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Transcript
CHAPTER 17 Electrical Energy and Current Conservative Forces: Work done on an object depends only on its initial and final position. The path from the initial to the final position is not important Example (Gravitational Force) Consider the work done on an object against gravity. W = Fg d = mgh Work is converted to gravitational potential energy The Electrostatic Force (Fe) is also a conservative force. W = Fe d = qEd (qE = k q21 q = Fe) r Work done by gravity (falling object) = -mgh Work done by electrostatic force = -qEd W = q E d Potential Difference: The change in potential energy (Electric Potential) of a charge, q1 divided by q1. Potential Difference Change in potential energy Potential Difference (V) (Electric Potential) V = PE q • PE occurs in a uniform electric field • q is a charge that changes position in the uniform field • V is a scalar quantity • V units = Joules/Coulomb 1 Volt (V) = 1 Joule/Coulomb Example Problem (moving a positive charge against an electric field) 15cm q E = 250 N/C q = +400C d = 15cm Work done on the charge = qEd = (400x10-6C)(250N/C)(15x10-2m) = 1.5 x 10-2 Joules Work done increases the potential energy of the charge. V = PE = qEd = Ed q q V = (250N/C)(15x10-2m) V = 38 Nm/C V = 38 J/C NOTE: [E] = V/m = N/C V = 38V Summary Direction of Movement Relative to E-Field + – Opposite (against) Opposite (against) Sign of PE + – + – Same direction (with) Same direction (with) – + Sign of Charge Electric Field Between Parallel Plates What is the Electric Potential Differenced in the above diagram if E=25x102N/C and d=15cm? Electric Potential Difference = V = VB – VA = -Ed (Potential Difference) = V = -(25x102 N/C)(15x10-2 m) = V = -375 Nm/C = V = -375 J/C V = -375V If the charge consisted of a proton: m = 1.67x10-27kg q=1.60x10-19C V= 38J/C What would be its change in potential energy and with what velocity would it be moving at “B” if it was at rest at point “A”? Strategy Calculate change in potential energy and convert potential energy to kinetic energy. Solve for v. PE = qV = (1.60x10-19C)(38J/C) PE = 6.1x10-19J PE = KE = KEB (KEA = 0) 6.1x10-18J = (1.67x10-27kg) v2 v = 6.0x104 m/s 2 Electric Potential Associated with Point Charges Between parallel plates E is uniform. E associated with a point charge is not uniform E= kq = kq r=d 2 2 r d V = Ed for only small d values Calculus to the rescue! dv = Edd dv = k 2q dd d d= r kq = d2 d= dv v= dd d=r -1 k q d d= v= k q r v= kq r Scalar Quantity Electric Potential Caused by Point Change Electric Potential Caused by 2 Point Charges When Analyzing Multiple Point Charges: • The principle of superposition applies. (Just like with calculating E-field due to multiple charges.) • However “v” is a scalar quantity (J/C) and “E” was a vector quantity (N/C) • Scalars are much easier to add than vectors because with scalars… we have no direction. Example Problem (Electric Potential : linear) Two point charges 20cm apart each with a charge of +50C are established. What is the electric potential 10.cm from each (midpoint)? What is the electric field at this point? Electric Potential (V) kq v= = k q1 + k q 2 r r1 r2 v = (9.0x109Nm2/C2)(+50x10-6C) (10x10-2m) v = 4.5x106 Nm/C 6 Volts 6 v = 4.5x10 v = 4.5x10 J/C Electric Field (F) ET = E Vectors 9Nm2/C2)(50x10-6C) (9.0x10 k q 1 = E1 = (10x10-2m)2 r2 E1 = 4.5x107 N/C Similarly E2 = 4.5x107 N/C directed away from q1 directed away from q2 E2 E1 q1 10cm ET = 0 N/C q2 10cm Example Problem (Electric Potential : 2 Dimensions) EA2 A 30cm EAl 60cm 52cm q2 = +50 C q1 = -50 C Calculate the electric potential of point A v= k( v= kq r q1 q2 r1 + r2 ) ( -6C + 50x10-6C -50x10 v = (9.0x109Nm2/C2) 60x10-2m 30x10-2m ) v = (9.0x109Nm2/C2)(-8.3x10-5C/m + 1.66x10-4C/m) v = (9.0x109Nm2/C2)(8.3x10-5C/m) v = 7.5x105Nm/C v = 7.5x105J/C v = 7.5x105 V Equipotential Lines (Surfaces) Gravitational Gravitational Field Lines Lines of Equipotential Earth Line of Equipotential is merely a line (surface in 3-dimensional system) where potential energy remains constant as an object moves along the line Lines of Equipotential are perpendicular to force field lines so that no work is done when the object moves W = F d cos Equipotential Lines (Surfaces) Electrostatic Equipotential Lines (surfaces) • encircle the charged particle • perpendicular to field lines • never cross each other because field lines never cross each other Electric Field Lines • directed away from a positive charge (i.e. direction a positive “test charge” would move) • closer together indicates greater E-field • exit perpendicular to the surface • never cross each other Test Yourself Draw E-fields and lines (surfaces) of Equipotential for the following situations. a) -q b) +q +q +q -q c) Current and Resistance (Electric) Current (I): The rate at which charge is flowing (through a wire). I= Q t Ampere (A): SI unit for current 1 Ampere = 1 Amp = 1 A = 1 Coulomb/sec Conventional Current: The flow of positive charge. If conventional current is flowing to the right, then in reality, electrons are flowing to the left. Positive charge, protons, don’t move. Resistance: Ohm (): A measure of what must be overcome to make charge flow SI unit for resistance 1 ohm = 1 = 1 Volt/Amp V R= I If a large current results from a small V, then the resistance must be small. If a small current results from a large V, then the resistance must be large. Example Problem A flow of 24 coulombs of charge passes through a wire in 2.1 seconds where a voltage of 37 volts is applied across the wire. Calculate a) how many charges moved through the wire, b) the current in the wire, and c) the resistance of the wire. Strategy: Extract the data. Q = 24 coulombs t = 2.1 seconds V = 37 V Apply Proper Formulas a) 24 coulombs 1 electrons 20electrons 1.5x10 1.60x10-19 coulombs Q 24 coulombs b) I = = t 2.1 seconds 11 amps I = 11 coulombs/sec V 37 Volts c) R = = I 11 amps R = 3.4 Volts/amp 3.4 ohms Resistance What factors might affect the resistance of a wire? l A R l A R= l A = resistivity of the material = ohms -meter Resistivity is another physical property of a material Resistors and Energy Loss It is really friction in the wire that results in resistance to flow of charge. Friction causes heat. A resistor should get hot when voltage causes a current passes through it (Ex: filament in a light bulb) Derivation: Voltage = Joules/coulomb Current = Coulomb/sec Joules x Coulomb Voltage x Current = Coulomb sec V x I = Joules/sec = Watts Power has units of Joules/sec or Watts P = IV P = I2R P =? V = IR