Download Physics 114 Exam 1 Fall 2016

Physics 114
Exam 1 Fall 2016
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For grading purposes (do not write here):
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 and touches a small metal ball
suspended by a thin thread. (a) Is there any movement of charge – if so, how?. (b) What does the
sphere do, if anything?
Now, the rod is taken away from the sphere and then brought back towards the sphere, but does
not touch it. (c) What do charges in the sphere do at this point?
(d) What does the sphere do (move towards or away or not move), why?
[Hint – think about the bell experiment]
Solution: (a) Negative charge transfers from the rod to the ball. The ball becomes negatively
charged – charged by conduction.
(b) As the rod approaches, the ball moves towards it due to polarization [this not really relevant
to the question at hand]. Once it touches and both are negatively charged the ball will be
repelled.
(c) The charges will polarize, even though there is a net charge on the sphere, more of these
would be on the far side of the rod.
(d) The ball will most likely be repelled. There is some chance, depending on the size of the
sphere, that the charges in it will be polarized enough so there is a net positive charge on the
nearer side to the rod that is attracted to the rod more strongly than the negative charges on the
far side are repelled. In that case it would move towards the bar.
Question 2. (10 points) Consider point A located an arbitrary distance from two positive point
charges in otherwise empty space as shown below. (a) Is it possible for an electric field to exist
at point A in empty space? Explain. (b) Does charge exist at this point? Explain. (c) Does a force
exist at this point? Explain.
Solution: (a) Yes, it does in fact exist there.
(b) No, only the two charges are present that are shown.
(c) No, there is no charge there; there is no force
Question 3. (10 points) Consider a point charge q1 = 8.85 C inside of a
cylindrical surface and another charge q2 = - 8.85 C outside the cylinder.
(a) Can Gauss’ law be used to find the total flux on the surface of the
cylinder? Why or why not? If one can, calculate it.
(b) Can Gauss’ law be used to find the electric field on the surface of
the cylinder? Why or why not? If one can, calculate it.
(c) Say that q2 were taken away and the size of cylinder reduced so
that its surface area was half of what it was originally? Would
your answers to parts (a) and (b) change – if so how?
q2
q1
 
Solution: (a) Yes. The flux just depends on the enclosed charge, q1.  0    0  E. dA  q enc
 = q1/0 = -8.85 × 10-6 C/8.85 × 10-12 (C2/N-m2) = -1 × 106 N-m2/C
(b) No, there is not enough symmetry.
(c) No the answers would not change. The flux is the same (same number of net field lines
penetrating surface). You would still not be able to use Gauss law to get the electric field. Your
know the field at the surface would be greater, but you could not use Gauss’ law to calculate it.
Problem 1. (15 points) A + 10 C charge (q1) is placed at the origin and a – 10 C charge (q2) is
placed along the x-axis at = 1 m. Note that q2 is negative.
(a) What is the electric field at the point x = 1m, y = 1m?
(b) If a + 10 nC charge (q3) is placed at this point (x = 1m, y =1m), what is the force on this
charge?
Solution: (a) The electric field due to q1 points
upward to the right 45 degrees above the x-axis
(1,1)
and that due to q2 points downward along y.
q1
=10 C
q2
= - 10 C



E = E1cos(45) i + (E1sin(45) + E2sin(270)) j
(9  109 )(10  10 6 ) 
(9  109 )(10  10 6 ) (9  109 )(10  10 6 ) 
= 2 2
i ( 2 2
)j
(
1 1
1 1
1



4.5  10 4 i  (4.5  10 4 j  9  10 4 j
 2
2





E = 3.2  10 4 i - 5.8  10 4 j N/C
(b) The force is just obtained by multiplying the charge and field:




F = qE = 10  10 9 (3.2  10 4 i - 5.8  10 4 j)


=3.2  10 4 i - 5.8  10 4 j
Problem 2. (15 points) Find the net electric flux (a) through a square surface
of area A = 2 m2



that is parallel to the x-z plane that is in a constant magnetic field given by E = 10 i + 10 j N/C,
(b) the closed spherical surface with R = 2 m in a uniform electric field of magnitude 10 N/C
shown below, (c) the closed cylindrical surface shown below where the magnitude of the electric
field is 10 N/C and points out of the surface on both sides as shown below. R is 2 m for the
cylinder and let its length be 5 m. (d) What can you conclude about the charges, if any, inside the
cylindrical surface?
Solution: (a) We can solve this using Cartesian coordinates or in terms of the magnitudes of the
vectors and the angles between them. The angle is 45 degrees. The magnitude of the electric
field is 14.14 N/C.
 
  E.A EA cos(45) = (14.14)(2)(0.707)=20 Nm2/C
 
  E.A  2ˆj  (10iˆ  10ˆj) = 20 Nm2/C
=
(b) No net flux goes through this surface as all lines going in also come out. The net flux is zero.
(c) No flux goes through the top and bottom. The sides have positive flux of equal magnitude
through each as the electric field are parallel to the normal (theta is zero). Thus
 = 2EA = (2)(10)()(22) = 80 = 251 Nm2/C
= 2.22 nC
(d) Have more than one charge and the net charge is positive. qenc   0
[This question only asked what you can conclude about the charges, it was stated vaguely so you
did not need to calculate the charge for full credit]
Problem 3. (15 points) A solid insulating sphere of radius 10.00 cm has a charge of 10.00 C
uniformly distributed throughout its volume. A conducting spherical shell of inner radius 20.00
cm and outer radius 30.00 cm is concentric with the solid sphere and has no net charge. Find the
electric field at (a) r = 5.00 cm, (b) r = 15.00 cm, (c) r =25 cm, and (d) r = 50 cm from the center
of this charge configuration. In addition (e), determine the total charge on the inner surface of the
conducting shell (at r = 20.00 cm) and on the outer surface of the shell (at r = 30.00 cm).
Solution: (a) This is inside the insulating sphere. Let the insulating sphere be that in yellow
below and so draw a Gaussian surface with r = 5 cm (shown in red).
4  r3
kQ
r3
3
ˆ
qin   0 E , Q
  0 AE  n , Q 3   0 4 r 2 E , E  Qr a 3 4 0  e 3 r ,
4  a3
a
a
3
E = (9 × 109)(10 × 10-6)(0.05)/(0.1)3 = 4.5 × 106 N/C
(b) Here the field is just that of the charge inside a Gaussian sphere of
radius 15 cm which is the entire 10 C. We have EA = q/0.
E4r2 = q/0; E = keq/ r2 = (9 × 109)(10 × 10-6)/(0.15)2 = 4.0 × 106 N/C
(c) This is inside the metallic shell. E = 0.
(d) We just have a situation like part b.
E = keq/ r2 = (9 × 109)(10 × 10-6)/(0.5)2 = 3.6 × 105 N/C
(e) If I consider a spherical Gaussian surface of radius 25 cm, we have E = 0, so q inside
must be zero. Thus the inner surface must have a charge of -10 C to balance that of the
sphere. Since the inner shell has -10 C and the net charge is zero, the outer surface must
have a charge of + 10 C.
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