Version 061 – Test #1 – Antoniewicz – (57030)
This print-out should have 22 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
1
8. Ic, IIc
9. Ib, IIc
Explanation:
The net electric field inside the conductor
has to be zero. Since this is a conductor,
charges will try to move to the inner surface
of the cavity, thus leaving a net +q charge on
the outer surface of the cavity. Since it creates
a net electric field inside the conductor, which
is not allowed, this will induce an equal and
opposite charge of −q on the inner surface of
the conductor. The answer is Ia.
This will in turn induce a charge +q on
the outer most surface of the conductor. The
answer is IIb.
And: Ia, IIb
002 (part 1 of 2) 5.0 points
A neutral conducting sphere contains a
spherical cavity. We put a charge +q inside the captivity. Choose the correct pair of
statements:
The total surface charge in the interior surface of the cavity is:
Ia. −q
Ib. +q
Ic. 0
The total surface charge in the outer most
surface of the conductor is:
IIa. −q
IIb. +q
IIc. 0
1. Ia, IIb correct
2. Ib, IIa
3. Ib, IIb
A charged particle with a negative charge
q1 is at a distance r from a neutral atom, as
shown in the following figure.
q1
Charge
r
Neutral atom
Determine the following two directions.
I. The direction of field which generates the
induced dipole is:
a. to the right
b. 0
c. to the left
II. The direction of the induced dipole is:
a. to the right
b. 0
c. to the left
4. Ia, IIc
1. Ib, IIb
5. Ic, IIa
2. Ic, IIb
6. Ia, IIa
3. Ib, IIc
7. Ic, IIb
4. Ia, IIc
5. Ia, IIb
Version 061 – Test #1 – Antoniewicz – (57030)
6. Ib, IIa
7. Ic, IIa
8. Ic, IIc correct
9. Ia, IIa
Explanation:
Since q1 < 0, at the neutral atom the
field and the induced dipole moment are both
pointing to the left. This is due to the fact
that the negative source charge plays the role
of a sink for the electric field lines so the field
! and
line is pointing to the left. Since !p = αE
α is positive, it follows that the direction of
! At the neutral
!p is the same as that of E.
!
atom, E is pointing to the left and hence the
induced dipole moment !p will be pointing to
the left as well.
Ans: Ic, IIc
2
between the two rings?
1. 88631.0
2. 54982.0
3. 50764.0
4. 41088.0
5. 39340.0
6. 86522.0
7. 32565.0
8. 75470.0
9. 42542.0
10. 48388.0
Correct answer: 50764 N/C.
Explanation:
The electric field vector is given by
#Ex , Ey , Ez $. However, the y and z components of the vector are zero. The x-component
of the electric field due to the ring is given by
E1 =
1
qx
4π$0 (R2 + x2 )3/2
here we note that
003 (part 2 of 2) 5.0 points
Determine the direction of the field at q1 generated by the induced dipole.
1. to the left correct
2. 0
3. to the right
Explanation:
Since the direction of the induced dipole
moment of the neutral atom is to the left, the
field due to the induced dipole at q1 is also
pointing to the left.
004 (part 1 of 3) 4.0 points
Two rings of radius R = 0.04 m are d =
0.134 m apart and concentric with a common
horizontal x axis (the +x axis is towards the
right). The ring on the left carries a uniformly
distributed charge of +40nC, and the ring
on the right carries a uniformly distributed
charge of −40nC.
The value of k is 9 × 109 Nm2 /C2 .
What is the magnitude of the electric field
due to the right ring at a location midway
x = 0.067 m
R = 0.04 m
q = 4 × 10−8 C
k=
1
= 9 × 109 Nm2 /C2
4 π $0
E1 = (9 × 109 Nm2 /C2 )
(4 × 10−8 C)(0.067 m)
((0.04 m)2 + (0.067 m)2 )
= 50764 N/C
005 (part 2 of 3) 3.0 points
Denote the magnitude of the electric field at
the middle due to the left ring by E1 , that due
to the right ring by E2 . Compare E1 and E2 .
Which of the following relations represents
the correct choice?
1. E1 > E2
2. E1 = E2 correct
3. E1 < E2
Explanation:
The electric field due to the left ring will
be of the same magnitude as that of the right
3/2
Version 061 – Test #1 – Antoniewicz – (57030)
ring since the parameters are the same. The
radial vector points in the opposite direction
but it is compensated by the charge, which is
of the opposite sign. Its direction will also be
along the same direction. Hence, the answer
is E1 = E2 .
006 (part 3 of 3) 3.0 points
What is the net electric field at a location
midway between the two rings?
1. Enet = E1 − E2
Let : me = 9.11 × 10−31 kg ,
qe = 1.6 × 10−19 C ,
v = 2 × 107 m/s ,
d = 4 cm .
4. Enet = E1 + E2 correct
Explanation:
The net electric field is the sum of the first
two electric fields (because both of them point
in the +x̂ direction) and the net field is given
by Enet = E1 + E2 .
007 10.0 points
The electron gun in a television tube is used to
accelerate electrons (mass of 9.11 × 10−31 kg
and charge of −1.6 × 10−19 C) from rest to
2 × 107 m/s within a distance of 4 cm.
What electric field is required?
1. 6943.6
2. 28468.8
3. 17253.8
4. 183013.0
5. 16267.9
6. 3847.13
7. 44175.6
8. 150717.0
9. 32432.8
10. 15599.3
F = qe E = me a
qe E
a=
me
The final velocity is
vf2 = vi2 + 2 a d = 2 a d
since vi = 0, so
2 d qe E
me
2
v me
E=
2dq
! e
"#
$
2 × 107 m/s2 9.11 × 10−31 kg
=
2 (4 cm) (1.6 × 10−19 C)
v2 =
= 28468.8 N/C .
008 10.0 points
1) Two uncharged metal balls, Y and X, each
stand on a glass rod and are touching.
X
Y
2) A third ball carrying a negative charge, is
brought near the first two.
−
X
Y
3) While the positions of these balls are fixed,
ball Y is connected to ground.
Correct answer: 28468.8 N/C.
−
Explanation:
and
The magnitude of the force is
2. Enet = −E1 − E2
3. Enet = −E1 + E2
3
X
Y
Version 061 – Test #1 – Antoniewicz – (57030)
4) Then the ground wire is disconnected.
−
X
Y
5) While Y and X remain in touch, the ball
carring the negative charge is removed.
X
Y
6) Then ball Y and X are separated.
X
Y
After these procedures, the signs of the
charge qY on Y and qX on X are
1. qY is positive and qX is positive. correct
2.
qY is positive and qX is neutral.
3.
qY is neutral and qX is negative.
4.
qY is negative and qX is positive.
5.
qY is neutral and qX is positive.
6.
qY is positive and qX is negative.
7.
qY is neutral and qX is neutral.
8.
qY is negative and qX is negative.
9.
qY is negative and qX is neutral.
Explanation:
When the ball with negative charge is
brought nearby, the free charges inside Y and
X rearrange themselves. The positive charges
are attracted and go to the right (i.e. move to
X), whereas the negative charges are repelled
and collect in the left hand side of the system
Y Y, i.e., in Y.
4
When we ground Y, the negative charges
which have collected in Y are allowed to escape (they strive to the left), whereas the
positive charges in X 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 X is redistributed in
the system Y Y, i.e. they share the positive
charge (equally if identical).
Finally we separate Y and X. The signs
of the charge on Y and that on X are both
positive.
009 (part 1 of 2) 5.0 points
Consider a solid conducting sphere with a
radius a and charge Q1 on it. There is a
conducting spherical shell concentric to the
sphere. The shell has an inner radius b (with
b > a) and outer radius c and a net charge
Q2 on the shell. Denote the charge on the
inner surface of the shell by Q"2 and that on
the outer surface of the shell by Q""2 .
b , Q"2
Q1 , a
P
Q""2 , c
Find the charge Q""2 .
1. Q""2 =
(Q1 + Q2 )2
Q1 − Q2
2. Q""2 = 2 (Q1 − Q2 )
3. Q""2 = 2 (Q1 + Q2 )
4. Q""2 = Q2 − Q1
5. Q""2 = 2 (Q2 − Q1 )
Q2 − Q1
2
Q
−
Q2
1
7. Q""2 =
2
6. Q""2 =
Q2
Version 061 – Test #1 – Antoniewicz – (57030)
5
8. Q""2 = Q1 + Q2 correct
5. EP = 0
9. Q""2 = Q1 − Q2
Q1 + Q2
10. Q""2 =
2
Explanation:
Sketch a concentric Gaussian surface S
(dashed line) within the shell.
r
6. EP =
7. EP =
8. EP =
9. EP =
10. EP =
Since the electrostatic field in a conducting
medium is zero, according to Gauss’s Law,
2 ke (Q1 − Q2 )
(a + b)2
2 ke (Q1 + Q2 )
(a + b)2
2 ke Q 2
(a + b)2
4 ke Q 2
(a + b)2
4 ke (Q1 − Q2 )
(a + b)2
Explanation:
Choose as your Gaussian surface the spherical surface S concentric with the centers of
the spheres, which passes through P . Thus
Q1
$0
Q1
EP =
4 π $0 r 2
ke Q 1
=
r2
4 ke Q 1
=
.
(a + b)2
4 π r 2 EP =
Q1 + Q"2
ΦS =
$0
=0
"
Q2 = −Q1
But the net charge on the shell is
Q2 = Q"2 + Q""2 ,
so the charge on the outer surface of the shell
is
Q""2 = Q2 − Q"2
= Q2 + Q1 .
010 (part 2 of 2) 5.0 points
Find the !magnitude of
" the electric field at
!
point P %EP % ≡ EP , where the distance
a+b
from P to the center is r =
.
2
4 ke Q 1
1. EP =
correct
(a + b)2
2 ke Q 1
2. EP =
(a + b)2
2 ke Q 1 a
3. EP =
(a + b)3
4 ke (Q1 + Q2 )
4. EP =
(a + b)2
011 (part 1 of 3) 4.0 points
A hollow ball with radius 2 cm has a charge
of −4 nC spread uniformly over its surface, as
shown by the following figure:
P #0, 5, 0$ cm
−4 nC
#−5.25, 0, 0$ cm
B
R = 2 cm
+6 nC
C
#1.5, 0, 0$ cm
The center of the ball is at
!r B = #−5.25, 0, 0$ cm.
A point charge of 6 nC is located at
!r C = #1.5, 0, 0$ cm.
Version 061 – Test #1 – Antoniewicz – (57030)
You will find the net electric field at the point
P, where
!r P = #0, 5, 0$ cm.
Since the points in the figure only have x
and y components, the electric field vector at
point P will be of the form
! ) = #Ex , Ey , 0$.
E(P
Find the x component, Ex (P ). Answer in
N/C.
1. -12218.0
2. -10999.9
3. -10639.1
4. -8161.51
5. -11720.6
6. -11598.2
7. -11433.5
8. -9876.54
9. -10875.6
10. -11858.7
%
% &
% CP %
%!r % = (1.5 cm)2 + (5 cm)2
√
= 27.25 cm2
= 5.22015 cm .
The vector pointing from the isolated
charge to the point P is given by
!r CP = #−1.5, 5, 0$ cm,
so the unit vector point in the same direction
would be
!r CP
|!r CP |
#−1.5, 5, 0$ cm
=
5.22015 cm
= #−0.287348, 0.957826, 0$ .
r̂ CP =
Correct answer: −10639.1 N/C.
Explanation:
The electric field at an arbitrary point 1
due to some charge q at a point 2 is given by
! 21 =
E
1
q
r̂ 21 .
4π$0 |!r 21 |2
To find the net electric field at point P, we
will superpose the contributions from the ball
(B) and the isolated charge (C):
! )=E
! BP + E
! CP .
E(P
However, we are only interested in the x
component right now:
Ex (P ) = ExBP + ExCP .
We will find the two contributions individually, and then add them together. Let’s start
with the isolated charge. The equation for the
field looks like
1
qC CP
r̂ ,
4π$0 |!r CP |2
%
%
% CP %
so we need to find the distance %!r % and the
unit vector in the direction from the isolated
charge to point P . The distance is given by
6
Now we have everything we need to solve
for ExCP . Plugging in and taking the x component of the unit vector, we obtain
6 nC
1
(−0.287348)
4π$0 (5.22015 cm)2
6 × 10−9 C
1
(−0.287348)
=
4π$0 (0.0522015 m)2
= −5686.36 N/C .
ExCP =
Now we need to apply the same procedure
to the hollow ball to find its contribution to
Ex (P ). Remember, since point P is outside
the ball, the ball’s charge can be thought of as
if it were all concentrated at the ball’s center.
So this is really no different from the isolated
charge in this case. Following the same steps
from above, we end up with:
ExCP =
%
% &
% BP %
!
r
%
% = (−5.25 cm)2 + (5 cm)2
√
= 52.5625 cm2
= 7.25 cm .
Version 061 – Test #1 – Antoniewicz – (57030)
r̂
BP
!r BP
= BP
|!r |
#−(−5.25), 5, 0$ cm
=
7.25 cm
= #0.724138, 0.689655, 0$ .
1
qB BP
r
BP
4π$0 |!r |2 unit,x
1
−4 nC
=
(0.724138)
4π$0 (7.25 cm)2
1 −4 × 10−9 C
(0.724138)
=
4π$0 (0.0725 m)2
= −4952.75 N/C .
6 × 10−9 C
1
(0.957826)
4π$0 (0.0522015 m)2
= 18954.5 N/C .
=
−4 nC
1
(0.689655)
4π$0 (7.25 cm)2
1 −4 × 10−9 C
=
(0.689655)
4π$0 (0.0725 m)2
= −4716.91 N/C .
EyBP =
ExBP =
Now we just add the two contributions:
Ex (P ) = ExCP + ExBP
= −5686.36 N/C + (−4952.75 N/C)
= −10639.1 N/C .
Summing the contributions, we find that
Ey (P ) = = EyCP + EyBP
= 18954.5 N/C + (−4716.91 N/C)
= 14237.6 N/C .
013 (part 3 of 3) 3.0 points
Which direction most closely matches the direction of the net electric field at point P ?
012 (part 2 of 3) 3.0 points
Find Ey (P ). Answer in N/C.
1. 9537.34
2. 14237.6
3. 14074.5
4. 16274.9
5. 17305.2
6. 13202.1
7. 15620.8
8. 12295.0
9. 11251.3
10. 11005.0
1. H correct
Correct answer: 14237.6 N/C.
2. C
Explanation:
This process is very much the same as
above. The charges are the same, as well as
the distances involved; we just need to multiply by the y components of the unit vectors
this time.
H
EyCP
A
B
G
C
F
D
E
3. F
4. B
5. A
6. D
1
6 nC
=
(0.957826)
4π$0 (5.22015 cm)2
7
7. G
Version 061 – Test #1 – Antoniewicz – (57030)
8. E
Explanation:
Ex (P ) is negative, so the x component of
the field points along the negative x axis. But
Ey (P ) is positive, so that component points
upward. Combining the two into the total
electric field vector, we see that the general
direction of the field at point P is upward and
to the left.
014 10.0 points
Two charges, +3q and −q, are fixed at the
positions shown in the figure.
8
charges have the same sign and attractive if
the signs are opposite. In region (a) the force
on +q because of the charge +3q is to the left,
while the force on +q because of −q is to the
right, which means they might cancel. However, the larger charge +3q is closer than the
charge −q, so the force exerted from the interaction with −q is always smaller; hence in
region (a) the forces cannot cancel. In region
(b), the forces on +q from both +3q and −q is
to the right, so the forces cannot cancel. In region (c), charge +3q exerts a force to the right
on +q, while the force exerted by −q on +q is
to the left. While +3q has a larger charge, the
force from −q can balance that by 3q because
the charge +q can be moved closer to −q.
015
10.0 points
In which of the three regions (a), (b),
and/or (c) could one place a third charge,
+q, such that it it experiences no force at
some point in the region?
1. (a) and (b)
2. (a)
3. Not enough information
4. (a) and (c)
5. (b)
6. (b) and (c)
7. None
8. All
9. (c) correct
Explanation:
The electric force between two charged objects is proportional to the magnitude of the
charges and proportional to the inverse of the
distance squared. The force is repulsive if the
A water molecule is a permanent dipole
with a known dipole moment p = qs. There
is a water molecule in the air a very short distance x from the midpoint of a long glass
rod of length L carrying a uniformly distributed positive charge Q. The axis of the
dipole is perpendicular to the rod. Note that
s << x << L. You may neglect the small
change in the dipole moment of the water
molecule induced by the rod.
Choose the answer that correctly expresses
Version 061 – Test #1 – Antoniewicz – (57030)
the magnitude and direction (along the xaxis) of the electric force on the water
molecule. Your f inal result must be expressed only in terms of k, Q, p, L, s and
x and any constant numerical factors.
Qp
Ls2
Qp
2. k 2
Lx
2Qp
3. k 2
Lx
2Qp
4. −k 2
xL
2Qp
5. k 2
Ls
2Qp
6. −k 2 correct
Lx
Qp
7. −k 2
Lx
2Qp
8. −k 2
Ls
Qp
9. k 2
Ls
2Qp
10. k 2
xL
Explanation:
Since we are given the condition that x <<
L, we may use the approximate equation for
the E-field of a long rod
1. −k
E≈
1 2(Q/L)
4π$0
r
Calculate the force by considering each charge
independently:
F = −
1 2Qq
4π$0 L
F =
1
x−

1 2Qq 

4π$0 L 
F =
s
2
+
1 2Qq
4π$0 L
1
x+
1 2Qq
4π$0 L
s
2
−
−s
1
x−
s2
x2 −
4
1
x+



s
2
F≈ −
9
1 2Qqs
4π$0 Lx2
F ≈ −k
where the approximation is justified since
s << x.
016 (part 1 of 3) 4.0 points
Consider a nonconducting semicircular arc
with radius r. The total charge Q on the
arc is negative and is distributed uniformly.
The charge on a small segment with angle ∆θ
is labeled ∆q.
y
y
∆θ
−− A
II
I
−−
−
x
r
θ −−
III IV
−
−
x
−
O
−
−
−−
−−
−−
B
What is ∆q?
1. ∆q = 2 π Q
Q
2π
Q ∆θ
3. ∆q =
2π
2 Q ∆θ
4. ∆q =
π
2. ∆q =
5. ∆q = π Q
6. None of these
s
2
2Qp
Lx2
2Q
π
Q ∆θ
8. ∆q =
correct
π
Q
9. ∆q =
π
7. ∆q =
10. ∆q = Q
Explanation:
Version 061 – Test #1 – Antoniewicz – (57030)
The angle of a semicircle is π, so
∆q
Q
= .
∆θ
π
017 (part 2 of 3) 3.0 points
What is the magnitude of the x-component of
the electric field at the center due to ∆q?
k |∆q| cos θ
correct
r2
k |∆q|
2. ∆Ex =
r2
10
5. 335529.0
6. 8232170.0
7. 889667.0
8. 1475340.0
9. 307847.0
10. 50696.1
Correct answer: 8.23217 × 106 N/C.
Explanation:
1. ∆Ex =
3. ∆Ex = k |∆q| r 2
4. ∆Ex = k |∆q| (sin θ) r
5. ∆Ex = k |∆q| (cos θ) r 2
k |∆q| sin θ
r
k |∆q| sin θ
7. ∆Ex =
r2
6. ∆Ex =
By symmetry of the semicircle, the ycomponent of the electric field at the center is
Ey = 0 . Combining part 1 and part 2,
∆Ex =
8. ∆Ex = k |∆q| (cos θ) r
9. ∆Ex = k |∆q| (sin θ) r 2
k |∆q| cos θ
10. ∆Ex =
r
Explanation:
Negative charge attracts a positive test
charge. At O, ∆E points toward ∆q . According to the sketch, the vector ∆Ex is pointing
along the negative x axis. The magnitude of
the ∆Ex is given by
∆Ex = ∆E cos θ =
Let : Q = −28.2 µC = −2.82 × 10−5 C ,
r = 14 cm = 0.14 m , and
k = 8.98755 × 109 N · m2 /C2 .
k |∆q|
cos θ .
r2
018 (part 3 of 3) 3.0 points
Determine the magnitude of the electric field
at O . The total charge is −28.2 µC, the radius
of the semicircle is 14 cm, and the Coulomb
constant is 8.98755 × 109 N · m2 /C2 .
1. 6707700.0
2. 1079560.0
3. 358494.0
4. 498208.0
k |∆q| cos θ
k |Q|
=
cos θ ∆θ ,
2
r
π r2
so the magnitude of the electric field at the
center is
π/2
k |Q|
2 k |Q|
E = Ex =
cos θ dθ =
2
π r2
−π/2 π r
#
$
2 8.98755 × 109 N · m2 /C2
=
π (0.14 m)2
× |(−2.82 × 10−5 C)|
-
= 8.23217 × 106 N/C .
The direction is along negative x axis.
y
y
∆θ
−− A
II
I
−−
−
x
r
θ −−
III IV
− E
−
x
−
O
−
−
−−
−−
−−
B
019 (part 1 of 2) 5.0 points
Version 061 – Test #1 – Antoniewicz – (57030)
11
|!p|l
X4
|!p|s
8. 3k 3
X
|!p|s
9. k 4
X
|!p|
10. 3k 4
X
7. 6k
What is the direction, if any, of the electric
field due to this configuration at a point X >>
l >> s?
1. −x
Explanation:
Using the formula for the electric field of a
dipole along its perpendicular axis,
! dip⊥ = −k !p
E
r3
2. +y correct
!
3. E(X)
=0
4. +x
5. −y
Explanation:
Since the dipole to the right of the origin is
closer to point X, its electric field will dominate; since the moment of the dipole points
in the − − y direction, the electric field at X
points in the +y direction.
020 (part 2 of 2) 5.0 points
Using a procedure similar to that used to
calculate the electric field of a dipole, find an
!
approximate algebraic expression for |E(X)|,
the magnitude of the electric field due to the
configuration at this point X, X >> l >> s.
|!p|l
correct
X4
|!p|ls
2. 3k 5
X
|!p|
3. 6k 3
X
|!p|l
4. 3k 3
X
|!p|s
5. 6k 4
X
1. 3k
6. 0
and expressing the distance to the left and
right dipoles as X + l/2 and X − l/2 respectively, we apply the superposition principle at
X:
|!p|
|!p|
(−ŷ)
+
k
ŷ
(X + l/2)3
(X − l/2)3
/
.
1
1
!
−
ŷ
Enet (X) = k|!p|
(X − l/2)3 (X + l/2)3
0.
/−3 .
/−3 1
|!
p
|
l
l
! net (X) = k
E
1−
− 1+
ŷ
X3
2X
2X
! net (X) = k
E
Making the small argument approximation
(since l << X) and taking the magnitude we
! net (X)|:
obtain an approximate value for |E
..
.
//
.
.
///
l
|!p|
l
1−3 −
− 1−3
k 3
X
2X
2X
therefore,
! net (X) ≈ k |!p| 6l
E
X 3 2X
! net (X) ≈ 3k |!p|l
E
X4
021
10.0 points
Consider symmetrically placed rectangular
insulators with uniformly charged distributions of equal magnitude as shown.
Version 061 – Test #1 – Antoniewicz – (57030)
y
++
++
−−
−−
x
! net at the origin is
The net field E
1. aligned with the positive x-axis.
2. zero and the direction is undefined.
3. aligned with the negative y-axis.
4. non-zero and is not aligned with either
the x- or y-axis.
5. aligned with the negative x-axis. correct
6. aligned with the positive y-axis.
Explanation:
At the origin, the positive slab of charge
produces an electric field pointed into quadrant III (away from the positively charged
slab). The negatively charged slab produces
an electric field of equal magnitude (as the
positively charged slab) but pointing into
quadrant II (toward the negatively charged
slab). The x-components of the two fields add
(producing Ex < 0), while the y-components
cancel, so the electric field is along the negative x-axis.
022 10.0 points
An electron 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 electron, there is a force F between the
electron and the carbon atom. If we change r
F"
where F "
to 2.6r, what will be the ratio of
F
is the new force between the two? Hint: To
find the r-dependence of the force, first find
the induced polarization (dipole moment) of
the atom as a function of r. Then find the
force exerted by this induced dipole on the
electron.
1. 0.0058105
12
2. 0.0022009
3. 0.0029802
4. 0.019404
5. 0.012559
6. 0.0041152
7. 0.0012621
8. 0.0084165
9. 0.0016538
10. 0.03125
Correct answer: 0.0084165.
Explanation:
The force will be proportional to r −5 and
F"
hence the correct answer will be
=
F
0.0084165 . The calculation is as follows.
The magnitude of electric field due to the
electron at the location of the carbon atom
will be given by
Ee =
1 e
4π$0 r 2
From this, the magnitude of the induced
dipole moment of the carbon atom can be
written down as p = αEe . This dipole moment will be pointing towards the electron,
and hence the electron will lie on the axis of
this dipole. The magnitude of the electric
field due to this dipole at the location of the
electron will be given by
Ep =
1 2p
4π$0 r 3
Here, we have used the fact that r >> d. The
magnitude of force on the electron due to the
dipole (and vice versa) can be obtained as
F = Ep e. Combining all the equations, we
get
.
/2
1 2αEe e
1
2αe2
F = Ep e =
=
4π$0 r 3
4π$0
r5
This implies that when r to 2.6r, the electric
field field, and consequently the force, is decreased by a factor of (2.6)−5 = 0.0084165,
and hence the answer is
F"
= (2.6)−5 = 0.0084165
F