Presentation on theme: "Ch. 24 Electric Flux Gauss's Law"— Presentation transcript:
1Ch. 24 Electric Flux Gauss's Law Electric flux is the amount of electric field going across a surfaceIt is defined in terms of a direction, or normal unit vector, perpendicular to the surfaceFor a constant electric field, and a flat surface, it is easy to calculateDenoted by EUnits of Nm2/CWhen the surface is flat, and the fields are constant, you can just use multiplication to get the fluxWhen the surface is curved, or the fields are not constant, you have to perform an integrationHere use A X normal unit vector for A vector of text.Really when everything in integral is constant, you can just multiply.So, if E dot n is constant along surface – you are OK.
6Total Flux Out of Various Shapes A point charge q is at the “center” of a (a) sphere (b) joined hemispheres (c) cylinder (d) cube. What is the total electric flux out of the shape?baqqaq
7Electric Flux (Hard example) A point charge q is at the center of a cylinder of radius a and height 2b. What is the electric flux out of (a) each end and (b) the lateral surface?topsbbrrzaConsider a ring of radius s and thickness dsqbalateral surface
8Gauss’s LawNo matter what shape you use, the total electric flux out of a region containing a point charge q is 4keq = q/0.Why is this true?Electric flux is just measuring how many field lines come out of a given regionNo matter how you distort the shape, the field lines come out somewhereIf you have multiple charges inside the region their effects addHowever, charges outside the region do not contributeqq4ϕ E = E ∙d A = q in ϵ 0q3q1q2
10Using Gauss’s Law to find total charge A cube of side a has an electric field of constant magnitude |E| = E pointing directly out on two opposite faces and directly in on the remaining four faces. What is the total charge inside the cube?A) 6Ea20 B) – 6Ea20 C) 2Ea20 D) – 2Ea20 E) None of the aboveaaaDo easy problem first – point charge,And problem 4 points on page 696.
13Using Gauss’s Law to find flux A very long box has the shape of a regular pentagonal prism. Inscribed in the box is a sphere of radius R with surface charge density . What is the electric flux out of one lateral side of the box?Perspective viewThe flux out of the end caps is negligibleBecause it is a regular pentagon, the flux from the other five sides must be the sameEnd view
14Using Gauss’s Law to find E-field A sphere of radius a has uniform charge density throughout. What is the direction and magnitude of the electric field everywhere?Clearly, all directions are created equal in this problemCertainly the electric field will point away from the sphere at all pointsThe electric field must depend only on the distanceDraw a sphere of radius r around this chargeNow use Gauss’s Law with this sphereDo point charge firstThen do as point charge with q = PVraIs this the electric field everywhere?
15Using Gauss’s Law to find E-field (2) A sphere of radius a has uniform charge density throughout. What is the direction and magnitude of the electric field everywhere? [Like example 24.3]When computing the flux for a Gaussian surface, only include the electric charges inside the surfacerar/a
16Electric Field From a Line Charge What is the electric field from an infinite line with linear charge density ?LrElectric field must point away from the line charge, and depends only on distanceAdd a cylindrical Gaussian surface with radius r and length LUse Gauss’s LawThe ends of the cylinder don’t contributeOn the side, the electric field and the normal are parallel
17Electric Field From a Plane Charge What is the electric field from an infinite plane with surface charge density ?Electric field must point away from the surface, and depends only on distance d from the surfaceAdd a box shaped Gaussian surface of size 2d L WUse Gauss’s LawThe sides don’t contributeOn the top and bottom, the electric field and the normal are parallelText uses cylinder. If truly infinite, never far from it. Consider 45 degrees area pf charge increaases as r squared as does 1/E.
21Conductors and Gauss’s Law Conductors are materials where charges are free to flow in response to electric forcesThe charges flow until the electric field is neutralized in the conductorInside a conductor, E = 0Draw any Gaussian surface inside the conductorIn the interior of a conductor, there is no chargeDo all Warmup04Conductors in equilibrium: (1) E zero inside, (2) charge on surface, (3) outside is sigma/eo, (4) greatest at pointThe charge all flows to the surface
23Electric Field at Surface of a Conductor Because charge accumulates on the surface of a conductor, there can be electric field just outside the conductorWill be perpendicular to surfaceWe can calculate it from Gauss’s LawDraw a small box that slightly penetrates the surfaceThe lateral sides are small and have no flux through themThe bottom side is inside the conductor and has no electric fieldThe top side has area A and has flux through it, pg 700The charge inside the box is due to the surface charge We can use Gauss’s Law to relate these
24Where does the charge go? A hollow conducting sphere of outer radius 2 cm and inner radius 1 cm has q = +80 nC of charge put on it. What is the surface charge density on the inner surface? On the outer surface?A) 20 nC/cm2 B) 5 nC/cm2 C) 4 nC/cm2 D) 0 E) None of the above80 nCThe Gaussian surface is entirely contained in the conductor; therefore E = 0 and electric flux = 0Therefore, there can’t be any charge on the inner surface1 cmFrom the symmetry of the problem, the charge will be uniformly spread over the outer surface2 cmPoint charge q/4Pi eps0 w/ q = 80 pi. Note at r = 2 cm, sigma/eps0 is same: 20/4 cm^2cutaway viewThe electric field:The electric field in the cavity and in the conductor is zeroThe electric field outside the conductor can be found from Gauss’s Law