# Version 025 – MT1 v3 – yao – (57465) 1

```Version 025 – MT1 v3 – yao – (57465)
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1
1
−
= (2q)2pk
3
(2r − s)
(2r + s)3
&quot;
#
4pkq
1
1
=
3 −
3
8r 3
1− s
1+ s
2r
001
10.0 points
1
2r
Using the small ǫ approximation, where ǫ =
s/2r, we obtain
due to p
≈
Fon
p1
pkq
3 p2
&times;
6ǫ
=
k .
2r 3
2 r4
The figure shows the set up of problem
15.P.44 in the M&amp;I text, which concerns the
interaction of two identical permanent dipoles
of moment p = qs separated by a distance r.
Let us consider a new situation: replace
the left dipole with a new dipole of moment
p1 = Qs′ , where s′ = 2s and Q = 2q. Let the
distance between the two dipoles be increased
to r ′ = 2r. The right dipole, denoted p,
remains unchanged. Find R, the ratio of the
force on p1 due to p to the force on the original
left dipole p0 due to p:
R=
due to p
Fon
p1
due to p
Fon
p0
R=
=
002
due to p
Fon
p1
due to p
Fon
p0
2
3 p
2 k r4
2
6k pr4
=
1
.
4
10.0 points
.
due to p
is given in the figure.
Note that Fon
p0
1. 1/4 correct
2. 1/2
3. 2
4. 1
5. 4
Explanation:
p
p
due to p
+ (−Q)EA
= QEB
Fon
p1
2pk
2pk
=Q&middot;
−Q&middot;
′ 3
′ 3
s
r′ − 2
r ′ + s2
A water molecule is a permanent dipole
with a known dipole moment p = qs. There
is a water molecule in the air a very small
distance d from the midpoint of a charged rod
of length L carrying a uniformly distributed
negative charge −Q. The axis of the dipole
lies along a line drawn through the center
of the rod. Note that s ≪ d ≪ L. You may
neglect the small change in the dipole moment
of the water molecule induced by the rod.
Version 025 – MT1 v3 – yao – (57465)
Choose the answer that correctly expresses
the magnitude and direction (along the xaxis) of the electric force on the water
molecule, FHrod
. Your f inal result must be
2O
expressed only in terms of k, Q, q, s, p, d
and/or L and any constant numerical factors.
1. −2
kQp
correct
Ld2
2. 0
kQp
Ld2
kQp
4. 2 3
d
kQp
5. − 3
d
kQp
6. − 2
Ld
kQq
7. 2
d
kQp
8. 2 2
Ld
kQp
9. 3
d
kQp
10. −2 3
d
Explanation:
We are told that L ≫ d, so we may use
the approximate field of the rod. From the
figure, we see that the dipole moment points
to the left, meaning that the positive dipole
charge is closer to the rod. Combining these
two facts, we obtain
k(−Q)
q
q
rod
FH2 O = 2
−
L
d − 2s
d + 2s
1
kQq
1
= −2
.
s −
s
Ld 1 − 2d
1 + 2d
3.
Letting ǫ = s/2d and applying the small ǫ
approximation,
kQq
[(1 − ǫ)−1 − (1 + ǫ)−1 ]
Ld
kQq
[1 + (−1)(−ǫ) − {1 + (−1)(ǫ)}]
= −2
Ld
Qq s
kQq
&middot; 2ǫ = 2k
&middot;
= −2
Ld
Ld d
FHrod
= −2
2O
= −2
2
kQp
.
Ld2
003 10.0 points
Two large, parallel, insulating plates are
charged uniformly throughout the plates.
Each plate has the same amount and sign
of charge per unit area +σ.
What is the magnitude of the resultant electric field E?
1.
σ
between the plates, zero outside
2 ǫ0
2. zero everywhere
3. zero between the plates,
4.
σ
everywhere
ǫ0
σ
outside
2 ǫ0
2σ
outside
ǫ0
2σ
2σ
6.
between the plates,
outside
ǫ0
ǫ0
2σ
7.
between the plates, zero outside
ǫ0
σ
between the plates, zero outside
8.
ǫ0
σ
9. zero between the plates,
outside corǫ0
rect
5. zero between the plates,
10.
σ
σ
between the plates,
outside
2 ǫ0
2 ǫ0
Explanation:
Each plate produces a constant electric field
σ
of E =
directed away from the plate for
2 ǫ0
positive charge density, and toward the plate
for negative charge density. Between the two
plates, the two fields cancel each other so that
Enet = 0. Outside of the two plates, the fields
Enet =
σ
.
ǫ0
004 (part 1 of 3) 10.0 points
A solid conducting sphere of radius R1 and
Version 025 – MT1 v3 – yao – (57465)
total charge q1 is enclosed by a conducting
R3 and total charge q2 . OA = a and OC = c.
1. ΦS =
2. ΦS =
q2
R1
q1
3. ΦS =
S
4. ΦS =
O
R2
5. ΦS =
A
C
R3
Find the electric field at C.
1. EP =
2. EP =
3. EP =
4. EP =
5. EP =
6. EP =
k (q1 + q2 )
c
k q1
c2
k q2
c2
k q1 k q2
+ 2
R12
R3
k (q1 + q2 )
R32
k q2
c
7. EP = 0
k (q1 + q2 )
correct
c2
Explanation:
Consider a spherical Gaussian surface
through the point C. C is outside of the
shell, so the enclosed charge qencl = q1 + q2
can be treated as a point charge and
qencl
q1 + q2
EA (4 π c2 ) =
=
ǫ0
ǫ0
(q1 + q2 )
1 q1 + q2
=k
.
EA =
2
4 π ǫ0
c
c2
8. EP =
005 (part 2 of 3) 10.0 points
Find the total flux emanating through the
Gaussian surface S drawn through the point
A.
6. ΦS =
7. ΦS =
8. ΦS =
3
−q2
ǫ0
q1
correct
ǫ0
q2
ǫ0
q2 − q1
ǫ0
−(q1 + q2 )
ǫ0
q1 − q2
ǫ0
q1 + q2
ǫ0
−q1
ǫ0
Explanation:
The enclosed charge is q1 , so the flux is
ΦS =
q1
.
ǫ0
006 (part 3 of 3) 10.0 points
Find the surface charge density on the inner
surface of the shell (i.e, at R2 ).
q2
4 π R22
q2 − q1
=
4 π R22
q1
correct
=−
4 π R22
q1
=−
4 π R12
q1
=
4 π R12
q2
=−
4 π R22
q1
=
4 π R22
q1 − q2
=
4 π R22
1. σ =
2. σ
3. σ
4. σ
5. σ
6. σ
7. σ
8. σ
Explanation:
Draw a gaussian surface inside the shell —
since this surface is inside a conductor, we
know immediately that ΦE = 0. From this
Version 025 – MT1 v3 – yao – (57465)
we conclude that the charge enclosed by this
surface is zero; the surface encloses q1 and the
charge on the inner surface of the shell, which
must be −q1 . Since the inner surface of the
shell is at R2 , the charge density there is
σ=−
q1
.
4 π R22
007 10.0 points
A lithium-7 nucleus (3 protons and 4 neutrons) and a neutral carbon atom of polarizability α are at a distance r apart (r is much
greater than the diameter d of the atom). Due
to polarization of the atom by the lithium nucleus, there is a force F between the 7 Li nucleus and the carbon atom. If we change r to
F′
5.3r, what will be the ratio of
where F ′ is
F
the new force between the two?
1. 0.0029802
2. 0.024485
3. 0.00021779
4. 0.00054192
5. 0.00019869
6. 0.00015236
7. 0.0012621
8. 0.00043602
9. 0.00013987
10. 0.00023912
4
axis of this dipole. The magnitude of the electric field due to this dipole at the location of
the 7 Li nucleus will be given by
Ep =
1 2p
4πǫ0 r 3
Here, we have used the fact that r ≫ d. The
magnitude of force on the nucleus due to the
dipole (and vice versa) can be obtained as
F = Ep &middot; 3e. Combining all the equations, we
get
F = Ep &middot; 3e =
=
1
4πǫ0
2
1 2(αEe )3e
4πǫ0
r3
18αe2
.
r5
This implies that when r to 5.3r, the electric
field field, and consequently the force, is decreased by a factor of (5.3)−5 = 0.00023912,
F′
= (5.3)−5 = 0.00023912
F
008
10.0 points
Explanation:
The force will be proportional to r −5 and
F′
=
hence the correct answer will be
F
0.00023912 . The calculation is as follows.
The magnitude of electric field due to the
7
Li nucleus at the location of the carbon atom
will be given by
E7 Li =
1 3e
4πǫ0 r 2
From this, the magnitude of the induced
dipole moment of the carbon atom can be
written down as p = αE7 Li . This dipole moment will be pointing away from the 7 Li nucleus, and hence the 7 Li nucleus will lie on the
The center of the spherical metal ball of
radius R, carrying a negative charge −Q, is
located a distance r from the center of a short,
thin, neutral copper wire of length L. Assume
that L , R ≪ r. The induced charges at the
two ends of the wire is &plusmn;q. Determine the polarizability of the wire segment which has a
length L = 0.0115 m. The units of polarizability are C2 m/N.
1. 1.013e-17
2. 3.811e-17
3. 1.389e-17
4. 1.736e-18
Version 025 – MT1 v3 – yao – (57465)
5. 3.814e-18
6. 2.112e-17
7. 3.417e-17
8. 4.764e-18
9. 4.234e-17
10. 2.4e-17
Correct answer: 2.112 &times; 10−17 .
Explanation:
First, we must determine the magnitude
of the induced charge. Since the wire is a
conductor, we must have E = 0 inside; using
this fact, we can calculate q:
EQ (A) = Edip (A)
1 Q
1
q
=2&middot;
2
2
4πǫ0 r
4πǫ0 L
2
Q
8q
= 2
2
r
L
2
Q L
.
q=
8 r
The polarizability is given by
α =
p
qL
=
kQ
E
r2
Using the value of q given above
(Q/8)(L/r)2 L
α =
kQ
r2
α =
C2 m
L3
= 2.112 &times; 10−17
8k
N
009 (part 1 of 2) 10.0 points
Two plastic balls are equally but oppositely
charged and held in place by insulating
threads, as shown in the following figure.
+
5
Neutral
Metal
~
F
−
~
F
Before
metal
ball
They attract each other with an electric
force of magnitude F . Then, an uncharged
metal ball is held in place by insulating
threads between the balls, closer to the left
ball. What happens to the magnitudes of the
net electric forces on the left and right charged
balls?
1. The magnitudes of the net electric forces
on both balls are increased. correct
2. The magnitude of the net electric force on
the right ball is reduced, and that on the left
ball is increased.
3. The magnitudes net electric forces on
both balls are reduced.
4. The neutral metal ball has no effect on
the magnitudes of the net electric forces.
5. The magnitude of the net electric force on
the left ball is reduced, and that on the right
ball is increased.
Explanation:
When placed between the two oppositely
charged balls, the metal sphere polarizes. The
electric field of both the positive and negative
charge point to the right at the position of the
neutral ball, polarizing it such that negative
charge collects on the left and positive charge
collects on the right.
The negative charge on the left side of the
metal sphere exerts a (small) rightward force
on the left positively charged ball, adding to
the pull from the negatively charged ball on
the right, meaning the net force increases.
Similarly, the excess positive charge on the
Version 025 – MT1 v3 – yao – (57465)
6
right side of the metal sphere exerts a (small)
leftward force on the negatively charged ball
on the right, increasing the net force on that
ball as well.
See the explanation for part 1. Since both
balls generate a field that points to the right
at the location of the neutral sphere, it must
polarize like option VII.
010 (part 2 of 2) 10.0 points
In the following diagram, choose the image (IIX) that best represents what happens to the
neutral metal ball when it is placed between
the two oppositely charged balls. (Option IX
means there is no change.)
011 10.0 points
A positive charge q and a dipole of moment ~p
contribute to a net field at location A that is
zero, as shown in the following figure. Both
~p and q are a distance a from point A, with
a ≫ dipole separation.
I
+
+
II
+
+
III
+
++
+
+
+
IV
−
−
+
+
−
−
+
−
−
−
−
+ +
++
+ +
a
+
q
a
+
+
VI
−
−−
−
VII
A
+
V
−
−
+ +
++
+ +
b
+
+
−
−
VIII
− −
−−
− −
−
−
−
−
−
IX
−
−
−
~p
If ~p points to the right, we denote it +p; if
~p points to the left, we denote it −p. Choose
the relationship between p and q that results
in a net field of 0 at A.
1. p = qa
2. p = −qa correct
qa
2
qa
4. p =
2
Explanation:
At A, the field due to the point charge
points to the left. In order to get cancellation of E at A, the field due to the dipole
must point to the right. Recall that, on the
axis perpendicular to the dipole moment, the
dipole field points in the direction opposite
the dipole moment, so the dipole moment
must point to the left: ~p = −p.
To determine the magnitude of the relationship between p and q, we simply set the
point charge field at A and the dipole field at
A equal to each other:
3. p = −
1. VIII
2. VI
3. I
4. V
5. II
6. IV
7. VII correct
8. III
9. IX
Explanation:
|Edip⊥ | = |Eq |
kq
k|~p|
= 2
3
a
a
Version 025 – MT1 v3 – yao – (57465)
|~p| = qa .
Combining the two results, our answer must
be: p = −qa.
7
Two vectors can define a plane. When these
two vectors are plotted in this plane, we have
012 10.0 points
~ has components
Vector A
Ax = −2.1,
Ay = 6.2,
Az = 6.7,
B
Bz = 5.
What is the angle θAB between these vectors? (Answer between 0◦ and 180◦ .)
1. 121.865
2. 105.72
3. 113.517
4. 140.101
5. 102.412
6. 118.162
7. 111.323
8. 142.246
9. 157.02
10. 125.71
5. 7
By = −7.7,
10
Bx = 7.7,
2◦
~ has components
while vector B
A
013 10.0 points
The diagrams in Figure 1 show a sequence of
events described below, which involves a small
lightweight solid metal ball that is suspended
Explanation:
~ =
kAk
q
A2x
+
A2y
+
A2z
q
= (−2.1)2 + 6.22 + 6.72
= 9.36696 and
q
q
2
2
2
~
kBk = Bx + By + Bz = 7.72 + (−7.7)2 + 52
= 11.9825 ,
so using
~ &middot;B
~ = Ax B x + Ay B y + Az B z
A
= (−2.1) 7.7 + 6.2 (−7.7) + 6.7 (5)
= −30.41 ,
cos θAB
θAB
~ &middot;B
~
A
−30.41
=
=
~ kBk
~
(9.36696) (11.9825)
kAk
= −0.270938
= arccos(−0.270938)
= 105.72◦ .
(a) You touch the ball briefly with your
fingers then release it.
(b) A block of metal that is known to be
charged is now moved near the ball.
Version 025 – MT1 v3 – yao – (57465)
(c) The ball briefly touches the charged
metal block. Then the ball swings away from
the block and hangs motionless at an angle as
shown in diagram 5 of Figure 1.
(d) Finally the block is moved far away.
A positively charged rod is brought near the
ball. The ball is repelled by the charged rod
shown in diagram 6 of Figure 1.
Among distributions E through N shown in
Figure 2, which of the diagrams best shows
the distribution of charge in the ball in slide 6
of Figure 1?
8
sion implies that either the ball is positively
charged (this rules out I-L) or the effective
positive charge in the ball is closer to the rod
which produces a net repulsion (this rules out
M).
Since the positive charge of the rod attracts
electrons in the positively charged ball, the far
side of the ball should appear more positively
charged. This condition rules out E,F,N.
G is the correct answer. Positive charge
on the ball leads to repulsion. It has more
positive charge on the far side, i.e. on the side
away from the charged rod. This is expected
since the negative charge on the surface of
the balls repositions to be as close to the
positively charged rod as possible.
014 10.0 points
Consider a thin glass rod of length L lying
along the x-axis with its left end at the origin. The rod carries a uniformly distributed
positive charge Q.
2. M
L
O
1. H
−a
+Q
3. L
4. J
5. I
6. G correct
7. F
What is the electric field due to the rod at
position −a on the x-axis, a &gt; L, as shown in
the figure above? Let the negative direction
point to the left.
{Hint: If you are unsure of your result, evaluate your integral to see if the result makes
physical sense.}
8. E
9. N
10. K
Explanation:
For slide 6 of Figure1, it is stated that the
rod is positively charged. The observed repul-
1 Q
1. E(−a) =
4 π ǫ0 L
ZL
dx
(x − a)2
0
1 Q
2. E(−a) = −
4 π ǫ0 L
ZL
0
dx
x2
Version 025 – MT1 v3 – yao – (57465)
1 Q
3. E(−a) =
4 π ǫ0 L
ZL
9
~
E
dx
x2
0
ZL
1 Q
4. E(−a) = −
4 π ǫ0 L
B F
~
dx
cor(x + a)2
0
rect
A α
5. E(−a) =
1 Q
4 π ǫ0 L
ZL
dx
(x + a)2
0
1 Q
6. E(−a) = −
4 π ǫ0 L
ZL
dx
(x − a)2
0
Explanation:
The charges of the rod are distributed between 0 and L. The distance from −a to a
point on the rod is given by x − (−a) = x + a
and the integration limits must be 0 and L.
Looking at the figure, since the charge on the
rod is positive, the field at −a must point to
the left and our answer must be negative.
1
∆Q
∆E = −
4 π ǫ0 (x + a)2
Q
L
∆x
1
=−
4 π ǫ0 (x + a)2
ZL
dx
1 Q
E=−
4 π ǫ0 L
(x + a)2
Now imagine that a lithium nucleus, containing three protons and four neutrons, is
placed at point B. What is the value of the
electric field at location B due to the alpha
particle? (Remember that e represents the
fundamental unit of charge, 1.6 &times; 10−19 C.
Also, r BA is the position vector pointing from
point A to point B.)
~
2. E
~
3. E
~
4. E
~
5. E
~
6. E
0
Digression: to use the hint, you must realize
that the electric field of the rod must be less
than that of a point charge Q at the origin
but greater than that of a point charge Q at
L:
|EQL (−a)| &lt; |Erod (−a)| &lt; |EQ0 (−a)| .
Determining bounds on the possible values
015 (part 1 of 4) 10.0 points
The following figure shows an alpha particle
(two protons and two neutrons) at location
~ at point B.
A creating an electric field E
Any charged particle placed at location B
~.
experiences a force F
1
2e BA
r̂
correct
4πǫ0 |~r BA |2
3e BA
1
=
r̂
4πǫ0 |~r BA |
1
2e BA
=−
r̂
4πǫ0 |~r BA |
1
11e BA
r̂
=
4πǫ0 |~r BA |2
5e BA
1
=−
r̂
4πǫ0 |~r BA |2
e
1
=
r̂ BA
4πǫ0 |~r BA |2
e2
1
=−
r̂ BA
4πǫ0 |~r BA |2
5e2 BA
1
r̂
=
4πǫ0 |~r BA |2
~ =
1. E
~
7. E
~
8. E
Explanation:
The electric field at point B is not related to
the charge at point B, only the one at point
A, so the answers with multiples of e2 are
incorrect. Recalling the form of the electric
field and remembering that the alpha particle
has charge 2e,
~ =
E
1 q
r̂,
4πǫ0 |~r |2
we see that it has the distance squared in the
denominator. Therefore the correct choice is
Version 025 – MT1 v3 – yao – (57465)
~ =
E
1
2e BA
r̂ .
4πǫ0 |~r BA |2
016 (part 2 of 4) 10.0 points
What is the force on the lithium nucleus?
11e2
~ = −1
r̂ BA
1. F
2
BA
4πǫ0 |~r |
3e BA
~ = 1
r̂
2. F
4πǫ0 |~r BA |
−1 2e2 BA
~
3. F =
r̂
4πǫ0 |~r BA |
6e2 BA
~ = 1
4. F
r̂
correct
4πǫ0 |~r BA |2
~ = −1 5e r̂ BA
5. F
4πǫ0 |~r BA |2
2e2 BA
~ = 1
6. F
r̂
4πǫ0 |~r BA |2
5e2 BA
~ = 1
7. F
r̂
4πǫ0 |~r BA |2
e2
~ = 1
r̂ BA
8. F
4πǫ0 |~r BA |2
Explanation:
Force involves both particles, and has the
general form
1 Q1 Q2
r̂,
4πǫ0 |~r |2
where ~r is the vector pointing between the two
charges. In this case, Q1 = 2e and Q2 = 3e,
so we end up with
~ =
F
~ =
F
6e2 BA
1
r̂ .
4πǫ0 |~r BA |2
017 (part 3 of 4) 10.0 points
The lithium nucleus is removed and an electron is placed at location B. Now what is the
value of the electric field at location B?
1
11e BA
r̂
4πǫ0 |~r BA |2
1
5e BA
=−
r̂
4πǫ0 |~r BA |2
1
e2
r̂ BA
=−
4πǫ0 |~r BA |2
~ =
1. E
~
2. E
~
3. E
10
1
e
r̂ BA
2
BA
4πǫ0 |~r |
3e BA
1
r̂
=
4πǫ0 |~r BA |
1
2e BA
=
r̂
correct
4πǫ0 |~r BA |2
5e2 BA
1
r̂
=
4πǫ0 |~r BA |2
1
2e BA
=−
r̂
4πǫ0 |~r BA |
~ =
4. E
~
5. E
~
6. E
~
7. E
~
8. E
Explanation:
See the explanation for part 1. The electric
field does not depend on the particle at point
B, so in this case the answer is the same as
for part 1.
018 (part 4 of 4) 10.0 points
What is the the force on the electron?
~ =
1. F
~ =
2. F
~ =
3. F
~ =
4. F
~ =
5. F
~ =
6. F
~ =
7. F
~ =
8. F
1
2e BA
r̂
4πǫ0 |~r BA |
−1 2e2 BA
r̂
4πǫ0 |~r BA |
−1 e2
r̂ BA
4πǫ0 |~r BA |2
−1 3e2 BA
r̂
4πǫ0 |~r BA |2
3e2 BA
1
r̂
4πǫ0 |~r BA |2
1
e2
r̂ BA
4πǫ0 |~r BA |2
−1 2e2 BA
r̂
correct
4πǫ0 |~r BA |2
1
2e2 BA
r̂
4πǫ0 |~r BA |2
Explanation:
Since the electric field of the alpha particle is now acting on a negative charge, the
force will be attractive, so we need an answer
proportional to −r̂ BA . Since the force is proportional to Q1 Q2 , we need an answer with
2e2 . As always, distance dependence goes like
1/r 2 . The only correct option is
Version 025 – MT1 v3 – yao – (57465)
11
VIII
2e2
~ = −1
F
r̂ BA .
2
BA
4πǫ0 |~r |
I
VII
II
III
VI
IV
V
019 (part 1 of 2) 10.0 points
Two dipoles consisting of two oppositely
charged balls connected by a wooden stick
are located as shown in the following figure.
A block of plastic is located between them,
as shown. Assume each of the molecules
throughout the block has the same polarizability. Locations B, C, and D all lie on a
line perpendicular to the axis of the dipole,
passing through the midpoint of the dipoles.
Plastic block
+ −
D
b
C
b
8. VII
+
+
+
−
I
−
−
+
H
−
− +
G
D
− +
F
4. I
b
C
− +
3. V
7. VI
− +
B
2. III correct
6. IV
Before selecting answers to the following
questions, you might want to draw your own
diagram of this situation, showing all the
fields and charge distributions requested.
Answer the following questions by selecting
either a direction (I-VIII) or an orientation of
a polarized molecule (A-J) from the following
two diagrams.
A
1. II
5. VIII
− +
B
Which of the arrows (I-VIII) best indicates
the direction of the electric field at location D
due only to the dipoles?
E
J
Other
Explanation:
The electric field along the perpendicular
bisector of a dipole is opposite its dipole moment. Since the upper dipole is closer to
point D, it’s field dominates that of the other
dipole. Since the dipole moment of the upper
dipole points left, its field at D points right.
Direction III is the correct answer.
020 (part 2 of 2) 10.0 points
Which of the diagrams (A-J) best indicates
the polarization of a molecule of plastic at
location D?
1. Diagram D
2. Diagram B correct
3. Diagram I
Version 025 – MT1 v3 – yao – (57465)
4. Diagram E
5. Diagram F
6. Diagram C
7. Diagram H
8. Diagram A
9. Diagram G
10. Diagram J
Explanation:
Since the field at D points to the right, the
dipole moment of a molecule at D must also
point to the right. Choice B shows the correct
polarization for a rightward dipole moment.
12
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