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Transcript
Today 2/28 Read Ch 19.3 Practice exam posted Electric Potential Energy, PEE Electric Potential, V Electric Potential Difference, V (watch out foe +/- signs!!!) More than one charge HW: “Charge Assembly” Due Monday 3/5 Problem: Two identical point charges of 0.01g are placed 1m apart. The right-hand charge is released. Find its velocity when it is 10cm farther away. v=0 v=? Two opposite point charges of 0.01g are placed 1m apart. The right-hand charge is released. Find its velocity when it is 10cm closer. v=? v=0 Problem: Two identical point charges of mass of 0.01g are placed 1m apart. The right-hand charge is released. Find its velocity when it is 10cm farther away. E Field and Force are not the same everywhere so Fnet = ma requiresv calculus. =0 v=? Also need a system to handle direction. Two opposite point charges of 0.01g are placed 1m apart. The right-hand charge is released. Find its velocity when it is 10cm closer. v=? v=0 Energy buckets are the way! Problem: Two identical point charges og mass .01g are placed 1m apart. The right-hand charge is released. Find its velocity when it is 10cm farther away. A Initial PE at A? Initial KE at A? B v=0 v=? 0 PEE KE PEE KE PEA,B = - KEA,B Final PE at B? (More or less than before?) Subscripts tell us from A to B Final KE at B? Equals the PE lost! “Potential Energy Difference” and “Potential Difference” Potential Energy Difference PEA,B is the change in PE the particular charge feels when it is moved from one location to another. Potential Difference VA,B is the change in PE a positive 1C charge would feel if it were moved from one location to another. VA,B = +108 Volts, and q = +1C PEA,B = +100J VA,B = -108 Volts, and q = +1C PEA,B = -100J Which way does the E Field point? PEA,B = +100J, and q = +1C What is VA,B? Remember: From A to B!!! VA,B = 108 Volts Higher Potential E field A Higher Potential B PEA,B = E field +100J, and q = VA,B = -108 Volts -1C What is VA,B ? “downhill” with respect to potential difference E Fields point What about Work, WA,B? PEA,B = +100J, and q = +1C What is VA,B? Remember: From A to B!!! VA,B = 108 Volts +Work Higher Potential E field A Higher Potential B +Work PEA,B = E field +100J, and q = VA,B = -108 Volts -1C What is VA,B ? “downhill” with respect to potential difference E Fields point Calculating VA,B V , A Q kQ r A Q is the “source charge.” A is a location near the source charge. r is the distance from the source charge to A. VA,B = V,B - V,A B V A V , A Example kQ r What is “the potential (VA)” at A, 5cm from a 4 x 10-9 C point charge? +4 x 10-9 C Q V , A 5cm A 9 (9 10 )( 4 10 ) 720V 0.05 9 V x V , x Example kQ r But it is the change in potential that is important. What is the potential difference between point A and point B? B is 10cm from the point charge. +4 x 10-9 C 5cm Q A B V , A 720V V ,B 360V VAB = -360V (the potential at B is less) V x V , x Example kQ r A 4g particle with charge q = +6C is released from rest at A. What is its speed at B? +4 x 10-9 C Q A B VAB = -360V PEAB = qV = (6 C)(-360V ) = -2.2 x 10-3 J KEAB = -PEAB = +2.2 x 10-3 J = mv2 v = 1m/s at location B More than one source What is the potential difference VAB? All distances are 5cm. Q1 = +4 x 10-9 C A B Q2 = +10 x 10-9 C VA = kq1/r1A + kq2/r2A Sum potentials at A VA = (9x109)(4x10-9)/.05 + (9x109)(10x10-9)/.10 VA = 720V + 900V = 1620V VB = (9x109)(4x10-9)/.10 + (9x109)(10x10-9)/.05 VB = 360V + 1800V = 2160V VAB = +540V potential is higher at B More than one source A 4g particle with charge +6C is released from rest at B. What is its speed at A? Q1 = +4 x 10-9 C A B q = +6C Q2 = +10 x 10-9 C VAB = +540V VBA = -540V PEBA = qV = (6 C)(-540V ) = -3.2 x 10-3 J KEBA = -PEBA = +3.2 x 10-3 J = 1/2 mv2 v = 1.3 m/s at location A More than one source What is the potential at point A midway between these two charges? (different charges than before) Q1 = +4 x 10-9 C A Enet Q2 = -4 x 10-9 C VA = 0 Two contributions add to zero Is there an E-field at A? Yes, Enet points right. Two contributions add as vectors, yet the potential is zero! The potential is negative just right of A and positive just left of A. There is E if V changes. More than one source How much work do I have to do to bring a 6C to point A from very far away? Q1 = +4 x 10-9 C VA = ? A Q2 = -4 x 10-9 C VA = 0 The work equals zero also since V = 0. Depending on the particular path we chose there will be + and - work done along the way but the net work done will always be zero for any path from far away to point A.