Download Old Miterm1 Exam with Solution

General Physics – PH 213
Midterm I (Ch 25 - 27)
October 24, 2008
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Name:____________
Exam is closed book, closed notes. Use only your note card.
Write all work and answers in the color papers provided.
Show all your work and explain your reasoning (except for true/false or multiple choices ones)
Partial credit will be given. No credit will be given if no work is shown (except for true/false or
multiple choices ones)
Part I – True or False: For questions 1 – 8, state whether each statement is true or false.
1. (True) The electric field of an infinite sheet of charge is the same everywhere above the sheet.
2. (True) Any charge q can be expressed as q   ne where n is an integer and e is the charge of an
electron.
3. (True) The An electron moving in +x direction is slowed down when it enters the field E  E0 iˆ .
4. (False) If the net charge on a conductor is zero, its surface charge density must be uniform.
5. (False) If the net flux through a closed surface is zero, then there can be no charge within that
surface.
6. (True) Units of electric fields are newtons per coulomb, N/C.
7. (False) Inside a conductor the electric field is always zero.
8. (True) A charged insulator and uncharged metal always attract each other electrostatically when
brought near each other.
Part II – Multiple choice questions: For questions 9 – 16 select the one alternative that best completes the
statement or answers the question.
9. The units of electric dipole moments are
A) N/m
B) C2m
C) N/C
D) C2/m
E) mC
10. A positive point charge is placed at the center of a spherical
conducting shell with radii Ri and Ro as shown. Which of the
following statements is true for the electric field vector?
A) E  0 for r < Ri and E = 0 for r > Ro .
B) E  0 for r > Ro and E = 0 for r < Ri .
C) E  0 for r < Ri and for r > Ro .
D) E  0 for Ri < r < Ro .
E) The E is zero everywhere.
11. When a negatively charged conductor touches a neutral conductor, the neutral conductor will
A) gain protons.
D) lose electrons.
B) lose protons.
E) gain electrons.
C) stay neutral
Ph 213 Midterm1 LBCC F08
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12. Particle A of charge +2q is located at the origin, and particle B of charge +q is located on the
positive x-axis. The electric force on the particle B due to particle A is represented by vector F.
Which vector represents the force on particle A due to particle B?
+2q
+q
F
A
B
A)
B)
C)
D)
F
-F
2F
-2F
E) 1/2 F
F) -1/2 F
G) The force is zero.
13. A charged particle of mass m is suspended between the horizontal plates of a charged capacitor.
Which of the following statements is not true?
A) The magnitude of the electric force on the
particle is equal to its weight.
––––––––––––––––
B) The electric field between the plates
point downward.
C) The plates are at different potentials.
D) The particle is charged positively.
+++++++++++++++++++++
14. A uniform spherical shell of charge of radius R surrounds a point charge at its center. The point
charge has value Q and the shell has total charge -Q. The electric field at a distance R/2 from the
center
A) is zero
B) does not depend on the charge of the spherical shell
C) is half of what it would be if only the point charge were present
D) is directed inward if Q>0.
15. A conducting rectangular plate is connected to ground through a switch. The switch is initially
closed. A negative charge Q is brought near the plate as shown, but the charge is not brought in
contact with the plate. While the charge is near the plate, the switch is opened and then the
charge is removed. What is now the charge state of the plate?
A) It is negatively charged
B) It is positively charged.
C) It is uncharged.
D) It can be any of the above depending
–Q
on the initial charge on the plate before
the charge –Q was brought nearby.
16. A charge Q is uniformly distributed throughout a nonconducting sphere of radius R. The charge
density in the sphere is
A)
Q
4R
B)
3Q
4R 3
Ph 213 Midterm1 LBCC F08
Q
4R 2
4Q
D)
4R 2
C)
E) Not given
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Part III – Questions and Problems
Please explain or show your work clearly for each of the following questions or problem.
17. The figure shows four spheres, each
with charge Q uniformly distributed
through its volume.
A) Rank the spheres according to
their volume charge density,
greatest first.
Uniform volume charge density is
given by:
Q
  , and since charge is the same for all spheres, the larger the volume, the smaller the
V
charge density Therefore:
a > b > c > d
B) The figure also shows a point P for each sphere, all at the same distance from the center of
the sphere. Rank the spheres according to the magnitude of the electric field they produce at
point P, greatest first
Take a spherical Gaussian surface passing over point P for each sphere (r = OP), which will
be the same size for all spheres. The electric field value is related to the enclosed charge
inside the Gaussian surface. For a and b, the enclosed charge is the same (all of Q), and c
contains more charge than d (higher charge density) but less than b or a. So: Ea = Eb > Ec >
Ea
18. A positively charged rod is held near, but not touching a
neutral metal sphere.
A) Add plusses and minuses to the figure to show the
charge distribution on the sphere.
The positive charge will induce charges on the
sphere as shown.
B) Does the sphere experience a net force? If so, in
which direction. Explain
Yes it does. There is an attractive force between the rod and front side of the sphere, and
repulsion with the far end. Since the attractive force is larger (due to shorter distance), there
is a net charge on sphere to the right.
19. A point charge +Q is placed at the center of the conductors. Find the induced charges for all
surfaces (i.e inside1, I1, outside1, O1, and inside2, I2 and outside2, O2).
Since inside the conductor, E= 0, the electric flux through a Gaussian surface taken inside the first
conductor is zero. This in turn means the enclosed charge by this surface must add up to zero.
Therefore: qenc = +Q + QI1 = 0  QI1 = –Q. Now since inside conductor is neutral, total charge on its
inside and outside surfaces must add up to neutral.
Hence: QI1 + QO1 = 0  QO1 + (– Q) = 0,  QO1 = +Q
Using another Gaussian surface inside the 2nd conductor, and using same
arguments as above, we conclude: QI2 = – Q, QO2 = +Q
Ph 213 Midterm1 LBCC F08
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20. An 8.85  10-12 C point charge sits at the origin of a coordinate
system. What is the flux of electric field through a 10 cm tall right
circular cylinder of diameter 6.0 cm centered on the origin?
+q
E
  q
   E  dA  enc
0
We do not need to calculate the integral for the flux, since it can be calculated from the right side
as well:

qenc
0

8.85  10 12 C
C2
12
8.85  10
N  m2
so
  1 .0
N  m2
C
21. Five +1.0 nC point charges are equally spaced around a semicircular arc
of 10 cm radius as shown. An electron sits at the center of the
2
semicircle.
A) What is the electric field at the center of the semicircle?
1
E5
E4
E4cos 45
A) The electric vector due to each charge is shown:
Since the distance of each charge to field point is the
same, the magnitude of all E-fields are equal, and let it be
E0.
3
E2cos 45
E1
E3
E2
4
2
q
1.0  10 9 C
9 N .m
Where E0  k 2  9  10

R
C2
0.01m 2
5
 E0 = 9.0  102 N/C
From the diagram:
Etotal = E1 + E2 + E3 + E4 + E5
Etotal = –E0 j + E0 cos45 i – E0 cos45 j + E0 i + E0 cos45 i + E0 cos45 j + E0 j

N
2
N
Etotal  9.0  10 2 (2 
 1) iˆ  2173 N / C iˆ  2200 iˆ
C
2
C

A) Etotal  2200
Nˆ
i
C
B) What force does the electron experience?


B) F  eE  1.6  10
Ph 213 Midterm1 LBCC F08
19
C  2200
Nˆ
i  3.5  10 16 iˆ
C
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22. An insulating sphere of radius R 
2a
has a volume charge density  (r )  ar  br 2 , where a and
b
b are constants.
A) What is the total charge contained in the volume of sphere?
 (r ) 
dq
dV
 dq   (r )dV
dV  4 r dr 
2
 dq    (r )dV

R
 dq   (ar  br )4 r
2
2
dr
0
4ar 4 4br 5 R  2ba
Q

|
4
5 0
32a 5
4b 5
16a 4
b
 Q  a 4 
b
5
48a 5
Q 4
5b
B) What is the electric field at a distance of r from the center of the
sphere?
A spherical Gaussian surface of radius r is shown inside the
charged sphere:
  q
   E  d A  enc
r
2R
0
Electric field everywhere on the surface of the radius r is uniform (not a function of the area) so:
r
 (ar  br )4 r
2
E  d A cos0 
0
0
2
dr
ar 4 (5  br)
 EA 
5 0
ar 4 (5  4br)
5 0
ar 2 (5  4br)
E (r ) 
5 0
E (r )4r 2 
C) What is the electric field at a distance of 2R from the center of the sphere.
Repeat part A, except that the Gaussian surface is of radius 2R and encloses all of the charge Q of
the sphere:
  q
   E  d A  enc
E (2 R)  

0
48a 5
 2a 
5b 4 4   
 b 
4
Ph 213 Midterm1 LBCC F08
48a 5
E 4 (2 R)    4
0
5b  0
2
2
Q
3a 3
 E (2 R)  
2 0b 2
5
D) At what radius does the electric field vanish?
E)
ar 2 (5  4br)
From part B: E (r ) 
 0  ar 2 (5  4br)  0
5 0
Therefore E(r) = 0 at r = 0 or r = 5/4b
23. Two curved plastic rods with uniform linear
charge densities form a semicircle as shown.
One rod carries a total charge +Q and the
other –Q uniformly distributed over their lengths.
Find the magnitude and direction of the electric
field at point P, the center of the semicircle.
The electric fields of an infinitesimal charge – dq
and infinitesimal charge +dq are shown on the diagram. The sum these two
infinitesimal fields cancel their x-component. Therefore by symmetry the net electric
field due to both rods at point P will be only in the y-direction.
dE1 =
1 dq
rˆ1
4πε0 R 2
and
dE 2 =
1 dq
rˆ2
4πε0 R 2
by symmetry dE 1 = dE 2 and cancel each other's x - component
 dE 1x = dE 2x
dE = dE1 + dE 2
dE = dE1y + dE 2y
dE = dE1 sinθ + dE 2 sinθ
 1 dq

dE = 2 
sinθ 
2
 4πε0 R

But
dq = λdl
where
λ=
Q
π
2
dand
dl = Rdθ
R
 1 λRdθ

dE = 2 
sinθ 
2
 4πε0 R

2 λ π2
 dE = 4πε0 R 0 sinθdθ
E=
2 λ
π
(-cosθ) 0
4πε0 R
Replacing λ =
Ph 213 Midterm1 LBCC F08
Q
π
2
2
 E=
R
Q
π ε0 R 2
2
in y  direction
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24. (Bonus Problem) An electron is initially traveling at 3.0 x 106 m/s upward when it enters a
uniform electric field of 200 N/C directed to the left. What is the velocity of the electron after it
has traveled to a point that has a vertical coordinate that is 10.0 cm upward of where the electron
entered the field?
y
v0  3.0 106 m/s ˆj
E= -200 mN iˆ
E
(x,10)
As the electron enters the electric field it will accelerate in the
opposite direction of the field (+x direction), but will continue its y
direction at its original constant speed:
Since the field is constant, so will be the acceleration of the
0
x
electron, hence we can use kinematics eqs:
y = yo + voy t
0.10m = 0 + 3.0  106 m/s t  t = 3.3  10–8sec.
So it will take the electron 3.3  10–8sec to reach the point (x , 0.10) in the field.
vx = vox t + ½ ax t2
vx = 0 + ½ (eE/m)t2
vx = ½ (1.6 10–19C  200 N/C /9.11  10-31kg) (3.3  10–8sec)2
vx = 0.020 m/s
Compared to y-component of the velocity this x – component is very small and insignificant.
Hence velocity of the electron is at the position of (x , 0.10m) is:
v = 0.020 m/s i + 3.0  106 m/s
magnitude of v is about 3.0  106 m/s and is moving still about y direction.
Ph 213 Midterm1 LBCC F08
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