• PHYS 2212 FINAL EXAM REVIEW
PHYS 2212 final exam review
Final Exam Review
23 Electric Fields
23.1 Electric Fields and Electric Field Lines
23.1.1 Definition of an Electric Field
An electric field is said to exist in the region of space around a charged object, the source charge.
When another charged object (the test charge) enters this electric field, an electric force acts on it. As an
example, we consider a small positive test charge qo placed near a second object carrying a much greater
positive charge Q. We define the electric field due to the source charge at the location of the test charge to
be the electric force on the test charge per unit change, or, to be more specific, the electrical field vector E →
at a point in space is defined as the electrical force F→E acting on a positive test charge qo placed at that
point divided by the test charge,
→
E→ = FE
qo
→
where the vector E has units of [ N/ C] . The direction of E → as shown on page 539 is in the direction of the
force a positive test charge experiences when placed in the field. Note that E → is the field produced by some
charge or charge distribution seperate from the test charge, it is not the eld produced by the source. The
presence of charge is not necessary for the field to exist. The test charge serves as a detector of the electric
field, an electric field exists at a point if a test charge at that point experiences an electrical force.
We can rearrange the equation above to give us
→ E = qE
F →
and this equation gives us the force on a charged particle q placed in the field. If q is positive, the force is
in the same direction as the field. If q is negative, the force and field are in opposite directions.
To determine the direction of an electric field, we consider a point charge q as a source charge. This charge
creates an electric field in all points of space surrounding it. Now suppose that a test charge qo is placed at
a point P (which is a distance r from the source charge). We imagine using the test charge to determine the
direction of the electric force and therefore that of the electric field. According to Coulomb's law, the force
exerted by q on the test charge is
→ = k qqo
F
E e 2 r^
r
where r^ is a unit vector directed from q to qo . The force is directed away from the source charge q. Becasue
→ =F
the electric field at P , the position of the test charge , is defined by E → E /qo the electric field at P created
by q is
→ = k q r^ .
E
e 2
r
1
,• PHYS 2212 FINAL EXAM REVIEW
PHYS 2212 final exam review
23.1.2 Electric Field Lines
A convenient way of visualizing electric field patterns is to draw electric feild lines. These electrical feild
lines are related to the electric feild in a region of space by
1. The electric field vector E→ is tangent to the electric feild line at each point. The line has a direction,
which is indicated by an arrow. The direction of the line is that of the force on a positive test charge.
2. The number of lines per unit area thorugh a surface perpendicular to the lines is proportional to the
magnitude of the electric field in that region. Therfore the lines are close together where the electric field is
strong and far apart when it is weak.
23.1.3 The Rules for Drawing Electric FIeld Lines
1. The lines must begin on a positive charge and terminate on a negative charge. In the case of an excess
of one type of charge, some lines will begin and end infinitley far away.
2. The number of lines drawn leaving a positive charge or approach a negitive charge is proportional to the
magnitude of the charge.
3. No two field lines can cross.
2
,• PHYS 2212 FINAL EXAM REVIEW
PHYS 2212 final exam review
23.2 Electrical Forces From Point Charges Using Coulomb³s Law
We can use the experiement performed by Coulomb to generalize two properties of electrical forces. We use
the term point charge to refer to a charged particle of zero size.
1. The Magnitude of the electrical force between two point charges is Coulomb³s Law
|q1| |q2|
FE = ke
r2
where ke is called Coulomb³s constant.
2. The value of the Coulomb constant depends on the choice of units. The SI unit of charge is the
Coulomb [ C]
m2
ke = 8.99 × 10 9 N 2
C
This constant can be written in the form 1
ke =
4πoo
where the constant oo is known as the permittivity of free space and has the value
C2
oo×= 8.85 10—12
N m2
fhe smallest unit of free charge e knomn in nature, the charge on an electron ( —e) or a proton (+e) , has a
magnitude
e = 1.60 × 10—19 C
Therfore one C of charge is equal to the charge of 6.24 × 1018 electrons or protons.
Exercise 1
We consider 3 point charges located at the corners of a triangle. We have q1 = q3 = 5.00µ C,
q2 = —2.00µ C, and a = 0.100 m. We need to find the resultant force exerted on q3.
SOLUTION:
We are trying to find the resultant force on q3. We can see from the figure that there is an electron to the
right of q3 which will create a attractive force, and we also see a proton located 45 ○ below q3 which will
create a repulsive force.
We define the force F→23 to be the force exerted by q2 on q3 and we define the force F→13 to be the force experted
by q1 on q3. We note the since q2 and q3 are opposite in sign, we know that the force will be attractive. This
means that the direction of F→23 will be to the left (toward q2 ). We can use this same thought process to see
that since q1 and q3 have the same sign, the force F→13 will be away from q1 and more precisely 45○ above
the x axis.
The equation to describe the Magnitude Electrical force exerted by q2 on q3 can be written as
|q | |q |
F→23 = ke 2 3
a2
3
, • PHYS 2212 FINAL EXAM REVIEW
PHYS 2212 final exam review
we can numberically solve for F→23 and get
N m2 .—2.00 × 10—6 C. .5.00 × 10—6 C.
F→23 =
9
8.99 × 10
C2 (0.100 m)2
= 8. 99 N.
We will perform the same steps to describe the magnitude of the electrical force exerted by q1 on q2
F→ = k |q1| |q3|
13 e √ 2
a 2
numberically we have
N m2 .5.00 × 10—6 C. .5.00 × 10—6 C.
→ 9
0.100 m 2
F13 = 8.99 × 10 √ 2
2
= 11. 2 N. C
Now, since the force F→13 is at an angle of 45○ above the x axis, we can break up the force into its respected
x and y components. We have
F→x13 = F13 cos 45○ = 11.2 N cos 45○ = 7. 92 N
F→y13 = F13 sin 45○ = 11.2 N sin 45○ = 7. 92 N
Now, we need to write the components of the resultant force acting on q3, we have
Fx3 = Fx13 + Fx23
= 7.92 N + (—8.99 N)
= —1. 07 N
and
Fy3 = Fy13 + Fy23
= 7.92 N + 0
= 7.92 N
we can conclude that
F→3 = (—1.07^
s + 7.92(^ ) N.
Exercise 2
Three points lie along the x axis shown on page 537. The positive charge q1 = 15.0µ C is at
x = 2.00 m. The positive charge q2 = 6.00µ C is at the origin and the net force acting on q3 is
zero. What is the x coordinate of q3?
4
PHYS 2212 final exam review
Final Exam Review
23 Electric Fields
23.1 Electric Fields and Electric Field Lines
23.1.1 Definition of an Electric Field
An electric field is said to exist in the region of space around a charged object, the source charge.
When another charged object (the test charge) enters this electric field, an electric force acts on it. As an
example, we consider a small positive test charge qo placed near a second object carrying a much greater
positive charge Q. We define the electric field due to the source charge at the location of the test charge to
be the electric force on the test charge per unit change, or, to be more specific, the electrical field vector E →
at a point in space is defined as the electrical force F→E acting on a positive test charge qo placed at that
point divided by the test charge,
→
E→ = FE
qo
→
where the vector E has units of [ N/ C] . The direction of E → as shown on page 539 is in the direction of the
force a positive test charge experiences when placed in the field. Note that E → is the field produced by some
charge or charge distribution seperate from the test charge, it is not the eld produced by the source. The
presence of charge is not necessary for the field to exist. The test charge serves as a detector of the electric
field, an electric field exists at a point if a test charge at that point experiences an electrical force.
We can rearrange the equation above to give us
→ E = qE
F →
and this equation gives us the force on a charged particle q placed in the field. If q is positive, the force is
in the same direction as the field. If q is negative, the force and field are in opposite directions.
To determine the direction of an electric field, we consider a point charge q as a source charge. This charge
creates an electric field in all points of space surrounding it. Now suppose that a test charge qo is placed at
a point P (which is a distance r from the source charge). We imagine using the test charge to determine the
direction of the electric force and therefore that of the electric field. According to Coulomb's law, the force
exerted by q on the test charge is
→ = k qqo
F
E e 2 r^
r
where r^ is a unit vector directed from q to qo . The force is directed away from the source charge q. Becasue
→ =F
the electric field at P , the position of the test charge , is defined by E → E /qo the electric field at P created
by q is
→ = k q r^ .
E
e 2
r
1
,• PHYS 2212 FINAL EXAM REVIEW
PHYS 2212 final exam review
23.1.2 Electric Field Lines
A convenient way of visualizing electric field patterns is to draw electric feild lines. These electrical feild
lines are related to the electric feild in a region of space by
1. The electric field vector E→ is tangent to the electric feild line at each point. The line has a direction,
which is indicated by an arrow. The direction of the line is that of the force on a positive test charge.
2. The number of lines per unit area thorugh a surface perpendicular to the lines is proportional to the
magnitude of the electric field in that region. Therfore the lines are close together where the electric field is
strong and far apart when it is weak.
23.1.3 The Rules for Drawing Electric FIeld Lines
1. The lines must begin on a positive charge and terminate on a negative charge. In the case of an excess
of one type of charge, some lines will begin and end infinitley far away.
2. The number of lines drawn leaving a positive charge or approach a negitive charge is proportional to the
magnitude of the charge.
3. No two field lines can cross.
2
,• PHYS 2212 FINAL EXAM REVIEW
PHYS 2212 final exam review
23.2 Electrical Forces From Point Charges Using Coulomb³s Law
We can use the experiement performed by Coulomb to generalize two properties of electrical forces. We use
the term point charge to refer to a charged particle of zero size.
1. The Magnitude of the electrical force between two point charges is Coulomb³s Law
|q1| |q2|
FE = ke
r2
where ke is called Coulomb³s constant.
2. The value of the Coulomb constant depends on the choice of units. The SI unit of charge is the
Coulomb [ C]
m2
ke = 8.99 × 10 9 N 2
C
This constant can be written in the form 1
ke =
4πoo
where the constant oo is known as the permittivity of free space and has the value
C2
oo×= 8.85 10—12
N m2
fhe smallest unit of free charge e knomn in nature, the charge on an electron ( —e) or a proton (+e) , has a
magnitude
e = 1.60 × 10—19 C
Therfore one C of charge is equal to the charge of 6.24 × 1018 electrons or protons.
Exercise 1
We consider 3 point charges located at the corners of a triangle. We have q1 = q3 = 5.00µ C,
q2 = —2.00µ C, and a = 0.100 m. We need to find the resultant force exerted on q3.
SOLUTION:
We are trying to find the resultant force on q3. We can see from the figure that there is an electron to the
right of q3 which will create a attractive force, and we also see a proton located 45 ○ below q3 which will
create a repulsive force.
We define the force F→23 to be the force exerted by q2 on q3 and we define the force F→13 to be the force experted
by q1 on q3. We note the since q2 and q3 are opposite in sign, we know that the force will be attractive. This
means that the direction of F→23 will be to the left (toward q2 ). We can use this same thought process to see
that since q1 and q3 have the same sign, the force F→13 will be away from q1 and more precisely 45○ above
the x axis.
The equation to describe the Magnitude Electrical force exerted by q2 on q3 can be written as
|q | |q |
F→23 = ke 2 3
a2
3
, • PHYS 2212 FINAL EXAM REVIEW
PHYS 2212 final exam review
we can numberically solve for F→23 and get
N m2 .—2.00 × 10—6 C. .5.00 × 10—6 C.
F→23 =
9
8.99 × 10
C2 (0.100 m)2
= 8. 99 N.
We will perform the same steps to describe the magnitude of the electrical force exerted by q1 on q2
F→ = k |q1| |q3|
13 e √ 2
a 2
numberically we have
N m2 .5.00 × 10—6 C. .5.00 × 10—6 C.
→ 9
0.100 m 2
F13 = 8.99 × 10 √ 2
2
= 11. 2 N. C
Now, since the force F→13 is at an angle of 45○ above the x axis, we can break up the force into its respected
x and y components. We have
F→x13 = F13 cos 45○ = 11.2 N cos 45○ = 7. 92 N
F→y13 = F13 sin 45○ = 11.2 N sin 45○ = 7. 92 N
Now, we need to write the components of the resultant force acting on q3, we have
Fx3 = Fx13 + Fx23
= 7.92 N + (—8.99 N)
= —1. 07 N
and
Fy3 = Fy13 + Fy23
= 7.92 N + 0
= 7.92 N
we can conclude that
F→3 = (—1.07^
s + 7.92(^ ) N.
Exercise 2
Three points lie along the x axis shown on page 537. The positive charge q1 = 15.0µ C is at
x = 2.00 m. The positive charge q2 = 6.00µ C is at the origin and the net force acting on q3 is
zero. What is the x coordinate of q3?
4
