CHAPTR 4 GAUSS LAW LCTRIC FLUX lectic flux is a measue f the numbe f electic filed lines penetating sme suface in a diectin pependicula t that suface. Φ = A = A csθ with θ is the angle between the and A. The diectin f A, the aea f a suface, is always nmal t that suface and pints utwad f clsed sufaces, (by clsed suface we mean that suface which divides space int inside and utside egins). It is bvius fm the last euatin that Φ is a scala uantity with the SI unit f N.m /C. If the electic field is nt cnstant ve the suface in uestin, Φ= d A suface Nte that if is cnstant ve the suface it shuld be taken ut f the integatin and we ecve the fist euatin as expected.
Cnside nw the clsed suface in the figue, If is utwad and θ < 90, hence the flux thugh this element is psitive. If is enteing the suface, θ > 90, and s the flux f such an element is negative. Fm this agument, we can expect that if a field line enteing and leaving the same clsed suface the net electic flux thugh the clsed suface fm that line is ze. xample 4.1 What is the e.flux thugh a sphee that has a adius f 1.00 m and caies a chage f 1.00 µc at its cente. Slutin: The e.field at any pint n the suface f the sphee is
= k = 1 6 9 1.00 10 ( 9.00 10 ) = 9.0 10 N/C (1.00) The field pints adially utwad and s, Φ = A= A csθ = 9.0 10 = 1.1 10 5 N.m /C ( 4πR ) cs0 xample 4. A cube f edge l is iented in a unifm e.field, as shwn. Find the net e.flux thugh the suface f the cube. y da da 1 da x z da 4 Slutin: The net flux thugh the cube is the sum f the fluxes thugh the 6-faces f the cube: The 4-faces named in the figue and the the tw unnumbeed faces (the fth and the back faces), that is
Φ = da1 da da 1 da4 da da 4 fth back The flux thugh the unnumbeed faces and the faces and 4 is ze because and da ae pependicula Φ = da 1 d 1 A Nw the angle between and da 1 is 180, while the angle between and da is ze Φ = da1 cs180 1 = A A= 0 da cs0 GAUSS LAW the electic flux thugh any clsed suface is eual t the net chage inside that suface divided by, that is Φ = da = in whee in dentes the chage inside the suface and the cicle in the integal sign indicates that the integatin is ve a clsed suface. We veify Gauss law by cnsideing a psitive pint chage suunded by tw clsed sufaces: S 1 is spheical, wheeas S is iegula. Culmb s law tells us that the magnitude f the electic field is cnstant
S 1 S eveywhee n the spheical suface and given as = k. Since the filed diectin is adial, we can evaluate the flux thugh S 1 as ( π ) Φ = da = A= k 4 Knwing that = ( 4π ) 1 k we get Φ = Figue (a) shws that the numbe f field lines cssing S 1 is the same as that lines cssing S, that is, the flux thugh the tw sufaces ae eual and independent f thei shapes. If the chage exists utside a clsed suface, the electic field lines enteing the suface must leave that suface. Hence, the electic flux thugh that suface is ze. The pactical utility f Gauss law lies lagely in pviding a smat way t evaluate the electic filed f a chage distibutin. F this way t be as easy as pssible we must be able t chse a hypthetical clsed suface (Gaussian
suface) such that the electic filed ve its suface is cnstant. This can be attained if the fllwing emaks ae satisfied: 1- The chage distibutin must have a high degee f symmety (spheical, cylindical with infinite length, plane with infinite extends). - The Gaussian suface shuld have the same symmety as that f the chage distibutin. - The pint at which is t be evaluated shuld lie n that Gaussian suface. 4- If is paallel t the suface ze at evey pint, then d A = 0 5- If is pependicula t the suface at evey pint, and since is cnstant, then the integal simply educes t d A = A xample 4.4 Find the e.f a distance fm a pint chage. Gaussian suface Slutin: Since the chage distibutin is spheical, we chse the Gaussian suface as a sphee f adius. It is clea that and da ae paallel and is cnstant ve the suface da da= 4π = = 4π ( )
xample 4.6 A thin spheical shell f adius a has a ttal chage unifmly distibuted ve its suface. Find the magnitude f the electic field at a pint a) utside the shell a distance >R b) inside the shell a distance <R Gaussian suface R Slutin a) Since the chage distibutin is spheical, we chse a spheical Gaussian suface Figue.5 xample.. cncentic with the shell. It is clea that is nmal t the suface at evey pint in the Gaussian suface, that is and da ae paallel, s we wite da = in A= But A, the aea f the Gaussian suface is A= 4π, then we btain = 4π b) In this case the Gaussian suface is inside the shell. This means that thee is n chage inside it, that is 0 which yield = 0. in = Gaussian suface R
xample 4.5 An insulating sphee f adius a has a ttal chage unifmly distibuted thugh its vlume. Calculate the electic filed a) utside the shell a distance >R b) inside the shell a distance <R Slutin We will efe t the Figue f the pevius example but nw the sphee is slid. a) Again, and because the spheical symmety f the chage distibutin, we select a spheical Gaussian suface f adius, cncentic with the sphee. As f the case in the pevius example we wite in d A = A= = Substituting f A by 4π we get = 4π b) In this case we chse a spheical Gaussian suface f adius <R. T find the chage in within the Gaussian suface f vlume V in, we use the fact that = ρ in V in whee ρ is the vlume chage density. Knwing that V in = 4 π and ρ = 4 R π
we btain in = R Applying Gauss law we btain da = which yields = 4π R in ( 4π ) = = R xample 4.7 Find the electic field at a distance fm an infinite line chage f unifm density λ. Slutin: As a Gaussian suface we select a cicula cylinde f adius with height h and caxial with the line chage. Since the cylinde has thee sufaces, the integal in Gauss's law has t be split int thee pats: the cuved suface, and the tw bases. This means that da da da = b b c in h da b da b da c
Fm the symmety f the system, is paallel t bth bases. Futheme, it has a cnstant magnitude and diected adially utwad at evey pint n the cuved suface f the cylinde. This means that the fist tw integals vanish and the last ne becmes simply A, with A is the aea f the cuved suface (πh). The ttal chage inside the gaussian suface is λh. Nw we wite and λh A = ( πh) = λ = π Nte that if the wie is nt t lng its ends will be clsed t any Gaussian suface. Since the electic field at, and clsed t the ends is nt unifm it will be impssible t manage the integal f Gauss law. xample 4.8 Find the electic field due t a nncnducting, infinite plane with unifm suface chage density σ. Gaussian suface
Slutin: T slve this pblem we select as a Gaussian suface a small cylinde whse axis is pependicula t the plane and whse ends each has an aea A, (Figue.-). As we d in the pevius example we wite Gauss law as da da da = b b c in By symmety, the electic filed lines will be unifm and nmal t the plane as shwn in Figue.-. This means that is pependicula t da f the cuved suface f the cylinde. Futheme, is diected nmally utwad and has a cnstant magnitude at each pint n the tw ends f the cylinde. This means that the thid integal vanishes and the fist tw integals each educe t A. Nting that = σa, we get in σa A= σ = Nte that this esult agees with the esult f xample 1.9. It is left as an execise t shw that the pblem can be slved using a Gaussian suface in the shape f paallelepiped. CONDUCTORS IN LCTROSTATIC UILIBRIUM If a cnduct is chaged, chages will mve a way fm each the due t the epulsin fce between them. F the chages t be as fa a way fm each the as they can, they will mve t the ute suface f the cnduct. Cnducts
with n mtin f chages ae said t be in electstatic euilibium. Such cnducts have the fllwing ppety: Any excess chage will eside entiely n the ute suface f an islated cnduct. Beaing this ppety in mind, if a Gaussian suface is cnstucted inside such a cnduct (see the Figue), it will nt enclse any chage. Using Gauss s law, we cnclude that The electic field must be ze inside any cnduct in electstatic euilibium. A thid imptant ppety f cnducts in electstatic euilibium is the fllwing: Gaussian suface Figue.7 An insulated cnduct f abitay shape in electstatic euilibium. Any Gausian suface inside such a cnduct cntains n chage. The electic field just utside a cnduct is always pependicula t the suface f the cnducts and eual t σ. If this is nt the case, the fee chages will mve alng the suface and this vilate the cnditin f euilibium. Let us nw use Gauss s law t calculate the magnitude f the electic filed just utside a chaged cnduct. Figue.8 A Gaussian suface in the shape f small cylinde is used t calculate the electic field just utside a chaged cnduct.
T d s we daw a gaussian suface in the shape f small cylinde as shwn Figue.8. Gauss s law then gives da da da = b b c in But the base inside the cnduct has n flux thugh it since =0, and the flux thugh the cuved suface is ze since is nmal t the aea vect f this suface. Hence, the last tw integals vanish leaving us with in σa A = = whee A is the aea f the cylinde s base and σ is the suface chage density n the cnduct. The electic filed is nw σ = xample 4.10 A cnducting sphee f adius a has a net chage. Cncentic with this sphee is a cnducting spheical shell f inne adius b and ute adius c and has a net chage f -. a) Find the electic field in the egins inside the sphee, between the sphee and the shell, inside the shell, and utside the shell. b) Detemine the induced chage n the inne and ute sufaces f the shell. Slutin: Since the sphee is cnducting we cnclude that the electic field in the fist egin is ze, i.e., 1 =0.
Induced chages - - a c b In the secnd egin we select a spheical Gaussian suface with adius a< < b. Since in = and is cnstant in magnitude ve the Gaussian suface and nmal t it, we find fm Gauss law ( 4π ) = in = Slving f we btain = a< < b 4π Gaussian sufaces In the thid egin the electic filed is again must be ze since this egin is inside the shell which is a cnduct, i.e., =0.
In the last egin utside the shell we cnstuct a Gaussian suface with adius > c. This suface enclse a ttal chage f (-)=. Gauss law is then gives 4 = > c 4π b) We said that the electic field is ze. If we cnstuct a gaussian suface in that egin with adius c > > b the net chage inside that suface must be ze. We cnclude that a chage - is induced by the chage n the sphee. The induced chage n the ute suface will be. Theefe, the net chage n the ute suface will be.