Name ___________________________________ (print clearly)
PHY-2054 College Physics II
Dr. Dubey – Dr. Bindell
January 10, 2010.
EXAMINATION #1
INSTRUCTIONS: Answer all questions carefully. It is always best to use symbols to solve
problems and to then add the numbers. This makes it easier to follow your work
and easier for you to check your work if you have time. Read the questions
carefully. Draw a diagram wherever possible. Be sure to put your name at the top
of each page so that if the pages get separated during the grading process, they can all be
identified. Use the back of the page if you need more room. Constants are below. Good luck.
 k=9.0 x 109 Nm2/C2
e=1.6 x 10-19 C 
Problem (1a) (They should generally express the
following ideas.)
The drawing shows three charges, labeled q1, q2, and q3. A
Gaussian surface is drawn around q1 and q2.
(a) Which charges determine the net electric flux
leaving or entering the Gaussian surface? (Explain)
1 & 2 determine the net flux because they are the
only charges inside the Gaussian surface and the
flux leaving the volume only depends on the charges that are inside.
(b) Which charges produce the electric field at the point P? (Explain)
ALL of the charges contribute to the field at point P. That field is just the superposition
of the charges in the area. The Gaussian surface is just imaginary and doesn’t modify
the electric field anywhere.
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Problem 1(b)
The location marked P in the drawing lies midway between the point charges +q and -q. The
lines labeled A, B, and C are edge-on views of three planes. Which of these planes is an
equipotential surface?
Explain:
The charges create an electric field at surface B that is perpendicular to the surface everywhere.
Therefore no work is necessary to move a charge from one part of the surface to another. The
angle between the force and the displacement (on the surface) is 90 degrees and the cosine is
zero. Thus W=0.
Problem (1c)
The potential at a point in space has a certain value, which is not zero. Is the electric potential
energy the same for every charge that is placed at that point? State your answer and explain it.
The electric potential energy is the (charge) x (the potential). If two charges are different, the PE
will be different as well.
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Problem (2)
Charges of −q and +2q are fixed in place, with a distance of 2.00 m between them. A dashed line is drawn
through the negative charge, perpendicular to the line between the charges. On the dashed line, at a
distance L from the negative charge, there is at least one spot where the total potential is zero. Find L
(From solution manual)
We begin by noting that the distance between the positive charge +2q and the zero-potential spot is
L2   2.00 m  . With this in mind and using V = kq/r
2
given by the Pythagorean theorem as
(Equation 19.6), we write the total potential at the spot in question as follows:
VTotal 
k  q 

L
k  2 q 
L2   2.00 m 
2
0
or
2
1

L
L2   2.00 m 
Squaring both sides of this result gives
1
4

L2 L2   2.00 m 2
L2   2.00 m   4L2
2
or
Solving for L, we obtain
3L2   2.00 m 
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2
or
L
2.00 m
 1.15 m
3
2
Name ___________________________________ (print clearly)
Problem (3)
1. The drawings show three charges that have the same magnitude but different signs. In all
cases the distance d between the charges is the same. The magnitude of the charges is |q|= 4.0
µC, and the distance between them is d = 5.0 mm. Determine the magnitude of the net force
on charge 2 for each of the three drawings. (Be careful of units!)
From Solution Manual:
REASONING The electrical force that each charge exerts on charge 2 is shown in the following drawings.
F21 is the force exerted on 2 by 1, and F23 is the force exerted on 2 by 3. Each force has the same
magnitude, because the charges have the same magnitude and the distances are equal.
q
F21
+q
1
F23
2
(a)
+q
+q
F23 +q
3
1
2
F21
+q
+q
3
1
F21 q
2
F23
(b)
3
+q
(c)
The net electric force F that acts on charge 2 is shown in the following diagrams.
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F21
F23
F21
F23
F23
(a)
F21
F
F=0N
F
(b)
(c)
It can be seen from the diagrams that the largest electric force occurs in (a), followed by (c), and then by
(b).
SOLUTION The magnitude F21 of the force exerted on 2 by 1 is the same as the magnitude F23
of the force exerted on 2 by 3, since the magnitudes of the charges are the same and the distances are
the same. Coulomb’s law gives the magnitudes as
F21  F23 
kq q
r2
8.99  109 N  m 2 /C2  4.0  106 C  4.0  106 C 


 5.7  104 N
2
5.0  103 m 
In part (a) of the drawing showing the net electric force acting on charge 2, both F21 and F23 point to the
left, so the net force has a magnitude of


F  2F12  2 5.7  104 N  11.4  104 N
In part (b) of the drawing showing the net electric force acting on charge 2, F21 and F23 point in opposite
directions, so the net force has a magnitude of 0 N .
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In part (c) showing the net electric force acting on charge 2, the magnitude of the net force can be
obtained from the Pythagorean theorem:
2
2
F  F21
 F23

 4.6  104 N    4.6  104 N 
2
2
 6.5  104 N
______________________________________________________________________________
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Name ___________________________________ (print clearly)
Problem (4)
An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the
positive plate (see the drawing). The plates are separated by a distance of 2.0 cm, and the electric field
within the capacitor has a magnitude of 1.5 106 V/m. What is the kinetic energy of the electron just as
it reaches the positive plate?

Electric field


+
+
Electron

F

+
x
+x
+
Capacitor plates
SOLUTION The total energy of the electron is conserved, so its total energy at the positive
plate is equal to its total energy at the negative plate:
KE positive  EPE positive  KE negative  EPE negative
Total energy at
positive plate
Total energy at
negative plate
Since the electron starts from rest at the negative plate, KEnegative = 0 J. Thus, the kinetic
energy of the electron at the positive plate is KEpositive = (EPEpositive  EPEnegative). We
know from Equation 19.4 in the REASONING section that EPEpositive  EPEnegative =
(e)(Vpositive  Vnegative), so the kinetic energy can be written as
KEpositive = (EPEpositive  EPEnegative) = e (Vpositive  Vnegative)
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Since the potential difference is related to the electric field E and the displacement s by
Vpositive  Vnegative   Es (Equation 19.7a), we have that


KE positive = e Vpositive Vnegative = e(  E s)

6
15


= 1.50  10
C  1.5 10 V/m  0.02 m   4.8 10
J



In arriving at this result, we have used the fact that the electric field is negative, since it
points to the left in the drawing.

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19
 
