Chapter 23: Electrostatic Energy and Capacitance Capacitors and Capacitance Capacitor • Any two conductors separated by an insulator (or a vacuum) form a capacitor • In practice each conductor initially has zero net charge and electrons are transferred from one conductor to the other (charging the conductor) • Then two conductors have charge with equal magnitude and opposite sign, although the net charge is still zero • When a capacitor has or stores charge Q , the conductor with the higher potential has charge +Q and the other -Q if Q>0
Transcript
Chapter 23: Electrostatic Energy and Capacitance
Capacitors and CapacitanceCapacitor• Any two conductors separated by an insulator (or a vacuum) form acapacitor
• In practice each conductor initially has zero net charge and electronsare transferred from one conductor to the other (charging the conductor)
• Then two conductors have charge with equalmagnitude and opposite sign, although the netcharge is still zero
• When a capacitor has or stores charge Q , the conductorwith the higher potential has charge +Q and the other-Q if Q>0
Capacitors and CapacitanceCapacitance• One way to charge a capacitor is to connect these conductors to oppositeterminals of a battery, which gives a fixed potential difference Vab betweenconductors ( a-side for positive charge and b-side for negative charge). Thenonce the charge Q and –Q are established, the battery is disconnected.
• If the magnitude of the charge Q is doubled, the electric field becomestwice stronger and Vab is twice larger.
• Then the ratio Q/Vab is still constant and it is called the capacitance C.
Q-Q
ltcoulomb/vo 1C/V 1 farad 1F 1units ====abV
QC
• When a capacitor has or stores charge Q , the conductorwith the higher potential has charge +Q and the other-Q if Q>0
Calculating CapacitanceParallel-plate capacitor in vacuum
• Charge density:AQ
=σ
• Electric field:A
QE00 εε
σ==
AQdEdVab
0
1ε
==• Potential diff.:
• Capacitance:dA
VQCab
0ε==
• The capacitance depends only on thegeometry of the capacitor.
• It is proportional to the area A.• It is inversely proportional to the separation d• When matter is present between the plates, itsproperties affect the capacitance.
-Q+Q
+Q
-Q
∫ ∫ =⋅=⋅−=−=a
b
b
abaab EddEdEVVV lrr
lrr
Calculating CapacitanceUnits
1 F = 1 C2/N m (Note [ε0]=C2/N m2) 1 µF = 10-6 F, 1 pF = 10-12 Fε0 = 8.85 x 10-12 F/m
Example 24.1: Size of a 1-F capacitor
F 0.1 , mm 1 == Cd
2812
3
0
m 101.1F/m 1085.8
m) 100.1F)(0.1(×=
××
== −
−
εCdA
Calculating CapacitanceExample 24.2: Properties of a parallel capacitor
kV 10.0 V 000,10 , m 00.2 , mm 00.5in vacuumcapacitor palte-parallelA
2 ==== VAd
F 0.00354F 1054.3m1000.5
)m F/m)(2.00 1085.8(
5
3
212
0
µ
ε
=×=
××
==
−
−
−
dAC
C 35.4C1054.3
V) 10C/V)(1.00 1054.3(5
49
µ=×=
××==−
−abCVQ
N/C 1000.2
)m 00.2)(mN/C 1085.8(C 1054.3
6
22212
5
00
×=
⋅××
=== −
−
AQE
εεσ
Calculating CapacitanceExample 24.3: A spherical capacitor
rarb
-
--
--
-
-
-
+ ++
+ +
+
+
+
-Q
Qr
From Gauss’s law: ∫ =⋅0ε
enclQAdErr
surfaceGaussian a as sphere aon point every at toparallel and magnitudein constant is AdErr
200
2
4)4(
rQEQrE
πεεπ =→=
This form is the same as that for a point charge
rQV
04πε=
ba
ab
babaab rr
rrQr
Qr
QVVV −=−=−=
000 444 πεπεπε
ab
ba
ab rrrr
VQC
−== 04πε
Calculating CapacitanceExample 24.4: A cylindrical capacitor (length L)
Signal wire
Outer metal braid
r r
line charge density λ
Q -Q
23.10 Example from ln2
0
0 rrV
πελ
=
a
b
a
bab
rrL
rr
LVQC
ln
2
ln2
0
0
πε
πελ
λ===
Capacitors in Series and ParallelCapacitors in series
Capacitors in Series and ParallelCapacitors in series (cont’d)
a
b
cVVab =1VVac =
2VVcb =
Q+
Q+Q−
Q−
1C
2C
22
11 C
QVVCQVV cbac ====
⎟⎟⎠
⎞⎜⎜⎝
⎛+=+==
2121
11CC
QVVVVab
21
11CCQ
V+=
The equivalent capacitance Ceq of the series combination is defined asthe capacitance of a single capacitor for which the charge Q is the sameas for the combination, when the potential difference V is the same.
∑=⇒+=→==i
ieqeqeqeq CCCCCQ
VCV
QC 111111
21
Capacitors in Series and ParallelCapacitors in parallel
VCQVCQ 2211 ==a
b
VVab = 1Q1C 2Q VCCQQQ )( 2121 +=+=
2C
21 CCVQ
+=
The parallel combination is equivalent to a single capacitor with thesame total charge Q=Q1+Q2 and potential difference.
∑=⇒+=i ieqeq CCCCC 21
Capacitors in Series and ParallelCapacitor networks
Capacitors in Series and ParallelCapacitor networks (cont’d)
Capacitors in Series and ParallelCapacitor networks 2
C C C CCA
B
A
B
CCC
CC
C C31
C
CC C
A
B
C C
C C
C C34
AC
4115
B
Energy Storage and Electric-field Energy
Work done to charge a capacitor• Consider a process to charge a capacitor up to Q with the final potentialdifference V.
CQV =
• Let q and v be the charge and potential difference at an intermediate stageduring the charging process.
Cq
=υ
• At this stage the work dW required to transfer an additional element of charge dq is:
CqdqdqdW ==υ
• The total work needed to increase the capacitor charge q from zero to Q is:
∫∫ ===QW
CQqdq
CdWW
0
2
0 21
Energy Storage and Electric-field Energy
Potential energy of a charged capacitor
• Define the potential energy of an uncharged capacitor to be zero.
• Then W in the previous slide is equal to the potential energy U ofthe charged capacitor
QVCVC
QU21
21
22
2
===
The total work W required to charge the capacitor is equal to the totalcharge Q multiplied by the average potential difference (1/2)V duringthe charging process
Energy Storage and Electric-field Energy
Electric-field energy
• We can think of the above energy stored in the field in the region betweenthe plates.
• Define the energy density u to be the energy per unit volume
20
2
212
1
EAd
CVu ε==
dAC 0ε
=
field volume
This relation works for any electric field
Energy Storage and Electric-field Energy
Example 24.9: Two ways to calculate energy stored• Consider the spherical capacitor in Example 24.3.
• The energy stored in this capacitor is:ab
ba
rrrrC
−= 04πε
ba
ab
rrrrQ
CQU −
==0
22
82 πε
204 r
QEπε
=• The electric field between two conducting sphere:
• The electric field inside the inner sphere is zero.
• The electric field outside the inner surface of the outer sphere is zero.
40
2
22
20
02
0 32421
21
rQ
rQEu
εππεεε =⎟⎟
⎠
⎞⎜⎜⎝
⎛==
∫∫∫−
==⎟⎟⎠
⎞⎜⎜⎝
⎛== b
a
b
a
r
rba
abr
r rrrrQ
rdrQdrr
rQudVU
0
2
20
22
40
2
2
884
32 πεπεπ
επ
Energy Storage and Electric-field Energy
Example : Stored energy
RQdV
rQU
EurQE
R )4(241
21
21
41
0
2
4
22
00
202
0
πεπεε
επε
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
=→=
∫∞
Dielectrics
Dielectric materials• Experimentally it is found that when a non-conducting material(dielectrics) between the conducting plates of a capacitor, thecapacitance increases for the same stored charge Q.
• Define the dielectric constant κ (= K in the textbook) as:
0CC
=κ
• When the charge is constant, VVCCCVVCQ // 0000 =→==
κκ00 EEVV =→=
Material κ κMaterialvacuum 1air(1 atm) 1.00059Teflon 2.1Polyethelene 2.25
Mica 3-6Mylar 3.1Plexiglas 3.40Water 80.4
Dielectrics
Induced charge and polarization• Consider a two oppositely charged parallel plates with vacuumbetween the plates.
• Now insert a dielectric material of dielectric constant κ.constant is Qwhen /0 κEE =
• Source of change in the electric field is redistribution of positiveand negative charge within the dielectric material (net charge 0).This redistribution is called a polarization and it produces inducedcharge and field that partially cancels the original electric field.
κεσσ
εσ 0
000
EEEE ind =−
==
0ty permittivi thedefine and 11 κεεκ
σσ =⎟⎠⎞
⎜⎝⎛ −=ind
εσ
=E 22000 2
121 EEu
dA
dACC εκεεκεκ =====
Dielectrics
Molecular model of induced charge
Dielectrics
Molecular model of induced charge (cont’d)
Dielectrics
Why salt dissolves
Normally NaCl is in a rigid crystal structure, maintained by the electrostatic attraction between the Na+ and Cl- ions.
Water has a very high dielectric constant (78). This reduces the field between the atoms, hence their attraction to each other. The crystal lattice comes apart and dissolves.
Dielectrics
Gauss’s law in dielectrics
cond
ucto
r
diel
ectri
c
++
+++
-
-
σ
indσ−
Gauss’s law:0
)(εσσ AEA ind−
=
κσσσ
κσσ =−⎟
⎠⎞
⎜⎝⎛ −= indind or 11
00
or εσκ
κεσ AEAAEA ==
∫ −=⋅0ε
κ freeenclQAdErr enclosed free
charge
ExercisesProblem 1
(a) If the distance d is halved, how much doesthe capacitance changes?
(b) If the area is doubled, how much does the capacitance changes?(c) For a given stored charge Q, to double the amount of energy stored
how much should the distance d be changed?Now a metal slab of thickness a (< d) and of the same area A is insertedbetween the two plates in parallel to the plates as shown in the figure(the slab does not touch the plates). (d) What is the capacitance of this arrangement?(hint:serial connection)(e) Express the capacitance as a multiple of the capacitance C0 when
the metal slab is not present.
aAn air capacitor is made by using two flat plateseach with area A separated by a distance d. d
Problem 1 Solution(a) doubled. is C so ,0 d
AC ε=
(b) doubled. is C so ,0 dAC ε=
doubled. be should d and 2
so ,2
and 0
22
0 AdQU
CQU
dAC
εε ===(c)
(d)
adACCC
adAC
ad
eqeq −=⇒=
−=
−
0eq
0
/2/1 :is C ecapacitanc equivalent the
Therefore .2 ecapacitanc thehascapacitor two theseofEach
plates. ebetween th 2/)( of gap a has which ofeach series,in connectedcapacitorstwoofsystemabetoconsideredbecan t arrangemen This
ε
ε
000 therefore, Cad
dCdAC eq −
== ε(e)
Problem 2In this problem you try to measure dielectric constant of a material. Firsta parallel-plate capacitor with only air between the plates is charged byconnecting it to a battery. The capacitor is then disconnected from thebattery without any of the charge leaving the plates.
(a) Express the capacitance C0 in terms of the potential difference V0between the plates and the charge Q if air is between the plates.
(b) Express the dielectric constant κ in terms of the capacitance C0 (air gap)and the capacitance C with material of the dielectric constant κ).
(c) Using the results of (a) and (b), express the ratio of the potentialdifference V/V0 if Q is the same, where V is the potential differencebetween the plates and a dielectric material dielectric constant is κfills the space between them.
(d) A voltmeter reads 45.0 V when placed across the capacitor. Whendielectric material is inserted completely filling the space, the voltmeterreads 11.5 V. Find the dielectric constant of this material.
(e) What is the voltmeter read if the dielectric is now pulled partway out sothat it fills only one-third of the space between the plates?(Use the formula for the parallel connection of two capacitors.)
Problem 2
00 /VQC =
0/ CC=(a)
κ(b)
(c) κκ /1/// 000 =→== VVCCVV(d) From (c) 91.35.11/0.45/0 === VVκ(e) In the new configuration the equivalent capacitor is
where C1/3 is the contribution from the part that has the dielectric materialand C0,2/3 is the part that has air gap.because the capacitance is proportional to the area.
