# Lect05 by liwenting

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```									           Physics 212
Lecture 5
Today's Concept:
Electric Potential Energy
Defined as Minus Work Done by Electric Field

Physics 212 Lecture 5, Slide 1
Music
Who is the Artist?

A)   Tito Puente
B)   Buena Vista Social Club
C)   Louis Prima
D)   Freddie Omar con su banda
E)   Los Hombres Calientes
Cuban Jazz !!

Thanks to Ry Cooder for bringing these guys
to our attention !!
Why ??

Cuban Jazz at Krannert (Ellnora)
Friday night (10:30pm)
Marc Ribot y Cubanos Postizos
Remembering Arsenio Rodriguez

FREE
Physics 212 Lecture 5
“This really seems like a rehash of mechanics with electric          Right! Nothing
charges instead of masses.”                                            really new

“Please discuss in lecture about how the point charges will affect the
electric field in different situations where it has both same charge or
please go over the potential energy equation to refresh my memory.”
“When solving for the potential energy, does r1 get subtracted from
r2? or is it the other way around?”
“Homework problem style examples would be helpful, the
checkpoints felt too easy.”                                       Example today:
“Still generally confused on some questions about the           calculation for CP 3
potential energy, like the third checkpoint.”
Discussion Sections
“Had there been office hours this week I definitely would have this week should help
been there. I'm still not 100% sure about Gauss' Law.”               WORKED
EXAMPLES!

“Labor Day Weekend and Physics must definitely have the same charge because the
weekend kept pushing the homework and this prelecture from getting done.”
05
Physics 212 Lecture 5, Slide 3
r2
Recall from physics 211:
W   F dr          WTOT   K
r1

F                   Object speeds up ( K > 0 )
dr     W>0

F
dr
or
W<0          Object slows down ( K < 0 )
F
dr

F
W=0           Constant speed ( K = 0 )
dr
9
Physics 212 Lecture 5, Slide 4
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.

F
+                   +

x       Coulomb force does negative work
+         +                Potential energy increases

Physics 212 Lecture 5, Slide 5
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

“Since potential energy is negative, the charge will
try to increase its potential energy, bringing it to
zero..”

“It wants to go to a spot with less PE.”

“constant potential energy would require no work to
preform.”

It will move in the same direction as F
F
Work done by force is positive
x
U = -Work is negative
Nature wants things to move in such a way that PE decreases
34
Physics 212 Lecture 5, Slide 6
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             FH
dr
E

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)   WH > 0   and WE > 0
B)   WH > 0   and WE < 0
C)   WH < 0   and WE < 0
D)   WH < 0   and WE > 0
14
Physics 212 Lecture 5, Slide 7
Not a conservative force.
Conservative force: U = - WE          Does not have any U.

FE            FH
dr
E

B) WH > 0 and WE < 0

Is U positive or negative?

A) Positive
B) Negative

16
Physics 212 Lecture 5, Slide 8
Example: Getting the signs right

Case A
d
Case B
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 9
Example: Getting the signs right
q1q2 1
• As usual, choose U = 0 to be at infinity:       U (r ) 
4 0 r
q2 1
Case A                    UA 
d             4 0 d
Case B                                         q2 1
UB 
2d                       4 0 2d

U(r)

UA > UB
U(d)
U(2d)
0                                                                    r

23
Physics 212 Lecture 5, Slide 10
Example: Two Point Charges

Calculate the change in potential energy for two point charges
originally very far apart moved to a separation of “d”

d
q1q2
U    k 2 dr                       d

r12              q1                  q2

q1q2           1 q1q2
U  k             
d            4 0 d

Charged particles w/ same sign have an increase in potential
energy when brought closer together.

For point charges often choose r=infinity as “zero” potential energy.

19
Physics 212 Lecture 5, Slide 11
Checkpoint 1
A charge of +Q is fixed in space. A second charge of +q was first placed at a distance r1
away from +Q. Then it is moved to a new position at a distance R away from its starting
point on a straight path. The final location of +q is at a distance r2 from +Q.

What is the change in the potential energy of the
charge +q in the process?                                                  1 Qq                    1 Qq
A. kQq/R            B. kQqR/r12          C. kQqR/r22        U initial                U final 
D. kQq(1/r2 - 1/r1) E. kQq(1/r1 - 1/r2)                                   4 0 r1                4 0 r2

“It is inversely proportional to the first radius.”
Qq  1 1 
“Simple conservation of energy problem: final                 U  U f  U i             
potential minus initial potential should equal                                    4 0  r2 r1 
change. ”
Note: +q moves AWAY from +Q.
“1/r1 will be larger then 1/r2 and this must be       Its Potential energy MUST DECREASE
positive”                                                              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)                 Q1                         q

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.

A) 0                                        Q
2
Q   1
B) U  4
0 d
3 Q2
Q2 1
C) U  2 4 d                      d          d       W   Wi  
0
40 d
2
D) U  3 Q 1
4 0 d                                                 3 Q2
Q2 1                                           U  
E) U  6                                                        40 d
4 0 d              Q            d       Q

Work to bring in first charge:       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

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?
2
A) 0                                        Q

Q2 1
B) U  1 4 d                                                          1 Q2
0
d              d       W   Wi  
40 d
2
Q 1
C) U  1 4 d
0
2
D)U  2 Q 1                                                 U  
1 Q2
4 0 d
Q2 1                                                    40 d
E) U  2                 1
Q            d           Q
4 0 d

Work to bring in first charge:       W1 = 0
1 Q2
Work to bring in second charge : W2  
40 d
1 Q2   1 Q2
Work to bring in third charge :     W3                0
40 d 40 d
29
Physics 212 Lecture 5, Slide 15
Checkpoint 2
Two charges which are equal in magnitude, but opposite in sign, are place at equal distances
from point A.

If a third charge is added to the system and placed at point A, how does the electric potential
energy of the charge collection change?
A. Increases          B. decreases        C. doesn’t change
D. The answer depends on the sign of the third charge

“inserting another charge is going to increase the magnitude of the potential energy.”
“No matter what the sign is the potential energy would be a positive number minus a negative
number minus another negative number. ”
“the change in potential is found by adding another kqq/r term. this term is dependent on the
sign of the new charge.“
31
Physics 212 Lecture 5, Slide 16
Checkpoint 3
You start with two point charges separated by some distance. The charge of the first is
positive. The charge of the second is negative and its magnitude is twice as large as that of
the first.

Is it possible to find a place to which you can bring a third charge in from infinity without changing
the total potential energy of the system?
A. YES, as long as the third charge is positive       B. YES as long as the third charge is negative
C. YES, no matter what the third charge is            D. NO

“The positive third charge will cancel out the
negative charge.“

“It doesn't matter what the sign is. Place the new
charge twice the distance from the - as the +
charge. It doesn't matter what the sign is because
the U added to the system will now be zero.”

“Adding a third charge that is opposite to at least
one other charge will cause a change in potential
energy”

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?

q                   -2q
x
X=0                  X=d

A)   0
B)   1
C)   2
D)   3

37
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                   -2q
A               X=0     B     C                                D
x
X=d

A)   A&B
B)   A&C
C)   B&C          Let’s calculate the positions of A and B
D)   B&D
E)   A&D

40
Physics 212 Lecture 5, Slide 19
Lets work out where A is
r                  d

q                -2q
A          X=0
x
X=d

1 Qq   1 2Qq
U         
40 r 40 r  d

Set U = 0

1   2

r rd

Makes Sense!
rd       Q is twice as far from -2q as it is from +q

43
Physics 212 Lecture 5, Slide 20
Lets work out where B is
r          d-r

q                         -2q
x
X=0       B                X=d

1   2
Setting U = 0                      
r d r

2r  d  r

d
r
3
Makes Sense!
Q is twice as far from -2q as it is from +q   Physics 212 Lecture 5, Slide 21
46
Summary

For a pair of charges:                           r
q1q2
Just evaluate   U k                    q1                       q2
r
(We usually choose U = 0 to be where the charges are far apart)

For a collection of charges:
q1q2
Sum up     U k             for all pairs
r

Physics 212 Lecture 5, Slide 22

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