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Physics 11bLecture #5Electric PotentialS&J Chapter 25What We Did Last Time Introduced electric potential V Work required to move a charge q from A to B is E field is negative gradient of V A point charge q generates potential For multiple and continuous charge,BABAWqdqV→=−⋅=∆∫Es01()4qVrrπε=01()4iiiqVπε=−∑rrr01()4dqVrπε=∫r(,,) , ,xyzVVVEEExyz⎛⎞∂∂∂==−⎜⎟∂∂∂⎝⎠EToday’s Goals Calculate electric potential due to continuous charge distributions Work our way through Ring Æ Disk Æ Sphere Compare with Gauss’s Law result What about charged conductors? Shielding effect Millikan’s oil-drop experimentBasic Formulae Electric potential V and electric field E Charge(s) or charge distribution and potential VBBAAVV V d∆= − =− ⋅∫Es ,,VVVxyz⎛⎞∂∂∂=−⎜⎟∂∂∂⎝⎠E01()4qVrrπε=01()4iiiqVπε=−∑rrr01()4dqVrπε=∫rLet’s do some exercise on this oneCharged Ring Imagine a thin ring of radius a, charge Q What’s the potential V alongthe axis? Take a little piece as usual Charge dq Distance  Distance is same for the entire ring Æ Take it out!ax22xa+22xa+220ring1()4dqVxxaπε=+∫dq22 22ring001()44QVx dqxa xaπε πε==++∫Charged Disk Now we make it a thin disk Slice the disk into thin rings Charge on the ring is Potential due to the ring is Can we integrate this?axrdr22area of ring 2 2area of diskrdr Qrdrdq Q Qaaππ===22 2220042dq QrdrdVxr axrπε πε==++()22 222222 200000()222aaQrdr Q QVx xr xaxaaaxrπε πεπε⎡⎤==+=+−⎣⎦+∫Yes!Charged Sphere We try a hollow sphere Slice into rings again Charge on the ring is Potential due to the ring is Let’s try integrating…acosxaθ−dθax222sin sin42adQddq Qaπθθ θθπ==2200sin()82cosQdVxxa xaπθθπεθ=+−∫θsinaθ22204(cos) sindqdVxa aπεθθ=−+Charged Sphere Have no fear Outside (x > a): Inside (x < a):()220022000sin()82cos12cos88QdVxxa xaQxa xaxaQxa xaxaππθθπεθθπεπε=+−⎡⎤=+−⎢⎥⎣⎦=+−−∫()2222sin2cos2cosdxaxa xadxa xaθθθθ+− =+−Use(Don’t worry if you can’t figurethis one out yourself.)22 22()x a xa xa xa+−=−=−0()4QVxxπε=0()4QVxaπε=Identical to point chargeConstantHow Does It Look? The potential V is Flat inside Falling as 1/x outside How about E? Remember V varies only in thex-directionxa04Qaπε04Qxπε,,VVVxyz⎛⎞∂∂∂=−⎜⎟∂∂∂⎝⎠E2040xQxadVxEdxxaπε⎧>⎪=− =⎨⎪<⎩xa204Qaπε204Qxπε0()Vx()xExDoes It Fit Together? Apply Gauss’s Law to an imaginary sphere outside/insidethe charged sphere Outside sphere contains Q Inside sphere contains no chargeQ2out out out0()4 ()EQrrErπεΦ= =outrinra2in in in()4 ()0ErrErπΦ= =204()0QrarErraπε⎧>⎪=⎨⎪<⎩Conductors In Lecture #3, I showed you: there is no E field inside conductors there is no net charge density inside conductors all charge lives on the surface of the conductors Immediately obvious: potential V is constant over the entire volume of a conductorAB0BBAAVV d−=−⋅=∫Eszero-- ---Induced Surface Charge Let’s put a conductor in an external E field E = 0 inside the conductor How does it do this? Something must create“canceling” E field E is created by charges No charges inside Free electrons in theconductor re-arrangeto create just the right surface charge density to counter-act the external E field We call it induced charge∆VEE = 0++ +++---+++Induced Surface ChargeUneven surface charge distribution ensures no E field inside the conductorElectric Shielding A hollow conductor can keep out external E field Induced charges will cancel E Widely used to eliminateelectric interference Doesn’t have to be solidconductor – mesh works OK “Faraday cage” That’s why Cell phones don’t work well inside steel-framed buildings Microwave ovens have metallic mesh on the windowEE = 0+++++++−−−−−−−−Irregularly-Shaped Conductor For irregular (not spherical) shaped conductor, surface charge density depends on the local curvature Small convex points (= pointy spots)attract more charge More charge Î Stronger E field Sharp points and edges ofconductors attracts high electricfield Think lightning rods+++++++++++++++++E largeMillikan Oil-Drop Experiment Robert Millikan (1968-1953) measured electric charge of small droplets of oil So small that they fallslowly in the air due toair resistance They may or may notbe charged – one canfind out by applying anexternal E field Use a telescope towatch one droplet ata time∆V-qEmgqEField offField onMillikan Oil-Drop Experiment Turn E on and off Negatively charged droplets goup and down (slowly) Measure the velocity Droplet size Æ Mass Determine the charge –q Millikan found every dropletcontained an integer multipleof a small charge-qmgqE-qmgairdragairdrag( 1,2,3,...)qne n==elementary charge = 1.60×10-19CSummary Electric potential due to continuous charge distributions Use Electric field/potential due tospherical charge distribution Looks like a point charge from outside Zero inside Discussed conductors Electric shielding (Faraday cage) Millikan’s oil-drop


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HARVARD PHYS 11b - Lecture #5

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