Reading: Chapter 8 Q 4πε r o k Q r e For r > a Gauss s Law 1
Chapter 8 Gauss s Law
lectric Flux Definition: lectric flux is the product of the magnitude of the electric field and the surface area, A, perpendicular to the field Φ = A The field lines may make some angle θ with the perpendicular to the surface Then Φ = A cos θ normal A Acos 3
lectric Flux: Surface as a Vector Vector, corresponding to a Flat Surface of Area A, is determined by the following rules: the vector is orthogonal to the surface the magnitude of the vector is eual to the area A The first rule 90 oa normal the vector is orthogonal to the surface does not determine the direction of. There are still two possibilities: A Area = A 90 o A or A 90 o You can choose any of them 4
lectric Flux: Surface as a Vector If we consider more complicated surface then the directions of vectors should be adjusted, so the direction of vector is a smooth function of the surface point correct or A A 1 wrong or A 3 5
lectric Flux Definition: lectric flux is the scalar product of electric field and the vector A A A A Acos 0 or A A Acos 0 6
lectric Flux A 1 1 1 area A 1 area A A 90 o o A Acos(90 ) Asin A A 1 1 1 sin if A 1 A sin A A A cos A then 1 flux is negative 1 cos sin flux is positive A sin 1 A1 0 A A and are orthogonal 7
lectric Flux In the more general case, look at a small flat area element A cosθ A i i i i i In general, this becomes lim A da i i A 0 i surface The surface integral means the integral must be evaluated over the surface in uestion The units of electric flux will be N. m /C 8
lectric Flux: Closed Surface The vectors A i point in different directions At each point, they are perpendicular to the surface By convention, they point outward lim A da i i A 0 i surface 9
lectric Flux: Closed Surface Closed surface 1 3 4 5 6 6 is orthogonal to A 3, A 4, A 5, and A 6 3 A3 0 4 A4 0 A A Then 5 5 0 6 6 0 1 o 1 A1 A1cos(90 ) A1sin A A 1 A 3 90 o 1 3 A A 1 4 A 4 3 5 4 ( sin ) A A1 A A sin 0 1 A A Then but 1 10 (no charges inside closed surface)
lectric Flux: Closed Surface A positive point charge,, 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 = k e / r lectric field is perpendicular to the surface at every point, so A has the same direction as at every point. Spherical surface 11
lectric Flux: Closed Surface has the same direction as at every point. ke r A Then da da i i i i i 4 4 0 e Spherical surface 4 k e 0 A r r k r Gauss s Law does not depend on r ONLY BCAUS 1 r 1
lectric Flux: Closed Surface A and have opposite directions at every point. ke r Then da da i i i i i 4 4 0 e Spherical surface 4 k e 0 A r r k r Gauss s Law does not depend on r ONLY BCAUS 1 13 r
Gauss s Law The net flux through any closed surface surrounding a point charge,, is given by /εo and is independent of the shape of that surface The net electric flux through a closed surface that surrounds no charge is zero A i A i 0 0 14
Gauss s Law Gauss s law states da in is the net charge inside the surface ε in o is the total electric field and may have contributions from charges both inside and outside of the surface A i A i 0 0 15
Gauss s Law Gauss s law states da in is the net charge inside the surface ε in o is the total electric field and may have contributions from charges both inside and outside of the surface 5 3 1 1 3 4 0 4 6 7 16
Gauss s Law Gauss s law states da in is the net charge inside the surface ε in o is the total electric field and may have contributions from charges both inside and outside of the surface 5 1 0 3 4 6 7 17
Gauss s Law Gauss s law states da in is the net charge inside the surface ε in o is the total electric field and may have contributions from charges both inside and outside of the surface 5 0 4 6 7 18
Gauss s Law: Problem What is the flux through surface 1 1 0 A A 0 0 1 A0 A 1 A 0 1 A A 3 19
Chapter 8 Gauss s Law: Applications 0
Gauss s Law: Applications Although Gauss s law can, in theory, be solved to find for any charge configuration, in practice it is limited to symmetric situations 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 da 5 ε in o 1 3 4 0 6 3 1 4 1
Gauss s Law: Point Charge SYMMTRY: - direction - along the radius - depends only on radius, r i 0 - Gauss s Law da da A r Then i i i i 0 4 r 0 4 Gaussian Surface Sphere Only in this case the magnitude of electric field is constant on the Gaussian surface and the flux can be easily evaluated ke r - definition of the Flux
Gauss s Law: Applications da ε in o 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. da can be expressed as a simple algebraic product da because and da are parallel The dot product is 0 because and da are perpendicular The field can be argued to be zero over the surface correct Gaussian surface wrong Gaussian surface 3
Gauss s Law: Applications Spherically Symmetric Charge Distribution SYMMTRY: - direction - along the radius - depends only on radius, r The total charge is Q A Select a sphere as the gaussian surface For r >a Q 4πε r in da da 4πr ε o ε o o k Q r e Q The electric field is the same as for the point charge Q 4
Gauss s Law: Applications Spherically Symmetric Charge Distribution Q Q k e 4πεor r The electric field is the same as for the point charge Q!!!!! a Q For r > a Q For r > a 5
Gauss s Law: Applications Spherically Symmetric Charge Distribution SYMMTRY: - direction - along the radius - depends only on radius, r A Select a sphere as the gaussian surface, r < a Q 4 4 3 a 3 r a 3 in 3 r Q 3 Q 3 da da 4πr ε 3 in Qr 1 Q e 3 e 3 o in k k r 4πε r a r a o 6
Gauss s Law: Applications Spherically Symmetric Charge Distribution Inside the sphere, varies linearly with r 0as r 0 The field outside the sphere is euivalent to that of a point charge located at the center of the sphere 7
Gauss s Law: Applications Field due to a thin spherical shell Use spheres as the gaussian surfaces When r > a, the charge inside the surface is Q and = k e Q / r When r < a, the charge inside the surface is 0 and = 0 8
Gauss s Law: Applications Field due to a thin spherical shell When r < a, the charge inside the surface is 0 and = 0 A 1 1 A1 A r 1 A r 1 1 A r the same solid angle 1 r A r k k k k 1 1 1 1 e e e e r1 r1 r1 A r k k k k e e e e r r r 1 1 0 A Only because in Coulomb law 1 r 9
Gauss s Law: Applications Field from a line of charge Select a cylindrical Gaussian surface The cylinder has a radius of r and a length of l Symmetry: is constant in magnitude (depends only on radius r) and perpendicular to the surface at every point on the curved part of the surface The end view 30
Gauss s Law: Applications Field from a line of charge da The flux through this surface is 0 The flux through this surface: The end view in da da πr in λ λ πr εo λ k πεr o e λ r ε o 31
Gauss s Law: Applications Symmetry: Field due to a plane of charge must be perpendicular to the plane and must have the same magnitude at all points euidistant from the plane da Choose a small cylinder whose axis is perpendicular to the plane for the gaussian surface The flux through this surface is 0 3
Gauss s Law: Applications Field due to a plane of charge A A 4 h A 5 da A 3 h A 6 A 1 1 1 A1 A A A A AA A 1 1 The flux through this surface is 0 in A 0 0 A A 0 0 does not depend on h 33
Gauss s Law: Applications 0 e k r 34
Gauss s Law: Applications Q 4 3 3 πa ρ 4 3 πa ρ Q 4 k 3 e r k 3 e r πk 3 eρr a a 3 4 πke ρr 3 35
xample Find electric field inside the hole R a r 36
4 πke ρr 1 3 4 πke ρr 3 R a r R r r 1 r R r 1 r 4 πke ρr 3 a 4 4 4 4 πk eρr1 πkρr e πkρr e ( 1 r) πkρa e const 37 3 3 3 3
xample The sphere has a charge Q and radius a. The point charge Q/8 is placed at the center of the sphere. Find all points where electric field is zero. r 1 Q 1 ke r 3 a k Q 8r e r 1 0 Q ke r k a Q 8r 3 e r 3 a 8 3 r a 38
Chapter 8 Conductors in lectric Field 39
lectric Charges: Conductors and Isolators lectrical conductors are materials in which some of the electrons are free electrons These electrons can move relatively freely through the material xamples of good conductors include copper, aluminum and silver lectrical insulators are materials in which all of the electrons are bound to atoms These electrons can not move relatively freely through the material xamples of good insulators include glass, rubber and wood Semiconductors are somewhere between insulators and conductors 40
lectrostatic uilibrium Definition: when there is no net motion of charge within a conductor, the conductor is said to be in electrostatic euilibrium Because the electrons can move freely through the material no motion means that there are no electric forces no electric forces means that the electric field inside the conductor is 0 F If electric field inside the conductor is not 0, 0 then there is an electric force F and, from the second Newton s law, there is a motion of free electrons. 41
Conductor in lectrostatic uilibrium The electric field is zero everywhere inside the conductor 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 euals the magnitude of the external field There is a net field of zero inside the conductor 4
Conductor in lectrostatic uilibrium If an isolated conductor carries a charge, the charge resides on its surface lectric filed is 0, so the net flux through Gaussian surface is 0 da ε in o Then in 0 43
Conductor in lectrostatic uilibrium The electric field just outside a charged conductor is perpendicular to the surface and has a magnitude of σ/ε o Choose a cylinder as the gaussian surface The field must be perpendicular to the surface If there were a parallel component to, charges would experience a force and accelerate along the surface and it would not be in euilibrium The net flux through the gaussian surface is through only the flat face outside the conductor The field here is perpendicular to the surface σa Gauss s law: A and ε o σ ε o 44
Conductor in lectrostatic uilibrium σ ε o σ σ 0 σ σ σ 45
Conductor in lectrostatic uilibrium: xample Find electric field if the conductor spherical shell has zero charge r 0 r 1 conductor 0 46
Conductor in lectrostatic uilibrium: xample Find electric field if the conductor spherical shell has zero charge 0 r 0 r 1 The total charge inside this Gaussian surface is 0, so the electric field is 0 surface charge, total charge is <0 surface charge, total charge is >0 This is because the total charge of the conductor is 0!!! 47
Conductor in lectrostatic uilibrium: xample Find electric field if the conductor spherical shell has zero charge 0 r r 1 0 The total charge inside this Gaussian surface is, so the electric field is ke r surface charge, total charge is >0 This is because the total charge of the conductor is 0!!! total charge is <0 48
Conductor in lectrostatic uilibrium: xample Find electric field if the conductor spherical shell has zero charge 0 r 0 r 1 ke r ke r surface charge, total charge is >0 This is because the total charge of the conductor is 0!!! total charge is <0 49
Conductor in lectrostatic uilibrium: xample Find electric field if the charge of the conductor spherical shell is Q 0 r 0 r 1 k e Q r ke r surface charge, total charge is Q+>0 This is because the total charge of the conductor is Q!!! total charge is <0 50