- Electric fields can be modeled as continuous distributions when charge densities are high. The continuous charge distribution is divided into small elements and the total field is found by summing the contributions of each element. - Gauss' law relates the electric flux through a closed surface to the net charge enclosed. It can be used to calculate electric fields for symmetric charge distributions.

1. Electric Field – Continuous Charge
Distribution
• The distances between charges in a group of
charges may be much smaller than the distance
between the group and a point of interest
• In this situation, the system of charges can be
modeled as continuous
• The system of closely spaced charges is
equivalent to a total charge that is continuously
distributed along some line, over some surface,
or throughout some volume

2. Electric Field – Continuous
Charge Distribution, cont
• Procedure:
– Divide the charge
distribution into small
elements, each of which
contains Δq
– Calculate the electric
field due to one of these
elements at point P
– Evaluate the total field
by summing the
contributions of all the
charge elements

3. Electric Field – Continuous Charge
Distribution, equations
• For the individual charge elements
• Because the charge distribution is
continuous
2
ˆe
q
k
r
∆
∆ =E r
2 20
ˆ ˆlim
i
i
e i e
q
i i
q dq
k k
r r∆ →
∆
= =∑ ∫E r r

4. Charge Densities
• Volume charge density: when a charge is
distributed evenly throughout a volume
– ρ = Q / V
• Surface charge density: when a charge is
distributed evenly over a surface area
– σ = Q / A
• Linear charge density: when a charge is
distributed along a line
– λ = Q / ℓ

5. Amount of Charge in a Small
Volume
• For the volume: dq = ρ dV
• For the surface: dq = σ dA
• For the length element: dq = λ dℓ

6. Problem Solving Hints
• Units: when using the Coulomb constant, ke, the
charges must be in C and the distances in m
• Calculating the electric field of point charges:
use the superposition principle, find the fields
due to the individual charges at the point of
interest and then add them as vectors to find the
resultant field

7. Problem Solving Hints, cont.
• Continuous charge distributions: the vector
sums for evaluating the total electric field at
some point must be replaced with vector
integrals
– Divide the charge distribution into infinitesimal pieces,
calculate the vector sum by integrating over the entire
charge distribution
• Symmetry: take advantage of any symmetry to
simplify calculations

8. Campo eléctrico debido a una
distribución de carga continua
• Una barra de 14cm esta cargada uniformemente y tiene una
carga total de -22µc.Determina la magnitud y dirección del
campo eléctrico a lo largo del eje de la barra en un punto a 36
cm de su centro

9. Un anillo cargado uniformemente de 10cm de radio
tiene una carga total de 75µc Encuentre el campo
electrico sobre el eje del anillo de a)1cm b)5cm
c)30cm d)100cm

10. Un disco cargado de modo uniforme de 35cm de radio tiene una densidad
de carga de 7.9x10-3 C/m2.Calcule el campo electrico sobre el eje del
disco en a)5cm, b)10cm c)50cm y d)200cm del Centro del disco

11. Electric Field Lines
• Field lines give us a means of representing the
electric field pictorially
• The electric field vector E is tangent to the
electric field line at each point
– The line has a direction that is the same as that of the
electric field vector
• The number of lines per unit area through a
surface perpendicular to the lines is proportional
to the magnitude of the electric field in that
region

12. Electric Field Lines, General
• The density of lines
through surface A is
greater than through
surface B
• The magnitude of the
electric field is greater on
surface A than B
• The lines at different
locations point in different
directions
– This indicates the field is
non-uniform

13. Electric Field Lines, Positive
Point Charge
• The field lines radiate
outward in all directions
– In three dimensions, the
distribution is spherical
• The lines are directed
away from the source
charge
– A positive test charge would
be repelled away from the
positive source charge

14. Electric Field Lines, Negative
Point Charge
• The field lines radiate
inward in all directions
• The lines are directed
toward the source
charge
– A positive test charge
would be attracted
toward the negative
source charge

15. Electric Field Lines – Dipole
• The charges are
equal and opposite
• The number of field
lines leaving the
positive charge
equals the number
of lines terminating
on the negative
charge

16. Electric Field Lines – Like
Charges
• The charges are equal
and positive
• The same number of
lines leave each
charge since they are
equal in magnitude
• At a great distance,
the field is
approximately equal to
that of a single charge
of 2q

17. Electric Field Lines, Unequal
Charges
• The positive charge is
twice the magnitude of the
negative charge
• Two lines leave the
positive charge for each
line that terminates on the
negative charge
• At a great distance, the
field would be
approximately the same
as that due to a single
charge of +q

18. Electric Field Lines – Rules for
Drawing
• The lines must begin on a positive charge and
terminate on a negative charge
– In the case of an excess of one type of charge,
some lines will begin or end infinitely far away
• The number of lines drawn leaving a positive
charge or approaching a negative charge is
proportional to the magnitude of the charge
• No two field lines can cross

19. Motion of Charged Particles
• When a charged particle is placed in an
electric field, it experiences an electrical
force
• If this is the only force on the particle, it
must be the net force
• The net force will cause the particle to
accelerate according to Newton’s second
law

20. Motion of Particles, cont
• Fe = qE = ma
• If E is uniform, then a is constant
• If the particle has a positive charge, its
acceleration is in the direction of the field
• If the particle has a negative charge, its
acceleration is in the direction opposite the
electric field
• Since the acceleration is constant, the kinematic
equations can be used

21. Un electrón entra ala región de un campo eléctrico
uniforme E=200N/C como se muestra en la figura
con una velocidad inicial de 3x10 6 m/s la longitud
horizontal de las placas es 0.1m
Encontrar la aceleracion del electrron mientras se
encuentra en le campo electrico

22. The Cathode Ray Tube (CRT)
• A CRT is commonly used to obtain a
visual display of electronic information in
oscilloscopes, radar systems, televisions,
etc.
• The CRT is a vacuum tube in which a
beam of electrons is accelerated and
deflected under the influence of electric or
magnetic fields

23. CRT, cont
• The electrons are
deflected in various
directions by two sets
of plates
• The placing of charge
on the plates creates
the electric field
between the plates
and allows the beam
to be steered

24. Flux Through Closed Surface, final
• The net flux through the surface is
proportional to the net number of lines
leaving the surface
– This net number of lines is the number of lines
leaving the surface minus the number
entering the surface
• If En is the component of E perpendicular
to the surface, then
E nd E dAΦ = × =∫ ∫E AÑ Ñ

25. Gauss’s Law, Introduction
• Gauss’s law is an expression of the
general relationship between the net
electric flux through a closed surface and
the charge enclosed by the surface
– The closed surface is often called a gaussian
surface
• Gauss’s law is of fundamental importance
in the study of electric fields

26. Gauss’s Law – General
• A positive point
charge, q, is located
at the center of a
sphere of radius r
• The magnitude of
the electric field
everywhere on the
surface of the
sphere is
E = keq / r2

27. Gauss’s Law – General, cont.
• The field lines are directed radially
outward and are perpendicular to the
surface at every point
• This will be the net flux through the
gaussian surface, the sphere of radius r
• We know E = keq/r2
and Asphere = 4πr2
,
E d E dAΦ = × =∫ ∫E AÑ Ñ
4E e
o
q
πk q
ε
Φ = =

28. Gauss’s Law – General, notes
• The net flux through any closed surface surrounding
a point charge, q, is given by q/εo and is independent
of the shape of that surface
• The net electric flux through a closed surface that
surrounds no charge is zero
• Since the electric field due to many charges is the
vector sum of the electric fields produced by the
individual charges, the flux through any closed
surface can be expressed as
( )1 2d d× = + ×∫ ∫E A E E AKÑ Ñ

29. Gauss’s Law – Final
• Gauss’s law states
• qin is the net charge inside the surface
• E represents the electric field at any point on
the surface
– E is the total electric field and may have contributions
from charges both inside and outside of the surface
• Although Gauss’s law can, in theory, be solved
to find E for any charge configuration, in
practice it is limited to symmetric situations
E A in
E
o
q
d
ε
Φ = × =∫Ñ

30. Applying Gauss’s Law
• To use Gauss’s law, you want to choose a
gaussian surface over which the surface
integral can be simplified and the electric
field determined
• Take advantage of symmetry
• Remember, the gaussian surface is a
surface you choose, it does not have to
coincide with a real surface

31. Conditions for a Gaussian Surface
• Try to choose a surface that satisfies one or
more of these conditions:
– The value of the electric field can be argued from
symmetry to be constant over the surface
– The dot product of E.
dA can be expressed as a
simple algebraic product EdA because E and dA
are parallel
– The dot product is 0 because E and dA are
perpendicular
– The field can be argued to be zero over the surface

32. Field Due to a Point Charge
• Choose a sphere as the
gaussian surface
– E is parallel to dA at each
point on the surface
2
2 2
(4 )
4
E
o
e
o
q
d EdA
ε
E dA Eπr
q q
E k
πε r r
Φ = × = =
= =
= =
∫ ∫
∫
E AÑ Ñ
Ñ

33. Field Due to a Spherically
Symmetric Charge Distribution
• Select a sphere as the
gaussian surface
• For r >a
in
2 2
4
E
o
e
o
q
d EdA
ε
Q Q
E k
πε r r
Φ = × = =
= =
∫ ∫E AÑ Ñ

34. Spherically Symmetric, cont.
• Select a sphere as
the gaussian
surface, r < a
• qin < Q
• qin = r (4/3πr3
)
in
in
2 3
4
E
o
e
o
q
d EdA
ε
q Q
E k r
πε r a
Φ = × = =
= =
∫ ∫E AÑ Ñ

35. Spherically Symmetric
Distribution, final
• Inside the sphere, E
varies linearly with r
– E → 0 as r → 0
• The field outside the
sphere is equivalent
to that of a point
charge located at
the center of the
sphere

36. Field Due to a Thin Spherical
Shell
• Use spheres as the gaussian surfaces
• When r > a, the charge inside the surface is Q and
E = keQ / r2
• When r < a, the charge inside the surface is 0 and E = 0

37. Field at a Distance from a Line of
Charge
• Select a cylindrical
charge distribution
– The cylinder has a
radius of r and a length
of ℓ
• E is constant in
magnitude and
perpendicular to the
surface at every point
on the curved part of
the surface

38. Field Due to a Line of Charge,
cont.
• The end view
confirms the field is
perpendicular to the
curved surface
• The field through the
ends of the cylinder
is 0 since the field is
parallel to these
surfaces

39. Field Due to a Line of Charge, final
• Use Gauss’s law to find the field
( )
in
2
2
2
E
o
o
e
o
q
d EdA
ε
λ
Eπr
ε
λ λ
E k
πε r r
Φ = × = =
=
= =
∫ ∫E A
l
l
Ñ Ñ

40. Field Due to a Plane of Charge
• E must be
perpendicular to the
plane and must have
the same magnitude at
all points equidistant
from the plane
• Choose a small
cylinder whose axis is
perpendicular to the
plane for the gaussian
surface

41. Field Due to a Plane of Charge,
cont
• E is parallel to the curved surface and
there is no contribution to the surface area
from this curved part of the cylinder
• The flux through each end of the cylinder
is EA and so the total flux is 2EA

42. Field Due to a Plane of Charge,
final
• The total charge in the surface is σA
• Applying Gauss’s law
• Note, this does not depend on r
• Therefore, the field is uniform everywhere
2
2
E
o o
σA σ
EA and E
ε ε
Φ = = =

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

44. Properties of a Conductor in
Electrostatic Equilibrium
• The electric field is zero everywhere inside the
conductor
• If an isolated conductor carries a charge, the
charge resides on its surface
• The electric field just outside a charged
conductor is perpendicular to the surface and
has a magnitude of σ/εo
• On an irregularly shaped conductor, the surface
charge density is greatest at locations where the
radius of curvature is the smallest

45. Property 1: Einside = 0
• Consider a conducting slab in
an external field E
• If the field inside the conductor
were not zero, free electrons in
the conductor would
experience an electrical force
• These electrons would
accelerate
• These electrons would not be
in equilibrium
• Therefore, there cannot be a
field inside the conductor

46. Property 1: Einside = 0, cont.
• Before the external field is applied, free
electrons are distributed throughout the
conductor
• When the external field is applied, the electrons
redistribute until the magnitude of the internal
field equals the magnitude of the external field
• There is a net field of zero inside the conductor
• This redistribution takes about 10-15
s and can be
considered instantaneous

47. Property 2: Charge Resides on
the Surface
• Choose a gaussian surface
inside but close to the actual
surface
• The electric field inside is
zero (prop. 1)
• There is no net flux through
the gaussian surface
• Because the gaussian
surface can be as close to
the actual surface as
desired, there can be no
charge inside the surface

48. Property 2: Charge Resides on the
Surface, cont
• Since no net charge can be inside the
surface, any net charge must reside on
the surface
• Gauss’s law does not indicate the
distribution of these charges, only that it
must be on the surface of the conductor

49. Property 3: Field’s Magnitude and
Direction
• Choose a cylinder as
the gaussian surface
• The field must be
perpendicular to the
surface
– If there were a parallel
component to E,
charges would
experience a force and
accelerate along the
surface and it would
not be in equilibrium

50. Property 3: Field’s Magnitude and
Direction, cont.
• The net flux through the gaussian surface
is through only the flat face outside the
conductor
– The field here is perpendicular to the surface
• Applying Gauss’s law
E
o o
σA σ
EA and E
ε ε
Φ = = =

51. Conductors in Equilibrium,
example
• The field lines are
perpendicular to
both conductors
• There are no field
lines inside the
cylinder

52. Derivation of Gauss’s Law
• We will use a solid
angle, Ω
• A spherical surface
of radius r contains
an area element ΔA
• The solid angle
subtended at the
center of the sphere
is defined to be 2
A
r
∆
Ω =

53. Some Notes About Solid Angles
• A and r2
have the same units, so Ω is a
dimensionless ratio
• We give the name steradian to this
dimensionless ratio
• The total solid angle subtended by a
sphere is 4π steradians

54. Derivation of Gauss’s Law, cont.
• Consider a point
charge, q, surrounded
by a closed surface of
arbitrary shape
• The total flux through
this surface can be
found by evaluating
E.
ΔA for each small
area element and
summing over all the
elements

55. Derivation of Gauss’s Law, final
• The flux through each element is
• Relating to the solid angle
– where this is the solid angle subtended by ΔA
• The total flux is
( ) 2
cos
cosE e
Aθ
Eθ A k q
r
∆
Φ = ×∆ = ∆ =E A
2
cosAθ
r
∆
∆Ω =
2
cos
E e e
o
dAθ q
k q k q d
rε
Φ = = Ω =∫ ∫Ñ Ñ

56. Densidad de las líneas de campo
∆NSuperficie gaussiana
N
A
σ
∆
=
∆
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

57. Densidad de líneas y constante de
espaciamiento
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 aes proporcional a ∆∆N/N/∆∆AA y esy es
igual aigual a kq/rkq/r22
en cualquier punto.en cualquier punto.
2
;
N kq
E E
A r
∆
∝ =
∆
εεοο se define como constante dese define como constante de
espaciamiento. Entonces:espaciamiento. Entonces:
0
1
4 k
ε
π
=:esεDonde 00E
A
N
ε=
∆
∆

58. Permitividad del espacio libre
La constante de proporcionalidad para la densidad deLa constante de proporcionalidad para la densidad de
líneas se conoce comolíneas se conoce como permitividadpermitividad εεοο y se define como:y se define como:
2
-12
0 2
1 C
8.85 x 10
4 N mk
ε
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 A
A
ε ε
∆
= ∆ = ∆
∆
Sumar sobre toda el área ASumar sobre toda el área A
da las líneas totales como:da las líneas totales como: N = εoEAN = εoEA

59. Ejemplo 5. Escriba una ecuación para encontrar el
número total de líneas N que salen de una sola
carga positiva q.
Superficie gaussiana
Radio r
rr
Dibuje superficie gaussiana esférica:Dibuje superficie gaussiana esférica:
2
2 2
; A = 4 r
4
kq q
E
r r
π
π
= =
Sustituya E y A de:Sustituya E y A de:
2
0 0 2
(4 )
4
q
N EA r
r
ε ε π
π
 
= =  
 
N = εoqA = qN = 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 =∆=∆

60. Ley de Gauss
Ley de Gauss:Ley de Gauss: El número neto de líneas de campoEl número neto de líneas de campo
eléctrico que cruzan cualquier superficie cerrada eneléctrico que cruzan cualquier superficie cerrada en
una dirección hacia afuera es numéricamente igual a launa 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 campoEl número neto de líneas de campo
eléctrico que cruzan cualquier superficie cerrada eneléctrico que cruzan cualquier superficie cerrada en
una dirección hacia afuera es numéricamente igual a launa 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
SiSi qq se representa como lase representa como la cargacarga
positiva neta encerradapositiva neta encerrada, la ley de, la ley de
Gauss se puede rescribir como:Gauss se puede rescribir como: 0
q
EA
ε
Σ =

61. Ejemplo 6. ¿Cuántas líneas de campo eléctrico
pasan a través de la superficie gaussiana
dibujada abajo?
+
-q1
q4
q3
-
+q2
-4 µC
+5 µC
+8 µC
-1 µC
Superficie gaussiana
Primero encuentre la cargaPrimero encuentre la carga
NETANETA ΣΣqq encerrada por laencerrada por la
superficiesuperficie::
ΣΣq = (+8 –4 – 1) = +3q = (+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

62. Ejemplo 6. Una esfera sólida (R = 6 cm) con una carga neta de
+8 µC está adentro de un cascarón hueco (R = 8 cm) que tiene
una carga neta de–6 µC. ¿Cuál es el campo eléctrico a una
distancia de 12 cm desde el centro de la esfera sólida?
ΣΣq = (+8 – 6) = +2q = (+8 – 6) = +2 µµCC
0N EA qε= Σ = Σ
-6 µC
+8 µC-
-
-
-
-
-
- -
Dibuje una esfera gaussiana a unDibuje 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
0
0
;net
q
AE q E
A
ε
ε
Σ
= =
2
2
-6
2 -12 2Nm
0 C
2 x 10 C
(4 ) (8.85 x 10 )(4 )(0.12 m)
q
E
rε π π
Σ +
= =

63. Ejemplo 6 (Cont.) ¿Cuál es el campo eléctrico a una
distancia de 12 cm desde el centro de la esfera sólida?
Dibuje una esfera gaussiana a unDibuje una esfera gaussiana a un
radio de 12 cm para encontrar E.radio de 12 cm para encontrar E.
ΣΣq = (+8 – 6) = +2q = (+8 – 6) = +2 µµCC
0N EA qε= Σ = Σ
0
0
;net
q
AE q E
A
ε
ε
Σ
= =
6 N
C2
0
2 C
1.25 x 10
(4 )
E
r
µ
ε π
+
= =
-6 µC
+8 µC-
-
-
-
-
-
- -
8cm
6 cm
12 cm
Superficie gaussiana
E = 1.25 MN/CE = 1.25 MN/C

64. Carga sobre la superficie de un conductor
Conductor cargado
Superficie gaussiana justo
adentro del conductor
Dado que cargas igualesDado que cargas iguales
se repelen, se esperaríase repelen, se esperaría
que toda la carga seque toda la carga se
movería hasta llegar almovería hasta llegar al
reposo. Entonces, de lareposo. Entonces, de la
ley de Gauss. . .ley de Gauss. . .
Como las cargas están en reposo, E = 0 dentro delComo 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

65. Ejemplo 7. Use la ley de Gauss para encontrar el campo E just
afuera de la superficie de un conductor. Densidad de carga
superficial: σ = q/A.
ConsidereConsidere q adentro de la cajaq adentro de la caja..
Las líneas deLas líneas de EE a través dea través de
todas las áreas son haciatodas las áreas son hacia
afuera.afuera.
Densidad de carga superficial σ
++
+ +
+
+ +
+
+
+ +++
A
E2
E1
0 AE qεΣ =
Las líneas de E a través de losLas 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
q
E
A
σ
ε ε
= =

66. Ejemplo 7 (Cont.) Encuentre el campo justo
afuera de la superficie si σ = q/A = +2 C/m2
.
Densidad de carga superficial σ
++
+ +
+
+ +
+
+
+ +++
A
E2
E1 E3
E3 E3
E3
1
0 0
q
E
A
σ
ε ε
= =
Recuerde que los camposRecuerde que los campos
laterales se cancelan y ellaterales se cancelan y el
campo interior es cero, decampo interior es cero, de
modo quemodo que
2
2
-6 2
-12 Nm
C
2 x 10 C/m
8.85 x 10
E
+
= E = 226,000 N/CE = 226,000 N/C

67. Campo entre placas paralelas
Cargas iguales y opuestas.Cargas iguales y opuestas.
Dibuje cajas gaussianas enDibuje cajas gaussianas en
cada superficie interior.cada superficie interior.
+
+
+
+
+
Q1 Q2
-
-
-
-
-
Campos ECampos E11 y Ey E22 a la derecha.a la derecha.
E1
E2
E1
E2
La ley de Gauss para cualquierLa ley de Gauss para cualquier
caja da el mismo campo (Ecaja da el mismo campo (E11 = E= E22).).
0 AE qεΣ = Σ
0 0
q
E
A
σ
ε ε
= =

68. Línea de carga
r
E
2πr
L
q
L
λ =
A1
A
A2
0
q
; =
2 L
q
E
rL
λ
πε
=
02
E
r
λ
πε
=
Los campos debidosLos campos debidos
a Aa A11 y Ay A22 se cancelanse cancelan
debido a simetría.debido a simetría.
0
; (2 )
q
EA A r Lπ
ε
= =
0 AE qεΣ =

69. Ejemplo 8: El campo eléctrico a una distancia de 1.5 m
de una línea de carga es 5 x 104
N/C. ¿Cuál es la
densidad lineal de la línea?
r
EL
q
L
λ =
02
E
r
λ
πε
=
02 rEλ πε=
2
2
-12 4C
Nm
2 (8.85 x 10 )(1.5 m)(5 x 10 N/C)λ π=
EE = 5 x 10= 5 x 1044
N/CN/C r = 1.5 mr = 1.5 m
λ = 4.17 µC/m

70. Cilindros concéntricos
+ + +
+ + + +
+ +
+ + + + +
+ + + +
+ +
+ +
a
b
λa
λb
r1
r2
-6 µC
ra
rb
12 cm
Superficie gaussiana
λa
λb
Afuera es como un largoAfuera es como un largo
alambre cargado:alambre cargado:
Para
r >
rb
02
a b
E
r
λ λ
πε
+
=
Para
rb > r > ra 02
a
E
r
λ
πε
=

71. Ejemplo 9. Dos cilindros concéntricos de radios 3 y 6 cm. La densidad
de carga lineal interior es de +3 µC/m y la exterior es de -5 µC/m.
Encuentre E a una distancia de 4 cm 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 superficieDibuje una superficie
gaussiana entre los cilindros.gaussiana entre los cilindros.
02
b
E
r
λ
πε
=
0
3 C/m
2 (0.04 m)
E
µ
πε
+
=

72. 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 deGaussiana afuera de
ambos cilindros.ambos cilindros.
02
a b
E
r
λ λ
πε
+
=
0
( 3 5) C/m
2 (0.075 m)
E
µ
πε
+ −
=
Ejemplo 8 (Cont.) A continuación, encuentre E a una
distancia de 7.5 cm desde el centro (afuera de ambos
cilindros)

73. Resumen 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 ;
q
EA q
A
ε σΣ = Σ =
C
N
r
kQ
q
F
E esUnidad2
==
vectorialSuma2∑=
r
kQ
E

74. CONCLUSIÓN: Capítulo 24
El campo eléctrico