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CHAPTER 17 Electrical Energy and Current Conservative Forces: 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 Example (Gravitational Force) Consider the work done on an object against gravity. W = F g d = mg h Work is converted to gravitational potential energy The Electrostatic Force (F e ) is also a conservative force. (qE = = F e ) k q 1 q r 2 W = F e d = qEd Work done by gravity (falling object) = -mg h Work done by electrostatic force = -qEd W = q E d

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Potential Difference: Potential Difference:The change in potential energy (Electric Potential)of a charge, q 1 divided by q 1. Potential Difference Potential Difference Change in potential energy Potential Difference ( V) (Electric Potential) 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 V = PE q

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Example Problem Example Problem (moving a positive charge against an electric field) q 15cm E = 250 N/C q = +400 C d = 15cm Work done on the charge = qE d = (400x10 -6 C)(250N/C)(15x10 -2 m) = 1.5 x 10 -2 Joules Work done increases the potential energy of the charge. V = PE q = qE d q = E d V = (250N/C)(15x10 -2 m) V = 38 N m/C V = 38 J/C V = 38V NOTE: [E] = V/m = N/C

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Summary Sign of Charge Direction of Movement Relative to E-Field Sign of PE +–+–+–+– Opposite (against) Same direction (with) + – – +

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Electric Field Between Parallel Plates What is the Electric Potential Differenced in the above diagram if E=25x10 2 N/C and d=15cm? Electric Potential Difference = V = V B – V A = -E d (Potential Difference) = V = -(25x10 2 N/C)(15x10 -2 m) = V = -375 N m/C = V = -375 J/C V = -375V

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If the charge consisted of a proton: m = 1.67x10 -27 kg q=1.60x10 -19 C 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 = q V = (1.60x10 -19 C)(38J/C) PE = 6.1x10 -19 J PE = KE = KE B (KE A = 0) 6.1x10 -18 J = (1.67x10 -27 kg) v 2 2 v = 6.0x10 4 m/s

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Electric Potential Associated with Point Charges Between parallel plates E is uniform. E associated with a point charge is not uniform E = k q r 2 = k q d 2 r = d V = E d for only small d values Calculus to the rescue! dv = Edd dv = dd k q d 2 dv = dd k q d 2 d= r d= v = -1 k q d d=r d= v= k q r v = k q r Scalar Quantity Electric Potential Caused by Point Change

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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 Example Problem (Electric Potential : linear) Two point charges 20cm apart each with a charge of +50 C are established. What is the electric potential 10.cm from each (midpoint)? What is the electric field at this point? Electric Potential (V) v = = + k q r k q 1 r 1 k q 2 r 2 v = (9.0x10 9 N m 2 /C 2 )(+50x10 -6 C) (10x10 -2 m) v = 4.5x10 6 N m/C v = 4.5x10 6 J/C v = 4.5x10 6 Volts

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Electric Field (F) E T = EVectors E 1 = k q 1 r 2 = (9.0x10 9 N m 2 /C 2 )(50x10 -6 C) (10x10 -2 m) 2 E 1 = 4.5x10 7 N/Cdirected away from q 1 Similarly E 2 = 4.5x10 7 N/C directed away from q 2 q1q1 q2q2 10cm E2E2 E1E1 E T = 0 N/C

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Example Problem Example Problem (Electric Potential : 2 Dimensions) Calculate the electric potential of point A v = k q r v = k + ( ) q1r1q1r1 q2r2q2r2 v = (9.0x10 9 N m 2 /C 2 ) ( ) -50x10 -6 C + 50x10 -6 C 60x10 -2 m 30x10 -2 m v = (9.0x10 9 N m 2 /C 2 )(-8.3x10 -5 C/m + 1.66x10 -4 C/m) v = (9.0x10 9 N m 2 /C 2 )(8.3x10 -5 C/m) v = 7.5x10 5 N m/C v = 7.5x10 5 J/C v = 7.5x10 5 V 60cm 30cm 52cm A q 2 = +50 Cq 1 = -50 C E Al E A2

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Equipotential Lines (Surfaces) Gravitational 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 Earth Lines of Equipotential Gravitational Field Lines

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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

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Test Yourself Draw E-fields and lines (surfaces) of Equipotential for the following situations. c) +q-q a) -q b) +q

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Current and Resistance (Electric) Current (I): The rate at which charge is flowing (through a wire). 1 Ampere = 1 Amp = 1 A = 1 Coulomb/sec I = QtQt Conventional Current: 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. Ampere (A):SI unit for current

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Resistance: Resistance:A measure of what must be overcome to make charge flow Ohm ( ): Ohm ( ):SI unit for resistance 1 ohm = 1 = 1 Volt/Amp R = VIVI 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.

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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 1.60x10 -19 coulombs 1.5x10 20 electrons b) I = = Q 24 coulombs t 2.1 seconds I = 11 coulombs/sec 11 amps c) R = = V 37 Volts I 11 amps 3.4 ohms R = 3.4 Volts/amp

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Resistance What factors might affect the resistance of a wire? l A R R = l A = resistivity of the material = ohms -meter Resistivity is another physical property of a material l A

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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/coulombCurrent = Coulomb/sec Voltage x Current V x I = Joules/sec= Watts Power has units of Joules/sec or Watts P = IV P = I 2 R P =? V = IR Joules x Coulomb Coulomb sec =

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