Energía electrostática Capacitores Campo eléctrico en ... · capacitancia. de un dispositivo se...

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


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.


Diagram

Thundercloud h = 10 km

r = 10 km

volume = π r2 h


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 polarized molecules of the dielectric tend to align themselves parallel to the electric field created by the charges on the conductors

−σb +σb

- - - - - - - -

+ + + + + + + +


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.


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= +


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


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


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 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


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


An equivalent capacitor can be found that performs the same function as the series combination

The charges are all the same

Q1 = Q2 = Q


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= + + +


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


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.


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.


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


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?




-external field exerts F=qE on each charge -net torque about the dipole’s center -dipole rotates to “align” with the 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:


θτ sin2Fanet =

qEF = aqp 2=

θθτ sinsin2 pEaqE ==

Ep ×=τ

but,

and

Thus:


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.


As the dipole aligns with the field, the system’s potential energy goes down.


-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.


-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


θτddW =

θτ sinpE=

-To rotate the dipole through some small angle dθ, an amount dW of work must be done.

but,


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:


0=iU 90=iθ

Let’s define the zero potential energy as being when the dipole is at θ = 90,

when


0=iU 90=iθ


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


0=iU 90=iθ


when

θcospEU −=



ffi pEpEpE f

iθθθθ θ

θ cos)cos(cos]cos[ −=−=−=

But, we already know


0=iU 90=iθ


EpU •−=

when

θcospEU −=



ffi pEpEpE f

iθθθθ θ

θ cos)cos(cos]cos[ −=−=−=

But, we already know


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 −×=

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

Energía electrostática Capacitores Campo eléctrico en ... · capacitancia. de un dispositivo se...

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