Physics 212
Lecture 5
Electric Potential Energy
Physics 212 Lecture 5, Slide 1
Recall from physics 211:

r2 

W   F  dr

r1
F
dr
W>0
Object speeds up ( K > 0 )
W<0
Object slows down ( K < 0 )
F
dr
or
F
dr
F
dr
9
WTOT   K
W=0
Constant speed ( K = 0 )
Physics 212 Lecture 5, Slide 2
Potential Energy
U  Wconservative
If gravity does negative work, potential energy increases!
Same idea for Coulomb force… if Coulomb force does
negative work, potential energy increases.
+
+
x
+
+
F
Coulomb force does negative work
Potential energy increases
Physics 212 Lecture 5, Slide 3
Analogy to gravity
m
Fext
Exert force Fext = - Fg
over distance h.
h
m
Fg
Work by me = Fext h = - Fgh = mgh = U

h
0
Fext d r    Fg d r  U g  h  U g  0  U g
h
0
Change in
Potential energy
Work by me
- (Work by gravity)
Checkpoint 4
A charge is released from rest in a region of electric field. The charge will start to move
A) in a direction that makes its potential energy increase
B) in a direction that makes its potential energy decrease
C) along a path of constant potential energy
F
x
It will move in the same direction as F
Work done by E field is positive
U = -Work is negative
34
Physics 212 Lecture 5, Slide 5
Example: Charge in External Field
You hold a positively charged ball and walk due west in a
region that contains an electric field directed due east.
FE
E
FH
dr
WH is the work done by the hand on the ball
WE is the work done by the electric field on the ball
Which of the following statements is true:
A)
B)
C)
D)
14
WH > 0
WH > 0
WH < 0
WH < 0
and WE > 0
and WE < 0
and WE < 0
and WE > 0
Physics 212 Lecture 5, Slide 6
Conservative force: U = - WE
FE
Not a conservative force.
Does not have any U.
FH
E
dr
B) WH > 0 and WE < 0
Is U positive or negative?
A) Positive
B) Negative
16
Physics 212 Lecture 5, Slide 7
Move charge q2 from r1  r2. Find the change in potential energy.
Fel
Fel
d rˆ
q2
q1
q2
r 12
d rˆ
q1 q2
Fel 
rˆ12
4  0
q1
r2
r2
q1 q2 rˆ12
U  U  r2   U  r1     Fel  d r  
dr
2

4  0 r1 r12
r1
From the diagram,
r2
r̂12  dr  dr
r
q1 q2 d r
q1 q2  1  2 q1 q2
U  

  
2


4  0 r1 r
4  0  r  r1 4  0
so,
1 1
    kq1 q2
 r2 r1 
1 1
  
 r2 r1 
Calculate the change in potential energy for two point charges
originally very far apart, moved to a separation of d.
d
q1
q2
Charged particles w/ same sign have an increase in potential
energy when brought closer together.
d
q1q2
U    k 2 dr  kq1q2
r12

q1q2
1 q1q2
1 1
 d     k d  4  d


0
For point charges we often choose r =  as “zero” potential energy.
So the potential energy of this configuration is,
U  d   U    d   U     U    d   k
19
q1 q 2
d
Physics 212 Lecture 5, Slide 9
Example: Getting the signs right
Case A
Case B
d
2d
In case A two negative charges which are equal in magnitude are
separated by a distance d. In case B the same charges are separated by
a distance 2d. Which configuration has the highest potential energy?
A) Case A
B) Case B
22
Physics 212 Lecture 5, Slide 10
Example: Getting the signs right
• As usual, choose U = 0 to be at infinity:
Case A
q2 1
UA 
4 0 d
d
Case B
q1q2 1
U (r ) 
4 0 r
2d
q2 1
UB 
4 0 2d
U(r)
UA > UB
U(d)
U(2d)
r
0
23
Physics 212 Lecture 5, Slide 11
Checkpoint 1
A
B
C
D
E
U initial 
1 Qq
4 0 r1
U  U f  U i 
U final 
1 Qq
4 0 r2
Qq  1 1 
  
4 0  r2 r1 
Note: +q moves AWAY from +Q.
Its potential energy MUST DECREASE
U < 0
34
Physics 212 Lecture 5, Slide 12
Potential Energy of Many Charges
Two charges are separated by a distance d. What is the change in
potential energy when a third charge q is brought from far away to a
distance d from the original two charges?
Q2
d
d
qQ1 1 qQ2 1
U 

4 0 d 4 0 d
(superposition)
q
Q1
d
25
Physics 212 Lecture 5, Slide 13
Potential Energy of Many Charges
What is the total energy required to bring in three identical
charges, from infinitely far away to the points on an equilateral
triangle shown ?
Q
2
Q 1
4 0 d
Q2 1
U  2
4 0 d
Q2 1
U  3
4 0 d
Q2 1
U  6
4 0 d
U 
d
Q
Work (by me) to bring in first charge:
d
d
3 Q2
W (by me)   Wi 
4 0 d
Q
3 Q2
U  
40 d
W1 = 0
2
1
Q
Work (by me) to bring in second charge : W2 
4 0 d
1 Q2
1 Q2
2 Q2
Work (by me) to bring in third charge : W3 


4 0 d 4 0 d 4 0 d
27
Physics 212 Lecture 5, Slide 14
Potential Energy of Many Charges
Suppose one of the charges is negative. Now what is the total
energy required to bring the three charges in infinitely far away?
A) 0
Q2 1
B) U  1 4 d
0
2
Q 1
C) U  1 4 d
0
2
D)U  2 Q 1
4 0 d
Q2 1
E) U  2
4 0 d
-Q
d
d
d
Q
1
Work to bring in first charge:
3
1 Q2
W  by me    Wi  
4 0 d
1 Q2
U  
40 d
Q
2
W1 = 0
1 Q2
Work to bring in second charge : W2  
40 d
1 Q2
1 Q2
2 Q2
Work to bring in third charge : W3  


4 0 d 4 0 d
4 0 d
29
Physics 212 Lecture 5, Slide 15
Checkpoint 2
A
31
B
C
D
Physics 212 Lecture 5, Slide 16
Checkpoint 3
LET’S DO THE CALCULATION !!
34
Physics 212 Lecture 5, Slide 17
Example: Potential Energy Changes
A positive charge q is placed at x=0 and a negative charge -2q is placed
at x=d. At how many different places along the x axis could another
positive charge be placed without changing the total potential energy of
the system?
A)
B)
C)
D)
37
q
-2q
X=0
X=d
x
0
1
2
3
Physics 212 Lecture 5, Slide 18
Example: Potential Energy Changes
At which two places can a positive charge be placed without changing the
total potential energy of the system?
q
A
X=0
A)
B)
C)
D)
E)
40
A&B
A&C
B&C
B&D
A&D
B
C
-2q
X=d
D
x
Let’s calculate the positions of A and B
Physics 212 Lecture 5, Slide 19
Lets work out where A is
d
r
A
q
-2q
X=0
X=d
x
1 Qq
1 2Qq
U  

40 r 40 r  d
Set U = 0
1
2

r rd
rd
43
Makes Sense!
Q is twice as far from -2q as it is from +q
Physics 212 Lecture 5, Slide 20
Lets work out where B is
d-r
r
q
X=0
Setting U = 0
B
-2q
x
X=d
1
2

r d r
2r  d  r
r
46
d
3
Makes Sense!
Q is twice as far from -2q as it is from +q
Physics 212 Lecture 5, Slide 21
Summary
For a pair of charges:
Just evaluate
q1q2
U k
r
r
q1
q2
(We usually choose U = 0 to be where the charges are far apart)
For a collection of charges:
Sum up
q1q2
U k
r
for all pairs
Physics 212 Lecture 5, Slide 22