Electric Field Continuous Charge DistributionThe distances between charges in a group of charges may be much smaller than the distance between the group and a point of interestIn this situation, the system of charges can be modeled as continuousThe 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, contProcedure:Divide the charge distribution into small elements, each of which contains qCalculate the electric field due to one of these elements at point PEvaluate the total field by summing the contributions of all the charge elements

Electric Field Continuous Charge Distribution, equationsFor the individual charge elements

Because the charge distribution is continuous

Charge DensitiesVolume charge density: when a charge is distributed evenly throughout a volume = Q / VSurface charge density: when a charge is distributed evenly over a surface area = Q / ALinear charge density: when a charge is distributed along a line = Q /

Amount of Charge in a Small VolumeFor the volume: dq = dVFor the surface: dq = dAFor the length element: dq = d

Problem Solving HintsUnits: when using the Coulomb constant, ke, the charges must be in C and the distances in mCalculating 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 integralsDivide the charge distribution into infinitesimal pieces, calculate the vector sum by integrating over the entire charge distributionSymmetry: take advantage of any symmetry to simplify calculations

Campo elctrico debido a una distribucin de carga continuaUna barra de 14cm esta cargada uniformemente y tiene una carga total de -22c.Determina la magnitud y direccin del campo elctrico 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 75c 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 LinesField lines give us a means of representing the electric field pictoriallyThe electric field vector E is tangent to the electric field line at each pointThe line has a direction that is the same as that of the electric field vectorThe 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, GeneralThe density of lines through surface A is greater than through surface BThe magnitude of the electric field is greater on surface A than BThe lines at different locations point in different directionsThis indicates the field is non-uniform

Electric Field Lines, Positive Point ChargeThe field lines radiate outward in all directionsIn three dimensions, the distribution is sphericalThe lines are directed away from the source chargeA positive test charge would be repelled away from the positive source charge

Electric Field Lines, Negative Point ChargeThe field lines radiate inward in all directionsThe lines are directed toward the source chargeA positive test charge would be attracted toward the negative source charge

Electric Field Lines Dipole The charges are equal and oppositeThe number of field lines leaving the positive charge equals the number of lines terminating on the negative charge

Electric Field Lines Like ChargesThe charges are equal and positiveThe same number of lines leave each charge since they are equal in magnitudeAt a great distance, the field is approximately equal to that of a single charge of 2q

Electric Field Lines, Unequal ChargesThe positive charge is twice the magnitude of the negative chargeTwo lines leave the positive charge for each line that terminates on the negative chargeAt a great distance, the field would be approximately the same as that due to a single charge of +q

Electric Field Lines Rules for DrawingThe lines must begin on a positive charge and terminate on a negative chargeIn the case of an excess of one type of charge, some lines will begin or end infinitely far awayThe number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the chargeNo two field lines can cross

Motion of Charged ParticlesWhen a charged particle is placed in an electric field, it experiences an electrical forceIf this is the only force on the particle, it must be the net forceThe net force will cause the particle to accelerate according to Newtons second law

Motion of Particles, contFe = qE = maIf E is uniform, then a is constantIf the particle has a positive charge, its acceleration is in the direction of the fieldIf the particle has a negative charge, its acceleration is in the direction opposite the electric fieldSince the acceleration is constant, the kinematic equations can be used

Un electrn entra ala regin de un campo elctrico 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, contThe electrons are deflected in various directions by two sets of platesThe placing of charge on the plates creates the electric field between the plates and allows the beam to be steered

Flux Through Closed Surface, finalThe net flux through the surface is proportional to the net number of lines leaving the surfaceThis net number of lines is the number of lines leaving the surface minus the number entering the surfaceIf En is the component of E perpendicular to the surface, then

Gausss Law, IntroductionGausss law is an expression of the general relationship between the net electric flux through a closed surface and the charge enclosed by the surfaceThe closed surface is often called a gaussian surfaceGausss law is of fundamental importance in the study of electric fields

Gausss Law General A positive point charge, q, is located at the center of a sphere of radius rThe magnitude of the electric field everywhere on the surface of the sphere is E = keq / r2

Gausss 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 rWe know E = keq/r2 and Asphere = 4r2,

Gausss Law General, notesThe net flux through any closed surface surrounding a point charge, q, is given by q/o and is independent of the shape of that surfaceThe net electric flux through a closed surface that surrounds no charge is zeroSince 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

Gausss Law FinalGausss law states

qin is the net charge inside the surfaceE represents the electric field at any point on the surfaceE is the total electric field and may have contributions from charges both inside and outside of the surfaceAlthough Gausss law can, in theory, be solved to find E for any charge configuration, in practice it is limited to symmetric situations

Applying Gausss LawTo use Gausss law, you want to choose a gaussian surface over which the surface integral can be simplified and the electric field determinedTake advantage of symmetryRemember, the gaussian surface is a surface you choose, it does not have to coincide with a real surface

Conditions for a Gaussian SurfaceTry 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 surfaceThe dot product of E.dA can be expressed as a simple algebraic product EdA because E and dA are parallelThe dot product is 0 because E and dA are perpendicularThe field can be argued to be zero over the surface

Field Due to a Point ChargeChoose a sphere as the gaussian surfaceE is parallel to dA at each point on the surface

Field Due to a Spherically Symmetric Charge DistributionSelect a sphere as the gaussian surfaceFor r >a

Spherically Symmetric, cont.Select a sphere as the gaussian surface, r < aqin < Qqin = r (4/3r3)

Spherically Symmetric Distribution, finalInside the sphere, E varies linearly with rE 0 as r 0The 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 ShellUse spheres as the gaussian surfacesWhen r > a, the charge inside the surface is Q and E = keQ / r2When r < a, the charge inside the surface is 0 and E = 0

Field at a Distance from a Line of ChargeSelect a cylindrical charge distribution The cylinder has a radius of r and a length of 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 surfaceThe field through the ends of the cylinder is 0 since the field is parallel to these surfaces

Field Due to a Line of Charge, finalUse Gausss law to find the field

Field Due to a Plane of ChargeE must be perpendicular to the plane and must have the same magnitude at all points equidistant from the planeChoose a small cylinder whose axis is perpendicular to the plane for the gaussian surface

Field Due to a Plane of Charge, contE is parallel to the curved surface and there is no contribution to the surface area from this curved part of the cylinderThe flux through each end of the cylinder is EA and so the total flux is 2EA

Field Due to a Plane of Charge, finalThe total charge in the surface is AApplying Gausss law

Note, this does not depend on rTherefore, the field is uniform everywhere

Electrostatic EquilibriumWhen 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 EquilibriumThe electric field is zero everywhere inside the conductorIf an isolated conductor carries a charge, the charge resides on its surfaceThe electric field just outside a charged conductor is perpendicular to the surface and has a magnitude of /oOn an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature is the smallest

Property 1: Einside = 0Consider a conducting slab in an external field EIf the field inside the conductor were not zero, free electrons in the conductor would experience an electrical forceThese electrons would accelerateThese electrons would not be in equilibriumTherefore, 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 conductorWhen the external field is applied, the electrons redistribute until the magnitude of the internal field equals the magnitude of the external fieldThere is a net field of zero inside the conductorThis redistribution takes about 10-15s and can be considered instantaneous

Property 2: Charge Resides on the SurfaceChoose a gaussian surface inside but close to the actual surfaceThe electric field inside is zero (prop. 1)There is no net flux through the gaussian surfaceBecause 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, contSince no net charge can be inside the surface, any net charge must reside on the surfaceGausss law does not indicate the distribution of these charges, only that it must be on the surface of the conductor

Property 3: Fields Magnitude and DirectionChoose a cylinder as the gaussian surfaceThe field must be perpendicular to the surfaceIf 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: Fields Magnitude and Direction, cont.The net flux through the gaussian surface is through only the flat face outside the conductorThe field here is perpendicular to the surfaceApplying Gausss law

Conductors in Equilibrium, exampleThe field lines are perpendicular to both conductorsThere are no field lines inside the cylinder

Derivation of Gausss LawWe will use a solid angle, A spherical surface of radius r contains an area element AThe solid angle subtended at the center of the sphere is defined to be

Some Notes About Solid AnglesA and r2 have the same units, so is a dimensionless ratioWe give the name steradian to this dimensionless ratioThe total solid angle subtended by a sphere is 4 steradians

Derivation of Gausss Law, cont.Consider a point charge, q, surrounded by a closed surface of arbitrary shapeThe total flux through this surface can be found by evaluating E.A for each small area element and summing over all the elements

Derivation of Gausss Law, finalThe flux through each element is

Relating to the solid angle

where this is the solid angle subtended by AThe total flux is

Densidad de las lneas de campoDNSuperficie gaussianaDensidad de lneas sLey de Gauss: El campo E en cualquier punto en el espacio es proporcional a la densidad de lneas s en dicho punto. DA

Densidad de lneas y constante de espaciamientoConsidere el campo cerca de una carga positiva q:Superficie gaussianaRadio rLuego, imagine una superficie (radio r) que rodea a q.E es proporcional a DN/DA y es igual a kq/r2 en cualquier punto.eo se define como constante de espaciamiento. Entonces:

Permitividad del espacio libreLa constante de proporcionalidad para la densidad de lneas se conoce como permitividad eo y se define como:Al recordar la relacin con la densidad de lneas se tiene:Sumar sobre toda el rea A da las lneas totales como:N = eoEA

Ejemplo 5. Escriba una ecuacin para encontrar el nmero total de lneas N que salen de una sola carga positiva q.Dibuje superficie gaussiana esfrica:Sustituya E y A de: N = eoqA = qEl nmero total de lneas es igual a la carga encerrada q.

Ley de GaussLey de Gauss: El nmero neto de lneas de campo elctrico que cruzan cualquier superficie cerrada en una direccin hacia afuera es numricamente igual a la carga neta total dentro de dicha superficie.Si q se representa como la carga positiva neta encerrada, la ley de Gauss se puede rescribir como:

Ejemplo 6. Cuntas lneas de campo elctrico pasan a travs de la superficie gaussiana dibujada abajo?Primero encuentre la carga NETA Sq encerrada por la superficie:Sq = (+8 4 1) = +3 mCN = +3 mC = +3 x 10-6 lneas

Ejemplo 6. Una esfera slida (R = 6 cm) con una carga neta de +8 mC est adentro de un cascarn hueco (R = 8 cm) que tiene una carga neta de6 mC. Cul es el campo elctrico a una distancia de 12 cm desde el centro de la esfera slida?Sq = (+8 6) = +2 mCDibuje una esfera gaussiana a un radio de 12 cm para encontrar E.8cm12 cm

Ejemplo 6 (Cont.) Cul es el campo elctrico a una distancia de 12 cm desde el centro de la esfera slida?Dibuje una esfera gaussiana a un radio de 12 cm para encontrar E.E = 1.25 MN/C

Carga sobre la superficie de un conductorDado que cargas iguales se repelen, se esperara que toda la carga se movera hasta llegar al reposo. Entonces, de la ley de Gauss. . .Como las cargas estn en reposo, E = 0 dentro del conductor, por tanto:Toda la carga est 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: s = q/A.Considere q adentro de la caja. Las lneas de E a travs de todas las reas son hacia afuera.++Las lneas de E a travs de los lados se cancelan por simetra.eoE1A + eoE2A = qEl campo es cero dentro del conductor, as que E2 = 0

Ejemplo 7 (Cont.) Encuentre el campo justo afuera de la superficie si s = q/A = +2 C/m2.Recuerde que los campos laterales se cancelan y el campo interior es cero, de modo que E = 226,000 N/C

Campo entre placas paralelasCargas iguales y opuestas.Dibuje cajas gaussianas en cada superficie interior.Campos E1 y E2 a la derecha.La ley de Gauss para cualquier caja da el mismo campo (E1 = E2).

Lnea de cargaLos campos debidos a A1 y A2 se cancelan debido a simetra.

Ejemplo 8: El campo elctrico a una distancia de 1.5 m de una lnea de carga es 5 x 104 N/C. Cul es la densidad lineal de la lnea?E = 5 x 104 N/Cr = 1.5 ml = 4.17 mC/m

Cilindros concntricosAfuera es como un largo alambre cargado:

Ejemplo 9. Dos cilindros concntricos de radios 3 y 6 cm. La densidad de carga lineal interior es de +3 mC/m y la exterior es de -5 mC/m. Encuentre E a una distancia de 4 cm desde el centro.E = 1.38 x 106 N/C, radialmente hacia afueraDibuje una superficie gaussiana entre los cilindros.

E = 5.00 x 105 N/C, radialmente hacia adentroGaussiana afuera de ambos cilindros.Ejemplo 8 (Cont.) A continuacin, encuentre E a una distancia de 7.5 cm desde el centro (afuera de ambos cilindros)

Resumen de frmulasIntensidad de campo elctrico E:Campo elctrico cerca de muchas cargas:Ley de Gauss para distribuciones de carga.

CONCLUSIN: Captulo 24El campo elctrico