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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
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
Electric Field – Continuous Charge Distribution, equations
• For the individual charge elements
• Because the charge distribution is continuous
2ˆ
e
qk
r
∆∆ =E r
2 20ˆ ˆlim
i
ie i eq
i i
q dqk k
r r∆ →
∆= =∑ ∫E r r
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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 eo
qπk q
εΦ = =
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
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 inE
o
qd
εΦ = × =∫Ñ
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
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
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
Eo
eo
qd EdA
ε
E dA Eπr
q qE k
πε r r
Φ = × = =
= =
= =
∫ ∫
∫
E AÑ Ñ
Ñ
Field Due to a Spherically Symmetric Charge Distribution
• Select a sphere as the gaussian surface
• For r >a
in
2 24
Eo
eo
qd EdA
ε
Q QE k
πε r r
Φ = × = =
= =
∫ ∫E AÑ Ñ
Spherically Symmetric, cont.
• Select a sphere as the gaussian surface, r < a
• qin < Q
• qin = r (4/3πr3)
in
in2 34
Eo
eo
qd EdA
ε
q QE k r
πε r a
Φ = × = =
= =
∫ ∫E AÑ Ñ
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
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
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
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
Field Due to a Line of Charge, final
• Use 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 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
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
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
22E
o o
σA σEA and E
ε εΦ = = =
Electrostatic Equilibrium
• When there is no net motion of charge within a conductor, the conductor is said to be in electrostatic equilibrium
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
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
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 10-15s and can be
considered instantaneous
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
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
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
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
Eo o
σA σEA and E
ε εΦ = = =
Conductors in Equilibrium, example
• The field lines are perpendicular to both conductors
• There are no field lines inside the cylinder
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 =
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
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
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
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 campo
∆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 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 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 00EA
N ε=∆∆
Permitividad 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. 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:
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 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. ¿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 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. 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) = +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.) ¿Cuál es el campo eléctrico a una distancia de 12 cm desde el 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 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. Use la ley de Gauss para encontrar el campo E justo afuera de la superficie de un conductor. Densidad de carga
superficial: σ = q/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.) 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
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 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 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: 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
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 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. 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 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.) A continuación, encuentre E a una distancia de 7.5 cm desde el centro (afuera de ambos
cilindros)
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 ;
qEA q
Aε σΣ = Σ =
CN
rkQ
qF
E es Unidad2==
vectorialSuma 2∑=rkQ
E
CONCLUSIÓN: Capítulo 24El campo eléctrico
