Lecture 8Electric Potential – Chapter 25Review• Electric Potential Energy, U –• Electric Potential, V -• Electrostatic force is conserved, work done by force is path independentWUUUif−=−=∆qWqUVVVif−=∆=−=∆Review• Equipotential surface –all points are at same potential • E field lines are ⊥ to the equipotential surface• If given equipotentialsurfaces can draw Efield linesReview• From and • Derived equation for finding a potential in a E field • Potential always decreases if move along a path in the direction of the E field qWV −=∆sdFdWrr•=∫•−=−fiifsdEVVrrElectric Potential (18)• Derive potential V around a charged particle• Imagine moving a + test charge from P to ∞∞∞∞• Path doesn’t matter so choose line radially with E∫•−=−fiifsdEVVrrElectric Potential (19)• Chose path so • Using radial path, rewrite• Use limits for i = R and f = ∞∞∞∞EdsdsEsdE ==•θcosrrdrds =∫∞∞−=−RREdrVVElectric Potential (20)•Use E for point charge• Define V∞∞∞∞= 02rqkE =∫∞−=−RdrrkqV210Electric Potential (21)• Finish integral• Letting R become any distance r from particle rqkV =RqkrkqVR−==−∞10Electric Potential (22)• Sign of V is same sign as q– + charge produces +V– - charge produces –V• V gets larger as r gets smaller–In fact V = ∞ when r = 0 (on top of charge)• From shell theorem this holds outside or on external surface of a spherical charge distributionrqkV =Electric Potential (23)• What is the force F, electric field E, and potential V, at a point P a distance r away from a point charge?20rqqkF =r2rqkE =rrqkV =Electric Potential (24)• Use superposition principle to find the potential due to n point charges• This is an algebraic sum, not a vector sum• Include the sign of the charge ∑∑====niiiniirqkVV11Electric Potential (25)• Checkpoint #4 – Rank a), b) and c) according to net electric potential V produced at point P by two protons. (Greatest first.) +=DqdqkVALL EQUALElectric Potential (26)• Replace one of the protons by an electron. Rank the arrangements now. +−=DqdqkVALL EQUALElectric Potential (27)• Potential due to a dipole•Sum V for 2 charges• Usually far away from dipole so r >>d+−=+−==+−−++−=∑rrrrkqrqrqkVVii21θcosdrr ≈−+−2rrr ≈+−Electric Potential (28)• For dipole • Measure θ from dipole axis to r=+−=+−−+2cosrqdkrrrrkqVθ2cosrpkVθ=qdp =Electric Potential (29)• How do we calculate E from V ?VqW ∆−=0dEqdFWrrrr•=•=dsEqdVqθcos00=−dsdVE −=θcosElectric Potential (30)• How do we calculate E from V ?• Component of E in direction of ds• Component of E in any direction is neg. rate of change of V with distance in that directionsVEs∂∂−=dsdVE −=θcosElectric Potential (31)• Take s axis to be x, y, or z axes• If E is uniform and s is ⊥ to equipotentialsurface zVEyVExVEzyx∂∂−=∂∂−=∂∂−= ,,sVE∆∆−=Electric Potential (32)• Checkpoint #6 – 3 pairs of parallel plates with same separation and V of each plate. E field is uniform between plates and ⊥ to the plates.• A) Rank (greatest first) magnitude of E between the platesElectric Potential (33)but asked for magnitude of E2, then 1 & 3sVE∆∆−=dE2001=dE2202=dE2003=Electric Potential (34)• Checkpoint #6 – b) For which pair does E point to the right?#3• C) If an electron is released midway between plates in (3) what does it do?Accelerate to the leftElectric Potential (35)• Define electric potential energy, U, of a system of charges as = to the W done by an external F to assemble the system, bringing each charge from ∞• Bring q1from ∞, W = 0 since no electric F yet• Bring q2from ∞, Wapp= q2V because q1exerts electrostatic F on q2during the moveElectric Potential (36)• Potential due to q1is • From definition of potential energy• Charges of like sign, W and U are +• Charges of opposite sign, W and U are -rqkV1=rqqkVqWU212===Electric Potential (37)• What is the potential energy when add an additional charge to system?• Move q1from ∞, W =U = 0• Move q2from ∞dqqkUW211212==Electric Potential (38)• Now bring in q3• Must also remember q2dqqkUW311313==dqqkUW322323==Electric Potential (39)• Total potential energy is the scalar sum231312UUUU ++=qqqqqq 2,4,321+=−=+=()( )()( )()()dqkdqqdqqdqqkU2102424−=+−++++−+=Electric Potential (40)• Using what we know about conductors– E = 0 inside– All excess charge is on surface• All points of a conductor – whether inside or on the surface – are at the same potential– A conductor is an equipotential
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