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Chapter 19 Electric Potential Energy and the Electric Potential Van de Graaff generator used as a particle accelerator. An electron is accelerated vertically starting from rest across a +100,000 V potential difference produced by a Van de Graaff. Find a) the change in electric potential energy of the electron, b) the kinetic energy gained by the electron (neglecting gravity) and c) the final speed of the electron. For fun, compare the result in a) with the change in the gravitational potential energy of the electron assuming the Van de Graaff is 1 m tall. a) EPEB - EPEA = qVB - qVA = q(VB - VA) = (-1.6 x 10-19)(105 - 0) = -1.6 x 10-14 J EPE decreases VB = +105 V b) Conservation of energy: KEB + EPEB = KEA + EPEA KEB = KEA - (EPEB - EPEA) = 0 - (-1.6 x 10-14) = 1.6 x 10-14 J = 105 eV = 100 KeV VA = 0 V electron q = -1.6 x 10-19 C , me = 9.11 x 10-31 kg vA = 0 m/s Van de Graaff example -- continued c) KEB = 1/2 mevB2 vB = [2KEB/me]1/2 = [2(1.6 x 10-14)/(9.11 x 10-31)]1/2 = 1.87 x 108 m/s The speed of light is 3.0 x 108 m/s, so this calculation would tell us that the electron is traveling at ~ 2/3 of the speed of light! To do part c) correctly, we really should use Einstein’s Theory of Special Relativity (gives 1.64 x 108 m/s) Fun calculation: compare EPEB - EPEA from a) with GPEB - GPEA for electron. GPEB - GPEA = meghB - meghA = meg(hB - hA) = (9.11 x 10-31)(9.8)(1 - 0) = 8.93 x 10-30 J From a), EPEB - EPEA = -1.6 x 10-14 J (1.6 x 10-14)/(8.93 x 10-30) = 1.79 x 1015 , GPE can be neglected for an electron!! LHC ring in Geneva area proton+proton collisions at 7 TeV CERN Large Hadron Collider ALICE setup ITS TOF FMD PMD HMPID TRD TPC PHOS MUON SPEC. 19.3 The Electric Potential Difference Created by Point Charges A positive test charge q0 is being repelled by another positive point charge q. Since the force on q0 is not constant between A and B, integral calculus must be used to find the work done by the force acting from A to B, WAB kqqo kqqo = − rA rB The potential difference between A and B is, − WAB kq kq VB − VA = = − qo rAB rAB If rB goes to infinity, and defining VB = 0 at infinity, and letting rA -> r and VA -> V, Potential of a point charge kq V= r 19.3 The Electric Potential Difference Created by Point Charges Example 5 The Potential of a Point Charge Using a zero reference potential at infinity, determine the amount by which a point charge of 4.0x10-8C alters the electric potential at a spot 1.2m away when the charge is (a) positive and (b) negative. 19.3 The Electric Potential Difference Created by Point Charges (a) kq = r 8.99 ×109 N ⋅ m 2 C 2 + 4.0 ×10 −8 C 1.2 m = +300 V V= ( )( ) A positive point charge raises the potential. (b) V = −300 V A negative point charge lowers the potential. 19.3 The Electric Potential Difference Created by Point Charges When two or more charges are present, the potential due to all of the charges is obtained by adding together the individual potentials. Example 6 The Total Electric Potential At locations A and B, find the total electric potential. 19.3 The Electric Potential Difference Created by Point Charges VA (8.99 ×10 = VB 9 )( ) ( )( ) N ⋅ m 2 C 2 + 8.0 ×10 −-98 C 8.99 ×109 N ⋅ m 2 C 2 − 8.0 ×10 −-98 C + = +240 V 0.20 m 0.60 m (8.99 ×10 = 9 )( ) ( )( ) N ⋅ m 2 C 2 + 8.0 ×10 −-98 C 8.99 ×109 N ⋅ m 2 C 2 − 8.0 ×10-9−8 C + =0V 0.40 m 0.40 m 19.3 The Electric Potential Difference Created by Point Charges Conceptual Example 7 Where is the Potential Zero? Two point charges are fixed in place. The positive charge is +2q and the negative charge is –q. On the line that passes through the charges, how many places are there at which the total potential is zero? 19.4 Equipotential Surfaces and Their Relation to the Electric Field An equipotential surface is a surface on which the electric potential is the same everywhere. kq V= r Equipotential surfaces for a point charge (concentric spheres centered on the charge) The net electric force does no work on a charge as it moves on an equipotential surface. 19.4 Equipotential Surfaces and Their Relation to the Electric Field The electric field created by any charge or group of charges is everywhere perpendicular to the associated equipotential surfaces and points in the direction of decreasing potential. In equilibrium, conductors are always equipotential surfaces. 19.4 Equipotential Surfaces and Their Relation to the Electric Field Lines of force and equipotential surfaces for a charge dipole. 19.4 Equipotential Surfaces and Their Relation to the Electric Field The electric field can be shown to be related to the electric potential. If you move a distance Δs and the electric potential changes by an amount ΔV, the component of the electric field in the direction of Δs is the negative of the ratio of ΔV to Δs, or more compactly, ΔV E=− Δs ΔV/Δs is called the potential gradient. 19.4 Equipotential Surfaces and Their Relation to the Electric Field Example 9 The Electric Field and Potential Are Related The plates of the capacitor are separated by a distance of 0.032 m, and the potential difference between them is VB-VA=-64V. Between the two equipotential surfaces shown in color, there is a potential difference of -3.0V. Find the spacing between the two colored surfaces. 19.4 Equipotential Surfaces and Their Relation to the Electric Field E=− Δs = − ΔV − 64 V = + = 2.0 ×103 V m Δs 0.032 m ΔV − 3.0 V −3 =− = 1 . 5 × 10 m 3 E 2.0 ×10 V m The gradient equation can be simplified for a parallel plate capacitor with voltage V across it and with plates separated by distance d as: E = V/d