1 1 W03D2 Work, Potential Energy and Electric Potential Today’s Reading Assignment: Course Notes: Sections 4.1-4.3 Announcements PS 3 due Week Three Tuesday at 9 pm in boxes outside 32-082 or 26-152 W03D3 Reading Assignment Course Notes: Sections 4.7-4.10 Exam One Thursday Feb 27 7:30-9:30 pm Room Assignments (See Stellar webpage announcements) 2 3 Outline Electrical Work Electric Potential Energy Electric Potential Difference Calculating Electric Potential Difference2 4 Electrical Work Work done by electrical force moving object 1 from A to B: W1=F12⋅ ds12AB∫Electrical force on object 1 due to interaction between charged objects 1 and 2: F12= keq1q2r122ˆr12PATH INTEGRAL 5 Concept Question: Sign of W Suppose a fixed positively charged object (charge qs > 0) is at the origin and we move a negatively charged object (charge q1 < 0) from A to B with rA < rB , where r is the distance from the origin. 1. Work done by the electrostatic force is positive and we do a positive amount of work 2. Work done by the electrostatic force is positive and we do a negative amount of work 3. Work done by the electrostatic force is negative and we do a positive amount of work 4. Work done by the electrostatic force is negative and we do a negative amount of work 6 Group Problem: Work Done by Electrical Force A point-like charged source object (charge qs) is held fixed. A second point-like charged object (charge q1)is initially at a distance rA from the fixed source and moves to a final distance rB from the fixed source. What is the work done by the electrical force on the moving object? Hint: What coordinate system is best suited for this problem?3 7 Sign of W: Negative Work Suppose a fixed positively charged source (charge qs > 0) is at the origin and a positively charged object (charge q1 > 0) moves from A to B with rA > rB , where r is the distance from the origin, then W < 0. rA> rB⇒1rB−1rA> 0 and qsq1> 0 ⇒W = −keqsq11rB−1rA⎛⎝⎜⎞⎠⎟< 08 Work and Change in Kinetic Energy W = ΔK9 Group Problem: Work-Kinetic Energy In a Uniform Electric field Consider two thin oppositely uniform charged thin plates separated by a distance d. The surface charge densities on the plates are uniform and equal in magnitude. An electron with charge –e and mass m is released from rest at the negative plate and moves to the positive plate. What is the speed of the electron when it reaches the positive plate?4 10 Potential Energy Difference ΔU ≡ UB−UA= −F12⋅ dsAB∫= −W = keqsq11rB−1rA⎛⎝⎜⎞⎠⎟ Suppose charged object 1 is fixed and located at the origin and charge object 2 moves from an initial position A a distance rA from the origin to a final position B, a distance rB from the origin. The potential energy difference due to the interaction is defined to be the negative of the work done object 2 in moving from A to B: 11 Potential Energy: Zero Point U (r) − U (∞) = U (r) =keqsq1r Choose the zero point for the potential energy at infinity. Then set rA = ∞ and rB = r . The potential energy difference between ∞ and any point on a circle of radius r is U (∞) ≡ 012 Concept Question: Motion of Charged Objects Two oppositely charged are released from rest in an electric field. 1. Both charged objects will move from lower to higher potential energy. 2. Both charged objects will move from higher to lower potential energy. 3. The positively charged object will move from higher to lower potential energy; the negatively charged object will move from lower to higher potential energy. 4. The positively charged object will move from lower to higher potential energy; the negatively charged object will move from higher to lower potential energy.5 13 Configuration Energy What is the potential energy stored in a configuration of charged objects? Start with all the charged objects at infinity. Choose (1) Bring in the first charged object. (2) Bring in the second charged object (3) Bring in the third charged object (4) Configuration energy ΔU2= U12= keq1q2/ r12 U (∞) ≡ 0 ΔU3= U23+ U13= keq2q3/ r23+ keq1q3/ r13 ΔU1= 0 ΔU = U12+ U23+ U13= keq1q2/ r12+ keq2q3/ r23+ keq1q3/ r1314 Group Problem: Build It How much energy does it take you to assemble the charges into the configuration at left, assuming they all started out an infinite distance apart? 15 Electric Potential Difference ΔV ≡ΔUqt= −FqtAB∫⋅ ds = −EAB∫⋅ dsUnits: Joules/Coulomb = Volts Change in potential energy per test object in moving the test object (charge qt) from A to B:6 16 Demonstration Van de Graaf D29 Breakdown of dry air 33 kV/cm Video of Tesla Coil http://www.youtube.com/watch?v=FY-AS13fl30 17 How Big is a Volt? AA Batteries 1.5 V High Voltage Transmission Lines 100 kV-700 kV Car Batteries 12 V Van der Graaf 300 kV US Outlet (AC) 120 V Tesla Coil 500 kV Distribution Power Lines 120 V- 70 kV Lightning 10-1000 MV 18 E Field and Potential: Effects F = qEIf you put a charged particle, (charge q), in a field: ΔU = qΔVTo move a charged particle, (charge q), in a field and the particle does not change its kinetic energy then:7 19 Concept Question: Motion of Charged Objects Two oppositely charged are released from rest in an electric field. 1. Both charged objects will move from lower to higher electric potential. 2. Both charged objects will move from higher to lower electric potential. 3. The positively charged object will move from higher to lower electric potential; the negatively charged object will move from lower to higher electric potential. 4. The positively charged object will move from lower to higher electric potential; the negatively charged object will move from higher to lower electric potential. 20 Potential & External Work Change in potential energy in moving the charged object (charge q) from A to B: ΔU = qΔVConservation of Energy Law: Wex t= ΔK + ΔU21 Demonstration Kelvin Water Drop in 26-152 Wimshurst Machine in 32-1528 22 Demonstration: Kelvin Water 23 Potential Created by Pt Charge ds = drˆr + r dφˆφ ΔV = VB− VA= −E ⋅ dsAB∫ E = kQˆrr2 = − kQˆrr2⋅ dsAB∫= −kQdrr2AB∫ = kQ1rB−1rA⎛⎝⎜⎞⎠⎟Take V = 0 at r = ∞: VPoint Charge(r) =kQr24 Concept Question: Two Point Charges The work done in moving a positively charged object that
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