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UW-Madison PHYSICS 208 - Lecture Notes

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PHYS208, SPRING 2008My office Hrs: Mon 12-1:30pm Ch4112•but if you email me we can find alternative times or I can answer to your question by email •[email protected]•We’ll talk about Antarctica and IceCube in Honor lecturesNo honors lecture this weekPHYS208, SPRING 2008MTE1, this afternoon•5:30-7pm: Ch 2103 Adam (301,310) Eli(302,311), Stephen (303,306)Science Hall 180 Amanda (305,307), Mike (304,309), Ye (308)•Alternate exams (you must have emailed us about your conflict): today at 2:30-4pm and 6-7:30pm Ch 3254•you can bring 8 1/2 x 11 handwritten note sheet (both sides) allowedCovers Chap. 21.5-7, 22-23,25-26+ lecture, lab, discussion, HWIt does not cover today’s lecture material From last time ! r E = kQr2ˆ r =Q4"#or2ˆ r k=9x109 N m2/C2 ! r E =r F qtest ! r E •Electric field is => superposition principle applies both to force and to•Electric field from a point charge:•From a set of point charges: ! r E net=r E ii"Calculating E-field for a continuous distribution requires some math... PHYS208, SPRING 2008Quick QuizWhat does an electric dipole in a uniform E-field do?A) accelerates leftB) accelerates rightC) rotates CW until aligned to ED) rotates CCW until aligned to -EE) accelerates up ! r p ! r " ! r " =r p #r E PHYS208, SPRING 2008Electric torque on dipolesU = - pE cos! = - p · EWhen the dipole is aligned to the field U is minimum: U = -pE equilibrium!! F+= qE, r = s/2F"= "qE, r = s/2Net force zerobut not aligned => dipole rotates ! r r + ! r r "Torque is a vector and RH rule tells direction of rotation. For a couple of forces ( = distance between lines along which forces act) ! r " ! l! p = qs ! r " = lF = sF sin#= qsE sin#= pE sin# ! r " =r p #r E PHYS208, SPRING 2008An example! dQ ="dxy+ + + + + + + + + + + + + + + + + + + ! dr E rx! r2+ x2! # ! r E = E r u y= 2k"dxr2+ x2cos#0$%= 2k"r ! r E ! cos"=rr2+ x2Infinite charged rodPHYS208, SPRING 2008SymmetryBut in situations of high symmetry we can use a smarter approach!Cylindrical symmetry Spherical symmetry Planar symmetryPHYS208, SPRING 2008Electric FluxElectric flux through a surface: (component of E-field 󲰥 to surface) x (surface area) ! "E proportional to number of E-field lines penetrating the surface! "EWhy perpendicular component?Which is the mathematical expression?PHYS208, SPRING 2008Flux equation! ˆ s This is equivalent to the scalar product: E uniform on A ! "E=r E #r A PHYS208, SPRING 2008Total Flux ! "r A i ! r E i ! "r A j ! r E jPHYS208, SPRING 2008Flux through a spherical surface! Asphere= dA = 4"r2#r2r2r2PHYS208, SPRING 2008GAUSS’ LAWPHYS208, SPRING 2008Quick QuizTwo spheres of radius R and 2R are centered on 2 positive charges of the same value q. The flux through the 2 spheres compares as:A) Flux(R) = Flux(2R)B) Flux(R) = 2Flux(2R)C) Flux(R) = 1/2 Flux(2R)D) Flux(R) = 1/4 Flux(2R)E) Flux(R) = 4 Flux(2R)E(R)x4!R2 = E(2R)x4!(2R)2! kqR24"R2= kq(2R)24"(2R)2PHYS208, SPRING 2008GAUSS’ LAW for any closed surfaceThe flux depends ONLY on the charge enclosed by the surface!!PHYS208, SPRING 2008How to use Gauss’ law?Apply to a uniformly charged sphere+QWhat is the field outside (r>a)?! "E= E # dA =$EdA$= E dA$= E 4%r2=Q&0E =Q4%&0r2= keQr2Same reasoning as for point charge!EdA1) Select a sphere as the gaussian surface thru which Given the symmetry, sphere of radius r is a natural choice2) Important: know E-field direction. Radially out.3) How its it oriented respect to dA? Parallel to E4) Flux thru surface: E4"r25) Enclosed charge + QPHYS208, SPRING 2008Where does it matter that the sphere is not point-like?•Field out of sphere and parallel to dA•charge = constant any in sphere element of volume (uniformly)•! in the gaussian surface (sphere with r<a) is the same as the charge sphere one+Q! dq ="dVvolumecharge density! "=Q43#a3=qin43#r3charge enclosed in gaussian surface! "=QVqin = Q (r3/a3)PHYS208, SPRING 2008Let’s plot E-field vs distance from centreof charge spherelinear with r!different from point-like chargePHYS208, SPRING 2008Quick Quiz•Which gaussian surface do you have to use to calculate E for an infinite line of charge?A) B) C) D) Noneof these ! dq ="dlFlux of E through the gaussian surfacecharge enclosedApply Gauss’ law to calculate E! ! "l ! E 2"rl ! E 2"rl =#l$0! E ="2#$0rPHYS208, SPRING 2008Plane of chargeE perpendicular to plane and same magnitude at all points equidistant from the planeS = Area of gaussian surface where Flux 2ESGauss’ law: 2ES = Q/$0 => E =%/(2$0)This is a uniform fieldS%=Q/S surface charge in C/m2PHYS208, SPRING 2008Conductors in equilibrium electrostatic equilibriumIn a conductor in electrostatic equilibrium there is no net motion of chargeProperty 1: E=0 everywhere inside the conductorConductor slab in an external field E: • if E-field not null inside the conductor, free electrons would be accelerated • These electrons would not be in equilibrium.• When the external field is applied, the electrons redistribute until the magnitude of the internal field equals the magnitude of the external field•The total field inside the conductor is zeroEtot = E+Ein= 0EinEtot =0PHYS208, SPRING 2008Property 2: Charge on only conductor surface• chose a gaussian surface inside the conductor (as close to the surface as desired)•There is no net flux through the gaussian surface (since E=0)•Any net charge must reside on the surface (cannot be inside!)E=0PHYS208, SPRING 2008Magnitude and direction of E-Field Field linesE-field perpendicular to the surface:parallel component to E would put force on chargescharges would accelerate along the surface NO equilibriumApplying Gauss’s lawPHYS208, SPRING 2008Questions for you •What is the field in all regions of space inside and outside a charged metal shell? •Can a uniformly charged sphere as the one considered in this lecture be made of metal? PHYS208, SPRING 2008The flux does not depend on the surface shape ! "#E=r E $ "r A = E cos%"A = kq"Acos%r2! "E= kqdAcos#r2=$kq d%$= 4&kq =q'0Surface elements infinitesimally small and sum (integral over surface)! "E= keqdAcos#r2= keq$d%$= ke4&q =q'0To derive Gauss’ lawwe use Coulomb’s law"Acos#The area element A subtends a solid angle (in steradians):The flux across the surface element is:Still true for any closed


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UW-Madison PHYSICS 208 - Lecture Notes

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