Transcript 1 farad Capacitor

Capacitors
Calculating capacitance
Energy stored in a capacitor
Capacitors with dielectric materials
25.2: Capacitance:
To store charge
To store energy
To control variation time
scales in a circuit
25.2: Capacitance:
When a capacitor is charged, its plates have charges of equal magnitudes but opposite signs: q+ and
q-. However, we refer to the charge of a capacitor as being q, the absolute value of these charges on
the plates.
The charge q and the potential difference V for a capacitor are proportional to each other:
The proportionality constant C is called the capacitance of the capacitor. Its value depends only on
the geometry of the plates and not on their charge or potential difference.
The SI unit is called the farad (F): 1 farad (1 F)= 1 coulomb per volt =1 C/V.
25.2: Charging a Capacitor:
The circuit shown is incomplete because switch S is open; that is, the switch does not electrically
connect the wires attached to it. When the switch is closed, electrically connecting those wires, the
circuit is complete and charge can then flow through the switch and the wires.
As the plates become oppositely charged, that potential difference increases until it equals the potential
difference V between the terminals of the battery. With the electric field zero, there is no further drive
of electrons. The capacitor is then said to be fully charged, with a potential difference V and charge q.
25.3: Calculating the Capacitance:
1 farad Capacitor
Example1: Given a 1 farad parallel plate capacitor
having a plate separation of 1mm. What is the area
of the plates?
A
C  0
d
A
Cd
0

1.0F 1.0  10 3 m 

8.85  10 12 F / m
 1.1  108 m 2
This corresponds to a square of about 10km on a side!
25.3: Calculating the Capacitance; A Cylindrical Capacitor :
As a Gaussian surface, we choose a cylinder of length L
and radius r, closed by end caps and placed as is shown. It
is coaxial with the cylinders and encloses the central
cylinder and thus also the charge q on that cylinder.
25.3: Calculating the Capacitance; A Spherical Capacitor:
25.3: Calculating the Capacitance; An Isolated Sphere:
We can assign a capacitance to a single isolated spherical conductor
of radius R by assuming that the “missing plate” is a conducting
sphere of infinite radius.
The field lines that leave the surface of a positively charged isolated
conductor must end somewhere; the walls of the room in which the
conductor is housed can serve effectively as our sphere of infinite
radius.
To find the capacitance of the conductor, we first rewrite the
capacitance as:
Now letting b→∞, and substituting R for a,
Example, Charging the Plates in a Parallel-Plate Capacitor:
25.4: Capacitors in Parallel:
25.4: Capacitors in Series:
Example, Capacitors in Parallel and in Series:
Example, Capacitors in Parallel and in Series:
Example, One Capacitor Charging up Another Capacitor:
25.5: Energy Stored in an Electric Field:
25.5: Energy Density:
Example, Potential Energy and Energy Density of an Electric Field:
26.5. Capacitors with Dielectrics
Dielectric is a non-conducting material, such as rubber, glass, or waxed paper.
When a dielectric is inserted between the plates of a capacitor, the capacitance
increases. If the dielectric completely fills the space between the plates, the
capacitance increases by a dimensionless factor k , which is called the dielectric
constant.
𝐶
𝐾=
𝐶0
26.6: Dielectrics, an Atomic View:
1. Polar dielectrics. The molecules of some dielectrics, like water, have permanent electric
dipole moments. In such materials (called polar dielectrics), the electric dipoles tend to
line up with an external electric field as in Fig. 25-14. Since the molecules are
continuously jostling each other as a result of their random thermal motion, this
alignment is not complete, but it becomes more complete as the magnitude of the applied
field is increased (or as the temperature, and thus the jostling, are decreased).The
alignment of the electric dipoles produces an electric field that is directed opposite the
applied field and is smaller in magnitude.
2. Nonpolar dielectrics. Regardless of whether they have permanent electric dipole
moments, molecules acquire dipole moments by induction when placed in an external
electric field. This occurs because the external field tends to “stretch” the molecules,
slightly separating the centers of negative and positive charge.
What Happens When You Insert a Dielectric?
 With battery attached, V=const, so
more charge flows to the capacitor
q  CV
q  CV

With battery disconnected, q=const, so
voltage (for given q) drops.
V
V 
q
C
q
C
October 10, 2007
What Does the Dielectric Do?
 A dielectric material is made of molecules.
 Polar dielectrics already have a dipole moment (like
the water molecule).
 Non-polar dielectrics are not naturally polar, but
actually stretch in an electric field, to become polar.
 The molecules of the dielectric align with the applied
electric field in a manner to oppose the electric field.
 This reduces the electric field, so that the net electric
field is less than it was for a given charge on the
plates.
 This lowers the potential (case b of the previous
slide).
 If the plates are attached to a battery (case a of the
previous slide), more charge has to flow onto the
October 10, 2007
𝐾=
𝐶0 =
𝐶=
𝐶
𝐶0
………1
𝑞
𝑞
=
𝑉0 𝐸0 d
…………2
𝑞
𝑞
=
𝑉 𝐸d
…………3
From 2 and 3
𝐶
𝑉0 𝐸0
=
=
=𝐾
𝐶0
𝑉
𝐸
……………4
From Gauss law
𝜎
𝜀0
𝜎
𝐸=
𝜀
𝐸0 =
…………5
… … … … … … . . .6
From 5 and 6
𝜀
𝐾=
𝜀0
…………………7
𝐶
𝑉0
0
𝑉
𝐾=𝐶 =
=
𝐸0
𝐸
𝜀
=𝜀
0
Dielectric and Gauss's Law
:
𝜙=
𝐸 𝑑𝐴 𝑐𝑜𝑠𝜃 =
𝑞𝑖𝑛
𝜀0
… … … … . .1
:
𝐸 = 𝐸0 − 𝐸𝑖
… … … … . .2
:
⟹
⟹ 𝐸0 =
𝐸0 𝑑𝐴 = 𝐸0 𝐴 =
𝑞
𝜎
=
𝜀0 𝐴 𝜀0
𝑞
𝜀0
… … … … . .3
𝑞
𝑞𝑖
⟹
𝐸 𝑑𝐴 𝑐𝑜𝑠𝜃 = 𝐸𝐴 =
𝐸=
⟹ 𝐸𝑖 =
𝑞 − 𝑞𝑖
𝜀0
𝑞
𝑞𝑖
−
𝜀0 𝐴 𝜀0 𝐴
𝑞𝑖
𝜎𝑖
=
𝜀0 𝐴 𝜀0
… … … … . .4
:
𝐾=
𝐸=
𝜀
𝜀0
𝑞
𝑞
=
𝜀𝐴 𝐾𝜀0 𝐴
… … … … . .5
:) (
⟹
𝑞
𝑞
𝑞𝑖
=
−
𝐾𝜀0 𝐴 𝜀0 𝐴 𝜀0 𝐴
… … … … . .6
⟹ 𝑞𝑖 = 𝑞 −
𝑞
1
=𝑞 1−
𝐾
𝐾
… … … … . .7
:
𝐸𝑐𝑜𝑠𝜃𝑑𝐴 =
𝑞 − 𝑞𝑖
𝜀0
𝐾𝐸𝑐𝑜𝑠𝜃𝑑𝐴 =
𝑜𝑟
𝐷𝑐𝑜𝑠𝜃𝑑𝐴 = 𝑞
𝑞
𝜀0
𝐾𝜀0 𝐸𝑐𝑜𝑠𝜃𝑑𝐴 = 𝑞
𝑜𝑟
𝐷 . 𝑑𝐴 = 𝑞
… … . .8
(electric displacement (
𝐷
:
𝐷 = 𝐾𝜀0 𝐸 = 𝜀𝐸
𝐶
𝑚2
𝑞
𝐷=
⟹ 𝑞 = 𝐷𝐴
𝐴
:D
𝐷 = 𝜀0 𝐸
… … … … . .9
Polarization and Electric Displacement
)polarization (
:
𝑝 = 𝑁𝑞𝑖 𝑑𝑙
… … … … . .10
𝑞𝑖
N
.
𝑞𝑖
𝑙
A
:
𝑝=
𝑞𝑖 𝑙
𝑞𝑖 𝑙 𝑞𝑖
= = 𝜎i
𝑙𝐴
A
𝐶
𝑚2
… … … … . .11
𝑙𝐴
:
) (
𝑞
𝑞
𝑞𝑖
=
+
𝜀0 𝐴 𝐾𝜀0 𝐴 𝜀0 𝐴
𝑞
𝑞
𝑞𝑖
= 𝜀0
+
𝐴
(𝐾𝜀0 𝐴) 𝐴
𝐷 = 𝜀0 𝐸 + 𝑃
p
.
… … … … . .12
D
E
( Electric Susceptibility (
:
)
(
D
𝐷 = 𝜀0 𝐸 + 𝑃
𝐾𝜀0 𝐸 = 𝜀0 𝐸 + 𝑃
𝑃 = 𝐾𝜀0 𝐸 − 𝜀0 𝐸
⟹ 𝑝 = 𝜀0 𝐸 𝐾 − 1 = 𝜀0 𝜒𝑒 𝐸
… … … … . .13
:
)
(
𝜎i = 𝜀0 𝜒𝑒 𝐸
… … … … . .14
𝜒𝑒
.
)
(
P
𝐷 = 𝜀0 𝐸 + 𝜀0 𝜒𝑒 𝐸
𝐷 = 𝜀0 𝐸 1 + 𝜒𝑒 = 𝜀0 𝐾𝐸
⟹ 𝜒𝑒 = 𝐾 − 1
𝐾=
𝜒𝑒 =
𝜀
𝜀0
𝜀
−1
𝜀0
… … … … . .15
. . . .16
… . .17
… . .18
:
𝑁𝑝2 𝐸
𝑝=
3𝐾𝐵 𝑇
… … … … . .19
:
)13 (
𝑁𝑝2
𝜒𝑒 =
3𝜀0 𝐾𝐵 𝑇
T
… … … … . .20
p
N
.
𝐽
𝐾𝐵 = 1.38 × 10−23 ( )
𝐾𝑔
𝐾𝐵
:
7
100 cm2
0.5 cm
1 cm
:
120 V
-
p,D
p,D
𝜀0 = 8.85 × 10−12
𝐶2
𝑁.𝑚 2
𝜀0 𝐴
8.85 × 10−12 (10−2 )
𝐶0 =
=
= 8.9 × 10−12 F
−2
𝑑
10
= 8.9 𝑝𝐹
𝑞 = 𝐶0 𝑉0 = 8.9 × 10−12 × 100
=
8.9 × 10−10 C
𝐸0 𝑑𝐴 = 𝐸0 𝐴 =
𝑞
𝜀0
𝑞
8.9 × 10−10
𝑉
4
⟹ 𝐸0 =
=
=
1
×
10
(
)
𝜀0 𝐴 8.85 × 10−12 × 10−2
𝑚
𝐸 𝑑𝐴 = 𝐸𝐴 =
𝐸=𝜀
𝑞
0
𝐸0
= =
𝐾𝐴 𝐾
1 ×10 4
7
𝑞
𝜀
𝑉
= 0.143 × 104 (𝑚 )
𝑉=−
𝐸 𝑑𝑙 = 𝐸0 𝑙1 + 𝐸 𝑙2
𝑞
𝐶=𝑉=-
4
-
𝐷 = 𝐾𝜀0 𝐸 = 7 × 8.85 × 10−12 × 0.143 × 10 =
8.9 × 10−8
𝐶
𝑚2
4
𝑝 = 𝜀0 𝐸 𝐾 − 1 = 8.85 × 10−12 × 0.143 × 10 7 − 1
−8 𝐶
= 7.6 × 10
𝑚2
4
𝐷0 = 𝜀0 𝐸0 = 8.85 × 10−12 × 1 × 10
−8
= 8.9 × 10
𝑝0 = 𝜀0 𝐸0 𝐾 − 1 = 0
𝐶
𝑚2
𝜒𝑒 = 𝐾 − 1 = 7 − 1 = 6
-
𝜎i = 𝜀0 𝜒𝑒 𝐸 =
=
4
8.85 × 10−12 × 6 × 0.143 × 10
Homework:
Problems 1, 2, 4, 6, 8,11,35,48
Serway Book