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Today 3/12 Plates if charge E-Field Potential HW: “Plate Potential” Due Friday, 3/14 Enet = 0 How big is E? E 2 0 (+) E 2 0 E net Enet = 0 (-) 0 Charged conducting plate 0 = Q/A A = Area of one side What’s wrong with this picture? Charged conducting plate L = 0/2 R = 0/2 Free charge always goes to surface of conductor. Charged conducting plate L = 0/2 EL = L/20 = 0/40 What is the electric field inside the conductor? R = 0/2 Charged conducting plate L = 0/2 R = 0/2 ER = R/20 = 0/40 What is the electric field inside the conductor? Charged conducting plate L = 0/2 What is the electric field inside the conductor? R = 0/2 The electric field is zero everywhere inside the conductor. Always, any conductor, no exceptions. What happens if I let it go? q=+1C Assume the particle gains 100 joules of kinetic energy as it moves from A to B. KE = 0 A KE = 100 J q=+1C B Now I stop it at B. How much work must I do to move it back to A? +100 J How does the potential energy change in moving from B to A? +100 J A q=+1C B DPEBA = +100 J DPEAB = -100 J What if q=+2C? How much work must I do to move it back to A? +200 J How does the potential energy change in moving from B to A? +200 J q=+2C A B DPEBA = +200 J DPEAB = -200 J Now we are back to our original definition. DVBA tells us how much PE changes when +1C is moved from B to A. A q=+1C B DVBA = +100 J/C DPEBA = (+100 J/C)x(q) DPEBA = DVBA q What if q = -1C? First I must turn my hand around. A q= -1C B What if q= -1C? How much work must I do to move it back to A? -100 J How does the potential energy change in moving from B to A? -100 J DVBA tells us how much PE changes when +1C is moved from B to A. A q= -1C B DPEBA = DVBA q DPEBA = (+100 J/C)x(q) DPEBA = (+100 J/C)x(-1) What if q= -1C? How much work must I do to move it back to A? -100 J How does the potential energy change in moving from B to A? -100 J DVBA does not depend on the sign of the point charge but DPEBA does!!!!! A q= -1C B DPEBA = DVBA q DPEBA = (+100 J/C)x(q) DPEBA = (+100 J/C)x(-1) A proton is released from rest at A. What is its speed when it reaches B? m= 1.7 x 10-27 kg q= 1.6 x 10-19 C What happens to the kinetic energy? DVAB = -100 J/C DVAB = -100 volts A B DPEAB = q DVAB DPEAB = q (-100 J/C) DPEAB = -1.6 x 10-17 J A proton is released from rest at A. What is its speed when it reaches B? DKEAB = +1.6 x 10-17 J m= 1.7 x 10-27 kg 1/2 mv2 = +1.6 x 10-17 J q= 1.6 x 10-19 C What happens to the kinetic energy? A B v = 1.4 x 105 m/s DPEAB = -1.6 x 10-17 J What direction is the force on an electron? F A E B An electron is released from rest at B. What is its speed when it reaches A? m= 9.1 x 10-31 kg q= -1.6 x 10-19 C DVAB = -100 J/C DVBA = +100 J/C A B DPEBA = q DVBA DPEBA = q (+100 J/C) What happens to the kinetic energy? DPEBA = -1.6 x 10-17 J An electron is released from rest at B. What is its speed when it reaches A? DKEBA = +1.6 x 10-17 J m= 9.1 x 10-31 kg 1/2 mv2 = +1.6 x 10-17 J q= -1.6 x 10-19 C What happens to the kinetic energy? A B v = 5.9 x 106 m/s DPEBA = -1.6 x 10-17 J DVAB = ? B A How does doubling the E-field affect DVAB = ? DVAB ? DVAB doubles B A How does moving point B affect DVAB = ? DVAB ? DVAB is halved Anything else? How does DVAB depend on E and D? D Bold Bnew A How much work must I do to move the charge from A to B? KEA=1/2 mv02 KEB=1/2 mv02 DKEAB = 0 D v0 v0 A B constant speed How much work must I do to move the charge from A to B? FHq B WAB = FHq x D WAB = FE x D WAB = qED FE = qE A WAB = qED What is the change in potential energy in going from A to B? PEAB = qED PEAB = qDVAB DVAB = EDAB WAB = qED FHq B FE = qE A Only applies when the field is uniform over the distance. DVAB‘s sign depends on the direction of E. In this case it’s positive.