Version 062 – Exam 1 – markert – (55305)
This print-out should have 20 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
Consider a charge q1 on a metallic ball with
radius R1 at the center, inside of a concentric
conducting shell with total charge q2 of inner
radius R2 and outer radius R3 .
1
treated as a point charge, and the electric
field is
q1 + q2
EA = k
.
a2
002 10.0 points
Consider a long, uniformly charged, cylindrical insulator of radius R with charge density
2 µC/m3 .
q2
R
R1
q1
3 cm
O
R2
R3
A
Find the electric field at A , where OA = a.
q1 − q2
2 a2
q1 − q2
2. EA = k
a2
q1 + q2
3. EA = k
R32
q1
4. EA = k 2
R3
1. EA = k
5. EA = 0
6. EA = k
7. EA = k
8. EA = k
9. EA = k
10. EA = k
q1
a2
q1 − q2
R32
q1 + q2
2 a2
q1
2 a2
q1 + q2
correct
a2
Explanation:
Consider a spherical Gaussian surface
through A. A is outside of the entire charge
distribution a distance a from the center, so
the enclosed charge Qencl = q1 + q2 can be
What is the magnitude of the electric field
inside the insulator at a distance 3 cm < R
from the axis? The permittivity of free space
is 8.8542 × 10−12 C2 /N · m2 and the volume
of a cylinder with radius r and length ℓ is
V = π r2 ℓ .
1. 621.174
2. 2032.93
3. 2710.58
4. 2371.76
5. 2936.46
6. 1982.11
7. 1626.35
8. 3388.22
9. 2055.52
10. 2783.99
Correct answer: 3388.22 N/C.
Explanation:
Let :
r = 3 cm = 0.03 m ,
ρ = 2 µC/m3 ,
= 2 × 10−6 C/m3 , and
ǫ0 = 8.8542 × 10−12 C2 /N · m2 .
Consider a cylindrical Gaussian surface of
radius r and length ℓ much less than the
length of the insulator so that the component of the electric field parallel to the axis is
negligible.
Version 062 – Exam 1 – markert – (55305)
2
kq
a2
√ kq
3. EO = 4 2 2 correct
a
√ kq
4. EO = 3 2 2
a
√ kq
5. EO = 2 2 2
a
kq
6. EO = 2
a
1 kq
7. EO = √
2 a2
1 kq
8. EO = √
2 2 a2
1 kq
9. EO = √
4 2 a2
1 kq
10. EO = √
3 2 a2
2. EO = 3
R
ℓ
r
The flux leaving the ends of the Gaussian
cylinder is negligible, and the only contribution to the flux is from the side of the cylinder.
Since the field is perpendicular to this surface,
the flux is
Φs = 2 π r ℓ E ,
and the charge enclosed by the surface is
Qencl = π r 2 ℓ ρ .
Using Gauss’ law,
Qenc
ǫ0
π r2 ℓ ρ
2πrℓE =
ǫ0
ρ
E=
r
2 ǫ0
2 × 10−6 C/m3 (0.03 m)
=
2 (8.8542 × 10−12 C2 /N · m2 )
Φs =
= 3388.22 N/C .
003 10.0 points
Consider a square with side a. Four charges
−q, +q, +q, and −q are placed at the corners
A, B, C, and D, respectively
A
B
−
+
O
a
−
+
D
C
What is the magnitude of the electric field
at the center O?
√ kq
1. EO = 2 2
a
Explanation:
The distance between each corner and the
a
center is √ , so the magnitude of each electric
2
field at D is
E=k q
a
√
2
2 = 2 k
q
a2
The two negative charges yield forces pointing away from them from O and the two positive charges yield forces pointing toward them
from O with the collinear charges adding algebraically:
~A + E
~ C k = kE
~B + E
~ Dk = 2 E = 4 k q .
kE
a2
EA + EC
E
EB + ED
Version 062 – Exam 1 – markert – (55305)
The Cartesian components of the two vectors with the origin at O are
q
1
1
~A + E
~B = 4k
E
− √ ı̂ + √ ̂
and
a2
2
2
q
1
1
~
~
EB + ED = 4 k 2 − √ ı̂ − √ ̂ , so
a
2
2
1
1
−√ − √
ı̂
2
2
1
1
+ √ −√
̂
2
2
√
q
= −4 2 k 2 ı̂ ,
a
√
q
with magnitude −4 2 k 2 .
a
~ = 4k q
E
a2
Correct answer: 6.55682 nN.
Explanation:
Let :
qo
qa
ra
qb
rb
ke
= 6 × 10−9 C ,
= 9 × 10−9 C ,
= 9 m,
= 4 × 10−9 C ,
= 9 m , and
= 8.98755 × 109 N · C2 /m2 .
Applying Coulomb’s Law for qo and qa ,
qo qa
2
roa
= −(8.98755 × 109 N · C2 /m2 )
(6 × 10−9 C) (9 × 10−9 C)
×
(9 m)2
= −5.9917 × 10−9 N
Foa = −ke
004 10.0 points
Three charges are arranged in the (x, y) plane
as shown.
y (m)
10
9
8
7
6
5
4
3
2
1
0
directed along the x-axis.
Applying Coulomb’s Law for qo and qb ,
4 nC
Fob = −ke
6 nC
3
9 nC
0 1 2 3 4 5 6 7 8 9 10
x
(m)
What is the magnitude of the resulting force on the 6 nC charge at
the origin?
The Coulomb constant is
9
8.98755 × 10 N · C2 /m2 .
1. 6.55682
2. 6.99354
3. 19.9255
4. 5.39282
5. 39.5464
6. 6.15879
7. 1.54088
8. 5.63272
9. 11.8312
10. 16.4304
qo qa
2
rob
= −(8.98755 × 109 N · C2 /m2 )
(6 × 10−9 C) (4 × 10−9 C)
×
(9 m)2
= −2.66298 × 10−9 N
directed along the y-axis.
The magnitude of the resultant force is
q
~ k = F2 + F2
kF
x
y
h
2
= −5.9917 × 10−9 N
2 i1/2
+ −2.66298 × 10−9 N
×
1 × 109 nN
1N
= 6.55682 nN .
005 10.0 points
The point charge Q shown is at the center of
Version 062 – Exam 1 – markert – (55305)
4
a metal box that is isolated, ungrounded, and
uncharged.
−
a
Q
+
(c)
−
+
(d)
For which situation(s) shown above is the
net force on the dipole equal to zero?
Which of the following is true?
1. The electric field outside the box is zero
everywhere.
1. (c) and (d) correct
2. (b), (c), and (d)
2. The electric field inside the box is the
same strength everywhere.
3. (b) and (d)
3. The electric field inside the box is zero.
4. (a) and (c)
4. The net charge on the outside surface of
the box is Q. correct
5. (a) and (d)
5. The electric field outside the box is the
same as if only the point charge (and not the
box) were there.
6. (c) only
7. (a) only
8. None of these
9. (a), (b), and (c)
10. Another combination
Explanation:
Consider a Gaussian surface between the
outside and inside surface of the box and apply Gauss’s law. The electric field on such
surface is zero, because the box is a conductor
and there is no current, so the charge on the
inside surface of the box is −Q. Since the box
is neutral, the charge on the outside surface
of the box is Q.
006 10.0 points
A dipole (electrically neutral) is placed in an
external field.
(a)
−
+
(b)
+
−
Explanation:
The force on a charge in the electric field is
~ = qE
~ and the torque is defined as
given by F
~
~
T = ~r × F .
X
~ = k ∆q r̂ and E
~ =
~i.
∆E
∆E
r2
Symmetry of the configuration will cause
some component of the electric field to be
zero.
Gauss’ law states
I
~ · dA
~ = Q.
ΦS = E
ǫ0
The electric dipole consists of two equal and
opposite charges separated by a distance. In
either situation (c) or (d), the electric field
is uniform and parallel everywhere. Thus,
the electric force on one charge is equal but
opposite to that on another so that the net
Version 062 – Exam 1 – markert – (55305)
force on the whole dipole is zero. By contrast,
electric fields are nonuniform for situations
both (a) and (b).
007 10.0 points
Consider a charged semicircular arc with radius 96 cm and total charge −37.6 µC distributed uniformly on the semicircle.
y
y
∆θ
−− A
II
I
−−
−
x
r
θ −−
III IV
−
−
x
−
O
−
−
−−
−−
−−
B
Find the magnitude of the electric field at
O . The value of the Coulomb constant is
8.98755 × 109 N · m2 /C2 .
1. 44929.4
2. 1546790.0
3. 233436.0
4. 2499300.0
5. 212034.0
6. 483712.0
7. 186688.0
8. 138471.0
9. 2056970.0
10. 76144.5
Correct answer: 2.33436 × 105 N/C.
Let : q = −37.6 µC = −3.76 × 10−5 C ,
r = 96 cm = 0.96 m , and
k = 8.98755 × 109 N · m2 /C2 .
y
∆θ
−− A
−−
−
r
θ −−
− E
−
−
O
−
−
−−
−−
−−
B
II
I
x
III
x
By symmetry of the semicircle, the ycomponent of the electric field at the center is Ey = 0 . We need consider only the
x-component of the electric field, so
ds = r dθ
q
q
∆q = λ ds = λ r dθ =
r dθ = dθ
πr
π
k |q|
k |∆q| cos θ
=
cos θ ∆θ ,
∆Ex =
2
r
π r2
and the magnitude of the electric field at the
center is
E = Ex =
Z
π/2
−π/2
=
k |q|
2 k |q|
cos θdθ =
2
πr
π r2
2 (8.98755 × 109 N · m2 /C2 )
π (0.96 m)2
× |(−3.76 × 10−5 C)|
= 2.33436 × 105 N/C
with direction along the negative x axis.
008 10.0 points
Two spheres, fastened to “pucks”, are riding on a frictionless airtrack. Sphere 1 is
charged with 3 nC, and sphere 2 is charged
with 15 nC. Both objects have the same mass.
1 nC is equal to 1 × 10−9 C.
As they repel,
1. sphere 2 accelerates 25 times as fast as
sphere 1.
Explanation:
y
5
2. they have the same magnitude of acceleration. correct
3. sphere 1 accelerates 5 times as fast as
sphere 2.
4. they do not accelerate at all, but rather
separate at constant velocity.
IV
5. sphere 2 accelerates 5 times as fast as
sphere 1.
6. sphere 1 accelerates 25 times as fast as
sphere 2.
Version 062 – Exam 1 – markert – (55305)
Explanation:
The force of repulsion exerted on each mass
is determined by
F =
1 Q1 Q2
= ma
4 π ǫ0 r 2
where r is the distance between the centers of
the two spheres. Since both spheres have the
same mass and are subject to the same force,
they have the same acceleration.
009 10.0 points
Two electrostatic point charges of +55.0 µC
and +58.0 µC exert a repulsive force on each
other of 195 N.
What is the distance between the two
charges? The value of the Coulomb constant
is 8.98755 × 109 N · m2 /C2 .
1. 0.330642
2. 0.441283
3. 0.411132
4. 0.39497
5. 0.383441
6. 0.379211
7. 0.422853
8. 0.400906
9. 0.357772
10. 0.432479
Correct answer: 0.383441 m.
Explanation:
010 10.0 points
A cubic box of side a, oriented as shown, contains an unknown charge. The vertically directed electric field has a uniform magnitude
E at the top surface and 2 E at the bottom
surface.
E
a
2E
How much charge Q is inside the box?
1. insufficient information
2. Qencl = 3 ǫ0 E a2
3. Qencl = 6 ǫ0 E a2
q1
q2
Fe
kC
= 55.0 µC ,
= 58.0 µC ,
= 195 N , and
= 8.98755 × 109 N · m2 /C2 .
q1 q2
Fe = kC 2
r r
kC q1 q2
r=
Fe
q
= 8.98755 × 109 N · m2 /C2
r
(5.5 × 10−5 C) (5.8 × 10−5 C)
×
195 N
= 0.383441 m .
E
ǫ0 a2
E
=2
ǫ0 a2
1
= ǫ0 E a2
2
4. Qencl = 3
5. Qencl
6. Qencl
Let :
6
7. Qencl = 2 ǫ0 E a2
8. Qencl = 0
9. Qencl = ǫ0 E a2 correct
10. Qencl =
E
ǫ0 a2
Explanation:
Electric flux through a surface S is, by convention, positive for electric field lines going
out of the surface S and negative for lines going in. No flux passes through the vertical
sides.
Version 062 – Exam 1 – markert – (55305)
The top receives Φtop = −E a2 (inward is
negative) and the bottom Φbottom = 2 E a2 ,
so the total electric flux is
ΦE = −E a2 + 2 E a2 = E a2 .
Using Gauss’s Law, the charge inside the
box is
Qencl = ǫ0 ΦE = ǫ0 E a2 .
7
y
++++
M
x
−−−−
Identify the configuration(s) where both xand y-components are non-zero.
1. Configuration M only correct
10.0 points
Consider non-zero components for total
electric field vectors in the following configurations.
y
+
+
+
+
S
−
++
++
++
9. Configuration L only
10. Configurations G and P only
X
k∆q
r̂
and
E
=
∆E .
r2
Symmetry of the configuration will cause
some component of the electric field to be
zero.
Configuration S is anti-symmetric about
the y-axis (opposite sign of charges):
y
+
+
+
S
+
∆E =
++
++
++
7. Configuration S only
Explanation:
y
x
−−−−−−
y
+++++
L
6. Configurations S, G and P only
+
+
+
+
+
+++++
−
−
−
− x
−
−
−
x
++++
P
5. Configurations S and G only
8. Configurations S, M and P only
++++
G
4. Configurations G, L and P only
−
−
y
3. Configuration G only
−
−
x
2. Configurations S and P only
−
011
x
so the electric field has no y-component.
Configuration G is symmetric by a rotation
of 180◦ :
Version 062 – Exam 1 – markert – (55305)
vanish.
y
++++
G
x
++++
so the electric fields generated by these two
pieces have opposite directions and the total
field is zero.
Configuration M is anti-symmetric by rotation of a 180◦ :
y
++++
M
x
−−−−
so the total field has non-zero components in
both x- and y-directions, just like the field
generated by just one piece of charge.
Configuration L is symmetric about the xaxis:
y
+++++
+
−
+
−
+
−
L
+
− x
+
−
+++++
++
++
++
++
++
++
so the y component of the total field must
vanish.
Configuration P is symmetric about the yaxis:
y
P
8
x
−−−−−−
so the x component of the total field must
012 10.0 points
How much positive charge is in 1.3 kg of oxygen?
The atomic weight (15.9994 g) of oxygen
contains Avogadro’s number of atoms, with
each atom having 8 protons and 8 electrons.
The elemental charge is 1.602 × 10−19 C and
Avogadro’s number is 6.023 × 1023 .
1. 57377500.0
2. 33754800.0
3. 72319300.0
4. 33363000.0
5. 62719900.0
6. 48246000.0
7. 44625100.0
8. 67509600.0
9. 27425300.0
10. 55673500.0
Correct answer: 6.27199 × 107 C.
Explanation:
Let : e = 1.602 × 10−19 C ,
m = 1.3 kg = 1300 g ,
NA = 6.023 × 1023 ,
AWC = 15.9994 g , and
npr = 8 protons/atom .
m npr NA
m
npr =
mA
AW
(1300 g) (6.023 × 1023 atoms)
=
15.9994 g
× (8 protons/atom)
= 3.9151 × 1026 protons and
Npr =
Q = e NA
= (1.602 × 10−19 C) (3.9151 × 1026 )
= 6.27199 × 107 C .
013 10.0 points
A 10.8 g piece of Styrofoam carries a net
Version 062 – Exam 1 – markert – (55305)
charge of −0.5 µC and floats above the center
of a very large horizontal sheet of plastic that
has a uniform charge density on its surface.
What is the charge per unit area on the
plastic sheet? The acceleration due to gravity
is 9.8 m/s2 and the permittivity of free space
is 8.85419 × 10−12 C2 /N/m2 .
1. -1.77013
2. -3.74851
3. -2.22134
4. -4.16501
5. -1.70071
6. -2.32546
7. -3.60968
8. -2.15192
9. -1.42305
10. -1.666
Correct answer: −3.74851 µC/m2 .
Explanation:
9
Two charged particles of equal magnitude
(+Q and +Q) are fixed at opposite corners of
a square that lies in a plane (see figure below).
A test charge −q is placed at a third corner.
+Q
−q
+Q
What is the direction of the force on the
test charge due to the two other charges?
1.
2.
3.
Let :
m = 10.8 g = 0.0108 kg ,
q = −0.5 µC = −5 × 10−7 C ,
g = 9.8 m/s2 , and
ǫ0 = 8.85419 × 10−12 C2 /N/m2 .
σ
The field E =
due to a nonconduct2 ǫ0
ing infinite sheet of charge is the same as that
very close to any plane uniform charge distribution, where σ is the surface charge density
(charge per unit area) of the plastic sheet.
The floating styrofoam must be in equilibrium, so the electric force must cancel the
force of gravity and
Fg = q E
σ
mg = q
2 ǫ0
mg
σ = 2 ǫ0
q
= 2 (8.85419 × 10−12 C2 /N/m2 )
(0.0108 kg) (9.8 m/s2 ) 106 µC
×
·
−5 × 10−7 C
1C
4.
correct
5.
6.
7.
8.
Explanation:
The force between charges of the same sign
is repulsive and the force between charges
with opposite signs is attractive.
+Q
= −3.74851 µC/m2 .
014
10.0 points
−q
+Q
Version 062 – Exam 1 – markert – (55305)
10
The resultant force is the sum of the two
vectors in the figure.
to the change in the kinetic energy of the
electrons, so
015 10.0 points
The electron gun in a television tube
is used to accelerate electrons with mass
9.109 × 10−31 kg from rest to 1 × 107 m/s
within a distance of 5 cm.
What electric field is required? The fundamental charge is 1.602 × 10−19 C .
1. 25845.5
2. 71075.2
3. 10934.6
4. 44889.6
5. 5686.02
6. 36553.0
7. 6048.95
8. 159919.0
9. 42118.6
10. 103382.0
Wf ield = Kf − Ki
1
qe E d = me vf2
2
vf2 me
E=
2 d qe
Correct answer: 5686.02 N/C.
Explanation:
Let : me = 9.109 × 10−31 kg ,
qe = −1.602 × 10−19 C ,
vf = 1 × 107 m/s , and
d = 5 cm .
The magnitude of the force is
F = qE = ma
qE
.
a=
m
Since vi = 0, the final velocity is
vf2 = vi2 + 2 a d = 2 a d =
E=
2 d |qe| E
me
vf2 me
2 d |qe|
(1 × 107 m/s)2 (9.109 × 10−31 kg)
=
2(0.05 m)| − 1.602 × 10−19 C|
= 5686.02 N/C .
Alternate Solution: Using conservation
of energy, the work done by the field is equal
016 10.0 points
A solid nonconducting sphere of radius R has
a charge Q uniformly distributed throughout
its volume. A Gaussian surface of radius r
with r < R is used to calculate the magnitude
of the electric field E at a distance r from the
center of the sphere.
Which equation results from a correct application of Gauss’s law for this situation?
Q r3
ǫ0 R3
Q 3 r3
2. E (4 π r 2 ) =
ǫ0 4 π R
Q
3. E (4 π r 2 ) =
ǫ0
1. E (4 π R2 ) =
4. E (4 π r 2 ) = 0
Q r3
correct
ǫ0 R3
Q
6. E (4 π R2 ) =
ǫ0
Q r3
4 3
7. E
πr =
3
ǫ0 R3
Q 3 r3
8. E (4 π R2 ) =
ǫ 4πR
0
Q r
4 3
πr =
9. E
3
ǫ0 R2
4 3
Q r2
10. E
πr =
3
ǫ0 R2
5. E (4 π r 2 ) =
Explanation:
Applying Gauss’s law,
I
S
~ · ~n dA =
E
I
S
E dA =
Qinside
.
ǫ0
Version 062 – Exam 1 – markert – (55305)
4π 3
V =
r ∝ r 3 ; because charge Q is uni3
formly distributed, the enclosed charges are
related by
Qsurf ace
r3
= 3
Qsphere
R
r 3 Qsphere
,
Qsurf ace =
R3
so Gauss’s law gives
E (4 π r 2 ) =
Q r3
.
ǫ0 R3
017 10.0 points
A charge of 5.4 µC is at the geometric center
of a cube.
What is the electric flux through one of
the faces? The permittivity of a vacuum is
8.85419 × 10−12 C2 /N · m2 .
1. 235294.0
2. 222117.0
3. 269176.0
4. 244705.0
5. 146823.0
6. 141176.0
7. 101647.0
8. 248470.0
9. 276705.0
10. 116706.0
Φ=
1
Φtot = 1.01647 × 105 N · m2 /C .
6
018 10.0 points
An electric field of magnitude 27000 N/C and
directed upward perpendicular to the Earth’s
surface exists on a day when a thunderstorm
is brewing. A truck that can be approximated
as a rectangle 6.6 m by 2 m is traveling along a
road that is inclined 8◦ relative to the ground.
Determine the electric flux through the bottom of the truck.
1. 352932.0
2. 217106.0
3. 110502.0
4. 730534.0
5. 281393.0
6. 134337.0
7. 196915.0
8. 208435.0
9. 430470.0
10. 137241.0
Correct answer: 3.52932 × 105 N · m2 /C.
Explanation:
Let :
E = 27000 N/C ,
ℓ = 6.6 m ,
w = 2 m , and
θ = 8◦ .
Correct answer: 1.01647 × 105 N · m2 /C.
Explanation:
By Gauss’ law,
q = 5.4 µC = 5.4 × 10−6 C and
ǫ0 = 8.85419 × 10−12 C2 /N · m2 .
Let :
~ · dA
~ = q .
E
ǫ0
The total flux through the cube is given by
Φ=
I
q
C
=
−12
ǫ0
8.85419 × 10
C2 /N · m2
= 6.09881 × 105 N · m2 /C ,
Φtot =
11
5.4 × 10−6
so the flux through one side of the cube is
~ ·A
~.
Φ=E
The flux through the bottom of the car is
Φ = E A cos θ = E ℓ w cos θ
= (27000 N/C)(6.6 m)(2 m) cos 8◦
= 3.52932 × 105 N · m2 /C .
019 10.0 points
Imagine a charge in the center of a conducting,
hollow sphere. There is no net charge on the
Version 062 – Exam 1 – markert – (55305)
sphere, and the sphere is not connected to
ground.
q
−
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Y
X
4) Then the ground wire is disconnected.
−
What will happen if the charge is moved a
little away from the center?
1. All of these can happen, depending on the
size of the charge.
Y
X
5) While X and Y remain in touch, the ball
carring the negative charge is removed.
2. The charge will move away from the center. correct
Y
X
3. The charge will remain stationary.
4. The charge will return to the center.
6) Then ball X and Y are separated.
5. There is not enough information to tell.
Explanation:
There will be an image-charge attracting it
toward any metal surface. It will move toward
the closest metallic surface because the closest
image-charge attraction will be stronger.
Any charge (free to move) will move toward
the closest conductor it can find.
020 10.0 points
1) Two uncharged metal balls, X and Y, each
stand on a glass rod and are touching.
Y
X
2) A third ball carrying a negative charge, is
brought near the first two.
−
Y
Y
X
After these procedures, what are the signs
of the charge qX on X and qY on Y?
1.
qX is positive and qY is positive correct
2.
qX is negative and qY is neutral
3.
qX is neutral and qY is negative
4.
qX is neutral and qY is positive
5.
qX is negative and qY is positive
6.
qX is neutral and qY is neutral
7.
qX is positive and qY is negative
8.
qX is positive and qY is neutral
9.
qX is negative and qY is negative
X
3) While the positions of these balls are fixed,
ball X is connected to ground.
Explanation:
When the ball with negative charge is
brought nearby, the free charges inside X and
Version 062 – Exam 1 – markert – (55305)
Y rearrange themselves. The positive charges
are attracted and go to the left (i.e., move
to Y), leaving negative charges on the right
hand side of the system X Y, i.e., in X.
When we ground X, electrons flow from the
ground to X (making it neutral), whereas the
positive charges in Y are still held enthralled
by the negative charge on the third ball. We
break the ground.
Now we remove the third ball with negative
charge. The charge on Y is redistributed in
the system X Y; i.e., they share the positive
charge (equally if identical).
Finally we separate X and Y. The signs
of the charge on X and that on Y are both
positive.
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