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Today 2/28
Read Ch 19.3
Practice exam posted
Electric Potential Energy, PEE
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.
v=0
v=?
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=?
v=0
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.
E Field and Force are not the same
everywhere so Fnet = ma requiresv calculus.
=0 v=?
Also need a system to handle direction.
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=?
v=0
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.
A
Initial PE at A?
Initial KE at A?
B
v=0 v=?
0
PEE KE
PEE KE
PEA,B = - KEA,B
Final PE at B? (More or less than before?) Subscripts tell us
from A to B
Final KE at B? Equals the PE lost!
“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.
Potential Difference VA,B is the change in PE a
positive 1C charge would feel if it were moved
from one location to another.
VA,B = +108 Volts,
and q = +1C
PEA,B = +100J
VA,B = -108 Volts,
and q = +1C
PEA,B = -100J
Which way does the E Field point?
PEA,B =
+100J, and q =
+1C What is
VA,B?
Remember:
From A to B!!!
VA,B = 108 Volts
Higher
Potential
E field
A
Higher
Potential
B
PEA,B =
E field
+100J, and q =
VA,B = -108 Volts
-1C What is
VA,B
? “downhill” with respect to potential difference
E Fields
point
What about Work, WA,B?
PEA,B =
+100J, and q =
+1C What is
VA,B?
Remember:
From A to B!!!
VA,B = 108 Volts
+Work
Higher
Potential
E field
A
Higher
Potential
B
+Work
PEA,B =
E field
+100J, and q =
VA,B = -108 Volts
-1C What is
VA,B
? “downhill” with respect to potential difference
E Fields
point
Calculating VA,B
V , A
kQ Q is the “source charge.”

r A is a location near the
source charge.
r is the distance from the
source charge to A.
Q
A
VA,B = V,B - V,A
B

V A  V , A
Example
kQ

r
What is “the potential (VA)” at A, 5cm from a
4 x 10-9 C point charge?
+4 x 10-9 C
Q
V , A
5cm
A
9
(9  10 )(4  10 )

 720V
0.05
9
V x  V , x
Example
kQ

r
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.
+4 x 10-9 C
5cm
Q
A
B
V,A  720V V,B  360V
VAB = -360V (the potential at B is less)
V x  V , x
Example
kQ

r
A 4g particle with charge q = +6C is released
from rest at A. What is its speed at B?
+4 x 10-9 C
Q
A
B
VAB = -360V
PEAB = qV = (6 C)(-360V ) = -2.2 x 10-3 J
KEAB = -PEAB = +2.2 x 10-3 J = mv2
v = 1m/s at location B
More than one source
What is the potential difference VAB? All
distances are 5cm.
Q1 = +4 x 10-9 C
A
B
Q2 = +10 x 10-9 C
VA = kq1/r1A + kq2/r2A Sum potentials at A
VA = (9x109)(4x10-9)/.05 + (9x109)(10x10-9)/.10
VA = 720V + 900V = 1620V
VB = (9x109)(4x10-9)/.10 + (9x109)(10x10-9)/.05
VB = 360V + 1800V = 2160V
VAB = +540V potential is higher at B
More than one source
A 4g particle with charge +6C is released from
rest at B. What is its speed at A?
Q1 = +4 x 10-9 C
A
B
q = +6C
Q2 = +10 x 10-9 C
VAB = +540V
VBA = -540V
PEBA = qV = (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
More than one source
What is the potential at point A midway between
these two charges? (different charges than before)
Q1 = +4 x 10-9 C
A
Enet
Q2 = -4 x 10-9 C
VA = 0 Two contributions add to zero
Is there an E-field at A?
Yes, Enet points right. Two contributions add as
vectors, yet the potential is zero!
The potential is negative just right of A and
positive just left of A. There is E if V changes.
More than one source
How much work do I have to do to bring a 6C to
point A from very far away?
Q1 = +4 x 10-9 C
VA = ?
A
Q2 = -4 x 10-9 C
VA = 0
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.