Download §13.1 Electrostatics and Electric Fields (part class) Def Electrostatics

§13.1 Electrostatics and Electric Fields (part class)
Def n Electrostatics deals with the study of electric charges that are stationary or very slow moving. It is
different from current electricity which examines charges that move more quickly.
E.g.: When we look at the reason why a balloon, which has been rubbed on your hair, sticks to a wall, we
are concerned with electrostatics.
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Def n Electrostatic forces are the forces involved in electrostatics.
E.g.: The force that causes the rubbed balloon to stick to the wall is an electrostatic force.
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Def n A field is a 3D region in which a property or quantity may be distributed (see p. 546, text).
Def n A force field is the 3D region in which a force acts.
E.g.: The region around a magnet where the magnetism exists is called its magnetic field.
E.g.: The region around Earth where gravity exists is called its gravitational field.
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§13.2 The Basis of Electric Charge – The Atom (part class)
Def n The basic unit of matter is called the atom. It consists of:
i.
A dense central nucleus which is made up of:
a. Protons with a positive electrical charge.
b. Neutrons with no electrical charge (neutral).
ii.
A mostly empty region around the nucleus which has moving electrons with a negative electrical
charge.
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If an object has the same number of protons and electrons, it is said to be electrically neutral. If the
number of electrons is greater than the number of protons, the atom (ion) is said to have an overall
negative electrical charge. If the number of protons is greater than the number of electrons, the atom (ion)
is said to have an overall positive electrical charge. However, since protons are held tightly in the
nucleus, an electrical charge is usually created on an object by removing or adding electrons. (See here
and here for animations of the atom)
Through experimentation, it can be shown that electric charges follow the Laws of Electric Charges.
1. Opposite electrical charges attract.
2. Like electrical charges repel.
3. Charged objects can attract some neutral objects.
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View MLO 1, CDLI
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§13.3 The Creation and Transfer of Electric Charge (1 class)
An object can be given an electric charge in 3 ways:
1. Friction (rubbing)
2. Conduction (contact, touching)
3. Induction
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Charging by Friction
If you look at the Electrostatic Series in Table 13.1 on p. 530, you should see that some materials hold
onto their electrons better than others. For example, rubber holds onto its electrons better than glass and
silk holds onto its electrons better than glass. This means that rubber has a high affinity for electrons
whereas glass has a low affinity for electrons.
If a glass rod is rubbed with a silk cloth, the silk cloth will take electrons away from the glass because it
can hold on to its electrons better than glass. This will leave the silk with an excess of electrons and a
negative charge. The glass rod will then have a deficiency of electrons and a positive charge (see Fig.
13.5, p. 530. You must be able to draw these diagrams.) Watch video here and animation here.
Draw diagrams here
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View MLO #2, CDLI
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Charging by Conduction
Once an object has an electric charge, it can induce (create) an electrical charge on a neutral object when
the objects touch.
Suppose a glass rod is rubbed with silk (charged by friction). The glass rod will take on a positive charge
because it will lose electrons to the silk. If the positively charged glass rod now touches an electrically
neutral third object which is not grounded, the rod will attract electrons from the third object and these
electrons will transfer from the third object to the glass rod until the charge on each object is the same.
When the objects are separated, the third object is left with a deficiency of electrons and a positive
charge. Note that the electrical charge on the third object is the same as the electrical charge on the glass
rod. (See Fig. 13.7b, p. 531. You must be able to draw these diagrams. Make sure you can explain how to
create a negative charge on a neutral object by conduction. See Fig. 13.7a, p. 531.)
Draw diagrams showing how to induce a positive charge by conduction.
Draw diagrams showing how to induce a negative charge by conduction.
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View MLO #3, CDLI
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Sample Exam Questions
1. What happens to a neutral electroscope when it is briefly touched by a positively charged glass rod?
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Charging by Induction
Charging by induction means inducing an electrical charge on a neutral object by bringing a charged
object near the neutral object, but NOT touching the neutral object. Grounding is used to create the
charge on the neutral object.
For example, if a negatively charged rod is brought near a neutral object, the electrons on the glass rod
will repel the electrons on the neutral object to the far side of the object. If the far side of the object is
grounded (touched by a finger), the electrons on the far side of the object will move out to the earth. If the
ground is removed and then the glass rod, the object will have a deficiency of electrons and a positive
charge. Note that the electrical charge on the newly charged object is the opposite of the electrical charge
on the glass rod. You must be able to draw these diagrams. Make sure you can explain how to create a
negative charge on a neutral object by induction. See Fig. 13.8, p. 532.) Watch video here.
Draw diagrams showing how to induce a positive charge by induction.
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Draw diagrams showing how to induce a negative charge by induction (see Fig. 13.8).
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View MLO’s #4-6, CDLI
Do #’s 35-40, p. 582
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§13.4 Coulomb’s Law (1 class)
The force of electrical attraction (or repulsion) between 2 point charges (so small they have no size) can
be found using a formula developed by Charles Coulomb.
****** Fe 
kq1q2
where
r2
Fe is the electric force acting on the point charges in Newtons (N).
q1 is the size of the electric charge on the first object in Coulombs (C).
q2 is the size of the electric charge on the second object in Coulombs (C).
r is the distance between the center of masses of the two objects in metres (m).
k is Coulomb’s constant which is 9.0 109
N  m2
.
C2
Note that this equation is very similar to the formula for gravitational force Fg 
Gm1m2
which was
r2
covered in Physics 2204.
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First we need to know what is meant by a Coulomb of charge. It is a collection of 6.242 1018 electrons
(e). This is similar to saying that a dozen is 12, a foot is 12 inches, or a metre is 100 centimetres.
If 1C  6.242 1018 e' s then
1C
6.242 1018 e' s

6.242 1018 e' s 6.242 1018 e' s
1.602 1019 C  1e
So the charge on one electron is 1.602 1019 C and therefore the charge on one proton is
1.602 1019 C .
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We can now calculate the electrical charge on an object by multiplying the number of excess electrons (or
protons) by the charge on one electron (or proton).
******q  Ne where
q is the charge on the object in C.
N is the number of electrons in deficit or excess.
e  1.602 1019 C for excess protons (deficiency of electrons) or e  1.602 1019 C for excess electrons.
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Example 3, p. 537
Fe  3.0 105 N
(a)
At a distance of r, Fe 
kq1q2
r2
At a distance of 3r,
kq1q2 1  kq1q2  1
 
  Fe at r
 3r  9r 2 9  r 2  9
So the electrostatic force between the objects when the distance between them is 3r would be
1
3.0 105 N   3.3 106 N .

9
Fe 
kq1q2
2

(b) For objects with charges of q1 and q2 , Fe 
For objects with charges
kq1q2
.
r2
1
q1 and 2q2 ,
4
1 
1

k  q1   2q2  k  q1q2 
1  kq q  1
4 
2
Fe  
  2    12 2   Fe with charges q and q .
2
1
2
r
r
2 r  2
So the electrostatic force between the objects when the charges are
1
q1 and 2q2 would be
4
1
3.0 105 N   1.5 105 N .

2
(c) At a distance of r and with charges q1 and q2 , Fe 
At a distance of 3r and for objects with charges
1 
k  q1   2q2 
4 
Fe  

2
 3r 
kq1q2
.
r2
1
q1 and 2q2 ,
4
1

k  q1q2 
2
  1  kq1q2   1 F with charges q and q at a distance r.
e
1
2


2
9r
18  r 2  18
So at a distance of 3r and for objects with charges
1
3.0 105 N   1.7 106 N .

18
7
1
q1 and 2q2 , the electrostatic charge would be
4
Example #4, p. 539
N  3.7 1024 , e  1.602 1019 C q  ?
q  Ne   3.7 1024  1.602 1019   5.9 105 C
The charge on the negative item is 5.9 105 C .
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Example #5, p. 540
q1  4.0 106 C, q2  3.0 106 C, r  0.20m, k  9.0 109 N  m2 /C2 , Fe  ?
9.0 10
kq q
Fe  12 2 
r
9
 4.0 10 3.0 10   2.7N
6
 0.20 
6
2
The electrostatic force between the 2 charges is 2.7N.
Note that the sign of the answer is positive. This indicates that the force is repulsive.
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Example #6, p. 541
Note that when the foam spheres are touched together, charge will transfer from one sphere to the other
until the charges on both spheres is equal. Therefore, if q is the original charge on the foam sphere before
1
they are touched, then the charge on each foam sphere after they touch must be q . So,
2
1
1
q1  q, q2  q, r  0.10m, k  9.0 109 N  m2 /C2 , Fe  6.4N, q  ?
2
2
Substituting into Fe 
kq1q2
gives
r2
1 
 9.0 10   12 q 
 q 
 2 
9
6.4 
 0.10 
2
4  0.10   6.4 
 q2
2
9.0 109
q  5.3 106 C
Since the charge on the original sphere was negative, q  5.3 106 C .
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*******Example #7, p. 542
q1  3.6 106 C, q2  2.7 106 C, q3  4.5 106 C, r1  0.30m, r2  0.20m,
k  9.0 109 N  m2 /C2 , F net  ?
There are two forces acting on q3 . The repulsive force of q1 on q3 and the attractive force of q2 on q3 .
Therefore,

 
F net3  F 1 on 3  F 2 on 3

 9.0 10  3.6 10  4.5 10   9.0 10  2.7 10  4.5 10 

9
6
 0.50 
6
2
6
9
 0.20 
6
2
 0.5832N[R]  2.724N[L]
 2.2N[L]
So the net force on q3 is 2.2N[L]
Note that the positive and negative signs DO NOT indicate the direction of the forces acting on q3 ,
they only indicate if the force is repulsive or attractive.
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Sample Exam Questions
1. How many excess electrons reside on a metal sphere with a charge of -0.150 C?
2. What is the force between a 1.50 105 C charge and a 1.03 105 C charge that are separated by a
distance of 12.0 m?
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3. Three charged objects are arranged as shown. Calculate the net electric force on object B due to
the presence of objects A and C. ANS: 0.9N[L]
4. Three charged spheres are arranged as shown. Calculate the magnitude of the net force on Y due
to the presence of X and Z. (June 07, 52a, Ans: 148N)
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View MLO #’s 7-10 CDLI
Do #’s 1, 2 ATC, p. 545
Do #’s 46, 52 Problems, pp. 582-583
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§13.5 Fields and Field-mapping Point Charges (part class)
Def n An electric field is the 3D region in space around a single charge or a collection of charges that can
exert an electrostatic force on another charge when it is in the region.
E.g.: If q1 attracts or repels q2 then q2 is in the electric field created by q1 . Likewise, q2 will also attract
or repel q1 , so q1 is in the electric field created by q2 .
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Def n An electric field map is a description of the forces exerted on any charge placed in an electric
field. The field is often represented by drawing electric field lines which show:
i.
The direction of the force on a charge placed in the field.
ii.
The strength of the force on the charge that is placed in the field. The closer together the field
lines, the stronger the force.
Scientists have agreed that the direction of the arrows on the field lines indicates the direction of the force
on a positive charge placed in the field.
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E.g.: Electric field map drawn to represent the electric field around a positive charge (2D view).
See http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html and
http://www.learnerstv.com/animation/animation.php?ani=86&cat=physics
E.g.: Electric field map drawn to represent the electric field around a negative charge (2D view).
See http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html
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There are places in the field where the force on a charge particle will be equal. We join these places
where the force or electric potential energy is the same with broken equipotential lines.
E.g.: Electric field map drawn to represent the electric field around a positive charge (2D view with
equipotential lines).
See http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html
Note that in 3D, equipotential lines become equipotential surfaces.
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Rules for Drawing Electric Field Lines
i.
ii.
The arrows on the field lines indicate the direction of the force on a positive charge.
The number of field lines is proportional to the size of the charge. See
http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html
iii. Equipotential lines are perpendicular to the field lines.
iv.
The shape of the electric field diagram depends on the shapes and positions of the charge(s) that
create the field.
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E.g.: Electric field map drawn to represent the electric field around two positive charges (2D view).
See http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html and
http://www.ph.biu.ac.il/~rapaport/java-apps/efield.html
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E.g.: Electric field map drawn to represent the electric field around two negative charges (2D view).
See http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html and
http://www.ph.biu.ac.il/~rapaport/java-apps/efield.html
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E.g.: Electric field map drawn to represent the electric field around a positive charge and a
negative charge (2D view).
See http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html and
http://www.ph.biu.ac.il/~rapaport/java-apps/efield.html
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E.g.: Electric field map drawn to represent the electric field around a pair of parallel plates with
opposite charges (2D view).
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Sample Exam Question
1. Which describes electric field lines?
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Do #1, p. 551
View MLO #11, 0-7:19 CDLI
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
§13.6 Field Strength  (1 class)
 
The size of the electric force F e on a test charge placed in an electric field depends on:
i.
The size of the test charge  qt  .
ii.
The strength of the electric field (field strength)  .

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Def n Electric field strength  is the force that would be exerted on a test charge with an electric

charge of 1C. Electric field strength can be calculated using the formula

Fe
where
qt
 is the electric field strength in N/C.
F e is the electric force at a point in the electric field in N.
qt is the size of electric charge placed in the electric field in C.
Fg
. Recall that the value a g does not change as long as the
m
distance above the surface of the earth stays the same. Similarly,  will remain the same along any
equipotential line.
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Example 9, p. 553
Note that this equation is similar to a g 
qt  1.5 106 C, F e  3.0N[L],   ?

Fe
3.0

 2.0 106 N/C[L]
6
qt 1.5 10
The electric field strength at that point is 2.0 106 N/C[L] .
14
kqt qm
where
r2
qm is the electric charge on the master charge in C. The master charge is the charge that creates the
electric field.
qt is the electric charge on the test charge in C. The test charge is the charge placed in the electric field.
We know that F e can be calculated using F e   qt . Recall from section 13.4, that F e 
Substituting for F e and solving for  gives
 qt 
k qt qm
r2
********  
kqm
r2
This formula can also be used to calculate the electric field strength.
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Example 10, p. 555
(a)
on A
 9.0 10  3.6 10   12900N/C[R]  12900N/C

on A
 9.0 10  2.7 10   607500N/C[L]  607500N/C

6
9
q
1
 0.50 
2
6
9
q
2
 0.20 
2
  net  12900  607500  477900  4.8 105 N/C
(b)
F e   qt   4.8 105 N/C  4.5 106 C   2.2N  2.2N[L]
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Sample Exam Questions
1. What force is experienced by a 2.50 106 C test charge placed in a 2.92 104 N/C electric field?
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2. What is the electric field strength at 6.7 m from a 6.0 106 C charged object?
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View MLO #12, CDLI (There is a test with answers at the end of the video.)
Do #1, 3, 4, ATC p. 557
Do #’s 59-64, 66 Problems, p. 584
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§13.7 Electric Potential  V  and Electric Potential Energy  E e  (1 class)
When we raise an object off the floor we give an object gravitational potential energy. The amount of
gravitational potential energy the object has is equal to the amount of work done raising he object (WorkEnergy Theorem).
Similarly, when a charged particle is moved from point A to point B in an electric field, the particle gains
(or loses) electric potential energy  E e  . The amount of electric potential energy it gains (or loses) equals
the amount of work done in moving the charged particle from point A to point B.
To find the amount of electric potential energy  E e  per unit charge (q) we divide E e by q. This gives
electric potential which we commonly call voltage  V  .
Def n The electric potential  V  is the quotient of the electric potential energy  E e  of a charged object
and the size of the charge (q) on that object.
Therefore,
V
Ee
where,
q
Ee is the electrical potential energy of the object in J.
q is the charge on the object in C.
V is the electric potential in J/C or volts (V)
From this equation, you should see that 1 volt is the electric potential at a point in an electric field when
1J of energy is needed to bring a charge of 1C from  to that point in the field.
q  1C
qm
Requires 1J of energy

q  1C
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Example 11, p. 560
W  2.5 105 J, q  1.0 106 C
(a) The change in electric potential energy is the same as the work done on the charge, so
Ee  2.5 105 J .
Ee 2.5 105

 25J/C  25V
q 1.0 106
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Recall that a Coulomb is a collection of a huge amount of charge (i.e. 6.242 1018 e' s ). Often we are
interested in the energy required to move a single electron through a potential difference of 1V. The
energy required to do this is called one electron-volt (eV).
(b) V 
Def n One electron-volt (eV) is the energy of 1 electron after it has moved through a potential of 1 volt.
Using Ee  Vq ,
1eV  1.00V   1.602 1019 C  1.602 1019 J
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#1, p. 561
q1  1.5 106 C, q2  5.0 106 C, r  10cm  0.10m
(a)
 9.0 10
kq q
Fe  12 2 
r
9
1.5 10  5.0 10   6.75N
6
6
 0.10 
W  F  d   6.75  0.10   6.8 101 J
2
(b)
V
Ee 6.8 101

 450000J/C  4.5 105 V
6
q
1.5 10
(c)
 9.0 10
kq q
Fe  12 2 
r
9
1.5 10  5.0 10   27N
6
 0.05
6
2
W  F  d   27  0.05   1.35J
V
Ee
1.35

 900000V  9.0 105 V
6
q 1.5 10
Therefore, the change in potential difference is 9.0 105   4.5 105   4.5 105 V
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Sample Exam Questions
1. How much work is done by a 9.00 V power supply in moving 8.50 1018 electrons?
2. Which refers to the work required to move a charge from one point to another in an electric field?
3.
What amount of energy is gained by a proton that moves through a potential difference of 1.0 V?
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1
Do # 2, ATC, p. 561 Remember that E k  mv 2
2
View MLO #11, 7:20-8:26
Do #’s 69-72, p. 584
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§13.8 Movement of Charged Particles in a Field – The Law of Conservation of Charge (1 class)
Recall that if you drop an object, the object accelerates towards Earth. The object’s potential energy
 E p  mgh  changes to kinetic energy  Ek  12 mv2  as the object falls since h gets smaller but v gets
larger. However, the total amount of mechanical energy  E k  E p  at any point during the object’s fall
remains the same (constant). This is due to the Law of Conservation of Energy.
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Law of Conservation of Energy – Energy is neither created nor destroyed, it is just changed from one
form (e.g.: E p ) to another (e.g.: Ek ).
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The energy of charged objects also follows the Law of Conservation of Energy.
E.g.: If a positively charged test charge  qt  is placed near a positive master charge  qm  , the test charge
will be repelled and will accelerate away from the master charge. As it moves away, its electric potential
energy  E e  is transformed into kinetic energy  Ek  . However, the total amount of energy at any point
along the path of the test charge remains constant. Note that if our charges are atomic particles, such as
18
electrons, protons, or neutrons, we will need their mass to calculate their kinetic energy. These masses are
found in the left margin on p. 566.
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Example #12, p. 563
From the margin on page 561, we know that positive charges move from areas of high electric potential
energy  High Ee  to areas of lower electric potential energy  Low Ee  .
(a) The charged particle will move toward position 2 because positive charges move from areas of
high electric potential energy  High Ee  to areas of lower electric potential energy  Low Ee  .
(b) The Law of Conservation of Energy states that the total amount of energy at position 1 and at
position 2 must be equal. In other words, ET1  ET2 .
Position 1
Position 2
v0
Ek 
v?
1 2
mv  0
2
Ek 
1
1.0 105  v 2  5.0 106 v 2

2
Ee  Vq   v  12  1.5 105 
Ee  Vq   v  1.5 105 
Etotal  Ek  Ee  0   v  12  1.5 105 
Etotal  Ek  Ee  5.0 106 v 2   v  1.5 105 
  v  12  1.5 105 
According to the Law of Conservation of Energy, E total1  E total2 so
 v  12  1.5 105   5.0 106 v 2   v  1.5 105 
v 1.5 105   12 1.5 105   5.0 106 v 2   v  1.5 105 
12 1.5 105   5.0 106 v 2
12 1.5 105 
5.0 106
v

5.0 106 v 2
5.0 106
12 1.5 105 
5.0 106
v  6.0m/s[R]
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A formula on p. 565 can be used to find the electric potential energy  E e  between 2 point charges
separated by a given distance.
*****Ee 
kq1q2
where
r
Ee is the electric potential energy in J.
k  9.0 109 N  m2 /C2
q1 is the charge on one point charge in C.
q2 is the charge on the other point charge in C.
r is the distance between the charges in m.
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Example #13, p. 566
Position 1
Position 2
v0
v  1.0 103 m/s
Ek 
2
1
9.111031 1.0 103   4.555 1025 J

2
 9.0 10  1.602 10  1.602 10 

19
9
Ee
Ek 
19
 2.310 10
 9.0 10  1.602 10  1.602 10 

19
9
Ee
1.0 103
25
1 2
mv  0
2
J

Etotal  Ek  Ee  4.6 1025  2.311025
2.310 10
r
r
28
J
Etotal  Ek  Ee  0 
 6.865 1025 J
19
2.310 1028 2.310 1028

r
r
According to the Law of Conservation of Energy, E total1  E total2 so
2.310 1028
6.865 10 
r
28
2.310 10
r
 3.5 104 m
6.865 1025
25
The moving electron will stop at a distance of 3.5 104 m from the stationary electron and then reverse
direction.
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20
The Electric Potential  V  around a Point Charge.
Recall that for a test charge, the electric potential is given by V 
master charge, Ee 
Ee
and that for a test charge near a
qt
kqt qm
. Substituting for Ee into the first equation gives
r
kqt qm
kq q
1 k qt qm kqm
or just
V r  t m 

qt
r
qt
r
qt r
***** V 
kqm
where
r
V is the electric potential in V.
k  9.0 109 N  m2 /C2
qm is the charge on the master charge in C.
r is the distance between the charges in m.
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Example #14, p. 568
r  4.0cm  0.040m, qm  3.20 1019 C, V  ?
V
9
19
kqm  9.0 10  3.20 10 
 7.2 108 V
r
0.040
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Example #15, p. 568
V  100V, q  2.5 107 C, W  Ee  ?
Using V 
Ee
gives
q
Ee
2.5 107
Ee  100   2.5 107   2.5 105 J
100 
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21
Sample Exam Questions
How much work is done by a 9.00 V power supply in moving 8.50 1018 electrons?
1.
2. An electron is placed between two oppositely charged parallel plates with an electric field strength of
2.7 104 N/C and accelerates horizontally toward one of the plates.
i.
Calculate the acceleration of the electron.
ii.
Calculate the speed of the electron after it has travelled 0.22 m.
22
3. What amount of energy is gained by a proton that moves through a potential difference of 1.0 V?
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Do #1, 3, 4, ATC p. 557
Do #’s 76, 77, Problems, p. 585
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