Download Physics 216 Sample Exam 2 Solutions

Physics 216
Sample Exam 2 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.
Constant values. For ease of calculation on this exam, use the values
Ke =
1
= 9 × 109 N m2 /C2
4πε0
and g = 10 m/s2
Circle your answers here. Do not detach this sheet from the test.
1.
a
b
c d
○
e
5.
a
b
c d
○
e
9.
2.
a
b
c d
○
e
6.
a
b
c d
○
e
a b
10. ○
3.
a
b
c
d e
○
7.
a
b c
○
d
e
11.
4.
a
b
c
d e
○
a b
8. ○
d
e
c
a
a
c d
○
e
c
d
e
b c
○
d
e
b
physics 216 sample exam 2 solns Copyright ©Wayne Hacker 2012. All rights reserved. 2
Gauss’s law
Problem 1. An imaginary disc with radius R is
immersed in a vector field ~v that points in the same
direction everywhere, perpendicular to the surface of
the disc. The magnitude of the field is v = kr, where
k > 0 is a constant and r is the distance from the center
of the disc. What is the magnitude of the flux Φ of the
field through the disc?
(a)
*(c)
(e)
4πkR3
3
2πkR3
Φ=
3
Φ=
(b)
(d)
R
r
dr- kR2
2
πkR2
Φ=
2
Φ=
None of these
Solution: The vector field ~v is perpendicular to the surface of the disc, so we don’t need
to worry about finding the normal component. We only need to integrate vdA over the
area of the disc.
Consider a ring in the disc with radius r and infinitesimal thickness dr, as shown in the
diagram. The increment of flux through the ring is
dΦ = vdA = (kr)(2πrdr)
To find the total flux through the disc, we integrate this from r = 0 to R.
Z
Z R
2πkR3
2πk 3 R
2
r r=0 =
Φ = dΦ = 2πk
r dr =
3
3
r=0
physics 216 sample exam 2 solns Copyright ©Wayne Hacker 2012. All rights reserved. 3
Problem 2. An imaginary cube is immersed in an electric field; there is no charge
enclosed inside the cube. The faces of the cube are numbered 1 through 6; the flux of
the electric field through face i is designated Φi . Which of the following statements are
true?
(i) If i and j are opposite faces of the cube, then Φi = −Φj .
(ii) If i and j are adjacent faces of the cube, then Φi = ±Φj .
(a) Statements (i) and (ii) are both true
(b) Statement (i) is true; statement (ii) is false
(c) Statement (i) is false; statement (ii) is true
*(d) Statements (i) and (ii) are both false
Solution: The best way to disprove statement (i) is to construct an electric field that contradicts it. Consider the situa~
d~s 6 E
r
tion shown at right. The imaginary cube (shown in dotted lines)
is viewed from the side, so we can’t see its third dimension. A
positive point charge Q is outside of the cube, in the plane of
~ is produced by the
the cube’s lower face. The electric field E
r r
~
charge Q. Consider the fluxes Φl and Φu through the lower and
Q
d~s ? E
the upper faces.
~ comes solely from the single positive point charge Q, it points straight away
Since E
~ is
from Q at all points. This means that at every point on the lower face of the cube, E
in the plane of the face; so it is perpendicular to the normal vector d~s to the face. Hence
~ · d~s = 0; so the total flux Φl = 0.
at every point on the lower face, E
~ makes an acute angle with d~s; so E
~ · d~s > 0. Hence
At every point on the upper face, E
for the upper face, Φu > 0.
Since Φu 6= −Φl , statement (i) is false.
We can use the same example to disprove statement (ii). Consider a face other than the
upper or lower face. Then that side much be adjacent to both the upper and lower faces.
Let Φi be the flux through that face. If statement (ii) were true, than Φi = ±Φl = 0,
and Φi = ±Φu 6= 0. Since this is a contradiction, statement (ii) must be false.
physics 216 sample exam 2 solns Copyright ©Wayne Hacker 2012. All rights reserved. 4
Problem 3. An infinite line of positive charge has linear charge density λ. A particle
with positive charge Q is held in place at a distance of R from the line of charge, then
released. How much work does the electric field do in moving the particle to a distance
of 2R from the line of charge?
λQ
λQR
(b) W =
(a) W =
πε0
4πR2 ε0
3λQR2
λQ
(c) W =
*(d) W =
ln 2
πε0
2πε0
(e)
None of these
Solution: To solve this, we need to know the force F (r) on the charge Q as a function
of its distance r from the line of charge. If we know that, then we can calculate the work
done by the force as
Z
2R
F (r)dr
W =
r=R
To calculate F (r), we need E(r); then F = EQ. To calculate E(r), consider a cylindrical
Gaussian surface with the line of charge running along its center, radius r, and length x.
~ points straight outward from the wire. Hence by Gauss’s law
At every point, E
Φε0 = 2πrxEε0 = λx
⇒
E=
λ
2πrε0
Substituting F = EQ and integrating from R to 2R yields
2R
Z 2R
dr
λQ
λQ
λQ
W =
=
ln r
=
ln 2
2πε0 r=R r
2πε0
2πε0
r=R
physics 216 sample exam 2 solns Copyright ©Wayne Hacker 2012. All rights reserved. 5
Problem 4. An infinite plane of positive charge has charge density σ. A particle with
positive charge qp and mass mp is held at a distance of R from the plane, then released
at time t = 0. What is the particle’s speed v as a function of time?
σqp t
2ε0
σqp t2
v=
ε0 (2R + t2 )
(a) v =
(c)
(e)
σqp t2
4ε0 mp
σqp t
*(d) v =
2ε0 mp
(b)
v =R+
None of these
Solution: We begin by using Gauss’s law to calculate the electric field strength E at
a distance R from the plane of charge. Consider a cylinder with its ends parallel to the
plane, each end with area A. By symmetry, the electric field is perpendicular to the
plane, so parallel to the sides of the cylinder; so there is no flux through the sides, and
the total flux through the cylinder is that through the two ends. The charge enclosed by
the cylinder is σA; so by Gauss’s law
Φε0 = 2AEε0 = σA
⇒
E=
σ
2ε0
Notice that this does not depend on R. The force on the particle is
F = Eqp =
σqp
2ε0
By Newton’s second law, the acceleration of the particle is
a=
F
σqp
=
mp
2ε0 mp
Since the particle is initially at rest, by the fundamental kinematic equations
v(t) = v0 + at = at =
σqp t
2ε0 mp
physics 216 sample exam 2 solns Copyright ©Wayne Hacker 2012. All rights reserved. 6
Problem 5. A small sphere with a mass of ms and a negative
charge of Q hangs from a thin insulating thread with length l.
The thread is attached to the ceiling at a horizontal distance of
x from a positively charged vertical wall, with a uniform charge
density of σ. What is the angle θ that the thread makes with
the vertical?
σQ
−1 4πε0 x − σQ
−1
(b) θ = sin
(a) θ = sin
2ε0 (l2 − x2 )
4πε0 l
σQ
−1
−1 2ε0 x − σQ
*(d) θ = tan
(c) θ = tan
2ε0 ms gl
2ε0 ms g
(e)
x
-
+
θ
+
+
l +
u
+
Q, ms
+
+
None of these
Solution: Sketch a force diagram for the sphere, as shown at
right. Three forces act on it: weight, pulling downward with a
force of ms g; the Coulomb force, pulling leftward with a force
of EQ; and the tension T in the string, pulling upward and to
the right. Since the sphere is in static equilibrium, the net force
on it is zero; so T sin θ = EQ and T cos θ = ms g. If we solve
this for θ, we get
EQ
sin θ
=
= tan θ
cos θ
ms g
EQ
θ
u
T
?ms g
We still need to find E. Consider a cylindrical Gaussian surface with its ends on either
side of and parallel to the charged wall and its sides perpendicular thereto; each end has
area A. By symmetry, the electric field from the charged wall is perpendicular to the
ends of the cylinder, and parallel to its sides. Hence the flux through the cylinder is
entirely through the ends. The total charge enclosed by the cylinder is σA; so by Gauss’s
law,
σ
Φε0 = 2AEε0 = σA ⇒ E =
2ε0
Substituting this into the preceding result, we get
σQ
σQ
−1
tan θ =
⇒ θ = tan
2ε0 ms g
2ε0 ms g
physics 216 sample exam 2 solns Copyright ©Wayne Hacker 2012. All rights reserved. 7
Electric potential energy and potential
Problem 6. (Similarity Problem) When two positive point charges are separated by
a distance of x1 , their electrical potential energy is U1 . When they are separated by a
distance of x2 , their electrical potential energy is U2 = U1 /2. What is the relationship
between x1 and x2 ?
√
(a) x2 = x1 /2
(b) x2 = x1 / 2
*(c) x2 = 2x1
(d) x2 = 4x1
(e) None of these
Solution: This is a similarity problem; the governing equation is the one for the electrical
potential energy between two charged particles:
U=
kqa qb
r
In the two experiments, the values of k, qa , and qb remain constant; the values of r and
U change. Separating the changing variables from the unchanging ones gives:
kqa qb = U r = U2 x2 = U1 x1
U1 x2
U2 =U1 /2
−−−−−→
= U 1 x1
2
2U1 x1
×2/U1
= 2x1
−−−→ x2 =
U1
Problem 7. Two metal spheres, both with radius R, are both given positive electrical
charges Q. The spheres are initially held with their centers at distance x from one
another; they are then pushed together until the spheres touch. How much work is done
in pushing the spheres together?
Q2
(a) W =
4πε0
Q2
*(c) W =
4πε0
(e)
x2 − R 2
·
R 2 x2
x − 2R
·
2Rx
Q2 x2 − 4R2
(b) W =
·
4πε0
4Rx
(d) W = ∞
None of these
Solution: To calculate work, we need to multiply force times distance. Since the force
varies with the distance, we need to integrate. When the centers of the spheres are
separated by a distance r, the increment of work done in pushing them closer together
by a distance dr is
1 Q2
dW = F dr = −
dr
4πε0 r2
(We need the negative sign because the motion and the force are in opposite directions.)
The centers of the spheres are initially separated by a distance x; when they touch, their
centers are separated by 2R. Hence
Z
W =
Q2
dW = −
4πε0
Z
2R
r=x
2R
dr
Q2 1
Q2
1
1
Q2 x − 2R
=
=
−
=
r2
4πε0 r r=x 4πε0 2R x
4πε0
2Rx
physics 216 sample exam 2 solns Copyright ©Wayne Hacker 2012. All rights reserved. 8
Problem 8. Consider an isolated positive point charge +q in interstellar space. Put
in a coordinate system and let the charge be located at the origin. It’s equipotential
surfaces are then spheres with center at the origin. Which of the following statements is
always true about the electric flux ΦE through the surface of any equipotential surface?
(a) ΦE = 0
(c) ΦE < 0
(b) ΦE > 0
*(d) ΦE ∝ q
Solution: Since the equipotential surface is a closed surface, we can apply Gauss’ Law
and take the equipotential surface as our gaussian surface. Gauss’ Law tells us that the
flux is proportional to the net charge enclosed by the sphere, which in this case is q.
+q
r
Problem 9. It takes W2 of work to move two positive point
J
charges of +q from infinity to a distance d from one another.
J
d
Jd
What is the work W3 necessary to move three charges of +q
J
from infinity so that they form the vertices of an equilateral
Jr
r
+q
+q
triangle with edge length d (see figure)?
d
(a) W3 = W2
(b) W3 = 2W2
*(c) W3 = 3W2
(d) W3 = 6W2
(e) None of these
Solution: See my solution to problem (7a) multiple-choice problems for chapter 28 on
the course website.
Problem 10. A positive point charge q1 and a negative point charge q2 = −2q1 are
placed at separate locations x1 and x2 on the x-axis. If the reference potential for each
point charge is Vi (∞) = 0 (for i = 1, 2), how many points x are there on the x-axis at
which Vnet (x) = 0? (Do not count x = ±∞.) Hint: For definiteness, let x1 = 0 and
x2 = a > 0.
(a) No such points
(b) One point
*(c) Two points
(d) Not enough information
Solution: Sketch the arrangement of charges, as at right. Let
Q
−2Q
x1 be the location of the positive charge q1 = +Q, and let x2
r
r
be the location of the negative charge q2 = −2Q. In our figure,
x1
x2
x1 < x2 ; but this obviously will not affect our final answer.
The potential at a point x is zero when
1
Q
2Q
V (x) =
−
= 0 ⇒ |x − x2 | = 2|x − x1 |
4πε0 |x − x1 | |x − x2 |
In the regions x < x1 and x > x2 , this becomes
x − x2 = 2x − 2x1
⇒
x = 2x1 − x2 = x1 − (x2 − x1 )
⇒
x=
In the region x1 < x < x2 , we get
x2 − x = 2x − 2x1
2x1 + x2
x 2 − x1
= x1 +
3
3
Hence there are two solutions for x: one to the left of x1 , and one between the two
charges, one-third of the way from x1 to x2 .
physics 216 sample exam 2 solns Copyright ©Wayne Hacker 2012. All rights reserved. 9
Problem 11. Consider a uniformly charged disk with a positive charge density +σ
located in the xy-plane with its center at the origin. Under this setup the z-axis is the
axis of symmetry for the electric field. In class we derived a formula for the electric field
and potential along the z-axis. The formula for the potential along the z-axis was found
to be:
σ √ 2
R + z 2 − |z| ,
Vdisk (z) =
2ε0
where R is the radius of the disk. Where is the location of the maximum potential along
the z-axis? There maybe more than one point.
(a) |z| = ∞
*(b) z = 0
(c) 0 < |z| < ∞
(d) both (a) and (b)
Advice: For the exam you should also know where the minimum of the potential occurs.
Solution: See my solution to problem (11a) multiple-choice problems for chapter 28 on
the course website.