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Today 3/12. Plates if charge E-Field Potential HW: “Plate Potential” Due Friday, 3/14 . (-). E net = 0. E net = 0. How big is E?. (+). Charged conducting plate. A = Area of one side. 0 = Q/A. What’s wrong with this picture?. Charged conducting plate. L = 0 /2.
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Today 3/12 • Plates if charge • E-Field • Potential • HW: “Plate Potential” Due Friday, 3/14
(-) Enet = 0 Enet = 0 How big is E? (+)
Charged conducting plate A = Area of one side 0 = Q/A 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 R = 0/2 EL = L/2e0 = 0/4e0 What is the electric field inside the conductor?
Charged conducting plate ER = R/2e0 = 0/4e0 L = 0/2 R = 0/2 What is the electric field inside the conductor?
Charged conducting plate L = 0/2 R = 0/2 The electric field is zero everywhere inside the conductor.Always, any conductor, no exceptions. What is the electric field inside the conductor?
Assume the particle gains 100 joules of kinetic energy as it moves from A to B. KE = 0 KE = 100 J q=+1C B A
Now I stop it at B. How much work must I do to move it back to A? q=+1C B A +100 J How does the potential energy change in moving from B to A? DPEBA = +100 J +100 J DPEAB = -100 J
What if q=+2C? How much work must I do to move it back to A? q=+2C B A +200 J How does the potential energy change in moving from B to A? DPEBA = +200 J +200 J DPEAB = -200 J
DPEBA = DVBA q DVBA tells us how much PE changes when +1C is moved from B to A. Now we are back to our original definition. q=+1C B A DVBA = +100 J/C DPEBA = (+100 J/C)x(q)
What if q = -1C? First I must turn my hand around. B A q= -1C
DVBA tells us how much PE changes when +1C is moved from B to A. What if q= -1C? How much work must I do to move it back to A? B A q= -1C -100 J How does the potential energy change in moving from B to A? DPEBA = DVBA q DPEBA = (+100 J/C)x(q) -100 J DPEBA = (+100 J/C)x(-1)
DVBAdoes not depend on the sign of the point charge but DPEBAdoes!!!!! What if q= -1C? How much work must I do to move it back to A? B A q= -1C -100 J How does the potential energy change in moving from B to A? DPEBA = DVBA q DPEBA = (+100 J/C)x(q) -100 J 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 10-27kg q= 1.6 x 10-19 C DVAB = -100 J/C DVAB = -100 volts B A DPEAB = q DVAB DPEAB = q (-100 J/C) DPEAB = -1.6 x 10-17 J What happens to the kinetic energy?
A proton is released from rest at A. What is its speed when it reaches B? m= 1.7 x 10-27kg q= 1.6 x 10-19 C DKEAB = +1.6 x 10-17 J B A 1/2 mv2 = +1.6 x 10-17 J v = 1.4 x 105 m/s What happens to the kinetic energy? DPEAB = -1.6 x 10-17 J
F E What direction is the force on an electron? B B A A
An electron is released from rest at B. What is its speed when it reaches A? m= 9.1 x 10-31kg q= -1.6 x 10-19 C DVAB = -100 J/C DVBA = +100 J/C B A DPEBA = q DVBA DPEBA = q (+100 J/C) DPEBA = -1.6 x 10-17 J What happens to the kinetic energy?
An electron is released from rest at B. What is its speed when it reaches A? m= 9.1 x 10-31kg q= -1.6 x 10-19 C DKEBA = +1.6 x 10-17 J B A 1/2 mv2 = +1.6 x 10-17 J v = 5.9 x 106 m/s What happens to the kinetic energy? DPEBA = -1.6 x 10-17 J
DVAB = ? A B
How does doubling the E-field affect DVAB ? DVAB = ? DVAB doubles A B
How does moving point B affect DVAB ? D DVAB = ? DVAB is halved Bold A Bnew Anything else? How does DVAB depend on E and D?
KEA=1/2 mv02 constantspeed v0 v0 How much work must I do to move the charge from A to B? KEB=1/2 mv02 DKEAB = 0 D A B
FHq FE = qE How much work must I do to move the charge from A to B? A B WAB = FE x D WAB = qED WAB = FHq x D
WAB = qED FHq FE = qE WAB = qED What is the change in potential energy in going from A to B? A B 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. PEAB = qED PEAB = qDVAB DVAB = EDAB