Download Physics 216 Sample Exam 1 Solutions

Physics 216
Sample Exam 1 Solutions
Print your name neatly. If you forget to write your name, or if the grader can’t read your
writing, you can lose up to 100 points. Answer all the questions that you can.
This exam will consist of 11 multiple-choice problems. You may not use calculators or
other electronic devices on this exam. The use of such a device will be regarded as an
attempt to cheat, and will be pursued accordingly. All diagrams and figures on this exam
are rough sketches: they are not generally drawn to scale.
No partial credit will be given for these problems. However, you can miss one of the 11
problems without penalty. Your grade will be based on your best 10 problems. You will
not receive extra credit for getting all 11 right.
Your grade on the exam will be based entirely on the answers that you circle on this
sheet. If you have no answer or a wrong answer there, the grader will not look at the
page with the problem to see if the right answer appears there. Illegible or ambiguous
answers will be graded as wrong. You are responsible for copying your answers clearly,
correctly, and in the right place.
Although there is no partial credit on this exam, you must show your work in the space
provided on the exam. There is additional scratch paper at the end of the exam: do
not use it unless you have filled all the scratch space provided on the page with the
problem. If you answer a difficult problem without doing any written work, the grader
will assume that you got the answer by guessing or by copying from someone else, and
will not give you credit for the problem even though you’ve indicated the correct solution
on the answer sheet.
Circle your answers here. Do not detach this sheet from the test.
1.
a
b
a b
2. ○
c d
○
e
a b
5. ○
c
d
e
9.
a
b
c d
○
e
c
d
e
6.
a
b
c d
○
e
10.
a
b
c d
○
e
11.
a
b c
○
3.
a
b c
○
d
e
7.
a
b c
○
d
e
4.
a
b
d e
○
8.
a
b c
○
d
e
c
Constant values. For ease of calculation on this exam, use the values
Ke =
1
= 9 × 109 N m2 /C2
4π0
and g = 10 m/s2
d
e
physics 216 sample exam 1 solns
Copyright ©Wayne Hacker 2012. All rights reserved. 2
Electric charge and Coulomb’s law
Problem 1. Three small objects are given nonzero electrical charges. Object 1 and
object 2 attract one another. Object 1 and object 3 repel one another. What can you
conclude about the signs of the charges on the three objects?
(a) All three charges have the same sign
(b) Charges 1 and 2 have the same sign; charge 3, the opposite sign
*(c) Charges 1 and 3 have the same sign; charge 2, the opposite sign
(d) Charges 2 and 3 have the same sign; charge 1, the opposite sign
(e) Not enough information
Solution: Since objects 1 and 3 repel one another, they have the same sign; since objects
1 and 2 attract one another, they have opposite signs.
Problem 2. In your physics lab, you conduct an experiment with three identical small
conducting spheres. The three spheres initially have no electrical charge. Two of the
spheres are brought into contact with one another and given a positive charge, then
placed at a distance of x from one another. At this distance, the force exerted by each
sphere on the other has magnitude F1 .
The three spheres are then brought together so that all three are in contact. One of
the spheres is carried out of the room; the other two are again placed at a distance of x
from one another. The force exerted by each sphere on the other is now F2 . What is the
relationship between F1 and F2 ?
4
2
*(a) F2 = F1
(b) F2 = F1
9
3
r
3
3
(c)
F2 =
F1
(d) F2 = F1
2
2
(e)
None of these
Solution: Let Q be the total charge initially placed on the two spheres. Since the spheres
are identical, conductive, and in contact with one another, they divide the charge equally;
so each has charge Q/2. After all three spheres have been brought together, the charge
on each one is Q/3. Then
1 (Q/2)2
1 (Q/3)2
and
F
=
2
4πε0 x2
4πε0 x2
F2
(Q/3)2
4
4
⇒
=
=
⇒ F2 = F1
2
F1
(Q/2)
9
9
F1 =
physics 216 sample exam 1 solns
Copyright ©Wayne Hacker 2012. All rights reserved. 3
Problem 3. An insulating string is fastened to an insulating
tabletop and stretched vertically upward. A bead with a positive
charge of Q1 = 2×10−5 C is fastened to the bottom of the string,
at y = 0. A second bead, with a mass of m2 = 0.09 kg and a
positive charge of Q2 = 3 × 10−5 C, is free to slide on the string,
as shown at right. At what height y2 will the second bead be in
static equilibrium?
r
√
3
(a) y2 =
m
*(b) y2 = 6 m
2
(c)
y2 = 5.4 m
(e)
None of these
(d)
Q = 3 × 10−5 C
u 2
6 m2
= 0.09 kg
y2
Q1 = 2 × 10−5 C
?u
y2 = 6 m
Solution: At y2 , the Coulomb force pushing the bead upward will equal the force of
gravity pulling it downward.
1/2
1 Q1 Q2
⇒ y2 =
4πε0 m2 g
1/2
√
(9 × 109 N m2 /C2 )(2 × 10−5 C)(3 × 10−5 C)
=
=
6m
(0.09 kg)(10 m/s2 )
1 Q1 Q2
= m2 g
4πε0 y22
Problem 4. Two identical small metal spheres are both given negative
charges, one of Q and the other of 2Q. The two spheres are then dropped
into a vertical non-conductive tube, as shown at right. When they reach
equilibrium, one sphere is at the bottom of the tube, at y = 0; the other
is at height y1 .
The spheres are pushed into contact with one another, then released.
When they again reach equilibrium, one sphere is again at the bottom of
the tube; the other is at height y2 . What is the relationship between y1
and y2 ?
r
r
2
3
(a) y2 =
y1
(b) y2 =
y1
3
2
2
3
(c) y2 = √ y1
*(d) y2 = √ y1
5
2 2
(e) None of these
w
y
6
w
Solution: At equilibrium, the Coulomb force pushing the upper sphere upward equals
the force of gravity pulling it downward.
1 q 1 q2
= mg
4πε0 y 2
Between the two situations, the values of q1 , q2 , and y change; everything else remains
constant. Initially, q1 = Q, q2 = 2Q, and y = y1 ; after the two spheres have been pushed
together, their charges are equal, so q1 = q2 = 3Q/2 and y = y2 .
q1 q2
(Q)(2Q)
(3Q/2)(3Q/2)
=
4πε
mg
=
=
0
y2
y12
y22
⇒
y22 =
9 2
y
8 1
⇒
3
y2 = √ y1
2 2
physics 216 sample exam 1 solns
Copyright ©Wayne Hacker 2012. All rights reserved. 4
Problem 5. Three small charged objects are placed in a line, as shown below. The
distance between q1 and q2 is twice the distance between objects q2 and q3 . If the net
electrical force on q3 is zero, what is the relationship between q1 and q2 ?
q1
u
q2
u
-
2x
*(a) q2 = −
(b) q2 = −
(c)
q1
9
√
q 2 = − 3 q1
(d) q2 = −3q1
(e)
None of these
q3
x
u
-
q1
3
Solution: The net electrical force on q3 is zero; so
1
q1 q 3
q 2 q3
+ 2 = 0 ⇒ q1 + 9q2 = 0
4πε0 (3x)2
x
⇒
Problem 6. Electrical charges are placed at the four corners
of a square, as shown at right: charges of q1 at two diagonally opposite corners, and charges of q2 at the other two. If
the net electrical force on the q1 charges is zero, what is the
relationship between q1 and q2 ?
√
√
(a) q2 = −2 2 q1
(b) q2 = − 2 q1
q1
*(c) q2 = − √
2 2
(e) None of these
q2 = −
q1
9
q2 u
uq1
q1 u
uq2
q1
(d) q2 = − √
2
Solution: By symmetry, we need only consider either the horizontal or vertical components of the forces acting on one of the q1 charges. We will consider the horizontal forces
acting on the q1 charge at upper right (UR). We can ignore the q2 charge at lower right,
since it only exerts a vertical force on UR. Then if the side of the square is s, the net
horizontal force on UR is
1
q1 q2
q1 q1
◦
+ √
cos 45 = 0
4πε0 s2
( 2 s)2
q1 1
q1
⇒ q2 + · √ = 0 ⇒ q2 = − √
2
2
2 2
physics 216 sample exam 1 solns
Copyright ©Wayne Hacker 2012. All rights reserved. 5
Electric fields
Problem 7. Two positive point charges, one of Q and one of 2Q, are
fixed in place along the y-axis as shown at right. Which of the following
statements is true about the electric field generated by the charges, both
on and off the y-axis?
(a)
*(b)
(c)
(d)
(e)
There are no points at which the electric field is zero
There is exactly one point at which the electric field is zero
There are exactly two points at which the electric field is zero
There are infinitely many points at which the electric field is
zero
None of these
6
2Q u
Q u
?
Solution: The electric field cannot be zero at any point that’s not on the y-axis. If,
for example, a point is to the left of the axis, then the electric fields generated by both
charges have components pointing leftward; so the electric field has a nonzero leftward
component at that point.
The electric field cannot be zero at any point on the y-axis above the upper charge, or
below the lower charge. At such a point, the electric fields from the two charges would
both point in the same direction, so the net electric field there would be nonzero.
There is only one point on the axis between the two charges where the electric field is
zero. At this point, the field pushing downward from the upper charge equals the field
pushing upward from the lower one. Above that point, the field from the upper charge
is stronger and the field from the lower charge is weaker; below that point, the field from
the lower charge is stronger than that from the upper charge.
physics 216 sample exam 1 solns
Copyright ©Wayne Hacker 2012. All rights reserved. 6
Problem 8. Two point charges, one of Q and one of −2Q, are
fixed in place along the y-axis as shown at right. The charge of Q
is at y = 0; the charge of 2Q is at y = 1. At what point along the
y-axis is the electric field equal to zero?
√
√
*(b) y = −1 − 2
(a) y = 1 − 2
√
√
−2 − 3
3−2
(c) y =
(d) y =
2
2
(e)
None of these
6
−2Q u y = 1
Q u y=0
?
Solution: We need to find a point at which the electric fields from the upper and
lower charge are equal in magnitude, and point in opposite directions. This point can’t
be between y = 0 and y = 1, because in that region both fields point in the positive
y-direction. It can’t be at any y > 1, because although the fields point in opposite
directions, such a point would be closer to the larger upper charge, so the field from that
charge would be stronger than the field from the smaller and more distant lower charge.
Hence we need to look in the region y < 0: there, the fields from the two charges point
in opposite directions, and there will be a point where the magnitude of the field from
the smaller but closer lower charge equals the magnitude of the field from the larger but
more distant upper charge.
We need to look for a point y where the magnitudes of the two electric fields are equal:
1 Q
2Q
1
=
4πε0 (y − 1)2
4πε0 y 2
⇒
2y 2 = (y − 1)2
⇒
y 2 + 2y − 1 = 0
√
√
This gives us solutions of y = −1 ± 2. One of these solutions, y = −1 + 2, is spurious:
it lies in the region 0 < y < 1. At this point, the electric fields produced by the two
√
charges are equal; but both point in the same direction. Only the solution y = −1 − 2
lies in the region y < 0, so is valid.
physics 216 sample exam 1 solns
Copyright ©Wayne Hacker 2012. All rights reserved. 7
Problem 9. Aboard a space station, where there is zero gravity, a metal sphere with a
mass of 60 g is given a charge of 4 × 10−4 C, and placed in a uniform electric field with
magnitude 300 N/C. The sphere is held in place, then released. How far will it travel in
the first 3 seconds after its release?
(a) 4.5 m
(b) 6 m
*(c) 9 m
(d) 12 m
(e) None of these
Solution: We know the sphere’s mass and charge, the strength of the electric field, and
the time that it moves. We want to calculate the distance that it travels in that time. If
we knew the acceleration, we could calculate the distance from the fundamental kinematic
equations. Since we know the mass, if we knew the force then we could calculate the
acceleration from Newton’s second law. We can calculate the force from the charge and
the electric field, F = qE.
1
1
x = x0 + v0 t + at2 = at2
2
2
qE
F
=
F = ma ⇒ a =
m
m
1 qE 2 (4 × 10−4 C)(300 N/C)(3 s)2
t =
=9m
x=
2m
2(0.06 kg)
⇒
Problem 10. A massless insulating rod with length l is
pivoted at its lower end, where it makes an angle of θ to the
horizontal. At the upper end of the rod is a small metal sphere
with mass ms , which carries a positive charge Q. A string is
fastened halfway along the rod and pulled horizontally. The
entire system is within a uniform electric field with magnitude
E, directed horizontally as shown at right. If the system is in
static equilibrium, what is the tension T in the string?
(a)
T = QEl sin θ + ms gl cos θ
*(c) T = 2QE +
(e)
2ms g
tan θ
(b)
E
w Q
T
?ms g
l
θ
ms
T = QEl cos θ + ms gl sin θ
(d) T =
2QEl 2ms gl
+
cos θ
sin θ
None of these
Solution: There are three torques about the lower end of the rod: τE , produced by the
Coulomb force on the sphere; τw , produced by the weight of the sphere; and τT , produced
by the tension in the string. When the system is in static equilibrium, the total torque
τ will be zero.
τ = τE + τw + τT = QEl sin θ + ms gl cos θ − T (l/2) sin θ = 0
⇒ T sin θ = 2(QE sin θ + ms g cos θ)
2ms g
⇒ T = 2QE +
tan θ
physics 216 sample exam 1 solns
Copyright ©Wayne Hacker 2012. All rights reserved. 8
Problem 11. A thin wire is bent into a ring with radius R
and given a nonzero electrical charge. The ring is placed in the
xy-plane, with its center at the origin. At what point along the
positive z-axis is the electric field at a maximum?
R
R
*(b) z = √
(a) z =
2
2
√
(d) z = 2R
(c) z = 2 R
(e)
6
z
θ
?
R -
None of these
~ as a function of z for
Solution: We will find the formula for E(z), the magnitude of E
points on the z-axis. We will then find the point of maximum E by differentiating and
setting dE/dz = 0.
By symmetry, the electric field at a point on the z-axis must point in the positive or
negative z-direction. Hence we need only consider the z-components of the electric field:
the x- and y-components from opposite sides of the ring cancel one another out.
The diagram above shows the ring viewed from the side, i.e. with our eyes in the xyplane. We use θ to indicate the angle between the plane and a vector that runs from a
point on the ring to the point (0, 0, z).
If φ is the measure of angle in the xy-plane, then the incremental charge on an infinitesimally short arc of the wire is
dq =
Qdφ
Q
(Rdφ) =
2πR
2π
The magnitude of the electric field produced by that small segment of charge is
1 Qdφ
1
2
4πε0 2π z + R2
and the z-component of the electric field produced by the small segment is
dE =
1 Qdφ
1
1 Q
z
sin θ =
dφ
2
2
2
4πε0 2π z + R
4πε0 2π (z + R2 )3/2
Then we can find E by integrating dE around the circle φ = 0 to 2π.
Z
Z 2π
1 Q
z
Q
z
E = dE =
dφ =
2
2
3/2
2
4πε0 2π (z + R )
4πε0 (z + R2 )3/2
φ=0
To find the point of maximum E, we differentiate w.r.t. z and set the derivative equal
to zero.
dE
Q d 2
Q
3 2
2 −3/2
2
2 −3/2
2 −5/2
=
z(z + R )
=
(z + R )
− z(z + R )
(2z)
dz
4πε0 dz
4πε0
2
Q (z 2 + R2 ) − 3z 2
Q R2 − 2z 2
=
=
=0
4πε0 (z 2 + R2 )5/2
4πε0 (z 2 + R2 )5/2
R
⇒ 2z 2 = R2 ⇒ z = ± √
2