Download Physics 114 Exam 1 Spring 2013

Physics 114
Exam 1 Spring 2013
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Question
Problem
1.
1.
2.
2.
3.
3.
Answer each of the following questions and each of the problems. Points for each
question and problem are indicated in red with the amount being spread equally among
parts (a,b,c etc). Be sure to show all your work.
Use the back of the pages if necessary.
Question 1. (10 points) A negatively charged rod is brought close to a small metal ball suspended
by a thin thread, but doesn’t touch it. (a) What does the sphere do, if anything. (b) What do the
charges in sphere do?
Now, still without the rod touching the sphere, but while holding the rod close to the sphere, you
ground the sphere by touching it. (c) What do charges in the sphere or elsewhere (like from the
ground) do at this point .
You take away your hand (the connection to ground). (d) What is the net charge of the sphere
now (positive, negative, or neutral)?
Solution
(a) The sphere is attracted to the rod and moves towards it
(b) They polarize; the negative charges move away from the rod so that the left side is
positive and right side is negative.
(c) Negative charges in the sphere flow into the ground.
(d) The charge on the sphere is positive.
Question 2. (10 points) Consider point A in the figure below located an arbitrary distance from
two positive point charges in otherwise empty space.
(a) Is it possible for an electric field to exist at point A in empty space?
(b) Does charge exist at this point?
(c) Does a force exist at this point?
Solution
(a) Yes. Electric field permeate space.
(b) No. It is empty space.
(c) No. There is no charge there. There is no force.
Question 3. (10 points). Consider the following image. (Take q1 = 3.00 nC and q2 = - 8.00nC.)
(a) Find the net electric flux through the cube shown in the figure above.
(b) Can you use Gauss's law to find the electric field on the surface of this cube? Explain why or
why not.
Solution
(a) The flux is just the charge inside divided by 0.
This is 3 × 10-9 C / 8.85 × 10-9 C2/N-m2
= 339 N-m2/C
(b) No. Gauss’s law only can be applied in cases of a symmetric charge distribution. In
addition, the symmetry of the Gaussian surface should match the field in the sense that
you will want the field to be constant over the surface. Here, the field due to q1 is not
constant over the surfaces. In addition, the field due to q2 also contributes to the field on
the surface. If you knew the positions of the charges relative to the cube, you could
calculate the field at the surface by integration, but you could not use Gauss’ law due to a
lack of symmetry.
Problem 1. (15 points) Three point charges with magnitude given by q = 2 C lie along a circle of
radius r= 3 m at angles of 30°, 150°, and 270° as shown in the figure below.
(a) Find the vector resultant electric field at the center of the circle.
(b) If a point charge Q = 10 nC is placed at the origin, what would the magnitude of the
net force be on that charge be due to the three charges shown in the Figure.
Solution
(a) This is solved by adding the vector electric fields due to all three charges. It is apparent
that the x-component of the field due to the two positive charges cancel each other out
and the negative charge creates no electric in the x-direction. Thus, there is only a ydirection of the resultant field. Note that the y-component of the field due to each charge
is in the negative direction (downward, imagine a positive test charge at the origin).
To do this properly, it could be useful to draw the E field vectors
y
x
Note that the electric field vectors from the positive charges point in directions minus 30
degrees below the positive and negative x-axis. That is the angles are 210 and 330
degrees.
 k

E= 2e  4sin(270)  2sin(210) + 2 sin(330)  j N
C
r
9

9  10
=
4
1
1)
j N




C
32

 6  109 j N
C
 
(b) To get the force on the charge we just apply E  F q . Here q = 10 nC = 1 × 10-8 C.
0



9
8
Thus, F=(  6 10 )(110 ) j N = -60 j N
Problem2. (15 points) The figure below represents the top view of a cubic gaussian
surface in a uniform electric field = 1 N/C oriented parallel to the top and bottom faces
of the cube. The field makes an angle θ with side , and the area of each face is A = 2
m2.
(a) If  = 0o (so the field would go straight up in the picture), what would be the flux
through side 1?
(b) If  = 0o, what would be the flux through side 2?
(c) If  = 30o what would be the flux through side 1?
(d) If  = 30o what would be the flux through side 2?
(e) What is the total flux through the cube (summing over all sides) in both cases ( = 0o
and  = 30o). Note there are no charges inside the surface.
Solution
   
(a) The Flux is given by    E.dA E. A , where the integral disappears since this is a
(b)
(c)
(d)
(e)
constant electric field and we have flat surfaces. The vector A points upward in the
diagram. Thus we have that the flux is just given by
= EA cos() = 1 N/C ×2 m2 × cos(0) = 2 N-m2/C
The vector A points to the right in figure. Thus,
= EA cos() = 1 N/C ×2 m2 × cos(90) = 0
= EA cos() = 1 N/C ×2 m2 × cos(30) = 1.73 N-m2/C
= EA cos() = 1 N/C ×2 m2 × cos(120) = - 1 N-m2/C
The total flux is zero in both cases – Gauss’ law says that the total flux will be zero if
there are no charges inside – and there are none.
Problem 3. (15 points) A point charge q = 5 mC is placed at the center of a spherical conducting
shell of inner radius r1 = 1 m and r2 = 2. M that has no net charge on it.
Calculate the electric field
(a) at r = 0.5 m
(b) at r = 1.5 m
(c) r = 3 m
(d) What is the total charge that is distributed on the inner surface of the spherical shell?
(e) What is the total charge that is distributed on the outer surface of the spherical shell
[Note that the total net charge on the spherical shell is zero].
r2
q
r1
Solution
 
(a) Gauss’s law states that    E.dA  qenc /  0 . The charge inside here is just q = 5 mC.
The field will be radial outward. Thus we have
E×4r2 = q/0
E ×42 = (5 × 10-3)/(8.85 × 10-12)
E = 1.80 × 108 N/C
(b) Here E = 0 since we are inside the conductor.
(c) This is like part (a) except r is different.
E×4r2 = q/0
E ×42 = (5 × 10-3)/(8.85 × 10-12)
E = 5.00 × 106 N/C
(d) Draw a Gaussian surface inside the sphere (in the metal).
The flux through that surface is zero as E is zero. Thus, the total charge is zero. Since
the point charge is q = 5 mC, the charge on the inner surface must be – 5 mC.
(e) Since the total, net charge on the conducting shell is 0, there must be + 5 mC on the outer
surface.
Possibly Useful Information
F
1 | q 1 || q 2 |
4  0 r 2
e = 1.6 X 10
 0  8.85 X 10 -12 ( C 2 / N  m 2 )
-19
C
| q|
E
4  0 r 2
1
ke 
 9 109 Nm 2 / C2
4 0
 
E  Fq
0
 
 0    0  E. dA  q enc

r = xi + yj + zk