Gauss s law and electic flux Gauss s Law Chapte 4 Gauss s law is based on the concept of flux: You can think of the flux though some suface as a measue of the numbe of field lines which pass though that suface. Flux depends on the stength of, on the suface aea, and on the elative oientation of the field and suface. A aea A Hee the flux is Φ = A Nomal to suface, magnitude A lectic flux The flux also depends on oientation: aea A lectic flux But what if the electic field is not constant? What if it vaies (possibly in both magnitude and diection) as a function of? This possibility is sketched hee fo the case of a closed suface. Angle (top of aea A is tilted fowad) aea A cos The numbe of field lines though the tilted suface equals the numbe though its pojection. Hence the flux though the tilted suface is simply given by the flux though its pojection: (Acos). Hee flux Φ = A cos = A Closed suface da Field lines How do you calculate the flux? Beak the suface into bits da. lectic flux But what if the electic field is not constant? What if it vaies (possibly in both magnitude and diection) as a function of? This possibility is sketched hee fo the case of a closed suface. lectic flux But what if the electic field is not constant? What if it vaies (possibly in both magnitude and diection) as a function of? This possibility is sketched hee fo the case of a closed suface. da is a small aea element, like A. Flux though da is dφ = da cos da dφ = da So how do we find the TOTAL Flux? We have to sum all the dφ s ove the entie suface: Φ = Σ dφ Fo accuacy the da s da and thus dφ s must be vey small.
lectic flux But what if the electic field is not constant? What if it vaies (possibly in both magnitude and diection) as a function of? This possibility is sketched hee fo the case of a closed suface. da Φ = lim = lim = d A dφ da 0 closed suface d A da 0 closed suface closed suface The loop means the integal is ove a closed suface. Gauss s Law lectic flux though any closed suface = (chage inside) / ε 0 Hence, Gauss Law states: Φ = dφ = da = inside ε 0 This is always tue. It s sometimes useless, but often a vey easy way to find the electic field (fo highly symmetic cases). q Apply Gauss s law to a point chage Conside a positive point chage q. Define a suface (i.e. a closed suface) which is a sphee of adius. By symmety, the lines of must be adially outwads, with magnitude depending only on. q = da = da = da ε q 0 using da = total aea = 4π 1 q gives = 4πε 0 Symmety Apply Gauss s Law to a point chage and what do you get? Answe: Coulomb s Law! We used the fact that a point chage in space is spheically symmetic. Gauss s Law is always tue, but is only useful fo poblems with usable symmety. Coulomb s Law! Is Gauss s Law moe fundamental than Coulomb s Law? Maybe? Hee we deived Coulomb s law fo a point chage fom Gauss s law. One can instead deive Gauss s law fo a geneal (even vey nasty) chage distibution fom Coulomb s law. The two laws ae equivalent. Gauss s law gives us an easy way to solve vey symmetic poblems in electostatics. Gauss s law also gives us geat insight into the electic fields in and on conductos and within voids inside metals. Gauss s law has applications in electicity, magnetism, and even gavity. Mathematically, it applies fundamentally to vecto fields and thei potentials. Mathematically, Gauss s law is vey fundamental. The total flux within a closed suface Gauss s Law Φ = da = is popotional to the enclosed chage. Q enclosed Gauss s Law is always tue, but is only useful fo poblems with usable symmety. ε 0
Symmety and the lectic Field Can we figue out how the field vaies with distance fom the field lines and the symmety? Symmety and the lectic Field Can we figue out how the field vaies with distance fom the field lines and the symmety? Look at a point chage: How do its field lines look? q This is a D pictue. Next we ll ty to get a pictue of a 3D piece. Field lines point out (o in) adially in all diections (3D) Symmety and the lectic Field Symmety and the lectic Field Can we figue out how the field vaies with distance fom the field lines and the symmety? Look at a point chage: How do its field lines look? q ecall that the magnitude of is elated to the density of field lines pe unit aea. Field lines point out (o in) adially in all diections (3D) The lines spead in diections. How does the numbe of field lines pe unit aea vay with distance? Invese squae law: 1 Symmety and the lectic Field Symmety and the lectic Field Line of chage: Line of chage: The lines spead in 1 diection. In this case only the vetical diection. Now how does flux density vay with distance? Now how does flux density vay with distance? 1
Symmety and the lectic Field Symmety and the lectic Field Sheet of chage: Sheet of chage: The lines don t spead at all. Now how does the field change with distance? Now how does field change with distance? Field is a constant! (If sheet is infinite.) Applications of Gauss s Law Poblem: Sphee of Chage Q A chage Q is unifomly distibuted though a sphee of adius. What is the electic field as a function of? Find at 1 and. Gauss s Law does what we just did above, but does it igoously. We ae now going to look at vaious chaged objects and use Gauss s law to find the field distibution. 1 Poblem: Sphee of Chage Q Poblem: Sphee of Chage Q A chage Q is unifomly distibuted though a sphee of adius. What is the electic field as a function of? Find at 1 and. ( ) 1 ( 1 ) Use symmety! This is spheically symmetic. That means that () is adially outwad, and that all points at a given adius ( =) have the same magnitude of field. Fist find () at a point outside the chaged sphee. Apply Gauss s law, using as the suface the sphee of adius pictued. & da What is the enclosed chage? Q
Poblem: Sphee of Chage Q Fist find () at a point outside the chaged sphee. Apply Gauss s law, using as the suface the sphee of adius pictued. & da What is the enclosed chage? Q What is the flux though this suface? Φ = d A = da = da = A = (4π ) xactly as though all the chage wee at the oigin! (fo >) Gauss: So Φ = Q enclosed / ε = Q/ o ε o Q/ε 0 =Φ=(4π )!" (!" 1 Q ) = 4πε o ˆ Poblem: Sphee of Chage Q Next find () at a point inside the sphee. Apply Gauss s law, using a little sphee of adius as a What is the enclosed chage? That takes a little effot. The little sphee has some faction of the total chage. What faction? That s given by volume atio: Q enc = 3 Q () 3 Again the flux is: Φ = A = (4π ) 3 3 ( / )Q settingφ = Q /ε enc o gives = 4πε o Q fo <, ( ) = ˆ 3 4πε o Poblem: Sphee of Chage Q Poblem: Sphee of Chage Q () Look close at these esults. The electic field at comes fom a sum ove the contibutions of all the little bits. Q > () is popotional to fo < () is popotional to 1/ fo > and () is continuous at It s obvious that the net at this point will be hoiontal. But the magnitude fom each bit is diffeent; and it s completely not obvious that the magnitude just depends on the distance fom the sphee s cente to the obsevation point. Doing this as a volume integal would be HAD. Gauss s law is ASY. Poblem: Sphee of Chage Q Poblem: Infinite chaged plane Now look at an obsevation point Q inside the sphee. < Because of the spheical symmety, the contibutions fom the bits outside the adius of exactly cancel one anothe! The field at is exactly what you would have if all the chage within the adius wee concentated to a point at the oigin. y Conside an infinite plane with a constant chage density (which is some numbe of Coulombs pe squae mete). What is at a point a distance above the plane? x Doing this as a volume integal would be HAD. Gauss s law is ASY.
Poblem: Infinite chaged plane Poblem: Infinite chaged plane Conside an infinite plane with a constant chage density (which is some numbe of Coulombs pe squae mete). What is at a point a distance above the plane? y So choose a suface which is a pillbox which has its top above the plane and its bottom below the plane, each a distance fom the plane. That way the obsevation point lies in the top. pillbox x Use symmety! The electic field must point staight away fom the plane (if >0). Maybe the magnitude depends on, but the diection is fixed. And is independent of x and y. Poblem: Infinite chaged plane Poblem: Infinite chaged plane Let the aea of the top and bottom be A. Let the aea of the top and bottom be A. pillbox pillbox Total chage enclosed by box = A Outwad flux though the top: A Outwad flux though the bottom: A Outwad flux though the sides: x (some aea) x cos(90 0 )=0 So the total flux is: A Poblem: Infinite chaged plane Poblem: Infinite chaged plane Let the aea of the top and bottom be A. pillbox Imagine doing this with an integal ove the chage distibution: beak the suface into little bits da. d Gauss s law then says that A/ε 0 =A so that =/ε 0, outwad. This is constant eveywhee in each halfspace! Notice that the aea A canceled: this is typical! Doing this as a suface integal would be HAD. Gauss s law is ASY.
Conductos Conductos A conducto is a mateial in which chages can move elatively feely. Usually these ae metals. In a static condition, the chages placed on a conducto will have moved as fa fom each othe as possible they epel each othe. In a static situation, the electic field is eo eveywhee inside a conducto. Why is =0 inside a conducto? Conductos Conductos Why is =0 inside a conducto? Because conductos ae full of fee electons, oughly one pe cubic Angstom. These ae fee to move. If is noneo in some egion, then the electons thee feel a foce e and stat to move. In an electostatics poblem, the electons adjust thei positions until the foce on evey electon is eo (o else it would move!). That means when equilibium is eached, =0 eveywhee inside a conducto. Because =0 inside, the inside is neutal. Suppose thee is an exta chage inside. Gauss s law fo the little spheical suface says thee would be a noneo neaby. But thee can t be, within a metal! Consequently the inteio of a metal is neutal. Any excess chage ends up on the suface. lectic field in conductos Poblem: Chaged coaxial cable This pictue is a coss section of an infinitely long line of chage suounded by an infinitely long cylindical conducto. Find. Q b This epesents the line of chage. Say it has a linea chage density of λ (some numbe of C/m ). a This is the cylindical conducto. It has inne adius a and oute adius b. Q Q Use symmety! Clealy points staight out, and its amplitude depends only on.
Poblem: Chaged coaxial cable Poblem: Chaged coaxial cable Fist find at positions in the space inside the cylinde (<a). L Fist find at positions in the space inside the cylinde (<a). L Choose as a suface a cylinde of adius and length L. What is the chage enclosed? λl What is the flux though the end caps? eo (cos90 0 ) What is the flux though the cuved face? x (aea) = (πl) Total flux = (πl) Gauss s law: (πl) = λl/ε 0 so () = λ/ πε 0 Poblem: Chaged coaxial cable Poblem: Chaged coaxial cable Now find at positions within the cylinde (a<<b). Thee s no wok to do: within a conducto =0. Still, we can lean something fom Gauss s law. Thee must be a net chage pe length λ attacted to the inne suface of the metal so that the total chage enclosed by this suface is eo. Make the same kind of cylindical suface. Now the cuved side is entiely within the conducto, whee =0; hence the flux is eo. Thus the total chage enclosed by this suface must be eo. And since the cylinde is neutal, these negative chages must have come fom the oute suface. So the oute suface has a chage density pe length of λ spead aound the oute peimete. So what is the field fo >b? asy! xample Poblem: Gauss Law fo Gavity Gauss law fo gavitation is Φ = g da = 4 g π Suface In which Φ g is the net flux of the gavitational field g though a gaussian suface that encloses a mass (m enclosed ). The field g is defined to be the acceleation of a test paticle on which m enclosed exets a gavitational foce. Calculate Newton s Law of Gavitation fom this. Gm enclosed Φ g Φ Φ xample Poblem: Gauss Law fo Gavity g g = g da = 4π Suface Gm enclosed = g da = g da Suface Suface = g da = g da Suface = ga = g ( 4π ) Suface m m F = m a = G 1 1 = 4 Gm enclosed 4 Gm Φ g g Φ = 4πGm g 4 ( π ) Gm accel. = g = = a