Energía electrostática Capacitores Campo eléctrico en materiales
Energía electrostática
Capacitores
Campo eléctrico en materiales
Capacitores
Capacitores
Un capacitor entonces acumula una carga –Q en un conductor y una carga +Q en el otro conductor
La carga acumulada crea un campo
eléctrico y por lo tanto, una diferencia de
potential entre dichos conductores.
–Q +Q
Un capacitor es un dispositivo que almacena carga eléctrica. Está formado por dos conductores próximos uno a otro, separados por un aislante, de tal modo que puedan estar cargados con el mismo valor, pero con signos contrarios.
4
Capacitancia
QCV
=
La capacitancia de un dispositivo se define como
La unidad es el faradio (F) = Culombio/Voltio (Remember: el culombio y por ende el faradio es una unidad enorme, de modo que en dispositivos ‘usuales’ las unidades que encontraremos son multiplos pequeños μF = 10-6 F o bien pF = 10-12 F
Q es la carga acumulada en un conductor V es la diferencia de potential entre los conductores
5
Capacitancia de un Conductor esférico
QV kR
=
El potencial en la superficie de un conductor esférico cargado con una carga Q y de radio R es
04Q RC RV k
πε= = =
De modo que su capacitancia es
6
Capacitancia de un capacitor de placas paralelas
Cuando dos placas conductoras de area A
están separadas por una distancia (pequeña
comparada con las dimensiones de las
placas) d el campo eláctrico entre ellas será
aproximadamente constante y de módulo
0 0/ /( )E Q Aσ ε ε= =
7
0 0( / ) /( )V E
Qd
d Adσ ε ε=
= =
Como el campo eléctrico es
constante, la diferencia de potencial
entre las placas es
entonces la capacitancia es
0CdAε
=
Capacitancia de un capacitor de placas paralelas
8
Capacitance Cylindrical Capacitors
0
12
2ln( / )R
CR
Lπε=
A coaxial cable of length L is an example of a cylindrical capacitor
R2
R1
Energía Electrostática
Electrostatic Energy
q1 q2
q3
Total work done 1 3 2 31 2
2 3kq q kq qkq qW W W
a a a= + = + +
2 2 1 21W q q kqV
a= =a
1 23 3 1 2 3( ) ( )W q q kq kqV V
a a= + = +
a a
Work is required to assemble a charge distribution
Electrostatic Energy
dW dqV=
W dW Vdq= =∫ ∫
dq
The work dW required to add an element of charge dq to an existing charge distribution is
where V is the potential at the final location of the charge element. The total work required is therefore
Since the electric is conservative, the work is stored as electrostatic energy, U.
Storage of Electrostatic Energy
Work must be done to move positive charge from a negatively charged conductor to one that is positively charged. Or to move negative charge in the reverse direction.
Storage of Electrostatic Energy
In moving charge dq, the electrostatic energy of the capacitor is increased by dU Vdq=
2
0
1212
Q q QU dqC C
QV
= =
=
∫Therefore,
212
U CV=
14
Energy Density of Electric Field
12
U VQ=
0/( )E Q Aε=
20 0
1 1( )( ) ( )2 2
EdE dA AU Eεε= =
V Ed=
Potential energy
Electric field
Electric potential
15
/( )Eenergyu Uvol
dume
A= =
The energy density uE
20
12Eu Eε=
This expression holds true for any electric field
Energy Density of Electric Field
16
Example – A Thunderstorm
How much electrical energy is stored in a typical thundercloud?
Assume a cloud of height h = 10 km, radius r = 10 km, with a uniform electric field E = 105 V/m.
http://redcrossggr.files.wordpress.com/2008/06/thunderstorm.jpg
Example – A Thunderstorm
Narrative The problem is about stored electric energy. Since the electric field, E, is uniform, so to is
the energy density uE = ½ ε0 E2 in the cloud. Therefore, the electric energy stored in the cloud is just the electric energy density times the volume of the cloud.
Example – A Thunderstorm
Diagram
Thundercloud h = 10 km
r = 10 km
volume = π r2 h
Example – A Thunderstorm
Calculation The electric energy density in the cloud is uE = ½ ε0 E2 = 4.4 x 10-2 J/m3. The volume of the cloud is, v = π r2 h, that is, v = 3.1 x 1012 m3. Therefore, the total electric energy stored in
the cloud is U = uEv = 140 GJ.
Using Capacitors
The Effect of Dielectrics
Michael Faraday 1791 – 1867
wikimedia
Michael Faraday discovered that the capacitance increases when the space between conductors is replaced by a dielectric. Today, we understand this to be a consequence of the polarization of molecules.
The Effect of Dielectrics
The polarized molecules of the dielectric tend to align themselves parallel to the electric field created by the charges on the conductors
−σb +σb
- - - - - - - -
+ + + + + + + +
The Effect of Dielectrics
The bound charge σb induced on the surface of the dielectric creates an electric field opposed to the electric field of the free charge σf on the conductors, thereby reducing the field between them.
Cuál es el mecanismo?
The Effect of Dielectrics The reduction in electric field strength from the initial field E0 to the reduced field E is quantified by the dielectric constant κ (kappa)
0EEκ
=
The dielectric increases the capacitance by the same factor κ.
The Effect of Dielectrics
For a parallel plate capacitor, with a dielectric between the plates, the electric field is
0ε κε= is called the permittivity
The product of the dielectric constant κ and the permittivity of free space ε0
0/( )E Q Aκε=
Capacitores en paralelo
At equilibrium, the potential across each capacitor is the same, namely, 12 V
same potential
same potential
The two capacitors are equivalent to a single capacitor with capacitance
1 2C C C= +
Capacitores en paralelo
The flow of charges ceases when the voltage across the capacitors equals that of the battery
The potential difference across the capacitors is the same And each is equal to the voltage of the battery ∆V1 = ∆V2 = ∆V
� ∆V is the battery terminal voltage The capacitors reach their maximum charge when the
flow of charge ceases The total charge is equal to the sum of the charges on the
capacitors Qtotal = Q1 + Q2
Capacitores en paralelo
The capacitors can be
replaced with one capacitor
with a capacitance of Ceq
The equivalent capacitor
must have exactly the same
external effect on the circuit
as the original capacitors
Capacitores en paralelo
Ceq = C1 + C2 + C3 + … The equivalent capacitance of a parallel
combination of capacitors is greater than any of the individual capacitors Essentially, the areas are combined
Use the active figure to vary the battery potential and the various capacitors and observe the resulting charges and voltages on the capacitors
Capacitores en paralelo
Capacitores en Serie
The sum of the potentials across both capacitors will be equal to 12 V
The potential V1 across C1 plus the potential V2 across C2 is equal to the potential difference V between points a and b: V = V1 + V2
1 21/ 1/ 1/C C C= +
Capacitores en serie
When a battery is connected to the circuit, electrons are transferred from the left plate of C1 to the right plate of C2 through the battery
Capacitores en serie
As this negative charge accumulates on the right plate of C2, an equivalent amount of negative charge is removed from the left plate of C2, leaving it with an excess positive charge
All of the right plates gain charges of –Q and all the left plates have charges of +Q
Capacitores en serie
An equivalent capacitor can be found that performs the same function as the series combination
The charges are all the same
Q1 = Q2 = Q
Capacitores en serie
The potential differences add up to the battery voltage ΔVtot = ∆V1 + ∆V2 + … The equivalent capacitance is
The equivalent capacitance of a series combination is always less than any individual capacitor in the combination
1 2 3
1 1 1 1
eqC C C C= + + +
Capacitores en serie
Energía almacenada en un Capacitor
Assume the capacitor is being charged and, at some point, has a charge q on it
The work needed to transfer a charge from one plate to the other is
The total work required is
qdW Vdq dqC
= ∆ =
2
0 2Q q QW dq
C C= =∫
Energy, cont
The work done in charging the capacitor appears as electric potential energy U:
This applies to a capacitor of any geometry The energy stored increases as the charge
increases and as the potential difference increases
In practice, there is a maximum voltage before discharge occurs between the plates
221 1 ( )
2 2 2QU Q V C VC
= = ∆ = ∆
Energy, final
The energy can be considered to be stored in the electric field
For a parallel-plate capacitor, the energy can be expressed in terms of the field as U = ½ (εoAd)E2
It can also be expressed in terms of the energy density (energy per unit volume)
uE = ½ εoE2
Resumen
Capacitancia C = Q / V (faradios) Placas paralelas C = ε0 A/d
Capacitores En paralelo C = C1 + C2
En serie 1/C = 1/C1 + 1/C2
Energia almacenada U = ½ QV Densidad de Energía uE = ½ ε0 E2
Efecto del dielectrico E = E0 / κ
TT: Efecto Piezoelectrico
https://www.youtube.com/watch?v=uNKMD8Z3-e4
Some Uses of Capacitors
Defibrillators When cardiac fibrillation occurs, the heart produces a
rapid, irregular pattern of beats A fast discharge of electrical energy through the heart
can return the organ to its normal beat pattern In general, capacitors act as energy reservoirs
that can be slowly charged and then discharged quickly to provide large amounts of energy in a short pulse
Dielectrics – An Atomic View
The molecules that make up the dielectric are modeled as dipoles
The molecules are randomly oriented in the absence of an electric field
Dielectrics – An Atomic View, 2
An external electric field is applied
This produces a torque on the molecules
The molecules partially align with the electric field
Dielectrics – An Atomic View, final An external field can
polarize the dielectric whether the molecules are polar or nonpolar
The charged edges of the dielectric act as a second pair of plates producing an induced electric field in the direction opposite the original electric field
Dielectrics – An Atomic View, 3 The degree of alignment of the molecules
with the field depends on temperature and the magnitude of the field
In general, the alignment increases with decreasing
temperature the alignment increases with increasing
field strength
Capacitors with Dielectrics
A dielectric is a nonconducting material that, when placed between the plates of a capacitor, increases the capacitance Dielectrics include rubber, glass, and waxed paper
With a dielectric, the capacitance becomes C = κCo The capacitance increases by the factor κ when the dielectric
completely fills the region between the plates κ is the dielectric constant of the material
Dielectrics, cont
For a parallel-plate capacitor, C = κεo(A/d) In theory, d could be made very small to create a
very large capacitance In practice, there is a limit to d
d is limited by the electric discharge that could occur though the dielectric medium separating the plates
For a given d, the maximum voltage that can be applied to a capacitor without causing a discharge depends on the dielectric strength of the material
Dielectrics, final
Dielectrics provide the following advantages: Increase in capacitance Increase the maximum operating voltage Possible mechanical support between the
plates • This allows the plates to be close
together without touching • This decreases d and increases C
Types of Capacitors – Tubular
Metallic foil may be interlaced with thin sheets of paraffin-impregnated paper or Mylar
The layers are rolled into a cylinder to form a small package for the capacitor
Types of Capacitors – Oil Filled
Common for high- voltage capacitors
A number of interwoven metallic plates are immersed in silicon oil
Types of Capacitors – Electrolytic
Used to store large amounts of charge at relatively low voltages
The electrolyte is a solution that conducts electricity by virtue of motion of ions contained in the solution
Types of Capacitors – Variable
Variable capacitors consist of two interwoven sets of metallic plates
One plate is fixed and the other is movable
These capacitors generally vary between 10 and 500 pF
Used in radio tuning circuits
Electric Dipole
An electric dipole consists of two charges of equal magnitude and opposite signs
The charges are separated by 2a
The electric dipole moment ( ) is directed along the line joining the charges from –q to +q
p
Thus, the total amount of work required to charge the capacitor from q = 0 to a final charge of q = Q is
∫ ∫ ===Q
Q
CQqdq
Cdq
CqW
00
2
21
But, in an isolated system with no non-conservative forces, total mechanical energy must be conserved.
Therefore, the work done to charge the capacitor must equal the change in the system’s potential energy.
Energy Stored in a Capacitor
( )22
21
21
2VCVQ
CQU ∆=∆==
Energy stored in a charged capacitor
Energy Stored in a Capacitor
EdV =∆
dAC 0ε
=
-It’s not obvious, but the potential energy stored in the capacitor actually resides in its electric field.
-This implies we should be able to solve the density of the energy stored in the field (J/m3).
-For a parallel plate capacitor, we already know:
-and, its capacitance is just:
-Substituting these into the purple equation,
( ) ( ) 20
220
21
21 EAddE
dAU εε
==
202
1 EuE ε=
-Dividing by the volume in between the plates of the capacitor (V=Ad), we get
Energy per unit volume in a capacitor (J/m3)
Energy Stored in a Capacitor ( )2
2
21
21
2VCVQ
CQU ∆=∆==
-We don’t attempt it here, but it can be shown that this result is valid for any electric field!
202
1 EuE ε= Energy per unit volume in an electric field.
-In a very real sense, electric fields “carry” energy.
Energy Stored in a Capacitor
Two capacitors, C1 and C2 (C1 > C2), are charged to the same initial potential difference, ΔVi. The charged capacitors are removed from the battery, and their plates are connected with opposite polarity, as shown. The switches, S1 and S2, are then closed.
(a) Find the final potential difference ΔVf between a and b after the switches are closed. (b) Find the total energy stored in the capacitors before and after the switches are closed and
determine the ratio of the final energy to the initial energy.
Rewiring two Charged Capacitors
What happens?
κ0VV ∆
=∆
Consider parallel-plate capacitor where ΔV0 = Q0/C0
Assume no battery is connected Q can’t change
When you stick a dielectric in between the plates
-where κ is a dimensionless constant called the “dielectric constant”
Capacitors with dielectrics
0
0
0
0
00
CC
VQ
VQ
VQC
κ
κ
κ=
∆=
∆=
∆=
-Q on the capacitor does not change
-Therefore:
-the capacitance is changed by a factor of κ. -as κ goes up, C goes up.
Capacitors with dielectrics
dAC
dAC
0
00
εκ
ε
=⇒
=
-For a parallel plate capacitor
To make capacitance ↑ -decrease d -increase A -increase κ
- Only limited by “dielectric strength” of the dielectric
Capacitors with dielectrics
Example values of dielectric constant
“Dielectric strength” is the maximum field in the dielectric before breakdown. (a spark or flow of charge)
max maxE V / d=
A parallel-plate capacitor is charged with a battery to a charge of Q0. The battery is then removed, and a slab of material that has a dielectric constant κ is inserted between the plates. Identify the system as the capacitor and the dielectric.
Find the energy stored in the system before and after the dielectric is inserted.
EG 26.5 – Energy stored before and after
0
20
0 2CQU =
CQU2
20
0 =
A parallel-plate capacitor is charged with a battery to a charge of Q0. The battery is then removed, and a slab of material that has a dielectric constant κ is inserted between the plates. Identify the system as the capacitor and the dielectric.
Find the energy stored in the system before and after the dielectric is inserted.
κκ0
0
20
0 2U
CQU ==
Before:
After:
Where did the energy go?
Rewiring two Charged Capacitors
The combination of two equal charges of opposite sign, +q and –q, separated by a distance 2a Every dipole can be characterized by it’s “dipole moment.” - vector which points from –q to +q -magnitude p = 2aq
1p 2p
1 2p p p= +
Electric Dipole in an Electric Field
What happens when we pop this baby in an external E-field?
Electric Dipole in an Electric Field
What happens when we pop this baby in an external E-field?
Electric Dipole in an Electric Field
-external field exerts F=qE on each charge -net torque about the dipole’s center -dipole rotates to “align” with the field
What happens when we pop this baby in an external E-field?
Electric Dipole in an Electric Field
-external field exerts F=qE on each charge -net torque about the dipole’s center -dipole rotates to “align” with the field
θτθτ
sinsin
FaFa
==
−
+
θτ sin2Fanet =
qEF = aqp 2=
θθτ sinsin2 pEaqE ==
but,
and
Thus:
Electric Dipole in an Electric Field
θτ sin2Fanet =
qEF = aqp 2=
θθτ sinsin2 pEaqE ==
Ep ×=τ
but,
and
Thus:
Electric Dipole in an Electric Field
The dipole and the external field are a system -electric force is an internal conservative force we can describe its work using a potential energy
In other words, different configurations of the dipole-field system have different potential energies.
Electric Dipole in an Electric Field
As the dipole aligns with the field, the system’s potential energy goes down.
Electric Dipole in an Electric Field
-Work must be done to “un-align” the dipole from the field. -in an isolated system, the work input must correspond to an increase in potential energy.
Electric Dipole in an Electric Field
-Work must be done to “un-align” the dipole from the field. -in an isolated system, the work input must correspond to an increase in potential energy. W = ΔK + ΔU
Electric Dipole in an Electric Field
θτddW =
θτ sinpE=
-To rotate the dipole through some small angle dθ, an amount dW of work must be done.
but,
Electric Dipole in an Electric Field
θτddW =
θτ sinpE=
-To rotate the dipole through some small angle dθ, an amount dW of work must be done.
∫∫∫ ===−f
i
f
i
f
i
dpEdpEdUU if
θ
θ
θ
θ
θ
θ
θθθθθτ sinsin
)cos(cos]cos[ fipEpE f
iθθθ θ
θ −=−=
but,
-so, to rotate the dipole from θi to θf, the change in potential energy is:
Electric Dipole in an Electric Field
0=iU 90=iθ
Let’s define the zero potential energy as being when the dipole is at θ = 90,
when
Electric Dipole in an Electric Field
0=iU 90=iθ
Let’s define the zero potential energy as being when the dipole is at θ = 90,
when
We’ll use this reference energy as an anchor point. At any time, we can write the system’s instantaneous potential energy, U, with respect to the zero-point potential energy.
090 −=−=−= = ffif UUUUUU θ
Electric Dipole in an Electric Field
0=iU 90=iθ
Let’s define the zero potential energy as being when the dipole is at θ = 90,
when
θcospEU −=
We’ll use this reference energy as an anchor point. At any time, we can write the system’s instantaneous potential energy, U, with respect to the zero-point potential energy.
090 −=−=−= = ffif UUUUUU θ
ffi pEpEpE f
iθθθθ θ
θ cos)cos(cos]cos[ −=−=−=
But, we already know
Electric Dipole in an Electric Field
0=iU 90=iθ
Let’s define the zero potential energy as being when the dipole is at θ = 90,
EpU •−=
when
θcospEU −=
We’ll use this reference energy as an anchor point. At any time, we can write the system’s instantaneous potential energy, U, with respect to the zero-point potential energy.
090 −=−=−= = ffif UUUUUU θ
ffi pEpEpE f
iθθθθ θ
θ cos)cos(cos]cos[ −=−=−=
But, we already know
Electric Dipole in an Electric Field
A water molecule has an electric dipole moment of 6.3x10-30 Cm. A sample contains 1021 water molecules. All of the dipoles are oriented in the direction of an external E-field, which has a magnitude of 2.5x105 N/C.
How much work is required to rotate all the dipoles from this orientation (θ = 0) to one in all the dipoles are perpendicular to the external field (θ = 90)?
The Water Molecule
A water molecule has an electric dipole moment of 6.3x10-30 Cm. A sample contains 1021 water molecules. All of the dipoles are oriented in the direction of an external E-field, which has a magnitude of 2.5x105 N/C.
How much work is required to rotate all the dipoles from this orientation (θ = 0) to one in all the dipoles are perpendicular to the external field (θ = 90)?
The Water Molecule
WU =∆)0cos()90cos(090 NpENpEUUW −−−=−= °°
)/105.2)(103.6)(10( 53021 CNmCNpE ×⋅×== −
J3106.1 −×=
A water molecule has an electric dipole moment of 6.3x10-30 Cm. A sample contains 1021 water molecules. All of the dipoles are oriented in the direction of an external E-field, which has a magnitude of 2.5x105 N/C.
How much work is required to rotate all the dipoles from this orientation (θ = 0) to one in all the dipoles are perpendicular to the external field (θ = 90)?
The Water Molecule
WU =∆)0cos()90cos(090 NpENpEUUW −−−=−= °°
)/105.2)(103.6)(10( 53021 CNmCNpE ×⋅×== −
J3106.1 −×=
Example P26.9
When a potential difference of 150 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of 30.0 nC/cm2. What is the spacing between the plates?
( )0 AQ V
d∈
= ∆
( ) ( ) ( )
( ) ( )12 2 2
09 2 4 2 2
8.85 10 C N m 150 V4.42 m
30.0 10 C cm 1.00 10 cm mV
d µσ
−
−
× ⋅∈ ∆= = =
× ×
Example P26.21 Four capacitors are connected as shown in
Figure P26.21. (a) Find the equivalent capacitance between
points a and b. (b) Calculate the charge on each capacitor if
ΔVab = 15.0 V.
1
2.50 F2.50 6.00 8.50 F
1 1 5.96 F8.50 F 20.0 F
s
p
eq
CC
C
µµ
µµ µ
−
== + =
= + =
1 1 115.0 3.00sC
= + ( ) ( )5.96 F 15.0 V 89.5 CQ C V µ µ= ∆ = =
( )( )
89.5 C 4.47 V20.0 F
15.0 4.47 10.53 V
6.00 F 10.53 V 63.2 C on 6.00 F
QVC
Q C V
µµ
µ µ µ
∆ = = =
− =
= ∆ = =
89.5 63.2 26.3 Cµ− =
Example P26.27
Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in Figure P26.27. Take C1 = 5.00 μF, C2 = 10.0 μF, and C3 = 2.00 μF.
( )( )
1
1
2
1
1 1 3.33 F5.00 10.02 3.33 2.00 8.66 F
2 10.0 20.0 F
1 1 6.04 F8.66 20.0
s
p
p
eq
C
C
C
C
µ
µ
µ
µ
−
−
= + =
= + =
= =
= + =
Example P26.35
A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled?
2 12d d= 2 112
C C= stored energy doubles,
. Therefore, the
Example P26.43
Determine (a) the capacitance and (b) the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.75 cm2 and plate separation of 0.040 0 mm.
( ) ( )12 4 2110
5
2.10 8.85 10 F m 1.75 10 m8.13 10 F 81.3 pF
4.00 10 mA
Cd
κ− −
−−
× ×∈= = = × =
×
( ) ( )6 5max max 60.0 10 V m 4.00 10 m 2.40 kVV E d −∆ = = × × =
Example P26.59
A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is 3.00 and whose dielectric strength is 2.00 × 108 V/m. The desired capacitance is 0.250 μF, and the capacitor must withstand a maximum potential difference of 4 000 V. Find the minimum area of the capacitor plates.
8 maxmax 2.00 10 V m
VE
d∆
= × =
60 0.250 10 FA
Cd
κ −∈= = ×
( ) ( )( ) ( )
62max
12 80 0 max
0.250 10 4 0000.188 m
3.00 8.85 10 2.00 10C VCdA
Eκ κ
−
−
×∆= = = =
∈ ∈ × ×
3.00κ =
Electric Circuits A circuit is a collection of objects usually containing a
source of electrical energy connected to elements that convert electrical energy to other forms
A circuit diagram – a simplified representation of an
actual circuit – is used to show the path of the real circuit
Circuit symbols are used to represent various elements