Capítulo 24 – Campo eléctricoCapítulo 24 – Campo eléctricoPresentación PowerPoint dePresentación PowerPoint de

Paul E. Tippens, Profesor de FísicaPaul E. Tippens, Profesor de Física

Southern Polytechnic State UniversitySouthern Polytechnic State University

© 2007

Objetivos: Después de terminar Objetivos: Después de terminar esta unidad deberá:esta unidad deberá:

• Definir el campo eléctrico y explicar qué determina su magnitud y dirección.

• Discutir las líneas de campo eléctrico y el significado de la permitividad del espacio.

• Escribir y aplicar fórmulas para la intensidad del campo eléctrico a distancias conocidas desde cargas puntuales.

• Escribir y aplicar la ley de Gauss para campos en torno a superficies con densidades de carga conocidas.

El concepto de campoEl concepto de campo

Un Un campocampo se define como una se define como una propiedad del espaciopropiedad del espacio en el que un objeto material experimenta una en el que un objeto material experimenta una fuerzafuerza..

.PSobre la Tierra, se dice que existe Sobre la Tierra, se dice que existe un un campo gravitacionalcampo gravitacional en en PP..

Puesto que Puesto que una masa una masa mm experimenta experimenta una una fuerzafuerza descendente en dicho descendente en dicho punto.punto.

¡No hay fuerza, no hay campo; no hay campo, no hay fuerza!

m

F

La La direccióndirección del campo está determinada por la del campo está determinada por la fuerzafuerza..

El campo gravitacionalEl campo gravitacional

Note que la fuerza Note que la fuerza F F es es realreal, pero el , pero el campo sólo es una forma campo sólo es una forma conveniente de conveniente de describir el espaciodescribir el espacio..

El campo en los puntos A o B se El campo en los puntos A o B se puede encontrar de:puede encontrar de:

Fg

m=

Si Si gg se conoce en se conoce en cada punto sobre la cada punto sobre la Tierra, entonces se Tierra, entonces se puede encontrar la puede encontrar la fuerza fuerza F F sobre una sobre una masa dada.masa dada.

La La magnitudmagnitud y y direccióndirección del del campo campo gg depende del peso, que depende del peso, que es la fuerza es la fuerza F.F.

•A• B Considere los puntos Considere los puntos AA y y BB sobre sobre

la superficie de la Tierra, sólo la superficie de la Tierra, sólo puntos en el puntos en el espacioespacio..

FF

FF

El campo eléctricoEl campo eléctrico1. Ahora, considere el punto 1. Ahora, considere el punto

PP a una distancia a una distancia rr de de +Q+Q..

2. 2. En En PP existe un campo eléctrico existe un campo eléctrico EE si una carga si una carga de pruebade prueba +q +q tiene tiene una fuerza una fuerza FF en dicho punto. en dicho punto.

3. La 3. La direccióndirección del del E E es igual es igual que la dirección de una que la dirección de una fuerzafuerza sobre la carga sobre la carga + (pos)+ (pos)..

E

4. La 4. La magnitudmagnitud de de EE está está dada por la fórmula:dada por la fórmula:

Campo eléctrico

++++ +

+++Q

.P

r

+qF

+

CN

qF

E unidades ;=

El campo es propiedad del El campo es propiedad del espacioespacio

E

Campo eléctrico

++++ +

+++Q

.

r

En un punto existe un campo En un punto existe un campo EE ya sea que en ya sea que en dicho punto haya o no una carga. La dicho punto haya o no una carga. La direccióndirección del campo es del campo es alejándosealejándose de la carga de la carga +Q+Q..

E

Campo eléctrico

++++ +

+++Q

.

r++q --q

F

F

La fuerza sobre La fuerza sobre +q+q está está en dirección del campo.en dirección del campo.

La fuerza sobre La fuerza sobre -q-q está contra la está contra la

dirección del campo.dirección del campo.

Campo cerca de una carga negativaCampo cerca de una carga negativa

Note que el campo Note que el campo EE en la vecindad de una en la vecindad de una carga carga negativa negativa –Q –Q es es hacia hacia la carga, la dirección en que se la carga, la dirección en que se movería una carga de prueba movería una carga de prueba +q+q..

La fuerza sobre La fuerza sobre +q+q está está en dirección del campo.en dirección del campo.

La fuerza sobre La fuerza sobre -q-q está contra la está contra la

dirección del campo.dirección del campo.

E

Campo eléctrico

.

r++q

F

---- -

----Q

E

Campo eléctrico

.

r

--qF

---- -

----Q

La magnitud del campo ELa magnitud del campo ELa La magnitudmagnitud de la intensidad del campo eléctrico en un de la intensidad del campo eléctrico en un punto en el espacio se define como la punto en el espacio se define como la fuerza por unidad fuerza por unidad de cargade carga (N/C)(N/C) que experimentaría cualquier carga de que experimentaría cualquier carga de prueba que se coloque en dicho punto.prueba que se coloque en dicho punto.

Intensidad de campo eléctrico E

Intensidad de campo eléctrico E

La La direccióndirección de de EE en un punto es la misma que la en un punto es la misma que la dirección en que se movería una carga dirección en que se movería una carga positivapositiva SISI se colocara en dicho punto.se colocara en dicho punto.

CN

qF

E unidades ;=

Ejemplo 1.Ejemplo 1. Una carga de Una carga de +2 nC+2 nC se se coloca a una distancia coloca a una distancia rr de una carga de una carga dede–8 –8 µµCC. Si la carga experimenta . Si la carga experimenta una fuerza de una fuerza de 4000 N4000 N, ¿cuál es la , ¿cuál es la intensidad del campo eléctrico E en intensidad del campo eléctrico E en dicho punto P?dicho punto P?

Campo eléctrico

.

---- -

----Q

P

Primero, note que la dirección de Primero, note que la dirección de E es hacia –Q (abajo).E es hacia –Q (abajo).

–8 µC

E

++q

E4000 N

-9

4000 N

2 x 10 C

FE

q= =

+2 nC

r

E = 2 x 1012 N/C hacia abajo

Nota: El campo Nota: El campo E E sería el sería el mismomismo para para cualquier cualquier carga que se carga que se coloque en el punto coloque en el punto PP. Es una propiedad de dicho . Es una propiedad de dicho espacioespacio..

Ejemplo 2.Ejemplo 2. Un campo constante Un campo constante E E de de 40,000 N/C40,000 N/C se mantiene entre las dos placas paralelas. se mantiene entre las dos placas paralelas. ¿Cuáles son la magnitud y dirección de la fuerza ¿Cuáles son la magnitud y dirección de la fuerza sobre un electrón que pasa horizontalmente entre sobre un electrón que pasa horizontalmente entre las placas?las placas?

E.FEl campo E es hacia abajo, y El campo E es hacia abajo, y

la fuerza sobre ela fuerza sobre e-- es arriba. es arriba.

; F

E F qEq

= =

-19 4(1.6 x 10 C)(4 x 10 )NCF qE= =

F = 6.40 x 10-15 N, hacia arribaF = 6.40 x 10-15 N, hacia arriba

+ + + + + + + + +

- - - - - - - - -

-ee-- -ee-- -ee--

Campo E a una distancia r Campo E a una distancia r desde una sola carga Qdesde una sola carga Q

++++ +

+++Q

.r

P

Considere una carga de prueba Considere una carga de prueba +q+q colocada en colocada en PP a una distancia a una distancia rr de de QQ..

La fuerza hacia afuera sobre +q La fuerza hacia afuera sobre +q es:es:

Por tanto, el campo eléctrico Por tanto, el campo eléctrico EE es: es:

2F kQq rE

q q= = 2

kQE

r=

++qF

2

kQqF

r=

++++

+

+++Q

.r

P

E

2

kQE

r=

Ejemplo 3.Ejemplo 3. ¿Cuál es la intensidad del ¿Cuál es la intensidad del campo eléctrico campo eléctrico EE en el punto en el punto PP, a una , a una distancia de distancia de 3 m3 m desde una carga negativa desde una carga negativa dede–8 nC–8 nC??

.r

P

-Q3 m

-8 nC

E = ? Primero, encuentre la magnitud:Primero, encuentre la magnitud:2

2

9 -9NmC

2 2

(9 x 10 )(8 x 10 C)

(3 m)

kQE

r= =

E = 8.00 N/CE = 8.00 N/C

La dirección es la misma que la fuerza sobre una La dirección es la misma que la fuerza sobre una carga positiva carga positiva sisi se colocase en el punto P: se colocase en el punto P: hacia –Qhacia –Q..

EE = 8.00 N, hacia -Q = 8.00 N, hacia -Q

El campo eléctrico resultanteEl campo eléctrico resultanteEl campo resultante El campo resultante EE en la vecindad de un número de en la vecindad de un número de cargas puntuales es igual a la cargas puntuales es igual a la suma vectorialsuma vectorial de los de los campos debidos a cada carga tomada individualmente.campos debidos a cada carga tomada individualmente.

Considere E para cada carga.Considere E para cada carga.

+

- •q1

q2q3

-

AE1

E3

E2

ERSuma vectorial:

E = E1 + E2 + E3

Suma vectorial:

E = E1 + E2 + E3

Las direcciones se basan en carga de prueba positiva.

Magnitudes a partir de:

2

kQE

r=

Ejemplo 4.Ejemplo 4. Encuentre el campo resultante en Encuentre el campo resultante en el punto el punto AA debido a las cargas de debido a las cargas de –3 nC–3 nC y y +6 +6 nCnC ordenadas como se muestra. ordenadas como se muestra.

+

-

•

q1

q24 cm

3 cm 5 cm

-3 nC

+6 nC

E para cada q se muestra E para cada q se muestra con la dirección dada.con la dirección dada.

E2

E1

1 21 22 2

1 2

; kq kq

E Er r

= =A

2

2

9 -9NmC

1 2

(9 x 10 )(3 x 10 C)

(3 m)E =

2

2

9 -9NmC

2 2

(9 x 10 )(6 x 10 C)

(4 m)E =

Los signos de las cargas sólo se usan para encontrar la dirección de ELos signos de las cargas sólo se usan para encontrar la dirección de E

Ejemplo 4.Ejemplo 4. (Cont.) (Cont.) Encuentre el campo Encuentre el campo resultante en el punto resultante en el punto AA. Las magnitudes son:. Las magnitudes son:

+

-

•

q1

q24 cm

3 cm 5 cm

-3 nC

+6 nC

E2

E1

A

2

2

9 -9NmC

1 2

(9 x 10 )(3 x 10 C)

(3 m)E =

2

2

9 -9NmC

2 2

(9 x 10 )(6 x 10 C)

(4 m)E =

EE11 = = 3.00 N, oeste3.00 N, oeste EE22 = = 3.38 N, norte3.38 N, norte

E2

E1

A continuación, encuentre el vector resultante EA continuación, encuentre el vector resultante ERR

ER

2 2 12 1

2

; tanR

EE E R

Eφ= + =

φ

Ejemplo 4.Ejemplo 4. (Cont.) (Cont.) Encuentre el campo resultante Encuentre el campo resultante en el punto en el punto AA con matemáticas vectoriales. con matemáticas vectoriales.

E1 = 3.00 N, oeste

E2 = 3.38 N, norte

Encuentre el vector resultante EEncuentre el vector resultante ERRE2

E1

ER

2 2(3.00 N) (3.38 N) 4.52 N;E = + =3.38 N

tan3.00 N

φ =

φφ = 48.4= 48.400 N de O; o N de O; o θθ = 131.6 = 131.600

Campo resultante: ER = 4.52 N; 131.60Campo resultante: ER = 4.52 N; 131.60

φ

Líneas de campo eléctricoLíneas de campo eléctrico

++++ +

+++Q --

-- -

----Q

Las Las líneas de campo eléctricolíneas de campo eléctrico son líneas imaginarias que se son líneas imaginarias que se dibujan de tal forma que su dirección en cualquier punto es dibujan de tal forma que su dirección en cualquier punto es la misma que la dirección del campo en dicho punto.la misma que la dirección del campo en dicho punto.

Las líneas de campo se Las líneas de campo se alejanalejan de las cargas de las cargas positivaspositivas y se y se acercanacercan a las cargas a las cargas negativasnegativas..

1. 1. La dirección de la línea de campo en cualquier punto La dirección de la línea de campo en cualquier punto es la misma que el movimiento de +q en dicho punto.es la misma que el movimiento de +q en dicho punto.

2. El espaciamiento de las líneas debe ser tal que 2. El espaciamiento de las líneas debe ser tal que estén cercanas donde el campo sea intenso y estén cercanas donde el campo sea intenso y separadas donde el campo sea débil.separadas donde el campo sea débil.

+ -qq11 qq22

EE11

EE22

EERR

Reglas para dibujar líneas de Reglas para dibujar líneas de campocampo

Reglas para dibujar líneas de Reglas para dibujar líneas de campocampo

Ejemplos de líneas de campo EEjemplos de líneas de campo EDos cargas iguales Dos cargas iguales pero pero opuestasopuestas..

Dos cargas Dos cargas idénticas idénticas (ambas +).(ambas +).

Note que las líneas Note que las líneas salen salen de las cargasde las cargas + + y y entran entran a las cargas a las cargas --..

Además, Además, EE es es más intensomás intenso donde las líneas de campo son donde las líneas de campo son más densasmás densas..

Electrical Force with Other Electrical Force with Other Forces, ExampleForces, Example

• The spheres are in The spheres are in equilibriumequilibrium

• Since they are separated, Since they are separated, they exert a repulsive they exert a repulsive force on each otherforce on each other

– Charges are like chargesCharges are like charges

• Proceed as usual with Proceed as usual with equilibrium problems, equilibrium problems, noting one force is an noting one force is an electrical forceelectrical force

Electrical Force with Other Electrical Force with Other Forces, Example cont.Forces, Example cont.

• The free body The free body diagram includes the diagram includes the components of the components of the tension, the electrical tension, the electrical force, and the weightforce, and the weight

• Solve for |Solve for |qq| |

• You cannot determine You cannot determine the sign of the sign of qq, only , only that they both have that they both have same signsame sign

Relationship Between Relationship Between FF and and EE

• FFee = = qqEE

– This is valid for a point charge onlyThis is valid for a point charge only

– One of zero sizeOne of zero size

– For larger objects, the field may vary over the For larger objects, the field may vary over the size of the objectsize of the object

• If If qq is positive, is positive, FF and and EE are in the same are in the same directiondirection

• If If qq is negative, is negative, FF and and EE are in opposite are in opposite directionsdirections

Electric Field Notes, FinalElectric Field Notes, Final

• The direction of The direction of EE is that is that of the force on a positive of the force on a positive test chargetest charge

• The SI units of The SI units of EE are N/C are N/C

• We can also say that an We can also say that an electric field exists at a electric field exists at a point if a test charge at point if a test charge at that point experiences an that point experiences an electric forceelectric force

Electric Field, Vector FormElectric Field, Vector Form

• Remember Coulomb’s law, between the Remember Coulomb’s law, between the source and test charges, can be source and test charges, can be expressed as expressed as

• Then, the electric field will be Then, the electric field will be

2ˆo

e e

qqkr

=F r

2ˆe

eo

qk

q r= =F

E r

More About ElectricMore About ElectricField DirectionField Direction

• a) a) qq is positive, is positive, FF is is directed away from directed away from qq

• b) The direction of b) The direction of EE is is also away from the also away from the positive source chargepositive source charge

• c) c) qq is negative, is negative, FF is is directed toward directed toward qq

• d) d) EE is also toward the is also toward the negative source chargenegative source charge

Superposition with Electric Superposition with Electric FieldsFields

• At any point At any point PP, the total electric field due , the total electric field due to a group of source charges equals the to a group of source charges equals the vector sum of electric fields of all the vector sum of electric fields of all the chargescharges

2ˆi

e ii i

qk

r= ∑E r

Superposition ExampleSuperposition Example

• Find the electric field due Find the electric field due to to qq11, , EE11

• Find the electric field due Find the electric field due to to qq22, , EE22

• EE = = EE11 + + EE22

– Remember, the fields add Remember, the fields add as vectorsas vectors

– The direction of the The direction of the individual fields is the individual fields is the direction of the force on a direction of the force on a positive test chargepositive test charge

Electric Field – Continuous Electric Field – Continuous Charge DistributionCharge Distribution

• The distances between charges in a group The distances between charges in a group of charges may be much smaller than the of charges may be much smaller than the distance between the group and a point of distance between the group and a point of interestinterest

• In this situation, the system of charges can In this situation, the system of charges can be modeled as continuousbe modeled as continuous

• The system of closely spaced charges is The system of closely spaced charges is equivalent to a total charge that is equivalent to a total charge that is continuously distributed along some line, continuously distributed along some line, over some surface, or throughout some over some surface, or throughout some volumevolume

Electric Field – Continuous Electric Field – Continuous Charge Distribution, contCharge Distribution, cont

• Procedure:Procedure:– Divide the charge Divide the charge

distribution into small distribution into small elements, each of which elements, each of which contains contains ΔΔqq

– Calculate the electric Calculate the electric field due to one of these field due to one of these elements at point elements at point PP

– Evaluate the total field Evaluate the total field by summing the by summing the contributions of all the contributions of all the charge elementscharge elements

Electric Field – Continuous Electric Field – Continuous Charge Distribution, Charge Distribution, equationsequations

• For the individual charge elementsFor the individual charge elements

• Because the charge distribution is Because the charge distribution is continuouscontinuous

2ˆ

e

qk

r

∆∆ =E r

2 20ˆ ˆlim

i

ie i eq

i i

q dqk k

r r∆ →

∆= =∑ ∫E r r

Charge DensitiesCharge Densities

• Volume charge densityVolume charge density: when a charge is : when a charge is distributed evenly throughout a volumedistributed evenly throughout a volume– ρρ = = QQ / / VV

• Surface charge densitySurface charge density: when a charge is : when a charge is distributed evenly over a surface areadistributed evenly over a surface area– σσ = = QQ / / AA

• Linear charge densityLinear charge density: when a charge is : when a charge is distributed along a linedistributed along a line– λλ = = QQ / / ℓℓ

Amount of Charge in a Small Amount of Charge in a Small VolumeVolume

• For the volume: For the volume: dqdq = = ρρ dVdV

• For the surface: For the surface: dqdq = = σσ dAdA

• For the length element: For the length element: dqdq = = λ λ ddℓℓ

Problem Solving HintsProblem Solving Hints

• UnitsUnits: when using the Coulomb constant, : when using the Coulomb constant, kkee, ,

the charges must be in C and the distances the charges must be in C and the distances in min m

• Calculating the electric field of point Calculating the electric field of point chargescharges: use the superposition principle, : use the superposition principle, find the fields due to the individual charges find the fields due to the individual charges at the point of interest and then add them as at the point of interest and then add them as vectors to find the resultant fieldvectors to find the resultant field

Problem Solving Hints, cont.Problem Solving Hints, cont.

• Continuous charge distributionsContinuous charge distributions: the : the vector sums for evaluating the total electric vector sums for evaluating the total electric field at some point must be replaced with field at some point must be replaced with vector integralsvector integrals

– Divide the charge distribution into infinitesimal Divide the charge distribution into infinitesimal pieces, calculate the vector sum by integrating pieces, calculate the vector sum by integrating over the entire charge distributionover the entire charge distribution

• SymmetrySymmetry: : take advantage of any take advantage of any symmetry to simplify calculationssymmetry to simplify calculations

Campo electrico debido a Campo electrico debido a una varilla cargadauna varilla cargada

Example – Charged DiskExample – Charged Disk

• The ring has a The ring has a radius radius RR and a and a uniform charge uniform charge density density σσ

• Choose Choose dqdq as a ring as a ring of radius of radius rr

• The ring has a The ring has a surface area 2surface area 2ππrr drdr

Electric Field LinesElectric Field Lines

• Field lines give us a means of representing Field lines give us a means of representing the electric field pictoriallythe electric field pictorially

• The electric field vector The electric field vector EE is tangent to the is tangent to the electric field line at each pointelectric field line at each point– The line has a direction that is the same as that The line has a direction that is the same as that

of the electric field vectorof the electric field vector

• The number of lines per unit area through a The number of lines per unit area through a surface perpendicular to the lines is surface perpendicular to the lines is proportional to the magnitude of the electric proportional to the magnitude of the electric field in that regionfield in that region

Electric Field Lines, GeneralElectric Field Lines, General

• The density of lines The density of lines through surface A is through surface A is greater than through greater than through surface Bsurface B

• The magnitude of the The magnitude of the electric field is greater on electric field is greater on surface A than Bsurface A than B

• The lines at different The lines at different locations point in different locations point in different directionsdirections– This indicates the field is This indicates the field is

non-uniformnon-uniform

Electric Field Lines, Positive Electric Field Lines, Positive Point ChargePoint Charge

• The field lines radiate The field lines radiate outward in all directionsoutward in all directions– In three dimensions, the In three dimensions, the

distribution is sphericaldistribution is spherical

• The lines are directed The lines are directed away from the source away from the source chargecharge– A positive test charge would A positive test charge would

be repelled away from the be repelled away from the positive source chargepositive source charge

Electric Field Lines, Negative Electric Field Lines, Negative Point ChargePoint Charge

• The field lines radiate The field lines radiate inward in all directionsinward in all directions

• The lines are directed The lines are directed toward the source toward the source chargecharge

– A positive test charge A positive test charge would be attracted would be attracted toward the negative toward the negative source chargesource charge

Electric Field Lines – Dipole Electric Field Lines – Dipole

• The charges are The charges are equal and oppositeequal and opposite

• The number of field The number of field lines leaving the lines leaving the positive charge positive charge equals the number equals the number of lines terminating of lines terminating on the negative on the negative chargecharge

Electric Field Lines – Like Electric Field Lines – Like ChargesCharges

• The charges are equal The charges are equal and positiveand positive

• The same number of The same number of lines leave each lines leave each charge since they are charge since they are equal in magnitudeequal in magnitude

• At a great distance, At a great distance, the field is the field is approximately equal to approximately equal to that of a single charge that of a single charge of 2of 2qq

Electric Field Lines, Unequal Electric Field Lines, Unequal ChargesCharges

• The positive charge is The positive charge is twice the magnitude of the twice the magnitude of the negative chargenegative charge

• Two lines leave the Two lines leave the positive charge for each positive charge for each line that terminates on the line that terminates on the negative chargenegative charge

• At a great distance, the At a great distance, the field would be field would be approximately the same approximately the same as that due to a single as that due to a single charge of +charge of +qq

Electric Field Lines – Rules Electric Field Lines – Rules for Drawingfor Drawing

• The lines must begin on a positive charge and The lines must begin on a positive charge and terminate on a negative chargeterminate on a negative charge

– In the case of an excess of one type of charge, In the case of an excess of one type of charge, some lines will begin or end infinitely far awaysome lines will begin or end infinitely far away

• The number of lines drawn leaving a positive The number of lines drawn leaving a positive charge or approaching a negative charge is charge or approaching a negative charge is proportional to the magnitude of the chargeproportional to the magnitude of the charge

• No two field lines can crossNo two field lines can cross

Motion of Charged ParticlesMotion of Charged Particles

• When a charged particle is placed in an When a charged particle is placed in an electric field, it experiences an electrical electric field, it experiences an electrical forceforce

• If this is the only force on the particle, it If this is the only force on the particle, it must be the net forcemust be the net force

• The net force will cause the particle to The net force will cause the particle to accelerate according to Newton’s second accelerate according to Newton’s second lawlaw

Motion of Particles, contMotion of Particles, cont

• FFee = = qqEE = = mmaa

• If If EE is uniform, then is uniform, then aa is constant is constant

• If the particle has a positive charge, its If the particle has a positive charge, its acceleration is in the direction of the fieldacceleration is in the direction of the field

• If the particle has a negative charge, its If the particle has a negative charge, its acceleration is in the direction opposite the acceleration is in the direction opposite the electric fieldelectric field

• Since the acceleration is constant, the Since the acceleration is constant, the kinematic equations can be usedkinematic equations can be used

Electron in a Uniform Field, Electron in a Uniform Field, ExampleExample

• The electron is projected The electron is projected horizontally into a uniform horizontally into a uniform electric fieldelectric field

• The electron undergoes a The electron undergoes a downward accelerationdownward acceleration

– It is negative, so the It is negative, so the acceleration is opposite acceleration is opposite EE

• Its motion is parabolic Its motion is parabolic while between the plateswhile between the plates

The Cathode Ray Tube The Cathode Ray Tube (CRT)(CRT)

• A CRT is commonly used to obtain a A CRT is commonly used to obtain a visual display of electronic information in visual display of electronic information in oscilloscopes, radar systems, televisions, oscilloscopes, radar systems, televisions, etc.etc.

• The CRT is a vacuum tube in which a The CRT is a vacuum tube in which a beam of electrons is accelerated and beam of electrons is accelerated and deflected under the influence of electric deflected under the influence of electric or magnetic fieldsor magnetic fields

CRT, contCRT, cont

• The electrons are The electrons are deflected in various deflected in various directions by two sets directions by two sets of platesof plates

• The placing of charge The placing of charge on the plates creates on the plates creates the electric field the electric field between the plates between the plates and allows the beam and allows the beam to be steeredto be steered

Flux Through Closed Surface, Flux Through Closed Surface, finalfinal

• The The netnet flux through the surface is flux through the surface is proportional to the net number of lines proportional to the net number of lines leaving the surfaceleaving the surface

– This net number of lines is the number of This net number of lines is the number of lines leaving the surface minus the number lines leaving the surface minus the number entering the surfaceentering the surface

• If If EEnn is the component of is the component of EE perpendicular perpendicular

to the surface, thento the surface, thenE nd E dAΦ = × =∫ ∫E AÑ Ñ

Gauss’s Law, IntroductionGauss’s Law, Introduction

• Gauss’s law is an expression of the Gauss’s law is an expression of the general relationship between the net general relationship between the net electric flux through a closed surface and electric flux through a closed surface and the charge enclosed by the surfacethe charge enclosed by the surface

– The closed surface is often called a The closed surface is often called a gaussian gaussian surfacesurface

• Gauss’s law is of fundamental Gauss’s law is of fundamental importance in the study of electric fieldsimportance in the study of electric fields

Gauss’s Law – General Gauss’s Law – General

• A positive point A positive point charge, charge, qq, is located , is located at the center of a at the center of a sphere of radius sphere of radius rr

• The magnitude of The magnitude of the electric field the electric field everywhere on the everywhere on the surface of the surface of the sphere is sphere is

EE = = kkeeqq / / rr22

Gauss’s Law – General, cont.Gauss’s Law – General, cont.

• The field lines are directed radially The field lines are directed radially outward and are perpendicular to the outward and are perpendicular to the surface at every pointsurface at every point

• This will be the net flux through the This will be the net flux through the gaussian surface, the sphere of radius gaussian surface, the sphere of radius rr

• We know We know EE = = kkeeqq//rr22 and and AAsphere sphere = 4= 4ππrr22,,

E d E dAΦ = × =∫ ∫E AÑ Ñ 4E eo

qπk q

εΦ = =

Gauss’s Law – General, Gauss’s Law – General, notesnotes

• The net flux through any closed surface The net flux through any closed surface surrounding a point charge, surrounding a point charge, qq, is given by , is given by qq//εεoo

and is independent of the shape of that surfaceand is independent of the shape of that surface

• The net electric flux through a closed surface The net electric flux through a closed surface that surrounds no charge is zerothat surrounds no charge is zero

• Since the electric field due to many charges is Since the electric field due to many charges is the vector sum of the electric fields produced by the vector sum of the electric fields produced by the individual charges, the flux through any the individual charges, the flux through any closed surface can be expressed as closed surface can be expressed as ( )1 2d d× = + ×∫ ∫E A E E AKÑ Ñ

Gauss’s Law – FinalGauss’s Law – Final

• Gauss’s law statesGauss’s law states

• qq inin is the net charge inside the surface is the net charge inside the surface

• EE represents the electric field at any point on represents the electric field at any point on the surfacethe surface– EE is the is the total electric fieldtotal electric field and may have contributions and may have contributions

from charges both inside and outside of the surfacefrom charges both inside and outside of the surface

• Although Gauss’s law can, in theory, be solved Although Gauss’s law can, in theory, be solved to find to find EE for any charge configuration, in for any charge configuration, in practice it is limited to symmetric situationspractice it is limited to symmetric situations

E A inE

o

qd

εΦ = × =∫Ñ

Applying Gauss’s LawApplying Gauss’s Law

• To use Gauss’s law, you want to choose To use Gauss’s law, you want to choose a gaussian surface over which the a gaussian surface over which the surface integral can be simplified and the surface integral can be simplified and the electric field determinedelectric field determined

• Take advantage of symmetryTake advantage of symmetry

• Remember, the gaussian surface is a Remember, the gaussian surface is a surface you choose, it does not have to surface you choose, it does not have to coincide with a real surfacecoincide with a real surface

Conditions for a Gaussian Conditions for a Gaussian SurfaceSurface

• Try to choose a surface that satisfies one or Try to choose a surface that satisfies one or more of these conditions:more of these conditions:

– The value of the electric field can be argued from The value of the electric field can be argued from symmetry to be constant over the surfacesymmetry to be constant over the surface

– The dot product of The dot product of EE..ddAA can be expressed as a can be expressed as a simple algebraic product simple algebraic product EdAEdA because because EE and and ddAA are parallelare parallel

– The dot product is 0 because The dot product is 0 because EE and and ddAA are are perpendicularperpendicular

– The field can be argued to be zero over the surfaceThe field can be argued to be zero over the surface

Field Due to a Point ChargeField Due to a Point Charge

• Choose a sphere as the Choose a sphere as the gaussian surfacegaussian surface

– EE is parallel to is parallel to ddAA at each at each point on the surfacepoint on the surface

2

2 2

(4 )

4

Eo

eo

qd EdA

ε

E dA Eπr

q qE k

πε r r

Φ = × = =

= =

= =

∫ ∫

∫

E AÑ Ñ

Ñ

Field Due to a Spherically Field Due to a Spherically Symmetric Charge DistributionSymmetric Charge Distribution

• Select a sphere as the Select a sphere as the gaussian surfacegaussian surface

• For For r r >>aain

2 24

Eo

eo

qd EdA

ε

Q QE k

πε r r

Φ = × = =

= =

∫ ∫E AÑ Ñ

Spherically Symmetric, cont.Spherically Symmetric, cont.

• Select a sphere as Select a sphere as the gaussian the gaussian surface, surface, rr < < aa

• qq inin < < QQ

• qq inin = = rr (4/3 (4/3ππrr33))in

in2 34

Eo

eo

qd EdA

ε

q QE k r

πε r a

Φ = × = =

= =

∫ ∫E AÑ Ñ

Spherically Symmetric Spherically Symmetric Distribution, finalDistribution, final

• Inside the sphere, Inside the sphere, EE varies linearly with varies linearly with rr

– EE →→ 0 as 0 as rr →→ 0 0

• The field outside the The field outside the sphere is equivalent sphere is equivalent to that of a point to that of a point charge located at charge located at the center of the the center of the spheresphere

Field Due to a Thin Spherical Field Due to a Thin Spherical ShellShell

• Use spheres as the gaussian surfacesUse spheres as the gaussian surfaces

• When When rr > > aa, the charge inside the surface is , the charge inside the surface is QQ and and

EE = = kkeeQQ / / rr22

• When When rr < < aa, the charge inside the surface is 0 and , the charge inside the surface is 0 and EE = 0 = 0

Field at a Distance from a Line Field at a Distance from a Line of Chargeof Charge

• Select a cylindrical Select a cylindrical charge distribution charge distribution – The cylinder has a The cylinder has a

radius of radius of rr and a length and a length of of ℓℓ

• EE is constant in is constant in magnitude and magnitude and perpendicular to the perpendicular to the surface at every point surface at every point on the curved part of on the curved part of the surfacethe surface

Field Due to a Line of Charge, Field Due to a Line of Charge, cont.cont.

• The end view The end view confirms the field is confirms the field is perpendicular to the perpendicular to the curved surfacecurved surface

• The field through the The field through the ends of the cylinder ends of the cylinder is 0 since the field is is 0 since the field is parallel to these parallel to these surfaces surfaces

Field Due to a Line of Charge, Field Due to a Line of Charge, finalfinal

• Use Gauss’s law to find the fieldUse Gauss’s law to find the field

( )

in

2

22

Eo

o

eo

qd EdA

ε

λEπr

ε

λ λE k

πε r r

Φ = × = =

=

= =

∫ ∫E A

ll

Ñ Ñ

Field Due to a Plane of ChargeField Due to a Plane of Charge

• EE must be must be perpendicular to the perpendicular to the plane and must have plane and must have the same magnitude at the same magnitude at all points equidistant all points equidistant from the planefrom the plane

• Choose a small Choose a small cylinder whose axis is cylinder whose axis is perpendicular to the perpendicular to the plane for the gaussian plane for the gaussian surfacesurface

Field Due to a Plane of Field Due to a Plane of Charge, contCharge, cont

• EE is parallel to the curved surface and is parallel to the curved surface and there is no contribution to the surface there is no contribution to the surface area from this curved part of the cylinderarea from this curved part of the cylinder

• The flux through each end of the cylinder The flux through each end of the cylinder is is EAEA and so the total flux is 2 and so the total flux is 2EAEA

Field Due to a Plane of Field Due to a Plane of Charge, finalCharge, final

• The total charge in the surface is The total charge in the surface is σσAA

• Applying Gauss’s lawApplying Gauss’s law

• Note, this does not depend on Note, this does not depend on rr

• Therefore, the field is uniform Therefore, the field is uniform everywhereeverywhere

22E

o o

σA σEA and E

ε εΦ = = =

Electrostatic EquilibriumElectrostatic Equilibrium

• When there is no net motion of charge When there is no net motion of charge within a conductor, the conductor is said to within a conductor, the conductor is said to be in be in electrostatic equilibriumelectrostatic equilibrium

Properties of a Conductor in Properties of a Conductor in Electrostatic EquilibriumElectrostatic Equilibrium

• The electric field is zero everywhere inside The electric field is zero everywhere inside the conductorthe conductor

• If an isolated conductor carries a charge, If an isolated conductor carries a charge, the charge resides on its surfacethe charge resides on its surface

• The electric field just outside a charged The electric field just outside a charged conductor is perpendicular to the surface conductor is perpendicular to the surface and has a magnitude of and has a magnitude of σσ//εεoo

• On an irregularly shaped conductor, the On an irregularly shaped conductor, the surface charge density is greatest at surface charge density is greatest at locations where the radius of curvature is locations where the radius of curvature is the smallestthe smallest

Property 1: Property 1: EEinsideinside = 0 = 0

• Consider a conducting slab in Consider a conducting slab in an external field an external field EE

• If the field inside the conductor If the field inside the conductor were not zero, free electrons in were not zero, free electrons in the conductor would the conductor would experience an electrical forceexperience an electrical force

• These electrons would These electrons would accelerateaccelerate

• These electrons would not be These electrons would not be in equilibriumin equilibrium

• Therefore, there cannot be a Therefore, there cannot be a field inside the conductorfield inside the conductor

Property 1: Property 1: EEinsideinside = 0, cont. = 0, cont.

• Before the external field is applied, free Before the external field is applied, free electrons are distributed throughout the electrons are distributed throughout the conductorconductor

• When the external field is applied, the When the external field is applied, the electrons redistribute until the magnitude of electrons redistribute until the magnitude of the internal field equals the magnitude of the the internal field equals the magnitude of the external fieldexternal field

• There is a net field of zero inside the There is a net field of zero inside the conductorconductor

• This redistribution takes about 10This redistribution takes about 10 -15-15s and can s and can be considered instantaneousbe considered instantaneous

Property 2: Charge Resides on Property 2: Charge Resides on the Surfacethe Surface

• Choose a gaussian surface Choose a gaussian surface inside but close to the actual inside but close to the actual surfacesurface

• The electric field inside is The electric field inside is zero (prop. 1)zero (prop. 1)

• There is no net flux through There is no net flux through the gaussian surfacethe gaussian surface

• Because the gaussian Because the gaussian surface can be as close to surface can be as close to the actual surface as the actual surface as desired, there can be no desired, there can be no charge inside the surfacecharge inside the surface

Property 2: Charge Resides Property 2: Charge Resides on the Surface, conton the Surface, cont

• Since no net charge can be inside the Since no net charge can be inside the surface, any net charge must reside surface, any net charge must reside onon the surfacethe surface

• Gauss’s law does not indicate the Gauss’s law does not indicate the distribution of these charges, only that it distribution of these charges, only that it must be on the surface of the conductormust be on the surface of the conductor

Property 3: Field’s Magnitude Property 3: Field’s Magnitude and Directionand Direction

• Choose a cylinder as Choose a cylinder as the gaussian surfacethe gaussian surface

• The field must be The field must be perpendicular to the perpendicular to the surfacesurface– If there were a parallel If there were a parallel

component to component to EE, , charges would charges would experience a force and experience a force and accelerate along the accelerate along the surface and it would surface and it would not be in equilibriumnot be in equilibrium

Property 3: Field’s Magnitude Property 3: Field’s Magnitude and Direction, cont.and Direction, cont.

• The net flux through the gaussian surface The net flux through the gaussian surface is through only the flat face outside the is through only the flat face outside the conductorconductor

– The field here is perpendicular to the surfaceThe field here is perpendicular to the surface

• Applying Gauss’s lawApplying Gauss’s law

Eo o

σA σEA and E

ε εΦ = = =

Conductors in Equilibrium, Conductors in Equilibrium, exampleexample

• The field lines are The field lines are perpendicular to perpendicular to both conductorsboth conductors

• There are no field There are no field lines inside the lines inside the cylindercylinder

Derivation of Gauss’s LawDerivation of Gauss’s Law

• We will use a solid We will use a solid angle, angle, ΩΩ

• A spherical surface A spherical surface of radius of radius r r contains contains an area element an area element ΔΔAA

• The solid angle The solid angle subtended at the subtended at the center of the sphere center of the sphere is defined to beis defined to be

2

A

r

∆Ω =

Some Notes About Solid Some Notes About Solid AnglesAngles

• AA and and rr22 have the same units, so have the same units, so ΩΩ is a is a dimensionless ratiodimensionless ratio

• We give the name We give the name steradiansteradian to this to this dimensionless ratiodimensionless ratio

• The total solid angle subtended by a The total solid angle subtended by a sphere is 4sphere is 4ππ steradians steradians

Derivation of Gauss’s Law, Derivation of Gauss’s Law, cont.cont.• Consider a point Consider a point

charge, charge, qq, surrounded , surrounded by a closed surface of by a closed surface of arbitrary shapearbitrary shape

• The total flux through The total flux through this surface can be this surface can be found by evaluating found by evaluating EE..ΔΔAA for each small for each small area element and area element and summing over all the summing over all the elementselements

Derivation of Gauss’s Law, Derivation of Gauss’s Law, finalfinal

• The flux through each element isThe flux through each element is

• Relating to the solid angleRelating to the solid angle

– where this is the solid angle subtended by where this is the solid angle subtended by ΔΔAA

• The total flux is The total flux is

( ) 2

coscosE e

AθEθ A k q

r

∆Φ = ×∆ = ∆ =E A

2

cosAθ

r

∆∆Ω =

2

cosE e e

o

dAθ qk q k q d

rεΦ = = Ω =∫ ∫Ñ Ñ

Densidad de las líneas de Densidad de las líneas de campocampo

∆NSuperficie gaussiana

N

Aσ ∆=

∆

Densidad de líneas σ

Ley de Gauss: El campo E en cualquier punto en el espacio es proporcional a la densidad de líneas σ en dicho punto.

Ley de Gauss: El campo E en cualquier punto en el espacio es proporcional a la densidad de líneas σ en dicho punto.

∆A

Radio r

rr

Densidad de líneas y constante de Densidad de líneas y constante de espaciamientoespaciamiento

Considere el campo cerca de una carga positiva q:Considere el campo cerca de una carga positiva q:

Superficie gaussiana

Radio r

rr

Luego, imagine una superficie (radio r) que rodea a q.Luego, imagine una superficie (radio r) que rodea a q.

EE es proporcional a es proporcional a ∆∆N/N/∆∆AA y es y es igual a igual a kq/rkq/r22 en cualquier punto. en cualquier punto.

2;

N kqE E

A r

∆ ∝ =∆

εεοο se define como constante de se define como constante de

espaciamiento. Entonces:espaciamiento. Entonces:

0

1

4 kε

π=:es ε Donde 00E

AN ε=

∆∆

Permitividad del espacio librePermitividad del espacio libreLa constante de proporcionalidad para la densidad de La constante de proporcionalidad para la densidad de líneas se conoce como líneas se conoce como permitividad permitividad εεοο y se define como:y se define como:

2-12

0 2

1 C8.85 x 10

4 N mkε

π= =

⋅

Al recordar la relación con la densidad de líneas se tiene:Al recordar la relación con la densidad de líneas se tiene:

0 0 N

E or N E AA

ε ε∆ = ∆ = ∆∆

Sumar sobre toda el área A Sumar sobre toda el área A da las líneas totales como:da las líneas totales como: N = εoEAN = εoEA

Ejemplo 5. Ejemplo 5. Escriba una ecuación para Escriba una ecuación para encontrar el número total de líneas encontrar el número total de líneas N N que que salen de una sola carga positiva salen de una sola carga positiva qq..

Superficie gaussiana

Radio r

rr

Dibuje superficie gaussiana esférica:Dibuje superficie gaussiana esférica:

22 2

; A = 4 r4

kq qE

r rπ

π= =

Sustituya E y A de:Sustituya E y A de:

20 0 2

(4 )4

qN EA r

rε ε π

π = =

N = εoqA = q N = εoqA = q

El número total de líneas es igual a la carga encerrada q.El número total de líneas es igual a la carga encerrada q.

EANAEN 00 y εε =∆=∆

Ley de GaussLey de GaussLey de Gauss:Ley de Gauss: El número neto de líneas de campo El número neto de líneas de campo eléctrico que cruzan cualquier superficie cerrada en eléctrico que cruzan cualquier superficie cerrada en una dirección hacia afuera es numéricamente igual a la una dirección hacia afuera es numéricamente igual a la carga neta total dentro de dicha superficie.carga neta total dentro de dicha superficie.

Ley de Gauss:Ley de Gauss: El número neto de líneas de campo El número neto de líneas de campo eléctrico que cruzan cualquier superficie cerrada en eléctrico que cruzan cualquier superficie cerrada en una dirección hacia afuera es numéricamente igual a la una dirección hacia afuera es numéricamente igual a la carga neta total dentro de dicha superficie.carga neta total dentro de dicha superficie.

0N EA qε= Σ = Σ

Si Si q q se representa como la se representa como la carga carga positiva neta encerradapositiva neta encerrada, la ley de , la ley de Gauss se puede rescribir como:Gauss se puede rescribir como: 0

qEA

εΣ =

Ejemplo 6.Ejemplo 6. ¿Cuántas líneas de campo ¿Cuántas líneas de campo eléctrico pasan a través de la eléctrico pasan a través de la superficie gaussiana dibujada abajo?superficie gaussiana dibujada abajo?

+

-q1

q4

q3-

+q2

-4 µC

+5 µC

+8 µC

-1 µC

Superficie gaussianaPrimero encuentre la carga Primero encuentre la carga NETA NETA ΣΣqq encerrada por la encerrada por la superficiesuperficie::

ΣΣq = (+8 –4 – 1) = +3 q = (+8 –4 – 1) = +3 µµCC

0N EA qε= Σ = Σ

N = +3 µC = +3 x 10-6 líneasN = +3 µC = +3 x 10-6 líneas

Ejemplo 6.Ejemplo 6. Una esfera sólida (R = 6 cm) con una Una esfera sólida (R = 6 cm) con una carga neta de +8 carga neta de +8 µµC está adentro de un cascarón C está adentro de un cascarón hueco (R = 8 cm) que tiene una carga neta de–6 hueco (R = 8 cm) que tiene una carga neta de–6 µµC. ¿Cuál es el campo eléctrico a una distancia C. ¿Cuál es el campo eléctrico a una distancia de 12 cm desde el centro de la esfera sólida?de 12 cm desde el centro de la esfera sólida?

ΣΣq = (+8 – 6) = +2 q = (+8 – 6) = +2 µµCC

0N EA qε= Σ = Σ-6 µC

+8 µC--

--

-

-- -

Dibuje una esfera gaussiana a un Dibuje una esfera gaussiana a un radio de 12 cm para encontrar E.radio de 12 cm para encontrar E.

8cm

6 cm

12 cm

Superficie gaussiana

00

; net

qAE q E

Aε

εΣ= =

2

2

-6

2 -12 2Nm0 C

2 x 10 C

(4 ) (8.85 x 10 )(4 )(0.12 m)

qE

rε π πΣ += =

Ejemplo 6 (Cont.)Ejemplo 6 (Cont.) ¿Cuál es el campo ¿Cuál es el campo eléctrico a una distancia de 12 cm desde el eléctrico a una distancia de 12 cm desde el centro de la esfera sólida?centro de la esfera sólida?

Dibuje una esfera gaussiana a un Dibuje una esfera gaussiana a un radio de 12 cm para encontrar E.radio de 12 cm para encontrar E.

ΣΣq = (+8 – 6) = +2 q = (+8 – 6) = +2 µµCC

0N EA qε= Σ = Σ

00

; net

qAE q E

Aε

εΣ= =

6 NC2

0

2 C1.25 x 10

(4 )E

r

µε π

+= =

-6 µC

+8 µC--

--

-

-- -

8cm

6 cm

12 cm

Superficie gaussiana

E = 1.25 MN/CE = 1.25 MN/C

Carga sobre la superficie de un conductorCarga sobre la superficie de un conductor

Conductor cargado

Superficie gaussiana justo adentro del conductor

Dado que cargas iguales Dado que cargas iguales se repelen, se esperaría se repelen, se esperaría que toda la carga se que toda la carga se movería hasta llegar al movería hasta llegar al reposo. Entonces, de la reposo. Entonces, de la ley de Gauss. . .ley de Gauss. . .

Como las cargas están en reposo, E = 0 dentro del Como las cargas están en reposo, E = 0 dentro del conductor, por tanto:conductor, por tanto:

0 or 0 = N EA q qε= Σ = Σ Σ

Toda la carga está sobre la superficie; nada dentro del conductorToda la carga está sobre la superficie; nada dentro del conductor

Ejemplo 7.Ejemplo 7. Use la ley de Gauss para encontrar el Use la ley de Gauss para encontrar el campo E justo afuera de la superficie de un campo E justo afuera de la superficie de un conductor. Densidad de carga superficial: conductor. Densidad de carga superficial: σ σ = = q/Aq/A..

Considere Considere q adentro de la cajaq adentro de la caja. . Las líneas de Las líneas de E E a través de a través de todas las áreas son hacia todas las áreas son hacia afuera.afuera.

Densidad de carga superficial σ

++

+ ++

+ ++

+

+ +++A

E2

E1

0 AE qεΣ =Las líneas de E a través de los Las líneas de E a través de los ladoslados se cancelan por simetría. se cancelan por simetría.

E3

E3 E3

E3

εεooEE11A + A + εεooEE22AA = = qq

El campo es cero dentro del conductor, así que EEl campo es cero dentro del conductor, así que E22 = 0 = 0

00

0 0

qE

A

σε ε

= =

Ejemplo 7 (Cont.)Ejemplo 7 (Cont.) Encuentre el campo justo Encuentre el campo justo afuera de la superficie si afuera de la superficie si σ σ = = q/A = q/A = +2 C/m+2 C/m22..

Densidad de carga superficial σ

++

+ ++

+ ++

+

+ +++A

E2

E1 E3

E3 E3

E3

10 0

qE

A

σε ε

= =

Recuerde que los campos Recuerde que los campos laterales se cancelan y el laterales se cancelan y el campo interior es cero, de campo interior es cero, de modo quemodo que

2

2

-6 2

-12 NmC

2 x 10 C/m

8.85 x 10E

+= E = 226,000 N/C E = 226,000 N/C

Campo entre placas paralelasCampo entre placas paralelasCargas iguales y opuestas.Cargas iguales y opuestas.

Dibuje cajas gaussianas en Dibuje cajas gaussianas en cada superficie interior.cada superficie interior.

+++++

Q1 Q2

-----

Campos ECampos E11 y E y E22 a la derecha. a la derecha.

E1

E2

E1

E2

La ley de Gauss para cualquier La ley de Gauss para cualquier caja da el mismo campo (Ecaja da el mismo campo (E11 = E = E22).).

0 AE qεΣ = Σ0 0

qE

A

σε ε

= =

Línea de cargaLínea de carga

r

E

2πr

L

q

Lλ =

A1

A

A2

0

q; =

2 L

qE

rLλ

πε=

02E

r

λπε

=

Los campos debidos Los campos debidos a Aa A11 y A y A2 2 se cancelan se cancelan debido a simetría.debido a simetría.

0

; (2 )q

EA A r Lπε

= =

0 AE qεΣ =

Ejemplo 8:Ejemplo 8: El campo eléctrico a una El campo eléctrico a una distancia de 1.5 m de una línea de carga es distancia de 1.5 m de una línea de carga es 5 x 105 x 1044 N/C. ¿Cuál es la densidad lineal de N/C. ¿Cuál es la densidad lineal de la línea?la línea?

r

EL

q

Lλ =

02E

r

λπε

=02 rEλ πε=

2

2

-12 4CNm

2 (8.85 x 10 )(1.5 m)(5 x 10 N/C)λ π=

E E = 5 x 10= 5 x 1044 N/CN/C r = 1.5 mr = 1.5 m

λ = 4.17 µC/m

Cilindros concéntricosCilindros concéntricos

+ + ++ + + +

+ +

+ + + + ++ + + +

+ +

+ + a

b

λa

λb

r1r2

-6 µCra

rb

12 cm

Superficie gaussiana

λa

λb

Afuera es como un largo Afuera es como un largo alambre cargado:alambre cargado:

Para r >

rb02

a bEr

λ λπε+= Para

rb > r > ra 02aE

r

λπε

=

Ejemplo 9.Ejemplo 9. Dos cilindros concéntricos de radios Dos cilindros concéntricos de radios 33 y y 6 cm6 cm. . La densidad de carga lineal interior es de La densidad de carga lineal interior es de +3 +3 µµC/mC/m y la y la exterior es de exterior es de -5 -5 µµC/mC/m. Encuentre E a una distancia de . Encuentre E a una distancia de 4 4 cmcm desde el centro. desde el centro.

+ + ++ + + +

+ +

+ + + + ++ + + +

+ +

+ + a = 3

cm

b=6 cm

-7 µC/m

+5 µC/m

E = 1.38 x 106 N/C, radialmente hacia afueraE = 1.38 x 106 N/C, radialmente hacia afuera

rr

Dibuje una superficie Dibuje una superficie gaussiana entre los cilindros.gaussiana entre los cilindros.

02bE

r

λπε

=

0

3 C/m

2 (0.04 m)E

µπε

+=

E = 5.00 x 105 N/C, radialmente hacia adentroE = 5.00 x 105 N/C, radialmente hacia adentro

+ + ++ + + +

+ +

+ + + + ++ + + +

+ +

+ + a = 3 cm

b=6 cm

-7 µC/m

+5 µC/m rr

Gaussiana afuera de Gaussiana afuera de ambos cilindros.ambos cilindros.

02a bE

r

λ λπε+=

0

( 3 5) C/m

2 (0.075 m)E

µπε+ −=

Ejemplo 8 (Cont.)Ejemplo 8 (Cont.) A continuación, encuentre E A continuación, encuentre E a una distancia de 7.5 cm desde el centro a una distancia de 7.5 cm desde el centro (afuera de ambos cilindros)(afuera de ambos cilindros)

Resumen de fórmulasResumen de fórmulas

Intensidad de campo eléctrico E:

Intensidad de campo eléctrico E:

Campo eléctrico cerca de muchas cargas:

Campo eléctrico cerca de muchas cargas:

Ley de Gauss para distribuciones de carga.

Ley de Gauss para distribuciones de carga. 0 ;

qEA q

Aε σΣ = Σ =

CN

rkQ

qF

E es Unidad2==

vectorialSuma 2∑=rkQ

E

CONCLUSIÓN: Capítulo 24CONCLUSIÓN: Capítulo 24El campo eléctricoEl campo eléctrico