1
Physics 4B
FINAL EXAM
Chapters 23  34
Fall 1996
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Solve the following problems in the space provided.
Use the back of the page if needed.
Each problem is worth
10 points.
You
must
show your work in a logical fashion starting with the correctly applied physical principles
which are on the last page.
Your score will be maximized if your work is easy to follow because partial credit will
be awarded.
1.
(a)Find the magnitude and direction of the electric force on the 3μC charge.
(b)Find the magnitude and direction of the electric field felt by the 3μC charge.
(a)The sizes of the forces can be found using Coulomb’s Rule,
F
5
=
(9.00x10
9
)
(3.00x10

6
)(5.00x10

6)
2
(0.300)
=
1.50N
F
4
=
(9.00x10
9
)
(3.00x10

6
)(4.00x10

6)
2
(0.300)
=
1.20N
Adding the vector components,
F
x
=
F
5
cos60˚
+
F
4
cos60˚
=
1.35N
and
F
y
=
F
5
sin60˚

F
4
sin60˚
=
0.260N
Find the magnitude using the Pythagorean Theorem,
F
=
F
x
2
+
F
y
2
=
(1.35)
2
+
(0.260)
2
=
1.37N
Find the direction using the defintion of tangent,
tan
θ =
F
y
F
x
⇒ θ =
arctan
F
y
F
x
=
arctan
0.260
1.35
=
10.9˚
(b)Using the defintion of EField
r
E
≡
r
F
q
E
=
F
q
=
1.37
3.00x10

6
=
4.57x10
5
N/C
Since the charge feeling the field is positive, the direction of the field is the
same
as
the
direction
of
the
force
.
2. State Gauss's Law in words (Don't just "write the equation" with words and don't use the word "flux" unless
you explain what it means.).
Explain why it works.
Discuss the conditions that must be met to use it to find the
electric field due to a charge distribution.
Gauss’s Law states that the total number of electric field lines coming out of any volume is proportional to the net
amount of charge contained in the volume.
This works because each unit of charge produces the same number of
field lines.
In order to use Gauss’s Law to find the electric field, the charge distribution must have a high degree
of symmetry.
+3μC
4μC
+5μC
30cm
30cm
30cm
60˚
60˚
F
5
F
4
x
y
F
θ
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3.
Find the potential at the origin due to a semicircular arc of radius R and total
charge Q.
The potential due to the point charge dq is:
dV
=
k
dq
R
There is no need to worry about components for potential because it isn't a
vector.
The total potential is,
V
=
k
dq
R
∫
=
k
1
R
dq
∫
Finally,
V
=
k
Q
R
4.
The charge on C
3
shown in the circuit at the right is 8.00μC. Find (a)the
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 Fall '09
 Staff
 Charge, Work, Magnetic Field, power supply, total charge Q., LC dt dt

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