Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
Answers are on the last page
Multiple choice questions [60 points]
Answer all of the following questions. Read each question carefully. Fill the correct
bubble on your scantron sheet. Each question has exactly one correct answer. All
questions are worth the same amount of points.
1.
The diagram shows a pair of heavily charged plastic cubes that attract
each other.
Cube 3 is a conductor and is uncharged. Which of the following
illustrates the forces between 1 and 3 and between 2 and 3?
A.
B.
C.
D.
E.
I
II
III
IV
V
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
2.
Consider the following procedural steps:
(1) ground the electroscope
(2) remove the ground from the electroscope
(3) touch a charged rod to the electroscope
(4) bring a charged rod near, but not touching, the electroscope
(5) remove the charged rod
To charge an electroscope by induction, use the sequence:
A.
B.
C.
D.
E.
3.
1,
4,
3,
4,
3,
4,
1,
1,
1,
5
5,
2,
2,
5,
2
5
5
2
Two identical conducting spheres A and B carry equal charge. They
are separated by a distance much larger than their diameters. A third
identical conducting sphere C is uncharged. Sphere C is first touched
to A, then to B, and finally removed. As a result, the electrostatic
force between A and B, which was originally F, becomes:
A.
B.
C.
D.
E.
F/2
F/4
3F/8
F/16
0
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
4.
Positive charge +Q is uniformly distributed on the upper half a rod
and a negative charge –Q is uniformly distributed on the lower half.
What is the direction of the electric field at point P, on the
perpendicular bisector of the rod?
A.
B.
C.
D.
E.
5.
Two point charges, q1 and q2, are placed a distance r apart. The
electric field is zero at a point P between the charges on the line
segment connecting them. We conclude that:
A. q1 and q2 must have the same magnitude and sign
B. P must be midway between q1 and q2
C. q1 and q2 must have opposite signs and may have different
magnitudes
D. q1 and q2 must have equal magnitudes and opposite signs
E. q1 and q2 must have the same sign but may have different
magnitudes
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
6.
The diagrams below depict four different charge distributions. The
charges are all the same distance from the origin. Rank the situations
according to the magnitude of the electric field at the origins, least to
greatest.
A.
B.
C.
D.
E.
7.
1
4
2
1
4
<
<
<
<
<
2
3
1
2
3
<
<
<
<
<
3
2
3
3
1
<
<
=
=
<
4
1
4
4
2
Which of the following graphs represents the magnitude of the
electric field as a function of the distance from the center of a solid
charged conducting sphere of radius R?
A.
B.
C.
D.
E.
I
II
III
IV
V
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
8.
Two large parallel plates carry charge of equal magnitude, one
positive and the other negative, that is distributed uniformly over their
inner surfaces. Rank the points 1 through 5 according to the
magnitude of the electric field at the points, least to greatest.
A.
B.
C.
D.
E.
9.
1
5
1
2
2
<
<
=
=
=
2
4
4
3
3
<
<
=
<
<
3
3
5
1
1
<
<
<
=
<
4
2
2
4
4
<
<
=
<
=
5
1
3
5
5
10 C of charge are placed on a spherical conducting shell. A –3 C
point charge is placed at the center of the cavity. The net charge in
coulombs on the outer surface of the shell is:
A.
B.
C.
D.
E.
-7
3
7
10
13
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
10. A particle with mass m and, charge –q is projected with speed v0 into the
region between two parallel plates as shown. The potential difference
between the two plates is V>0 and their separation is d. The change in
kinetic energy of the particle as it traverses this region is:
Neglect any fringe effect due to the holes in the plates
A.
B.
C.
D.
positive
negative
0
Can't tell. There is not enough information.
11. The above experiment is repeated with a particle of mass 2m. All the
other values are unchanged (same charge, same initial velocity). How
does the change of kinetic energy between the two plates of the particle
of mass 2m, ∆KE2m, compare to the change of kinetic energy between
the two plates of the particle of mass m, ∆KEm?
A.
B.
C.
D.
∆KE2m < ∆KEm
∆KE2m > ∆KEm
∆KE2m = ∆KEm
Can't tell. There is not enough information.
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
12. The particle of mass m and charge –q is now projected in the region
between the two plates at a 30° angle with an initial velocity v0. How
does its change of kinetic energy between the two plates compare to its
change of kinetic energy between the two plates when it is projected
horizontally as in question 10?
0
V
m
-q
v0
30°
d
A.
B.
C.
D.
same
greater
smaller
Can't tell. There is not enough information.
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
PROBLEM [40 points]
A very long cylindrical plastic rod of radius a=2.5 cm has electric charge per
unit length of density λ = +6.0 × 10-7 C/m distributed uniformly throughout
its volume. This rod is surrounded by a concentric cylindrical metal shell of
inner radius R1 = 7.5 cm and outer radius R2 = 10 cm, as shown. The metal
shell is initially uncharged.
A is a point on the rod (at r=a), B a point on the inner surface of the shell (at
r = R1), C is a point on the outer surface of the shell (at r = R2). P is a point
between the rod and the shell a distance r from the center of the rod (a < r <
R1). There is nothing (a vacuum) between the rod and the shell.
R1 a
R2
P
A
B
C
1). [5 pts] What is the surface charge density σ1 on the inner surface of the
shell? Explain
2). [5 pts] What is the surface charge density σ2 on the outer surface of the
shell? Explain
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
3). [10 pts] What is the electric field at point P, located a distance r
(a < r < R1) from the center of the rod? Give the magnitude (no numbers
just letters), direction and units. Explain.
4). [10 pts] What is the potential difference between point A on the rod and
point C on the outer shell? Give magnitude, sign and units. Explain
5). [5 pts] An electron (charge q = -1.6 × 10-19 C) is released at rest at point
B on the inner surface of the shell. What is the electron kinetic energy
when it reaches the rod?
6). [5 pts] The shell is now grounded. Does the value of the electric field at
P change? If so what is the new value? If not, why does it stay the same?
Explain clearly.
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
Answer key:
1.C, 2.B, 3.C, 4.A, 5.E, 6.B, 7.E, 8.C, 9.C, 10.A, 11.C, 12.A
Problem:
1) Take for a Gaussian surface a cylinder with a radius > R1 and < R2 and height h.
The flux through the surface is 0 (Since E is radial, there is no flux through the ends of the cylinder.
Since E = 0 in the conductor, there is no flux through the side of the cylinder).
The net charge inside the surface is therefore 0.
Thus
0 = λh + σ 1 2πR1 h
σ1 = −
λ
= −1.27 × 10 −6 C / m 2
2πR1
2) The net charge of the conductor is zero since it is initially uncharged. For a piece of the conductor of
height h:
σ 1 2πR1 h + σ 2 2πR2 h = 0
σ 2 = −σ 1
R1
= 9.55 × 10 −7 C / m 2
R2
3) Take a Gaussian cylinder of radius r and height h. By symmetry, E is radial. It is directed away from the
positively charged rod. The units are N/C.
The net flux through the Gaussian cylinder is
Φ = E 2πrh
Also using Gauss’s law
Φ=
λh
ε0
Thus
E =
λ
2π rε
0
4)
r r
R1
V A − VC = ∫ E • ds = ∫
a
λ
λ
R 
dr =
ln 1  = 11.9kV
2πrε 0
2πε 0  a 
Note that since E is 0 in the conductor the integral is taken from a to R1 (and not R2).
5) Use the work energy theorem
KE f − KE i = Wnet = q (VB − V A )
But KEi = 0 and VB = VC (two points of a conductor)
KE f = −1.6 × 10 −19 × −11851.8 = 1.90 × 10 −15 J
6) The computation in 3) depends only on the charge inside the Gaussian surface. The charge on the
conductor is irrelevant. The electric field at P doesn’t change.