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Chapter 19: Electric Potential Energy and Potential • Electric potential energy and potential • Equipotential surfaces, electric field • Omit 19.5 (capacitors and dielectrics) 1 Electrical potential energy • A potential energy can be defined for a conservative force. Example, gravity. • For the electrostatic force: – the work done in moving a charge from one place to another is independent of the path taken (no dissipation of energy by friction, etc) ⇒ The electrostatic force is conservative ⇒ An electrical potential energy can be defined... 2 Gravitational and Electrical Potential Energy Change in PE from B to A is work done in moving object from B to A ΔPE = mg(hA − hB) ΔPE = q0E(hA − hB) 3 Electric Potential Energy Release the charge q0 from point A. The electrostatic force does work, accelerates the charge: WAB = FΔs = q0EΔs = KEB − KEA Conservation of energy: Displacement = Δs KEA + PEA = KEB + PEB So, KEB − KEA = PEA − PEB = q0EΔs 4 WAB = PEA − PEB = q0EΔs Write: PEA = q0VA, PEB = q0VB VA,B = electrical potential (volts) Then, PEA − PEB = q0(VA −VB) = q0EΔs = WAB WAB (volts) q0 VA −VB ΔV E= =− Δs Δs VA −VB = Unit: N/C or V/m (equivalent) • If ΔV = 0, the potential energy of charge q0 is constant and E = 0 in the direction from A to B. • The line from A to B is then an “equipotential” line. 19.2 5 Prob. 19.2 An electric force moves a charge of q = 1.8 × 10-4 C from A to B and performs 5.8 × 10-3 J of work on the charge. a) What is PEA – PEB ? b) What is VA – VB ? c) Which point has the higher potential? 6 Equipotentials and electric field V1 All points between B and C on the blue line are at the same potential ⇒ a charge has constant PE on the blue line (same q0V1) V2 ⇒ no work is done in moving the charge from B to C along the equipotential ⇒ The electric field (red lines) points in a direction at right angles to the equipotentials. A conductor is an equipotential volume E = 0, V = constant throughout 7 Electric field and equipotential surface E must be perpendicular to equipotential! If the electric field is not perpendicular to the equipotential surface, then there is a component parallel to it, so work would have to be done to move a charge along the surface – which would no longer be an equipotential! 8 Equipotentials and Field Lines E E V V V=0 9 ΔV E =− Δs Capacitor 10 Prob. 19.56 The four outer points are all 6 mm from the origin. Find the electric field at the origin. 19.32 11 Potential Energy Potential energy of 1 electron charge at a potential of 1 V. PE = qV = eV = (1.6×10-19 C)×(1 V) = 1.6×10-19 J or, PE = 1 eV (electron-volt) 1 eV = 1.6×10-19 J As the charge moves, ΔKE = –ΔPE (i.e., KE + PE = constant) An electron charge gains a kinetic energy of 1 eV in falling through a potential drop of 1 V. 19.6 12