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Published bySerena Jersey Modified about 1 year ago

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Today 3/12 Plates if charge E-Field Potential HW:“Plate Potential” Due Friday, 3/14

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(+) How big is E? (-) E net = 0

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0 = Q/A Charged conducting plate What’s wrong with this picture? A = Area of one side

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L = 0 /2 Charged conducting plate Free charge always goes to surface of conductor. R = 0 /2

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L = 0 /2 Charged conducting plate R = 0 /2 What is the electric field inside the conductor? E L = L /2 0 = 0 /4 0

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L = 0 /2 Charged conducting plate R = 0 /2 What is the electric field inside the conductor? E R = R /2 0 = 0 /4 0

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L = 0 /2 Charged conducting plate R = 0 /2 What is the electric field inside the conductor? The electric field is zero everywhere inside the conductor. Always, any conductor, no exceptions.

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q=+1C What happens if I let it go?

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KE = 0KE = 100 J A B Assume the particle gains 100 joules of kinetic energy as it moves from A to B. q=+1C

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A 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 PE BA = +100 J PE AB = -100 J q=+1C

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A B PE BA = +200 J PE AB = -200 J What if q=+2C? How much work must I do to move it back to A? How does the potential energy change in moving from B to A? +200 J q=+2C

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A B Now we are back to our original definition. PE BA = (+100 J/C)x(q) V BA tells us how much PE changes when +1C is moved from B to A. PE BA = V BA q V BA = +100 J/C q=+1C

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A B What if q = -1C? First I must turn my hand around. q= -1C

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A 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 PE BA = (+100 J/C)x(q) PE BA = (+100 J/C)x(-1) V BA tells us how much PE changes when +1C is moved from B to A. PE BA = V BA q q= -1C

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A 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 PE BA = (+100 J/C)x(q) PE BA = (+100 J/C)x(-1) V BA does not depend on the sign of the point charge but PE BA does!!!!! PE BA = V BA q q= -1C

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A B V AB = -100 J/C V AB = -100 volts A proton is released from rest at A. What is its speed when it reaches B? m= 1.7 x kg q= 1.6 x C PE AB = q V AB PE AB = q -100 J/C) PE AB = -1.6 x J What happens to the kinetic energy?

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A B KE AB = +1.6 x J A proton is released from rest at A. What is its speed when it reaches B? m= 1.7 x kg q= 1.6 x C mv 2 = +1.6 x J v = 1.4 x 10 5 m/s What happens to the kinetic energy? PE AB = -1.6 x J

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A B A B What direction is the force on an electron? EF

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A B V AB = -100 J/C V BA = +100 J/C An electron is released from rest at B. What is its speed when it reaches A? m= 9.1 x kg q= -1.6 x C PE BA = q V BA PE BA = q +100 J/C) PE BA = -1.6 x J What happens to the kinetic energy?

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A B KE BA = +1.6 x J mv 2 = +1.6 x J v = 5.9 x 10 6 m/s An electron is released from rest at B. What is its speed when it reaches A? m= 9.1 x kg q= -1.6 x C What happens to the kinetic energy? PE BA = -1.6 x J

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B A V AB = ?

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B A How does doubling the E-field affect V AB ? V AB = ? V AB doubles

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B new A How does moving point B affect V AB ? V AB = ? How does V AB depend on E and D? D Anything else? V AB is halved B old

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B A KE AB = 0 constant speed v0v0 v0v0 KE A =1/2 mv 0 2 KE B =1/2 mv 0 2 How much work must I do to move the charge from A to B? D

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B A F E = qEF Hq W AB = F E x D W AB = qED W AB = F Hq x D

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B A What is the change in potential energy in going from A to B? F E = qE F Hq W AB = qED PE AB = qED PE AB = q V AB V AB = ED AB W AB = qED Only applies when the field is uniform over the distance. V AB ‘s sign depends on the direction of E. In this case it’s positive.

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