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UW-Madison PHYSICS 208 - Lect 11

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From last timeMore on electric potential and connection to E-fieldHow to calculate E-field from VCapacitors and CapacitanceToday:Gauss’ lawWork, potential energy Electric potentialconnection between E-field and potentialThis week Honor •Prof. S. Westheroff on THE HIGHEST ENERGY COSMIC RAYS with the Pierre Auger observatoryElectric Potential€ ΔU /q ≡ V =Electric potentialWork done by E-field produced by source charge Q on a test charge q to move it along a path: € Wel= −ΔU =r F Coulomb⋅ dr s = qr E ⋅∫∫dr s = qr E ⋅ dr s ∫Potential energy of 2 charges:Electric potential of a point charge Q:€ UQqr( )= keQqr€ Vqr( )=UQqr( )q= keQrSuperposition: electric potential of dipole+Q -Qx=+ax=-aSuperposition of• potential from +Q• potential from -Q+ =yEg: V(y=0) = kQ/a-kQ/a=0In general: for a group of point charges€ V = keqirii∑€ V =kQrElectric Potential of a continuous charge distributionConsider a small charge element dq. The potential at some point due to this charge element isTotal potential:This value of V used the reference that V = 0 at infinite distanceQuick QuizThe total work required for YOU to assemble the set ofcharges as shown below is:+Q+Q− Q5 m5 m5 m1. positive2. zero3. negativeWext = ΔU = Ufinal - Uinitial <0 => U decreases.123€ W1= 0W2= kQ(−Q)dW3= kQQd+ kQ(−Q)dTotal work = −kQ2d€ = QV12= 0Work to put Q at pos 1Work to put -Q at pos 2where V1 = kQ/r€ = −QV1Work to put Q at pos 3where V12 = 0Electric field from PotentialDifferential form:Spell out the vectors:Each component:One only symbol:Work done = qoVB-qoVA = € −r F Coulomb( )• dr l AB∫= -Wel € dWel=r F Coulomb• dr s = −dU = −qdV ⇒r F Coulomb⋅ dr s = qr E • dr s = −qdV € dV = −r E • dr s € dV = − Exdx + Eydy + Ezdz( )€ Ex= −dVdx, Ey= −dVdy, Ez= −dVdz € r E = −r ∇ V = −dVdx,dVdy,dVdz      Work to move a charge q0 from A to B and overcome the Coulomb repulsive force dure to Q:ΔV=VB-VApotential differenceQuick QuizSuppose the electric potential is constant everywhere. What is the electric field?A) PositiveB) NegativeC) IncreasingD) DecreasingE) Zero € dV = −r E • dr sExplain € r E V=Vo € dr l € V = Vo−r E dr l € V = Vo+r E dr l € dr l € dV = −r E • dr l E-field can be used to find changes in V € dr l € V = VoPotential changes largest in direction of E-fieldZero changes in direction perpendicular to E-fieldEquipotential linesLines of constant potential are perpendicular at any point to E-fieldQuick QuizHow does the electric potential outside a uniforminfinite sheet of positive charge vary with distancefrom the sheet?A. Is constantB. Increasing as (distance)1C. Decreasing as (distance)1D. Increasing as (distance)2E. Decreasing as (distance)2Notice the E field is uniform! € dV = −r E • dr l ABxdxE+++++++ExplainOnly one dimension => easy problem! Remember: € r E || dr s E cnst € Ex= E = −dVdx⇒ΔV = VB− VA= −r E • dr x AB∫= − EdxAB∫= −E (xB− xA) = −EdABxdxEConstant E-field corresponds to linearlydecreasing (in direction of E) potentialParticle gains kinetic energy equal to thepotential energy lost € dV = −r E • dr s ++++++++€ Wel= −ΔU = −qΔV = ΔKPotential of spherical conductor•Another 1D problem•charge is on surface•E-field is in radial directiondifficultpatheasypathEasy because is samedirection as E,€ E • ds = E dr = EdrAt any distance r =>Inside the sphere E = 0 so all points have the same potential of the surfaceV(R) = kQ/R+Q€ V (r) = kQdrr2= kQrr∞∫R € V (∞) − V (r) = −r E • dr s = −r∞∫Edrr∞∫Quick QuizTwo conducting spheres of diff radii connected by long conducting wire.What is approximately true of Q1, Q2?A) Q2>Q1B) Q2<Q1C) Q2=Q1R1R2Q1Q2Explain Since both must be at the same potential,€ kQ1R1=kQ2R2⇒Q1Q2=R1R2Surface charge densities? € σ=Q4πR2⇒σ1σ2=R2R1Charge proportionalto radiusSurface charge densityproportional to 1/RElectric field?Since , € E =σ2εoLocal E-field proportional to 1/R(1/radius of curvature)The surface chargedensity and the E-field are larger at sharp points!Varying E-fields outside a conductor Expect larger electric fields near the small end. Electric field approximately proportional to 1/(local radius of curvature). Large electric fields at sharp points because E-field proportional to charge density and charge density is larger for smaller radius of curvature Fields can be so strong that air ionized and ions accelerated.Any conductor quizConsider this conducting object. When it has totalcharge Qo, its electric potential is Vo. When it hascharge 2Qo, its electric potentialA. is VoB. is 2VoC. is 4VoD. depends on shapeCapacitance and capacitors Electric potential of any conducting objectproportional to its total charge.€ V =1CQA capacitor consists of two conductors:•Conductors generically called ‘plates’•Charge transferred between plates•Plates carry equal and opposite charges•Potential difference between plates ΔVproportional to charge transferred Q€ ΔV =1CQ The SI unit of capacitance is the farad (F) 1 Farad = 1 Coulomb / Volt This is a very large unit: typically use µF = 10-6 F, nF = 10-9 F, pF = 10-12 FParallel plate capacitorσ=Q/A surface charge densityA=surface area€ Eleft+ Eright=σ/2εo+σ/2εo=σ/εo-Q+Qd+++++++++++++++---------------EleftErightUniform E-fieldNotice outside the plates E = 0Work done = qoVB-qoVA = € −r F Coulomb( )• dr l AB∫Work to move q from positive to negative plate = € C =εoAdThis is a geometrical factor+ -E+E-E-E+E-E+ € q(V−−V+) = −qr E • dr s 0d∫= −qEd ⇒ ΔV = Ed =σε0d = Qdε0A      =QC€ ΔV = V+−


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UW-Madison PHYSICS 208 - Lect 11

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