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Physics 202, Lecture 5 Today’s Topics Announcements: Homework #3 on WebAssign by tonight Due (with Homework #2) on 9/24, 10 PM Review: (Ch. 25-Part I) Electric Potential Energy, Electric Potential Electric Potential (Ch. 25-Part II) Electric Potential For Various Charge Distributions Point charges, Continuous distributions (uniform ring, sphere, shell) More on Conductors in Electric Fields, Equipotentials Electric Potential Energy, Electric Potential Work required to move q0 from A to B in E field: potential energy B B ! ! ! ! ! ! !U = !W = " Fus ids = # " Fe ids = #q0 " Eids B A Point charges: kqq U(r) = e 0 r A A U = !! i< j j ke qi q j rij q r q0 (work required to assemble charge configuration) Electric potential difference: (independent of test charge q0) ! ! !U = " # E ids = VB " VA q0 A B !V = Example: uniform E field parallel to path !V = "Ed 1 Example: Uniform Electric Field In the uniform electric field shown: 1. Find potential at B,C,D,G 2. If a charge +q is placed at B, what is the potential energy UB? D 3. If now a –q is at B, what is UB? A VA=0 C B G d 4. If a -q is initially at rest at G, will it move to A or B? What is its final kinetic energy? Recall: Particles will move to minimize their final potential energy! Electric Potential: Point Charges Charge at origin: B rB A rA q (choose zero of potential at infinity) ! ! kq kq V (r) = ! " Eids = ! " e2 dr = e r r # r Potential difference: V (rB ) ! V (rA ) = General: q ! V(r ) = ke ! ! | r ! r" | Single charge q ! V(r ) = ke # ! i ! i | r ! ri" | ke q ke q ! rB rA ! r! q ! ! r ! r" ! r P Multiple charges 2 Electric Potential: Continuous Charge Distributions Finite charge distributions: usually set V=0 at infinity. If charge distribution is known: dq r Here: r = distance b/w source and obs. point V = ke ! Note: scalar integral! More precisely (see board): dq ! V(r ) = ke # ! ! | r ! r" | ! r : field (observation) point ! r ! : source point Calculating Electric Field, Electric Potential Three ways to calculate E field: Coulomb’s Law: (lecture 2) ! ! ! ! dq(r ! r " ) E(r ) = ke # ! ! 3 | r ! r" | Gauss’s Law: (lecture 3,4) ! ! q !e = " EidA = in #0 Potential: (lecture 4,today) dq ! V(r ) = ke # ! ! | r ! r" | ! ! V = ! " Eids ! ! dV = ! Eids Ex = ! "V , Ey = ! "V , E = ! "V "x "y z "z ! E = !"V 3 Example: Uniformly Charged Ring For a uniformly charged ring, show that the potential along the central axis is keQ V= x2 + a2 Solution: (also see board) V = ! = ! = k e dq r k e dq x2 + a2 ke ! dq x2 + a2 =Q Uniformly Charged Ring: Electric Field Find the electric field along the central axis. Approach 1: Superposition. (Eg. 23.18) k e dq x r2 r k xQ k xQ E x = " dE x = e 3 = 2 e 2 3 / 2 r (x + a ) dE x = dE cos # = E! = 0 due to symmetry Approach 2: derivative of potential V (x) = (x keQ 2 +a ) 2 1/ 2 Ex = " !V k xQ = 2 e 2 3/ 2 !x (x + a ) 4 Example: Uniformly Charged Spherical Shell For uniformly charged spherical shell. Again, use: B ! ! !V = " # Eids = VB " VA A Tip: V is the same inside E=0 region Er = E =0 keQ r2 r>R r<R Example: Uniformly Charged Sphere Show that the potential of a R uniformly charged sphere is: Hint: more convenient to use !V = " B Eids = V " V # B A A since from Gauss’s law: Er = keQ r2 r>R Er = keQr R3 r<R 5 Conductors (Lecture 3 Review) For conductors, electrons inside able to move freely Key points: Free electrons inside the conductor move under the influence of any applied electric field Electron redistribution: additional electric fields. Eventually (~10-16s), reach electrostatic equilibrium (E=0 inside conductor). +−+−+−+− +−+−+−+− +−+−+−+− No Applied Field − + − E=0 + − + In Applied Field Conductors in Electrostatic Equilibrium Regardless of shape : Electric field inside conductor is zero All net charges reside on the surface Electric field on surface of conductor always normal to the surface, magnitude σ/ε0 Electric field also zero inside any empty cavity within the conductor sharper edge larger field. Since E=0 inside conductor, Potential is the same throughout the conductor: Equipotential 6 Charge Distribution On Conductor |E| higher at smaller radius of curvature (more charge density) |E| lower (less charge density) Example: Charge Distribution On Conductors (I) The total charge on this conducting sphere is 5q. How is the charge distributed? Evenly distributed throughout the body Qsurface=5q, Qbody=0 None of above + + + + + + + + Note: Regardless of shape, charge resides only on the surface of a conductor 7 Example: Charge Distribution On Conductors (II) The total charge on this conducting shell is 5q, How is the charge distributed? (Router=2Rinner) QInner_surface = 2.5q, QOuter_surface=2.5q, Qbody=0 QInner_surface = q, QOuter_surface=4q, Qbody=0 QInner_surface = 0, QOuter_surface=5q, Qbody=0 + + + E=0 + + + + + Note: Regardless of shape, charge resides only on outer surface of a conductor if no charge inside cavity (Einside=0). Example: Charge Distribution On Conductors (III) The total charge on this conducting shell (Router=2Rinner) is +5q. A point charge of +q is placed at the center. How is the charge distributed? QInner_surface = - q, QOuter_surface=6q, Qbody=0 QInner_surface = q, QOuter_surface=4q, Qbody=0 QInner_surface = 0, QOuter_surface=5q, Qbody=0 + + + + - + - + q - + - + + + - - + - + - + - + + + 8 Example: Charge Distribution On Conductors (IV) Initially, the total charge on this shell is +5q. A point charge of +q is placed at the center, and the shell is then grounded. How is the charge distributed? (Router=2Rinner) QInner_surface = - q, QOuter_surface=6q, Qbody=0 QInner_surface = q, QOuter_surface=4q, Qbody=0 QInner_surface = -q, QOuter_surface=0, Qbody=0 Q=0 everywhere. - - q - - - - Charge Distribution on Conductors: Field lines and Equipotential Surfaces Note: equipotentials are normal to field lines 9 Equipotentials Defined as: The locus of points with the same potential. • Example: for a point charge, the equipotentials are spheres centered on the charge. The electric field is always perpendicular to an equipotential surface! ! ! VB ! VA = ! " Eids B A Along equipotential surface, no change in V ! ! ! " Eids = #V = 0 B A Therefore, ! ! Eids = 0 ! ! E ! ds Electric field perpendicular to surfaces of constant V 10