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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 PEA,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 VA,B is the change in PE a positive 1C charge would feel if it were moved from one location to another.
Calculating VA,B
r
kQV A,
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.
VA,B = V,B - V,AA
B
Q
VA,B is negative!
More than one sourceWhat is the potential difference VAB? All distances are 5cm.
A B
VA = kq1/r1A + kq2/r2A Sum potentials at A
Q1 = +4 x 10-9 C Q2 = +10 x 10-9 C
VA = (9x109)(4x10-9)/.05 + (9x109)(10x10-9)/.10
VA = 720V + 900V = 1620V
VB = (9x109)(4x10-9)/.10 + (9x109)(10x10-9)/.05VB = 360V + 1800V = 2160V
VAB = +540V potential is higher at B
More than one source
A BQ1 = +4 x 10-9 C Q2 = +10 x 10-9 C
VAB = +540V
A 4g particle with charge +6C is released from rest at B. What is its speed at A?
VBA = -540V
PEBA = qV = (6 C)(-540V ) = -3.2 x 10-3 J
KEBA = -PEBA = +3.2 x 10-3 J = 1/2 mv2
v = 1.3 m/s at location A
q = +6C
More than one source
AQ1 = +4 x 10-9 C Q2 = -4 x 10-9 C
VA = 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, Enet points right. Two contributions add as vectors, yet the potential is zero!
Enet
The potential is negative just right of A and positive just left of A. There is E if V changes.
More than one source
AQ1 = +4 x 10-9 C Q2 = -4 x 10-9 C
VA = ?
How much work do I have to do to bring a 6C 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.
VA = 0
How much energy is stored in the assembly of charges? All are +1 x 10-3C and 1g (1 x 10-3kg).
4m
5m
3mEnergy 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 r
kQV
0
V
VqPE
Energy stored
+1 x 10-3C and 1g
4m
5m
3m
First moves in for free. Work = PEr = 0 Upper left requires work. Work = 9/4 x 103 J
Upper right also requires work, two contributions.
Work = (9/3 + 9/5) x 103 J
Energy stored in the system is the sum of the work done.Energy Stored = 7 x 103 J
r
kQV
0
V
VqPE
+1 x 10-3C and 1g
4m
5m
3m
What happens if we let go of the charges?
Energy Stored = 7 x 103 J
They accelerate away from each other.
When they are very far away from each other the system will have 7 x 103 J of kinetic energy shared among these three charges. (energy is conserved)
Energy stored r
kQV
0
V
VqPE
+1 x 10-3C and 1g
4m
5m
3m
What happens if we let go the upper right charge?
PE after = 9/4 x 103 J
Buckets!
Energy stored
0PEE KEPEE KE
Before After
Total System PE Total System KE, all in the moving charge
7 x 103 J
PE after = ?
r
kQV
0
V
VqPE
PE = 19/4 x 103 J
KE = +19/4 x 103 J , + = speeds up
+ or - ?
v = 3082m/s
+1 x 10-3C 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 103 J
Each charge has 0.9 x 103 J of kinetic energy. (split 50/50 by symmetry)0.9 x 103 = 1/2mv2
v = 1340m/s
Energy stored r
kQV
0
V
VqPE
Both are 1C in magnitude.
5m
What about the case where one is now a negative charge?
Energy stored =
-1.8 x 10-3 J
Energy stored r
kQV
0
V
VqPE
-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.
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