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Today 2/28 Read Ch 19.3 Practice exam posted Electric Potential Energy, PE E Electric Potential, V Electric Potential Difference, V (watch out foe +/- signs!!!) More than one charge HW:“Charge Assembly” Due Monday 3/5

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Problem: Two identical point charges of 0.01g are placed 1m apart. The right-hand charge is released. Find its velocity when it is 10cm farther away. Two opposite point charges of 0.01g are placed 1m apart. The right-hand charge is released. Find its velocity when it is 10cm closer. v = 0 v = ?

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Problem: Two identical point charges of mass of 0.01g are placed 1m apart. The right-hand charge is released. Find its velocity when it is 10cm farther away. Two opposite point charges of 0.01g are placed 1m apart. The right-hand charge is released. Find its velocity when it is 10cm closer. v = 0 v = ? E Field and Force are not the same everywhere so F net = ma requires calculus. Also need a system to handle direction. Energy buckets are the way!

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Problem: Two identical point charges og mass.01g are placed 1m apart. The right-hand charge is released. Find its velocity when it is 10cm farther away. v = 0v = ? PE E KEPE E KE Initial PE at A? Initial KE at A? Final PE at B? (More or less than before?) Final KE at B? 0 Equals the PE lost! PE A,B = - KE A,B AB Subscripts tell us from A to B

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“Potential Energy Difference” and “Potential Difference” Potential Energy Difference PE A,B is the change in PE the particular charge feels when it is moved from one location to another. Potential Difference V A,B is the change in PE a positive 1C charge would feel if it were moved from one location to another. V A,B = +10 8 Volts, and q = +1 C PE A,B = +100J V A,B = -10 8 Volts, and q = +1 C PE A,B = -100J

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Which way does the E Field point? PE A,B = +100J, and q = +1 C What is V A,B ? V A,B = 10 8 Volts V A,B = -10 8 Volts PE A,B = +100J, and q = -1 C What is V A,B ? AB Higher Potential E Fields point “downhill” with respect to potential difference E field Remember: From A to B!!!

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What about Work, W A,B ? PE A,B = +100J, and q = +1 C What is V A,B ? V A,B = 10 8 Volts V A,B = -10 8 Volts PE A,B = +100J, and q = -1 C What is V A,B ? AB Higher Potential E Fields point “downhill” with respect to potential difference E field Remember: From A to B!!! +Work

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Calculating V A,B Q is the “source charge.” A is a location near the source charge. r is the distance from the source charge to A. V A,B = V ,B - V ,A A B Q

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Example What is “the potential (V A )” at A, 5cm from a 4 x 10 -9 C point charge? A+4 x 10 -9 C 5cm Q

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Example But it is the change in potential that is important. What is the potential difference between point A and point B? B is 10cm from the point charge. V AB = -360V (the potential at B is less) AB+4 x 10 -9 C 5cm Q

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Example A 4g particle with charge q = +6 C is released from rest at A. What is its speed at B? AB V AB = -360V PE AB = q V = (6 C)(-360V ) = -2.2 x 10 -3 J KE AB = - PE AB = +2.2 x 10 -3 J = mv 2 v = 1m/s at location B +4 x 10 -9 C Q

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More than one source What is the potential difference V AB ? All distances are 5cm. AB V A = kq 1 /r 1A + kq 2 /r 2A Sum potentials at A Q 1 = +4 x 10 -9 CQ 2 = +10 x 10 -9 C V A = (9x10 9 )(4x10 -9 )/.05 + (9x10 9 )(10x10 -9 )/.10 V A = 720V + 900V = 1620V V B = (9x10 9 )(4x10 -9 )/.10 + (9x10 9 )(10x10 -9 )/.05 V B = 360V + 1800V = 2160V V AB = +540V potential is higher at B

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More than one source AB Q 1 = +4 x 10 -9 CQ 2 = +10 x 10 -9 C V AB = +540V A 4g particle with charge +6 C is released from rest at B. What is its speed at A? V BA = -540V PE BA = q V = (6 C)(-540V ) = -3.2 x 10 -3 J KE BA = - PE BA = +3.2 x 10 -3 J = 1 / 2 mv 2 v = 1.3 m/s at location A q = +6 C

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More than one source A Q 1 = +4 x 10 -9 CQ 2 = -4 x 10 -9 C V A = 0 Two contributions add to zero What is the potential at point A midway between these two charges? (different charges than before) Is there an E-field at A? Yes, E net points right. Two contributions add as vectors, yet the potential is zero! E net The potential is negative just right of A and positive just left of A. There is E if V changes.

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More than one source A Q 1 = +4 x 10 -9 CQ 2 = -4 x 10 -9 C V A = ? How much work do I have to do to bring a 6 C to point A from very far away? The work equals zero also since V = 0. Depending on the particular path we chose there will be + and - work done along the way but the net work done will always be zero for any path from far away to point A. V A = 0

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