ESV =
20
How much energy is stored in the Earth’s atmospheric electric field?(Order of magnitude estimate)
atmosphere
h
R
Earth
20 km
R = 6x106 m
210100 == mVE
VolumeEU ⋅= 20
21 ε
hRVolume 24=318426 106.8)102()106(4 mVolume ×=××=
)106.8)(10)(10(21 318211
2
2
mU mV
NmC ×= −
JU 11103.4 ×=This energy is renewed daily by the sun. Is this a lot?
The total solar influx is 200 Watts/m2
dayJsJUsun211626 102102)106(14.3200 ×=×≅×⋅=
10102 −×≅sunUUOnly an infinitesimal fraction gets converted to electricity.
World consumes about 1018 J/day. This is 1/2000 of the solar flux.
21
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.
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.
VCCVCVCQQQ )( 212121 +=+=+=
2121 CCCCCV
Q+=⇒+=
VQC =
Q1Q2
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Series Combination of Capacitors
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:
)1
()11
(2121
21C
QCC
QC
Q
C
QVVV =+=+=+=
So 211
111
CCC+= Therefore,
21
21
CC
CCC
+=
Q QV1 V2
CQV
VQC
=
=
23
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.
21
2112
2112
111
CC
CCC
CCC +=⇒+=
FC 3.31550
510510
12 ==+×
=
C12 and C3 are in parallel.
FCCCeq 3.70.43.3312 =+=+=
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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
100100.4 633 ⋅×== −VCQ
c) What is the total energy stored in the circuit?
