Transcript Chapter 21: Electric Charge and Electric Field

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
Capacitors and Capacitance
 Capacitance
• One way to charge a capacitor is to connect these conductors to opposite
terminals of a battery, which gives a fixed potential difference Vab between
conductors ( a-side for positive charge and b-side for negative charge). Then
once the charge Q and –Q are established, the battery is disconnected.
• If the magnitude of the charge Q is doubled, the electric field becomes
twice stronger and Vab is twice larger.
• Then the ratio Q/Vab is still constant and it is called the capacitance C.
-Q
C
Q
Vab
units 1 F  1 farad  1 C/V  1 coulomb/volt
• When a capacitor has or stores charge Q , the conductor
with the higher potential has charge +Q and the other
-Q if Q>0
Q
Calculating Capacitance
 Parallel-plate
• Charge density:
• Electric field:
• Potential diff.:
capacitor in vacuum
Q
A

Q
E 
0 0 A

Vab  Ed 


a 
b 
Vab  Va  Vb   E  d    E  d   Ed
b
• Capacitance:
1 Qd
0 A
a
Q
A
C
 0
Vab
d
• The capacitance depends only on the
geometry 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, its
properties affect the capacitance.
+Q
-Q
+Q
-Q
Calculating Capacitance
 Units
1 F = 1 C2/N m (Note [0]C2/N m2)
0 = 8.85 x 10-12 F/m
 Example
1 mF = 10-6 F, 1 pF = 10-12 F
24.1: Size of a 1-F capacitor
d  1 mm , C  1.0 F
(1.0 F)(1.0 103 m)
8
2
A

 1.110 m
12
0
8.8510 F/m
Cd
Calculating Capacitance
 Example
24.2: Properties of a parallel capacitor
A parallel- paltecapacitorin vacuum
d  5.00 mm , A  2.00 m 2 , V  10,000 V  10.0kV
A (8.851012 F/m)(2.00m 2 )
C  0 
d
5.00103 m
 3.54105 F  0.00354mF
Q  CVab  (3.54109 C/V)(1.00104 V)
 3.54105 C  35.4mC

Q
3.54105 C
E


 0  0 A (8.851012 C 2 / N  m 2 )(2.00 m 2 )
 2.00106 N/C
Calculating Capacitance
 Example
-Q
-
-
+
+ ra +
+
+
r ++ +
Q
-
24.3: A spherical capacitor
  Qencl
From Gauss’s law:
 E  dA   0
-


E is constantin magnitudeand parallelto dA
rb-
at everypointon a sphereas a Gaussian surface
E (4 r 2 ) 
Q
0
E
Q
4 0 r 2
This form is the same as that for a point charge
Q
V
4 0 r
Vab  Va  Vb 
C
Q
4 0 ra
rr
Q
 4 0 a b
Vab
rb  ra

Q
4 0 rb

Q rb  ra
4 0 ra rb
Calculating Capacitance
 Example
24.4: A cylindrical capacitor (length L)
Q
-Q
Outer metal braid
r
r
V
r
l
ln 0 fromExample23.10
2 0 r
C
Q

Vab
lL
r
l
ln b
2 0 ra
Signal wire
line charge density l

2 0 L
r
ln b
ra
Capacitors in Series and Parallel

Capacitors in series
Capacitors in Series and Parallel

Capacitors in series (cont’d)
a
Q
Vab  V
Q
c
Q
Q
C1
Vac  V1
C2
Vcb  V2
Q
Q
Vac  V1 
Vcb  V2 
C1
C2
1
1 

Vab  V  V1  V2  Q  
 C1 C2 
V
1
1
 
Q C1 C2
b
The equivalent capacitance Ceq of the series combination is defined as
the capacitance of a single capacitor for which the charge Q is the same
as for the combination, when the potential difference V is the same.
Ceq 
Q
V
1 V
1
1
1
1
1
 
 

 i
Ceq Q
Ceq C1 C2
Ceq
Ci
Capacitors in Series and Parallel

Capacitors in parallel
Q1  C1V
a
Vab  V
b
Q1
C1 Q2
C2
Q2  C2V
Q  Q1  Q2  (C1  C2 )V
Q
 C1  C2
V
The parallel combination is equivalent to a single capacitor with the
same total charge Q=Q1+Q2 and potential difference.
Ceq  C1  C2  Ceq  i Ci
Capacitors in Series and Parallel

Capacitor networks
Capacitors in Series and Parallel

Capacitor networks (cont’d)
Capacitors in Series and Parallel

Capacitor networks 2
C
C
C
C
C
A
A
C
C
C
C
C
1
C
3
B
B
C
C
C
C
C
C
C
A
A
15
C
41
4
C
3
C
B
B
C
C
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 potential
difference V.
Q
V
C
• Let q and v be the charge and potential difference at an intermediate stage
during the charging process.
q

C
• At this stage the work dW required to transfer an additional element of
charge dq is:
qdq
dW  dq 
C
• The total work needed to increase the capacitor charge q from zero to Q is:
W
W 
0
1 Q
Q2
dW   qdq 
C 0
2C
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 of
the charged capacitor
Q2 1
1
U
 CV 2  QV
2C 2
2
The total work W required to charge the capacitor is equal to the total
charge Q multiplied by the average potential difference (1/2)V during
the 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 between
the plates.
• Define the energy density u to be the energy per unit volume
C
0 A
d
1
CV 2
1
2
u
 0E2
Ad
2
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.
C  4 0
ra rb
rb  ra
• The energy stored in this capacitor is:
Q2
Q 2 rb  ra
U

2C 8 0 ra rb
• The electric field between two conducting sphere: E 
Q
4 0 r 2
• The electric field inside the inner sphere is zero.
• The electric field outside the inner surface of the outer sphere is zero.
2
1
1  Q 
Q2
2
 
u   0 E   0 
2
2  4 0 r 2 
32 2 0 r 4
 Q2
 2
Q2
4r dr 
U   udV   
ra 32 2 r 4 
8 0
0


rb

rb
ra
dr Q 2 rb  ra

2
r
8 0 ra rb
Energy Storage and Electric-field Energy

Example : Stored energy
E
1
Q
1
 u  0E2
2
4 0 r
2
2
1  1  Q2
Q2
 4 dV 
U    0 
R 2
2(4 0 ) R
 4 0  r

Dielectrics

Dielectric materials
• Experimentally it is found that when a non-conducting material
(dielectrics) between the conducting plates of a capacitor, the
capacitance increases for the same stored charge Q.
• Define the dielectric constant k ( K in the textbook) as:
k
C
C0
• When the charge is constant, Q  C0V0  CV  C / C0  V0 / V
V
E
V  0 E 0
k
k
Material
vacuum
air(1 atm)
Teflon
Polyethelene
k
1
1.00059
2.1
2.25
Material
Mica
Mylar
Plexiglas
Water
k
3-6
3.1
3.40
80.4
Dielectrics

Induced charge and polarization
• Consider a two oppositely charged parallel plates with vacuum
between the plates.
• Now insert a dielectric material of dielectric constant k.
E  E0 / k when Q is constant
• Source of change in the electric field is redistribution of positive
and negative charge within the dielectric material (net charge 0).
This redistribution is called a polarization and it produces induced
charge and field that partially cancels the original electric field.

E0 
0

   ind
E
0
E
E0
k
1
 ind   1   and define thepermittivity   k0
 k

E

C  kC0  k 0
A
A
1
1

u  k 0 E 2  E 2
d
d
2
2
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
Gauss’s law:
+ -   ind
+

++
dielectric
conductor
+
EA 
 ind
(   ind ) A
0

 1
  1   or    ind 
k
 k
A
A
EA 
or kEA 
k0
0
  Qencl free
 kE  dA 
0
enclosed free
charge
Exercises
Problem 1
An air capacitor is made by using two flat plates
d
a
each with area A separated by a distance d.
(a) If the distance d is halved, how much does
the 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 inserted
between 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.
Problem 1 Solution
A
, so C is doubled.
d
A
(b) C   0 , so C is doubled.
d
A
Q2
Q2d
(c) C   0 andU  , so U 
and d should be doubled.
d
2C
2 0 A
(a) C   0
(d) T hisarrangement can be consideredt o be a syst emof t wo capacit ors
connect edin series, each of which has a gap of (d  a ) / 2 bet ween t he plat es.
2A
Each of t hese t wo capacit orhas t hecapacit ance C   0
. T herefore
d a
A
t heequivalentcapacit ance C eq is :1 / Ceq  2 / C  Ceq   0
d a
(e) C   A , therefore C  d C
0
0
eq
0
d
d a
Problem 2
In this problem you try to measure dielectric constant of a material. First
a parallel-plate capacitor with only air between the plates is charged by
connecting it to a battery. The capacitor is then disconnected from the
battery without any of the charge leaving the plates.
(a) Express the capacitance C0 in terms of the potential difference V0
between the plates and the charge Q if air is between the plates.
(b) Express the dielectric constant k in terms of the capacitance C0 (air gap)
and the capacitance C with material of the dielectric constant k).
(c) Using the results of (a) and (b), express the ratio of the potential
difference V/V0 if Q is the same, where V is the potential difference
between the plates and a dielectric material dielectric constant is k
fills the space between them.
(d) A voltmeter reads 45.0 V when placed across the capacitor. When
dielectric material is inserted completely filling the space, the voltmeter
reads 11.5 V. Find the dielectric constant of this material.
(e) What is the voltmeter read if the dielectric is now pulled partway out so
that it fills only one-third of the space between the plates?
(Use the formula for the parallel connection of two capacitors.)
Problem 2
(a)
(b)
(c)
(d)
(e)
C0  Q /V0
k  C / C0
V0 / V  C / C0  k  V / V0  1/ k
From (c) k  V0 / V  45.0 / 11.5  3.91
In the new configuration the equivalent capacitor is Ceq  C1/ 3  C0, 2 / 3
where C1/3 is the contribution from the part that has the dielectric material
and C0,2/3 is the part that has air gap. C1/ 3  (1/ 3)C and C0,2 / 3  (2/3)C0
because the capacitance is proportional to the area.
Ceq  C1/ 3  C0,2 / 3  (1/ 3)C  (2 / 3)C0  C0[(1/ 3)k  (2 / 3)]
Using the results from (c)
V0 / V  Ceq / C0  [(1 / 3)k  (2 / 3)] 
 3 
V  V0 /[(1 / 3)k  (2 / 3)]  (45.0 V)
  22.8 V
 5.91