# Today 2/28  Read Ch 19.3  Practice exam posted  Electric Potential Energy, PE E Electric Potential, V Electric Potential Difference,  V (watch out.

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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

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 = ?

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!

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

“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

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!!!

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

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

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

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

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

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

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

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.

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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