Transcript electrical potential energy

Unit 4
Electric Potential Energy and Electric Potential
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
1
Charges and the MVE – WS 4 #1
In Physics 210 we discussed energy and the MVE.
This equation still applies to charged particles even though they are so
small.
However, when we discuss charges, we will modify the MVE.
Gravitational Potential Energy has little impact on charges; therefore, we
will ignore it.
Elastic Potential Energy and Rotational Kinetic Energy will likewise be
ignored.
However, we will need to add a term for Electrical Potential Energy (U).
This type of energy acts like Gravitational Potential Energy and is caused
by the attraction between charged particles.
The MVE looks like the equation below when these changes are made.
KE1  GPE1  EPE1KE
 RKE
Woo  KE
KE
GPE
U2 2  EPE2  RKE2
1 U
1 1W
2 
2 
1
1 2 1
1
1 2 1
2
2
2
2
m v1  m gy1  kx1  I1  wother  m v2  m gy2  kx2  I 2
2
2
2
2
2
2
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
2
Electric Potential – WS 4 #2
In the picture below, the negative charge rest against the positively charged plate,
and the positive charge rests against the negatively charged plate.
Observe what happens as the switches in the picture below are closed.
Essentially, the electric field was reversed; therefore, the charges moved to the
opposite plates.
In order for the charges to move, there must be a force applied to them that caused
them to move through a distance.
As a result, there was work done on the charges.
On the next slide we will determine the work done by using the modified MVE.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
3
Work in a Uniform Electric Field – WS 4 #3
 In order for the charges in the previous slide to move, work must
be done on them.
 Recall that in our mechanical systems, work was determined using
the equation below where r is the distance an object is moved.
 From unit 13 we learned that the force on a charged particle in an
electric field is given by the following equation.
 As a result, the work done in moving a charge from one position
to another in an electric field would be as follows.
 Electrical systems, just as in mechanical systems, can have
positive and negative work.
 Consider the two charges below: in which direction do the charges
“wish” to move in the electric field shown?
 The negative charge does not wish to move because it is next to
the positively charged plate and is “happy.”
 The work done in moving this charge toward the negatively
charged plate would be negative because the force on the charge
due to the electric field is up whereas the displacement would be
down.
 What kind of work was done on the positive charge when we
moved it to the negatively charged plate ?
 Explain your answer.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
W  F r  Fr cos 
F  q0 E
W  q0 Er
4
GPE v. U – WS 4 #4
 When we dealt with gravitation potential energy (GPE),
the ground was considered to be our lowest point where
the height is equal to zero and subsequently the GPE = 0.
 When dealing with charges in uniform electric fields, our
reference point where U = 0 would be the charged plate
where the electric charge is the “happiest.”
 Is a positive charge “happy” when it is located near a
positively charged plate? Why?
 It is not happy because like charges repel each other;
therefore, it would have a high potential energy.
 What about a negative charge when it is located near a
positively charged plate? Why?
 A negative charge would very “happy” because opposite
charges attract; therefore, it would have a zero electric
potential energy.
 “Up” or “Down” are meaningless terms in these
electrical systems.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
5
Electric Potential Energy Derivation (Uniform Field) – WS 4 #5
 Using the MVE below, derive the equation for the amount of electric potential
energy acquired by the negative charge when work is done to move it to the
negatively charged plate.
 What are the initial and final kinetic energies of the negatively charged particle?
 What is the initial electric potential energy of the negative charge? Why?
 After eliminating the appropriate terms and making the appropriate substitutions,
we find the equation needed to calculate the potential energy of the negative
charge when it is located near the negatively charged plate.
KE1  U1  wother  KE2  U2
U 2  wother
W  q0 Er
U 2  q0 Er
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
6
Understanding Work – WS 4 # 6 – 8 (Homework)
 Three charges in an uniform electric field are released
from the same distance from a negatively charged metal
plate.
 Are they initially in an area of high potential energy or
low potential energy? Explain.
 They follow the paths shown.
 Which charge requires the most work in order to move the
charge?
 In the end are they in a higher or lower potential energy
state? Explain.
 Now consider the red charge.
 How much work is done in moving it from its initial to its
final position?
 Explain your answer.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
7
WS 4 # 9 (Homework)
 A charge (q = 8.3  10-8 C) held in place 1.5 cm from a positively charged plate is
pushed 0.87 cm closer to the negatively charged plate.
 The magnitude of the electric field is 5.35  105 N/C.
 Calculate the force (magnitude and direction) acting on the charge.
 What is the electric potential energy of the charge in its initial position?
 To find this electric potential energy, use the MVE from the position of zero
potential energy to the position it now holds.
 How much work is done in moving the charge as
specified above?
 What is the charges new potential energy?
KE1  U1  wother  KE2  U2
F  q0 E
q
Ek 2
r
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
W  q0 Er
qq0
U2  k
r
8
WS 4 # 10
 A beam of charged particles (q = -2.2  10-12 C, m = 8.92  10-15) initially
traveling at 125.0 m/s is fired through an electric field (E = 900 N/C) generated
by the two parallel plates below.
 The particles deflect by 4.5 mm as they pass through the plates (Note: we do
not say rise or fall).
 Draw in arrows representing the direction the field is acting in the figure below.
 What is the new speed of the charges as they leave the parallel plates?
KE1  U1  Wo  KE2  U2
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
9
WS 25 # 11 – 13 (Homework)
Two charges are positioned as below on different plates of two parallel plates below separated
by 1.0 cm.
The purple charge has a charge of |qp| = 2.2  10-12 C and a mass of 5.0  10-11 kg, and the green
charge has a charge of |qg| = 1.1  10-12 C and a mass of 5.0  10-11 kg.
The switches are closed reversing the direction of the electric field, and the charges move as
shown.
What are the charges (+ or -)of the green and purple charges? Explain your answer.
If the force acting on the purple charge is 1.50 N, then find the following: the magnitude of the
electric field, the force acting on the green charge, the work required to move the charges to the
opposite plates, and the electric potential energy of the charges immediately before they begin to
travel from one plate to the other.
Calculate the speed of the particles when they collide with the opposite plates.
W  q0 Er
F  q0 E
q
Ek 2
r
KE1  U1  wother
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
U 2  q0 Er
 KE2  U2
10
Electric Potential Energy Derivation (Two Charged Particles)
 On previous slides we derived the equation needed to
calculate the potential energy of a charged particle
moving in an uniform electric field.
 We will now consider the potential energy of one
charged particle in the vicinity of another charged
particle.
 Electrical Potential Energy is the energy a charged
particle has due to its position relative to an electric field.
 Watch as the test charge, q0, moves from position r1 to
position r2 in the electric field generated by the positive
charge.
 As this particle moves, its potential energy relative to the
large positive charge changes.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
q0
r1
r2
11
Electrical Potential Energy – WS 4 #14 (Homework)
Electrical Potential Energy is the energy a charged particle
has due to its position relative to an electric field.
Watch as the test charge, q0, moves from position r1 to
position r2.
As this particle moves, its potential energy relative to the
large positive charge changes.
The equation for the particle’s Electrical Potential energy
(U) at position 1 and position 2 is as shown to the right.
The magnitude (+ or -) of the potential energy will depend
on the charge of the test charge q0.
If q0 is positive, then the sign for U will be positive and
would increase as q0 moved closer to q1.
If q0 is negative, then the sign for U will be negative and
would decrease as q0 moved closer to q1.
How would these values change if we replaced q1 in the
center with a negative charge?
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
U1  k
qq0
r1
U2  k
qq0
r2
q0
r1
r2
12
Electric Potential Energy Derivation (Two Charged Particles) – WS 4 #15
 Using the MVE below, derive the equation for work required to move the test
charge closer to the positive charge.
 What are the initial and final kinetic energies of the test charge?
 Beginning with the MVE and making the appropriate substitutions, we derive the
following equation.
KE1  U1  wother  KE2  U2
q0
r1
wother  U2  U1
U k
wk
qq0
r
qq0
r2
k
qq0
r1
1 1
 k qq0   
 r2 r1 
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
r2
q0
13
WS 4 # 16 & 17
Two charges (|qbig| = 4.2  10-8 C, and |qsml| = 7.7  10-10 C) are in close proximity to one
another in as shown below.
The red charge is moved from its initial position (r1 = 1.25 cm) to the new position
(r2 = 0.17 cm) below.
Is its new position in an area of higher or lower potential energy? Explain.
Calculate the work required to move the charge to the new position.
Now suppose the charge moved along a different path to the same point as shown.
Without doing any calculations, determine the work needed to move this charge to
this same point along the new position.
Explain your answer.
KE1  U1  wother  KE2  U2
qq0
U k
r
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
14
WS 4 #18 & 19
 Charges q1 and q2 each have a charge of 9.0 
10-19 C.
 Each of the squares in the graph are 0.01 cm 




0.01 cm.
The third charge has a charge of q3 = 1.3  10-17 C.
The charge is moved from point a to point b as
shown.
Has this charge moved to an area of higher or
lower potential energy. Explain.
How much work is done moving this charge from
point a to point b?
n
V  k
i 1
qi
ri
b
q1
q2
q3
a
W  q0 Er
F  q0 E
q
Ek 2
r
U 2  q0 Er
KE1  U1  wother  KE2  U2
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
15
Electric Potential – WS 4 # 20-21
Electric Potential (V) is the electric potential energy (U) per unit of
charge.
The equation for electric potential is given below in two forms.
V
U
q0
U k
qq0
r
V k
q
r
The units for electric potential is the J/C which is defined as the volt.
We normally like to consider the Electric Potential Difference
(commonly known as voltage) between to different positions.
On the next slide we will consider the potential difference between
two parallel plates.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
16
Electric Potential – WS 4 #22
Determine the change in electric potential of the charges
below when moved to the opposite metal plate.
What are the initial and final kinetic energies of the
charges in the picture below?
Solving the equation for work, we get the following.
This work is just the change in the electrical potential
energy (U) of the charges.
Electric Potential (V) is the electric potential energy (U)
per unit of charge.
Substituting, we get the following equation.
As a result, the electric potential between the starting and
ending positions would be as follows.
This difference in electric potential is known as the
voltage between the starting and ending points of the
charges as they go from one position to another in the
presence of an electric field.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
KE1  U1  wother  KE2  U2
w  U2  U1
w U 2  U1

q0
q0
U
V
q0
W
 V2  V1
q0
W
V 
q0
17
Electric Potential
In the previous slide, we derived the equation needed to find the electric potential
when we move a charge from an area of zero electric potential energy to and area of
higher or from a high electric potential energy to a zero electric potential energy.
In this slide we derive the equations needed to determine the electric potential when
the charge travels from one non-zero potential energy to another.
In the animation below both charges move in this manner.
W
 V2  V1  V
q0
W  F r  Fr cos 
2
1
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
18
Electric Potential
Look at the two charges below.
The positive charge is very “happy” because it is near the negatively charged plate.
As a result, we say that the positive charge has U1 = 0 (V1 = 0) because it does not
want to move; however, the negative charge is very “unhappy” because it too is near
the negatively charged plate.
As a result it has a very high U1 > 0 and wants to move away.
Now lets move the two charges across to the next parallel plate.
The negative charge has U2 = 0 (V2 = 0); however, the positive charge has U2 > 0
(V2 > 0).
In order to move the charges, we had to do work on them.
The voltage difference that we had to overcome by the work can be found as
follows.
0
W
 V2  V1
q0
V
W
q0
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
19
Why Do “Happy” Charges Move?
Students often wonder why a charge would move when it is already “happy.”
Consider the charge below.
What is the charge on the plates in the figure below?
The bottom one is negative, and the top one is positive.
The electric field due to these plates is as shown.
Watch the charge.
Why do you think it moved?
There are many reasons why it might move.
One possibility is another electric field that
opposes the field established by the AA battery.
As the flux (electric field density) of this
field is higher, the electric field produced
by the D batteries does more work on the
charge than the field produced by the AA
battery.
As a result, the charge moves.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
20
WS 4 # 23-24 (Homework)
 Answer the questions below as they pertain to the charges
shown in the figure to the right.
 Which charges have an electric potential V = 0? Explain.
 Do the charges to the right move from a high potential to a
low potential or from a low potential to a high potential?
Explain.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
21
WS 4 # 25
 What is the change in electric potential of the charge in problem 9?
U
V
q0
V k
q
r
KE1  U1  wother  KE2  U2
F  q0 E
q
Ek 2
r
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
W  q0 Er
qq0
U2  k
r
22
Electric Potential of Two Point Charges – WS 4 #26
Derive the equation needed in order to calculate the
work done when moving the small charge along the
path shown.
KE1  U1  wother  KE2  U2
w  U2  U1
qq0
U k
r
wk
qq0
qq
k 0
r2
r1
V 
w U 2  U1

q0
q0
w
V 

q0
k
qq0
qq
k 0
r2
r1
q0
1 1
V  kq   
 r2 r1 
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
23
WS 4 # 27
 A point charge has a charge of -2.6  10-15 C.
 Find the distances ra, rb, and rc if the electric potentials at these
points are 30.0 V, 20.0 V, and 10.0 V respectively.
 What is the electric potential energy of this charge at each of
these three points?
ra
U
q
q
V k
r
V
rb
rc
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
24
WS 4 # 28-31
 A small charged body hangs from a string (l = 12.0 cm) in the
center between two metal plates that are 3.5 cm apart and
connected to a battery as shown below.
 The charge has a charge of 5.7  10-5 C.
 When the switch is closed, the body moves to the position shown
where it makes an angle of 15.0 with the vertical.
 What is the magnitude (+ or -) on the charge? Explain.
 How much work is done by the electric field in moving the charge
in the horizontal direction.
 What is the magnitude of the electric field?
 What is the change in electric potential of the charge?
U
V
q0
V k
q
r
KE1  U1  wother  KE2  U2
F  q0 E
q
Ek 2
r
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
W  q0 Er
qq0
U2  k
r
25
WS 4 # 32-34 (Homework)
 A small charged body (m = 0.95 g) hangs from a string in the
center between two metal plates that are 3.5 cm apart and
connected to a battery as shown below.
 The charge has a charge of 5.7  10-5 C.
 When the switch is closed, the body moves to the position
shown where it makes an angle of 15.0 with the vertical.
 What is the magnitude (+ or -) on the charge? Explain.
 How much work is done by the electric field in moving the
charge in the horizontal direction.
 What is the magnitude of the electric field?
 What is the change in electric potential of the charge?
U
V
q0
V k
q
r
KE1  U1  wother  KE2  U2
F  q0 E
q
Ek 2
r
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
W  q0 Er
qq0
U2  k
r
26
Electric Potential Due to a System of Point Charges WS 4 #35 (Homework)
 We may find the electric potential due to a system of point charges like the one
appearing below using the following equation.
 Recall that electric potential is a scalar quantity; therefore, only the distance, not the
direction, of the charges from a point in space is needed.
 In the figure below, the square has a length of l = 0.90 m.
 Calculate the electric potential at point P due to the five charges below given that the
magnitude of q1 = 15.0 nC and the magnitude of q2 = 22.5 nC.
n
n
qi
V  k Vi  k 
i 1
i 1 ri
2
2
l
l
l l
r2       
 0.636m r1   0.450m
2
2
2 2
n
 q2 q1 q2 q1 q1 
V  k Vi  k 

 
 
r2
r1
r2
r2 
i 1
 r2
 q2 q1 q2 
V  k Vi  k 

   80.66V
r2
r1 
i 1
 r2
n
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
q2
q1
r2
q1

P
r1
q2
q1
27
Electric Potential due to a Non-uniform Electric Field WS 4 #36
 An electric charge moves a distance of l from
point a to point b through a non-uniform
electric field as shown.
 Determine the electric potential difference
experienced by this charge.
a
W  F l  qE l
dl
dW  qE dl
b
W  q  E dl
b
a
V 
W
q
b
V   E dl
a
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
28
Equipotential Surfaces
Equipotential Surfaces indicate positions
in an electric field where the electric
potential is constant.
These equipotential surfaces within a
uniform electric field are indicated by the
blue lines in the figure to the right.
When a charge moves along the
equipotential surfaces, the electric field does
no work on the charge.
The electric field does work on a charge
only when it moves from one equipotential
surface to another.
The equipotential surfaces about an
isolated point charge would be as shown.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
29
Equipotential Surfaces for a Dipole WS 4 #37
The equipotential surfaces for a dipole
would look those appearing in the figure
to the right.
Remember, if a charge moves along
these surfaces, then the electric field does
no work on the charge.
Only when a charge moves from one
surface to the other does the electric field
do work on a charge.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
30
Small Angle Approximation WS 4 #38
 Consider the two points P1 and P2 separated by a distance
of l and located a distance of r1 and r2 from point P.
 If r1>>l and r2>>l, then we can approximate that r1  r2.
 As a result, the difference in the lengths of r1 and r2 may be
found as follows.
 Additionally, if l is very small, then the following is also
true.
P1
l
P2
r1
r2
l y  l cos
r2  r1  l y  r1  l cos
r2  r1  l cos 
P1
r1  r2
r1  r2  r
2
l
P2
 ly
lx
P
r1  r2  l cos 
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
31
Electric Potential due to a Dipole WS 4 #39
 Calculate the electric potential at point P due to the dipole
below given that the magnitude of q1 = q2 = 15.0 nC = q.
r1  r2  l cos 
q1
l
r1  r2  r 2
q2
n
qi
V  k
i 1 ri
 q q   r2 q r1q 
 r2  r1 
V k  

  q

r
r
r
r
r
r
r
r
 1 2  21 12
 12 
V 
r1
r2
ql cos 
r2
P
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
32
Electric Potential: Continuous Charge Distribution WS 4 #40
 When we have a long rod with a continuous charge
distribution, we need to integrate over the charge
distribution in order to find the electric potential at point P.
q
V k
r
q

l
V  k
L
0
dq
dV  k
r
dq
V  k
r
r  l  d
dq   dl
2
dq
 dl
 dl
 k

k

1
0 2 2 12
0
2
2
2
r
l  d 
l  d 
L
2

1
2
dl
r
L
L
l
P
d
V  k  ln l   l 2  d 2  2 

0
1
1
1
 


2
2
2
2
2
2
2
2
V  k  ln  L   L  d   0   0  d    k  ln  L   L  d  2   ln d 



 

 L  L2  d 2 12 


 A


ln A  ln B  ln  
V  k  ln


d
B


1
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
33
Electric Potential Due to a Charged Disk WS 4 #41
 Determine the electric potential at point P located a distance
of z from a uniformly charged disk of radius R.
q   A   x
V  k
2
dq
2 xdx
 k
2
2 1/ 2
r
x  z 
V  k   x  z
2

dq  2 xdx
  2xdx 
2 1/ 2
u 1/ 2 du   n  1 u  n1
P
r   x2  z 2 

1/ 2
r
z
x
R
dr
R
k  2
2 1/ 2 
V
x z  



0
2
k  2
2 1/ 2

V
R

z

z



2 
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
34
WS 25 # 32-34
 A
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
35
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36
q2
q1
r2
q1

P
r1
q2
a
q1
dl
b
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
37
c
a
d
b
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
e
38
Electric Potential Due to a Charged Disk
 Determine the electric potential at point P located a distance
of z from a uniformly charged disk of radius R.
q   A   x
V  k
2
dq
2 xdx
 k
2
2 1/ 2
r
x  z 
V  k   x  z
2

dq  2 xdx
  2xdx 
2 1/ 2
u 1/ 2 du   n  1 u  n1
P
r   x2  z 2 

1/ 2
r
z
x
r
R
R
k  2
2 1/ 2 
V
x z  



0
2
k  2
2 1/ 2

V
R

z

z



2 
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
39