Today 3/3  Read Ch 19.4  Practice exam posted  Energy of Assembly  Calculating Velocities  HW:“Charge Assembly 2” Due Thursday  Lab: “Mapping Equipotential Lines” (get E field lines from these)

“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. This is also the Work required to move the charge. 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.

Calculating  V A,B Q is the “source charge.” A is a location (field point) near the source charge. r is the distance from the source charge to point field point.  V A,B =  V ,B -  V ,A A  B Q  V A,B is negative!

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 CQ 2 = +10 x 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 V = 2160V  V AB = +540V potential is higher at B

More than one source AB Q 1 = +4 x CQ 2 = +10 x 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 J  KE BA = -  PE BA = +3.2 x 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 CQ 2 = -4 x 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 CQ 2 = -4 x 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

How much energy is stored in the assembly of charges? All are +1 x C and 1g (1 x kg). 4m 5m 3m Energy stored is the same as the work required to assemble the charges. Move them all very far away, bring them in one at a time, and total up the work done. Energy stored

+1 x C and 1g 4m 5m 3m First moves in for free. Work =  PE  r = 0 Upper left requires work.Work = 9/4 x 10 3 J Upper right also requires work, two contributions. Work = (9/3 + 9/5) x 10 3 J Energy stored in the system is the sum of the work done. Energy Stored = 7 x 10 3 J

+1 x C and 1g 4m 5m 3m What happens if we let go of the charges? Energy Stored = 7 x 10 3 J They accelerate away from each other. When they are very far away from each other the system will have 7 x 10 3 J of kinetic energy shared among these three charges. (energy is conserved) Energy stored

+1 x C and 1g 4m 5m 3m What happens if we let go the upper right charge? PE after = 9/4 x 10 3 J Buckets! Energy stored 0 PE E KEPE E KE BeforeAfter Total System PE Total System KE, all in the moving charge 7 x 10 3 J PE after = ?  PE = 19/4 x 10 3 J  KE = +19/4 x 10 3 J, + = speeds up + or - ? v = 3082m/s

+1 x C and 1g 5m How fast are these charges moving after they are released and they are very far away from each other? Energy stored = 1.8 x 10 3 J Each charge has 0.9 x 10 3 J of kinetic energy. (split 50/50 by symmetry) 0.9 x 10 3 = 1/2mv 2 v = 1340m/s Energy stored

Both are 1  C in magnitude. 5m What about the case where one is now a negative charge? Energy stored = -1.8 x J Energy stored -q +q When the energy is negative we call it the “binding energy.” Binding energy is the work required to separate the charges to infinity. Electrons are bound to protons in this way.