Download Physics 370 Midterm Exam #2 Solution Key Spring Semester 2014

1
Physics 370 Midterm Exam #2 Solution Key
Spring Semester 2014
Performance Notes: This exam had a total of 100 points possible (not including extra credit). The average score was 65.8 out of 100 with a standard
deviation of 12.3 points. That is a bit lower than I expected, sorry, this was
a second tougher than average test in a row! A summary of statistics on a
question by question basis is presented below.
Question
1a
1b
1c
2a
2b
2c
2d
2e
3a
3b
3c
4a
4b
4c
4d
4e (Extra Credit)
5a
5b
5c
5d
5e
TOTAL
Possible Average Std. Dev. Max Min
6
6
0
6
6
6
4
2.2
6
2
8
4.3
2
8
3
3
2.7
0.8
3
1
5
4.2
1.2
5
2
5
4.2
1.3
5
2
4
2.5
1.4
4
0
8
4.7
2
8
2
5
4.3
0.8
5
3
12
8.3
3
12
5
8
2
1.8
4
0
5
4.2
1.1
5
3
6
2.6
2.6
6
0
14
7.8
2.7
10
4
5
2.8
0.4
3
2
3
1
0
1
1
5
4
0
4
4
5
3
0
3
3
10
8
0
8
8
6
4
0
4
4
4
4
0
4
4
100
65.8
12.3
83
55
2
[Because I am an algebraically inept individual who can’t add
numbers properly, I accidentally made Problem #1 15 points instead of 20 points on the printed exam (so that the total is 100
points). Sorry about that.]
1. Easy in Concept? (20 points TOTAL) Answer the following conceptual problems:
(a) (6 Points) If you use separation of variables as a technique for
solving a particular boundary value problem (for example, the
electric potential inside some box, given the potential on the 6
sides of the box), what is a fundamental assumption you make?
In other words, what must be true about the electric potential
function V (x, y, z) for it to be considered “separable?”
For us to use separation of variables to solve a particular boundary value problem (such as the electric potential in a region given
the values of the potential at the boundaries of the region), we
must assume the potential we are solving for is separable, that is,
that the potential is a function of the various coordinate variables
in a way that can be written as the product of functions of single coordinates. More succinctly, V (x, y, z) must be writable as
X(x)Y (y)Z(z) in order to be considered separable.
(b) (6 Points) Is there any way a charged particle can move through a
uniform magnetic field (and no electric field) without experiencing
a net force? If so, describe an example of such a situation. If not,
explain why not.
A charged particle in an uniform magnetic field will experience a
~ If the charged particle moves in a
force in the direction ~v × B.
~ = 0 and the
direction parallel to the magnetic field, then ~v × B
total magnetic force on the charged particle is zero!
3
(c) (8 points) If I run a current through a (perfectly conducting)
metallic spring as illustrated here, will the spring compress, expand, or neither?
+
–
Does it matter if we reverse how the battery is hooked up? Full
credit requires you to clearly explain your reasoning. HINT: Consider just two adjacent loops of the spring.1
Consider just two adjacent loops of the spring as shown below.
I
I
In both of these loops, the current are moving parallel to one another. Therefore, using the right hand rule to get the direction of
the Biot-Savart law, we see the top loop is immersed in a magnetic field pointed out of the page from current in the bottom loop.
Now apply Lorentz’s Law and realize the direction of the magnetic
~ which is downward. Applying the
force on the top loop is I~ × B
same analysis to the bottom loop reveals the magnetic field of the
loop above it is into the page, so the bottom current experiences a
force upward. The net result is the spring gets compressed when
a current is run through it. It does not matter what direction the
current is running, any parallel currents will have a net attractive
force between them.
1
This question was borrowed from MIT’s OpenCourseWare AP Physics course notes!
4
z
P
R
y
x
2. Dielectric Dome (25 points TOTAL) A hemispherical dielectric of radius R is oriented “flat-face down” and has uniform polarization pointing in the z direction, P~ = P zˆ (as shown to the right).
(a) (3 points) Determine the bound volume charge density ρb .
~ · P~ . But since we are
The bound volume charge density ρb = −∇
told P~ is uniform, the divergence of P~ must be zero, and the bound
volume charge density ρb = 0 .
(b) (5 points) Determine the bound surface charge density σb on the
hemispherical surface.
The bound surface charge density σb = P~ · n
ˆ . For the top surface,
n
ˆ = rˆ, and so σb = P~ · rˆ = P zˆ · rˆ = P cos θ where θ is the polar
angle in spherical coordinates.
(c) (5 points) Determine the bound surface charge density σb on the
flat bottom.
At the bottom of the dielectric, n
ˆ = −ˆ
z , therefore
~
σb = P · (−ˆ
z ) = P zˆ · (−ˆ
z ) = −P .
(d) (4 points) Sketch rough features of the electric field using the
front-view outline below. Briefly explain your reasoning.
Given the accumulation of positive bound surface charge at the top
of the sphere and negative bound surface charge at the bottom, the
5
electric field inside the dielectric must point downward. My (not
terribly precise) sketch of the electric field is shown.
(e) (8 points) Compute the total bound surface charge and show that
it
expected. NOTE: It may be useful to recall
R takes on the value
cos θ sin θdθ = − 21 cos2 θ.
Since the only bound charges are surface charges, all I need to do
is integrate the bound surface charge over all the surfaces
Z
Z
Qb = σtop datop + σbottom dabottom
(1)
For the top surface, we know datop = Rdθ(R sin θdφ) = R2 sin θdθdφ
for θ = 0 → π/2 and φ = 0 → 2π in spherical
R coordinates. For
the bottom
surface, since σbottom is uniform, σbottom dabottom =
R
σbottom dabottom = σbottom πR2 . Putting this all together, equation
1 becomes
Z π/2
Z 2π
2
Qb =
(P cos θ)R sin θdθ
dφ + σbottom πR2
(2a)
θ=0
= 2πP R2
φ=0
Z
π/2
θ=0
cos θ sin θdθ − P πR2
(2b)
R
Via u-substitution, we could have shown cos θ sin θdθ = − 21 cos2 θ,
therefore:
π/2
1
2
2
Qb = 2πP R − cos θ
cos θ sin θdθ − P πR2
(3a)
2
θ=0
2 1
= 2πP R
cos θ sin θdθ − P πR2
(3b)
2
=0
(3c)
6
As expected via conservation of charge, the total bound charge is zero .
7
z
3q
2a
-q
-q
a
x
a
y
2a
-q
3. Multipole Madness in March (25 Points TOTAL) Consider the
above charge distribution. You are interested in the electric potential
far away from the charges (r ≫ a).
(a) (5 points) This problem of the electric potential far from the
charges lends itself to a multipole expansion . Explain why the
multipole expansion is an approximation and explain what order
of the multipole expansion you expect to give a good approximation in this case.
A multipole expansion is an infinite series representation of the potential that was built by assuming we√are far away from the charge
distribution (such that sepmag = r 1 + ǫ ≈ r where epsilon ≪
1). If we cut off the series at some term, it is typically an approximation of the potential, although since higher order terms drop
off more quickly with increasing r, the approximation is usually
quite good even if you stick to only the first non-zero term. In this
particular case, since the total charge Q is zero, we would expect
the monopole term would be zero, so the next most dominant term
will be the dipole term.
(b) (12 points) Determine the electric potential V (r, θ, φ) far from the
charges. Only keep the first non-zero term in the multipole expansion. Be clever to avoid extra work.
Since we rejected the monopole term in part (a), let’s consider the
dipole term. Given the handful of fixed charges, the easiest way to
compute the dipole terms is to use the dipole moment. The dipole
8
moment is
~p =
X
qi~ri′
(4)
i
= 3q(2a)ˆ
z − qaˆ
y − q(−2a)ˆ
z − q(−a)ˆ
y
= 8qaˆ
z
(5)
(6)
and applying this dipole moment to solve for the potential we have
(and noting zˆ · rˆ = cos θ where θ is the polar angle coordinate:
V (~r) =
p~ · rˆ
4πǫ0 r 2
V (r, θ, φ) =
(7)
8qa cos θ
4πǫ0 r 2
(8)
(c) (8 points) Determine the work done bringing moving a charge q
from ~ri = (xi , yi , zi ) = (0, 0, 40a) to ~rf = (xf , yf , zf ) = (0, 20a, 0)
assuming you never bring it close to this charge configuration.
The work done is just going to be the change in energy in the
charge q moving from ~ri to ~vf , that is
∆W = q∆V = q (V (~rf ) − V (~ri ))
(9)
where these positions in Cartesian coordinates become ~ri = (ri , θi , φi ) =
(40a, 0, 0) and ~rf = (rf , θf , φf ) = (20a, π2 , 0) so
8qa
−0
(10)
∆W = q
4πǫ0 (40a)2
q2
=
(11)
1600πǫ0 a
9
4. Infinite Images: (30 Points TOTAL) One of the Griffiths’ problems
I considered assigning states:
Two infinite parallel grounded conducting planes are held a
distance a apart. A point charge q is placed in the region
between them, a distance x from one plate. Find the force
on q.
a
q
x
A sketch of this situation is shown above with the dark dot representing
the real charge between two infinite conducting plates (the gray slabs).
(a) (5 Points) This problem can be solved via the method of images. If
you want to solve a given electrostatics problem using the method
of images, what are the requirements of the corresponding images
problem?
To solve a given electrostatics problem using the method of images requires you to set up an “image problem” that has the same
boundary values for the electric potential as the original electrostatics problem. By the uniqueness theorem, if you find a solution
to a boundary value problem that matches the boundary values,
than you have found the solution to that boundary value problem,
regardless of method used to solve the problem (including adding
virtual charges to the problem such that the boundary value potentials match). In this case, the key requirement will be that
V = 0 on the surfaces of the two conducting plates.
(b) (6 Points) To solve this problem, you have to consider a very
surprising equivalent image problem with an infinite number of
image charges in the configuration shown below.
q –q
a
q –q
–q
q
x
x
a-x
q –q
q –q
10
From the point of view of the method of images technique, explain why you need this charge configuration to solve the problem
of the force on the point charge? In the process explain (i) explain
why adding a single −q image charge alone doesn’t work in this
case [recall that it did work with a single conducting plane] and
(ii) explain why for each −q image charge you add, you have to
also add a +q image charge, explaining why they are placed where
they are placed. HINT: It may help to start by add one image
charge at a time to explain their ordering.
Imagine adding a single −q image charge distance x−a to the right
of the right conducting plane. It gives me V = 0 on the right conducting plane, but not on the left, so to get V = 0 on the left plate,
I have to place a q image charge distance a + (a − x) = 2a − x
to the left of the left conducting plate. But that now messes up
the potential at the right plate, so we have to add an image charge
−q at distance 2a − x + a = 3a − x to right of the right plate...
and so on. This explains half the charges. The other half are explained by trying to get v = 0 on the left plate first with image
charge −q distance x to the left of the left plate, which makes V
non-zero on the right plate. Trying to cancel that out leads to the
other charges. A infinite sequence of charges on both sides in the
pattern shown.
(c) (14 Points) So, write an expression for the force on charge q. As
you might imagine, this will be an infinite series, but try to simplify it a bit. HINT: Consider the contributions of just the q
image charges separately from the −q image charges.
This is just a bunch of point charges, so to figure out the total
force, we just have to add up the forces between all the point image charges and q.
• The q image charges are spaced at intervals of 2a to the right
and left of q, so their total force precisely cancels out to zero!
• The −q image charges to the right of the right conducting
plane are at distances 2(a − x) + n(2a) (where n = 0, 1, 2, ...)
and pull the charge q rightward. Assuming rightward is posi-
11
tive, these charges contribute force
Fright =
∞
1 X
q2
4πǫ0 n=0 (2(a − x) + n(2a))2
(12)
The −q image charges to the left of the left conducting plane
are at distances 2x + n(2a) (where n = 0, 1, 2, ...) pull the
charge leftward. Adding up their contributions we have
Flef t
∞
q2
1 X
=−
4πǫ0 n=0 (2x + n(2a))2
(13)
So the −q image charges contribute total force on q of
F = Fright +Flef t
∞
q2 X
1
1
=
−
.
2
4πǫ0 n=0 (2(a − x) + n(2a))
(2x + n(2a))2
(14)
(d) (5 Points) Check that your answer is correct in the special case
where the charge is exactly half-way between the conducting planes,
so x = a2 . NOTE: If you can’t check it, state what the answer
should be for partial credit!
If charge q is midway between the two plates, the total force should
be zero since it is equally attracted to the induced surface charges
on both plates. Evaluating equation 14 for x = a/2 we have
∞
q2 X
1
1
F =
− a
a
2
4πǫ0 n=0 (2(a − 2 ) + n(2a))
(2 2 + n(2a))2
=
∞
q2 X
1
1
−
4πǫ0 n=0 (a + n(2a))2 (a + n(2a))2
= 0!
(15)
(16)
(17)
(e) (3 Points Extra Credit) Check that your answer is correct in the
special case where one of the conducting planes is infinitely far
away, so a → ∞ NOTE: If you can’t check it, state what the
answer should be for partial credit!
12
For a → ∞, this is equal to a charge a distance x from a single
plate, which we know should experience an attractive force equal to
that of an negative image charge distance 2x from it. Evaluating
equation 14 for a → ∞, any terms with a in the denominator will
die away, leaving the only the n = 0 term of the second component
of equation 14, thus we find
F =−
q2
1
4πǫ0 (2x)2
which is the answer we expected.
(18)
13
5. I cannae change the laws of physics! (30 points TOTAL) Consider
the two current loops shown below.
I2
I1
dℓ1
dℓ2
(a) (5 Points) What is the magnitude and direction for the magnetic
~1
force dF~2 on the current element d~ℓ2 due to the magnetic field dB
~
created current element dℓ1 ? Clearly explain your reasoning and
write this force only in terms of the variables given here.
~
Given the Lorentz Force Law we know generally that F = I(d~ℓ×B)
~ 1 . However, dB
~ 1 will be into the
or in this case dF~2 = I2 d~ℓ2 × dB
page at d~ℓ2 (via the right hand rule a la Biot-Savart), therefore
the force dF~2 has magnitude I2 dℓ2 dB1 and points downward.
(b) (5 Points) Now consider the inverse problem of the magnitude and
direction for the magnetic force dF~1 on the current element d~ℓ1
~ 2 created current element d~ℓ2 ?Again,
due to the magnetic field dB
~ 2 at d~ℓ1 ?
clearly explain your reasoning. HINT: What is dB
~ r) =
Here, the trick is to realize that via the Biot-Savart Law B(~
R
µ0 I
d~
ℓ′ × ˆ
~
~
~
~
2 , which says dB2 at dℓ1 due to dℓ2 is zero, since dℓ2 is
4π
parallel to the separation vector ~ between them! Thus the force
dF~1 is zero!
Hopefully you have discovered that these two forces mentioned in
parts (a) and (b) are NOT equal and opposite, which would seemingly violate Newton’s Third Law of Motion! This problem arises
because the current elements d~ℓ1 and d~ℓ2 can’t actually exist in
isolation! In reality, we need to consider the total force on one
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entire current loop due to the other entire current loop.
(c) (10 Points) Defining the separation between a current element in
~ field) and another current element in loop
loop 1 (“feeling” the B
~ field) as ~ 12 = ~r1 − ~r2 , write an expression for
2 (“creating” the B
the total force dF~1 on current element d~ℓ1 due to the magnetic
field B2 created by all of current loop 2 (exploit Biot-Savart to get
~ 2 ). Then to get the total force F~1 on loop 1, integrate the exB
pression you find over loop 1. HINT: Since the shapes of loops 1
and 2 are not clearly defined,
H just indicate a closed integral around
any loop with the symbol n where n indicates the loop number.
From the Lorentz force law, I know the force on element d~ℓ1 will
be
~ 2 ),
dF12 = I1 (dℓ~1 × B
(19)
so I can’t evaluate the force on loop 1 by loop2 without figuring
out an expression for the magnetic field due to loop 2.
To find the magnetic force of the loop 2 on d~ℓ1 , I have to use
the Biot–Savart law and integrate over all of loop 2 to find the
magnetic field created by loop 2 at d~ℓ1 . This is at a location on
loop 1 (vecr1 ), the magnetic field due to the current in loop 2
should be
I ~
dℓ2 × ˆ 12
µ 0 I2
~
.
(20)
B2 (~r1 ) =
2
4π 2
12
Inserting equation 20 into our expression for the force on the current element in loop 1, I have
!
I ~
ˆ 12
µ
I
d
ℓ
×
0
2
2
(21)
dF~12 = I1 dℓ~1 ×
2
4π 2
12
and integrating over all of loop 1, we would have a total force of
loop 2 on loop 1 of
!
I
I ~
ˆ
µ
I
d
ℓ
×
0 2
2
12
F~12 = I1 dℓ~1 ×
.
(22)
2
4π
1
2
12
15
Collecting the constants outside the integral, I have
!
I
I ~
ˆ 12
d
ℓ
×
µ
I
I
2
0
1
2
dℓ~1 ×
.
F~12 =
2
4π
1
2
12
(23)
(d) (6 Points) Analogous to the previous problem, start by defining
~ field generating current (loop 1)
the separation between the B
and the current feeling the field (loop 2) as ~ 21 = ~r2 − ~r1 = −~ 12 .
Write an expression for the total force F2 on current loop 2 due
to the magnetic field B1 created by current loop 1.
The approach here is identical to what we did in the last problem,
except we consider the force due to the magnetic field created by
loop 1 at some location on loop 2. Via the Biot–Savart law, we
know:
I ~
dℓ1 × ˆ 21
µ 0 I1
~
B1 (~r2 ) =
.
(24)
2
4π 1
21
where the separation vector now goes from loop 1 to 2 (thus the
change in subscript). This expression for the magnetic field can
be inserted into the Lorentz force law applied to loop 2 to state:
!
I ~
I
ˆ 21
d
ℓ
×
µ
I
1
0
1
F~12 = I2 dℓ~2 ×
(25)
2
4π 1
2
21
!
I
I ~
ˆ 21
d
ℓ
×
µ 0 I1 I2
1
(26)
dℓ~2 ×
=
2
4π
2
1
21
!
I
I ~
ˆ 12
µ 0 I1 I2
d
ℓ
×
1
=−
(27)
dℓ~2 ×
2
4π
2
1
12
where for equation 27 I used the fact that ˆ 21 = − ˆ 12 to say
ˆ
−
12
2
12
ˆ
21
2
21
=
.
(e) (4 Points) While it is NOT trivial to show this is true, what relationship needs to exist between your last two derived expressions
in order for Newton’s Third Law to hold? Do not try to prove this
16
relationship for now!
For Newton’s Third Law to hold, F~12 = −F~21 , therefore, combining
equations 23 and 27, we have the condition that
µ 0 I1 I2
4π
I
1
I
1
F~12 = −F~21
(28)
!
!
I ~
I
I ~
ˆ 12
ˆ 12
d
ℓ
×
d
ℓ
×
µ
I
I
2
1
0
1
2
dℓ~1 ×
dℓ~2 ×
=
2
2
4π
2
2
1
12
12
(29)
!
!
I
I ~
I ~
dℓ2 × ˆ 12
dℓ1 × ˆ 12
~
~
=
d
ℓ
×
(30)
d ℓ1 ×
2
2
2
2
12
2
1
12
It does in fact turn out that you can show equation 30 is true, so
Newton’s Third Law is upheld.