Version 098 – QZ1 – ditmire – (58565)
This print-out should have 23 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 (part 1 of 2) 10.0 points
An electron with 9.109 × 10−31 kg is accelerated from rest for 1 × 10−9 s by a uniform electric field that exerts a force of 7.3 × 10−15 N
on the electron.
What is the magnitude of the electric field?
The fundamental charge is 1.602 × 10−19 C .
1. 28089.9
2. 46192.3
3. 21223.5
4. 45568.0
5. 33707.9
6. 39950.1
7. 19975.0
8. 20599.3
9. 43695.4
10. 48064.9
Correct answer: 45568 V/m.
Explanation:
Let : qe = −1.602 × 10−19 C and
F = 7.3 × 10−15 N .
The magnitude of the force is
F = |qe | E
F
7.3 × 10−15 N
E=
=
|qe |
| − 1.602 × 10−19 C|
= 45568 V/m .
002 (part 2 of 2) 10.0 points
What is the speed of the electron after it has
accelerated for 1 × 10−9 s?
1. 9770560.0
2. 7794490.0
3. 5159730.0
4. 5379300.0
5. 8014050.0
6. 6367330.0
7. 7355360.0
8. 3952140.0
1
9. 8782520.0
10. 7904270.0
Correct answer: 8.01405 × 106 m/s.
Explanation:
Let : me = 9.109 × 10−31 kg
t = 1 × 10
−9
and
s.
The force on the electron is F = me a and
its acceleration is
∆v
F
=
∆t
me
(7.3 × 10−15 N)(1 × 10−9 s)
F ∆t
=
∆v =
me
9.109 × 10−31 kg
a=
= 8.01405 × 106 m/s .
003 10.0 points
Consider a charged semicircular arc with radius 87 cm and total charge −74.5 µC distributed uniformly on the semicircle.
y
y
∆θ
−−− A
−
−
r
θ −−
−
−
−
O
−
−
−−
−−
−−
B
II
I
x
III
IV
x
Find the magnitude of the electric field at
O . The value of the Coulomb constant is
8.98755 × 109 N · m2 /C2 .
1. 402509.0
2. 563170.0
3. 810892.0
4. 605217.0
5. 890560.0
6. 51235.5
7. 184773.0
8. 618024.0
9. 530494.0
10. 275863.0
Version 098 – QZ1 – ditmire – (58565)
5
Correct answer: 5.6317 × 10 N/C.
Explanation:
Let : q = −74.5 µC = −7.45 × 10−5 C ,
r = 87 cm = 0.87 m , and
k = 8.98755 × 109 N · m2 /C2 .
y
y
∆θ
−−− A
−
−
r
θ −−
− E
−
−
O
−
−
−−
−−
−−
B
II
1. 0.128793
2. 0.174141
3. 0.0426897
4. 0.101429
5. 0.0577787
6. 0.0169779
7. 0.0360777
8. 0.151328
9. 0.0235921
10. 0.0302473
Correct answer: 0.151328 mJ.
I
x
III
IV
x
Explanation:
When the three capacitors are connected in
series,
1
1
1
1
+
+
C1 C2 C3
1
=
1
1
1
+
+
5.2 µF 1.6 µF 9.6 µF
= 1.08522 µF
Cs =
By symmetry of the semicircle, the ycomponent of the electric field at the center is Ey = 0 . We need consider only the
x-component of the electric field, so
ds = r dθ
q
q
∆q = λ ds = λ r dθ =
r dθ = dθ
πr
π
k |q|
k |∆q| cos θ
=
cos θ ∆θ ,
∆Ex =
r2
π r2
and the magnitude of the electric field at the
center is
Z π/2
k |q|
2 k |q|
E = Ex =
cos θdθ =
2
π r2
−π/2 π r
=
2
2 (8.98755 × 109 N · m2 /C2 )
π (0.87 m)2
× |(−7.45 × 10−5 C)|
= 5.6317 × 105 N/C
with direction along the negative x axis.
004 (part 1 of 2) 10.0 points
Three capacitors of 5.2 µF, 1.6 µF, and 9.6 µF
are connected to the terminals of a 16.7 V
battery.
How much energy does the battery supply
if the capacitors are connected in series?
and the energy supplied by the battery is
1
1
Cs V 2 = 1.08522 µF (16.7 V)2
2
2
= 0.151328 mJ .
U=
005 (part 2 of 2) 10.0 points
How much energy does the battery supply if
the capacitors are connected parallel?
1. 2.01228
2. 3.13681
3. 2.2869
4. 5.1779
5. 0.618164
6. 4.92134
7. 1.14109
8. 2.03913
9. 2.76479
10. 1.03824
Correct answer: 2.2869 mJ.
Explanation:
Version 098 – QZ1 – ditmire – (58565)
When the capacitors are connected in parallel,
Cp = C1 + C2 + C3
= 5.2 µF + 1.6 µF + 9.6 µF
= 16.4 µF
is directed in the positive x-direction. Suppose a 24 µC charge moves from the origin to
point A at the coordinates, (30 cm, 50 cm).
y
280 V/m
(30 cm, 50 cm)
A
and the energy supplied by the battery is
1
1
U = Cp V 2 = 16.4 µF (16.7 V)2
2
2
= 2.2869 mJ .
006 10.0 points
Imagine a charge in the center of a conducting,
hollow sphere. There is no net charge on the
sphere, and the sphere is not connected to
ground.
q
3
x
O
What is the absolute value of the change in
potential from the origin to point A?
1. 73.92
2. 93.24
3. 84.0
4. 65.78
5. 98.56
6. 91.31
7. 82.94
8. 77.91
9. 78.98
10. 101.16
Correct answer: 84 V.
Explanation:
Let :
1. The charge will return to the center.
2. All of these can happen, depending on the
size of the charge.
3. The charge will move away from the center. correct
4. There is not enough information to tell.
5. The charge will remain stationary.
Explanation:
There will be an image-charge attracting it
toward any metal surface. It will move toward
the closest metallic surface because the closest
image-charge attraction will be stronger.
Any charge (free to move) will move toward
the closest conductor it can find.
007 10.0 points
A uniform electric field of magnitude 280 V/m
x = 30 cm ,
y = 50 cm ,
and
~ = 280 V/m .
kEk
The potential difference from O to A is
defined as
Z A
~ · d~s .
∆V = VA − VO = −
E
O
~ = (280 V/m) ı̂ . We need
We know that E
to choose a path to integrate along. Because
the electric force is conservative, it doesn’t
matter which path we take; they all give the
same answer. There are two choices of path
for which the math is simple (see the figure
below.)
y
E
(x, y) A
II
O
I
I
What will happen if the charge is moved a
little away from the center?
B
x
Version 098 – QZ1 – ditmire – (58565)
Path I:
VA − VO = (VA − VB ) + (VB − VO ),
~ and d~s are both along the
From O to B, E
~ · d~s = E dx. From B to A, E
~ and
x-axis, so E
~
d~s are perpendicular, so E · d~s = 0.
Z B
Z A
~
~ · d~s
VA − VO = −
E · d~s −
E
ZOx
Z yB
=−
E dx −
0 dy
0
0
Z x
= −E
dx = −E ∆x
4
1. 63.6269
2. 304.128
3. 1110.63
4. 234.476
5. 928.122
6. 118.911
7. 814.993
8. 330.947
9. 566.352
10. 411.802
Correct answer: 1110.63 N m2 /C.
Explanation:
O
= −(280 V/m) (0.3 m)
= −84 V .
Let : s = 9.37 m ,
h = 4.23 m , and
E = 50.6 N/C .
The absolute value is
|∆V | = 84 V .
By Gauss’ law,
~ · d~s = E cos θ ds .
Path II: In this case, E
x
⇒ x = l cos θ .
where cos θ =
l
Z l
VA − VO = −E cos θ
ds
O
= −E l cos θ
= −E x .
which is the same as the result for the other
path.
008 10.0 points
A (9.37 m by 9.37 m) square base pyramid
with height of 4.23 m is placed in a uniform
vertical electric field of 50.6 N/C. The pyramid encloses no charge.
~ ·A
~.
Φ=E
Since there is no charge contained in the
pyramid, the net flux through the pyramid
must be 0 N/C. Since the field is vertical, the
total flux through the base of the pyramid is
equal and opposite to the total flux through
the four sides, so the total flux through the
base of the pyramid is
Φtotal = E A = E s2 = (50.6 N/C) (9.37 m)2
= 4442.52 N m2 /C ,
and the flux through one of the four triangular
sides is
Φside =
4.23 m
4442.52 N m2 /C
= 1110.63 N m2 /C .
4
009
10.0 points
b
9.37 m
50.6 N/C
Calculate the electric flux which goes out
through one of the pyramid’s four slanted
surfaces.
When the switch is in position a, an isolated
capacitor of unknown capacitance has been
charged to a potential difference of 114 V.
When the switch is moved to position b, this
charged capacitor is then connected parallel
to the uncharged 12 µF capacitor. The voltage across the combination becomes 27 V.
Version 098 – QZ1 – ditmire – (58565)
5
Consider a long, uniformly charged, cylindrical insulator of radius R with charge density
1.1 µC/m3 .
12 µF
C1
S b
114 V
a
R
3 cm
Calculate the unknown capacitance.
1. 1.47059
2. 2.64158
3. 5.34286
4. 8.26429
5. 2.45575
6. 3.72414
7. 3.25217
8. 5.85714
9. 2.17308
10. 8.7
Correct answer: 3.72414 µF.
Explanation:
Let :
V1 = 114 V ,
V2 = 27 V and
C2 = 12 µF .
Since the capacitor has been charged to a
potential difference of V1 , the charge in the
capacitor is Q = C1 V1 , where C1 is the unknown capacitance. When we connect the
second capacitor parallel with the first one,
the charge of the system is still Q, while the
total capacitance is Ctotal = C1 + C2 . Denoting the voltage across the combination as
V2 , we have Q = Ctotal V2 . Equating the two
expressions we obtained for Q, we get
C1 V1 = (C1 + C2 ) V2
C2 V2
(12 µF) (27 V)
C1 =
=
V1 − V2
114 V − 27 V
= 3.72414 µF .
010
10.0 points
What is the magnitude of the electric field
inside the insulator at a distance 3 cm < R
from the axis? The permittivity of free space
is 8.8542 × 10−12 C2 /N · m2 and the volume
of a cylinder with radius r and length ℓ is
V = π r2 ℓ .
1. 1106.82
2. 1863.52
3. 2591.99
4. 1535.99
5. 1490.82
6. 1982.11
7. 1688.46
8. 1185.88
9. 1219.76
10. 2078.11
Correct answer: 1863.52 N/C.
Explanation:
Let :
r = 3 cm = 0.03 m ,
ρ = 1.1 µC/m3 ,
= 1.1 × 10−6 C/m3 , and
ǫ0 = 8.8542 × 10−12 C2 /N · m2 .
Consider a cylindrical Gaussian surface of
radius r and length ℓ much less than the
length of the insulator so that the component of the electric field parallel to the axis is
negligible.
Version 098 – QZ1 – ditmire – (58565)
6
4. negative
ℓ
r
R
Explanation:
The field on the surface is directed toward
the inner shell wall, so the sign of the charge
must be positive. Displace the +q charge
from the center of the uncharged spherical
conducting shell.
The flux leaving the ends of the Gaussian
cylinder is negligible, and the only contribution to the flux is from the side of the cylinder.
Since the field is perpendicular to this surface,
the flux is
Φs = 2 π r ℓ E ,
+
+
-
+
-
-
+
+
and the charge enclosed by the surface is
+
- +
+
- - +
- +
+q +
+
Qencl = π r 2 ℓ ρ .
Using Gauss’ law,
Qenc
ǫ0
π r2 ℓ ρ
2πrℓE =
ǫ0
ρ
E=
r
2 ǫ0
1.1 × 10−6 C/m3 (0.03 m)
=
2 (8.8542 × 10−12 C2 /N · m2 )
Φs =
= 1863.52 N/C .
011 (part 1 of 4) 10.0 points
Consider an uncharged, conducting hollow
spherical shell. The electric field on the inside
surface of the spherical shell is measured to be
directed toward the inner shell wall and varies
in magnitude over the spherical shell’s inside
surface. It is known that the only charge in
the hollow cavity inside the spherical shell is
a point charge with magnitude q.
What is the sign of the charge q?
The field lines leave the point charge radially and arrive at the inside surface of the
uncharged spherical conductor perpendicular
to the inside surface (radial). There are as
many field lines touching the inside surface
as the outside surface (the spherical shell has
net charge of zero). The field lines leave the
outside surface radially directed and are uniformly distributed.
If this is not obvious, use Gauss’ law
Φc =
~ · dA
~ = Qencl = Q .
E
ǫ0
ǫ0
A
I
Since the normal is outward and the electric
field is outward as well, the flux is positive
and so is the charge.
012 (part 2 of 4) 10.0 points
What can you say about the position of the
charge?
1. A point charge cannot create a charge distribution on the inside surface of the spherical
shell which varies in magnitude.
1. cannot be determined
2. zero
3. positive correct
2. The charge is zero, so position doesn’t
matter.
3. The charge is inside the spherical shell
Version 098 – QZ1 – ditmire – (58565)
but not at the center of the spherical shell.
correct
7
4. The charge is outside of the spherical
shell.
014 (part 4 of 4) 10.0 points
What can you say about the charge distribution on the outside surface of the spherical
shell?
5. The charge is in the center of the spherical
shell.
1. The charge distribution is positive and is
uniform. correct
Explanation:
Since the electric field on the surface is
found to vary in magnitude, the charge is
not in the center of the spherical shell. This
follows from the fact that the strength of the
electric field varies with distance, so if the
electric field strength varies as we go over
the spherical shell’s surface, then so must the
distance to the charge. If the charge would
have been in the center, the distance from a
point on the surface to the charge would have
been constant over the surface.
013 (part 3 of 4) 10.0 points
What is the net electric flux through the
spherical shell?
1. Φ = 0 correct
2. Φ =
q
ǫ0
3. Φ = 2 q
4. Φ =
q
2
q
2 ǫ0
2q
6. Φ =
ǫ0
5. Φ =
7. Φ = q
Explanation:
The net charge inside the spherical shell is
zero; i.e., the charge on the inside surface of
the point charge is equal and opposite that of
the point charge. Again, according to Gauss’
law
I
~ · dA
~ = Qencl = 0 ,
E
Φc =
ǫ0
A
where E = 0 in the conductor, so the net
electric flux through the spherical shell is zero.
2. The charge distribution is zero.
3. The charge distribution is positive and is
not uniform.
4. The charge distribution is negative and is
uniform.
5. The charge distribution is negative and is
not uniform.
Explanation:
The induced charge on the inside of the
spherical shell is equal in magnitude but opposite in sign to that of the point charge.
The field just inside the outside surface of the
spherical shell is zero, so the charge distribution on the outside surface of the spherical
shell is symmetric and constant. The sign
and magnitude of the charge is the same as
the point charge inside the spherical shell.
If the field lines arriving at the inside surface or leaving the outside surface of the
spherical conductor were not radial, then
charge on the surface of the conductor would
experience a force and move (since there
would be an electric field along the surface
of the conductor).
015 10.0 points
Three point charges lie in a straight line along
the y-axis. A charge of −5.5 µC is at 5.0 m,
and a charge of −4.1 µC is at −4.0 m. The
net electric force on the third point charge is
zero.
Where along the y−axis is this charge located?
1. -2.88291
2. -0.14304
3. 0.170112
4. 1.71487
Version 098 – QZ1 – ditmire – (58565)
The third point charge is located at
(x3 , y3 ) = (0 m, 0.170112 m).
5. 0.675868
6. -0.82324
7. -0.306978
8. -1.06187
9. 0.127735
10. -0.506201
Correct answer: 0.170112 m.
Explanation:
= −5.5 µC = −5.5 × 10−6 C ,
= 5.0 m ,
= −4.1 µC = −4.1 × 10−6 C ,
= −4.0 m .
Let : q1
y1
q2
y2
8
and
016 (part 1 of 3) 10.0 points
Consider two “solid” conducting spheres with
radii r1 = 6 R and r2 = 7 R , separated by a
large distance so that the field and the potential at the surface of sphere #1 only depends
on the charge on #1 and the corresponding
quantities on #2 only depend on the charge
on #2. Place an equal amount of charge on
both spheres: q1 = q2 = Q .
r2
r1
r1,2 = |y1 | + |y2 | = 9 m .
The equilibrium position for q3 is between
q1 and q2 , so if r1,3 = x, then r2,3 = r1,2 − x.
q1 q2
Fe = kC
and the net electric force is
r2
zero, so
kC
F1,3
q1 q3
2
r1,3
q1
2
r1,3
q1
x2
= F2,3
q2 q3
= kC 2
r2,3
q2
= 2
r2,3
q2
=
(r1,2 − x)2
(r1,2 − x)2
q2
=
2
x
q
r1
r1,2 − x
q2
=
x
q1
r
q2
r1,2 − x = x
q1
r q2
x 1+
= r1,2
q1
r1,2
9m
r =
r
q2
−4.1 × 10−6 C
1+
1
+
q1
−5.5 × 10−6 C
= 4.82989 m
x=
#1
Q
#2
After the electrostatic equilibrium on each
sphere has been established, what is the ratio
V2
of the potentials at the “centers” of the
V1
two solid conducting spheres?
1.
2.
3.
4.
5.
6.
7.
8.
9.
V2
V1
V2
V1
V2
V1
V2
V1
V2
V1
V2
V1
V2
V1
V2
V1
V2
V1
49
72
36
=
49
49
=
36
=
=1
=
=
=
=
=
6
correct
7
7
12
7
3
49
18
7
6
below q1 .
y3 = y1 − x = 5 m − 4.82989 m
= 0.170112 m .
Q
Explanation:
Version 098 – QZ1 – ditmire – (58565)
For a solid conducting sphere of radius R
and charge Q, the constant potential throughout the sphere is
V (r) = k
Q
,
r
r ≤ R.
r2
7R
7
=
= , so
r1
6R
6
Q
r2
Q
k
r1
k
V2
=
V1
=
r1
6R
6
=
= .
r2
7R
7
017 (part 2 of 3) 10.0 points
E2
What is the ratio
of the electric fields at
E1
the “surfaces” of the two spheres?
1.
2.
3.
4.
5.
6.
7.
8.
9.
E2
E1
E2
E1
E2
E1
E2
E1
E2
E1
E2
E1
E2
E1
E2
E1
E2
E1
7
3
7
=
6
36
=
correct
49
7
=
12
=
Q
2 2
r22
r1
E2
6R
36
=
=
.
=
=
Q
E1
r2
7R
49
k 2
r1
k
018 (part 3 of 3) 10.0 points
Now “connect” the two spheres with a wire.
There will be a flow of charge through the
wire until equilibrium is established.
r2
r1
#1
q1
q2
#2
E2
of the electric fields at
E1
the “surfaces” of the two spheres once equilibrium has been reached?
What is the ratio
E2
=1
E1
E2
49
2.
=
E1
36
E2
6
3.
= correct
E1
7
E2
7
4.
=
E1
6
7
E2
=
5.
E1
3
E2
36
=
6.
E1
49
E2
49
7.
=
E1
18
E2
7
8.
=
E1
12
49
E2
9.
=
E1
72
Explanation:
From Gauss’ law, for a conducting sphere
of radius R with charge q,
1.
=1
49
72
6
=
7
49
=
18
49
=
36
=
Explanation:
For a conducting sphere of radius R, the
charge Q is uniformly distributed at the surface, so from Gauss’ law, the electric field on
the surface is
E(r) = k
9
Q
,
r2
and
q
, r ≥ R,
r2
q
V (r) = k , r ≤ R .
r
E(r) = k
and
Version 098 – QZ1 – ditmire – (58565)
After equilibrium has been established, the
ends of the wire are at the same potential, so
V2 = V1 and
q2
q2
k
2
r
E2
r2
= q12 =
q
E1
1
k 2
k
r1
r1
1
V2
r1
r2
=
=
1
r2
V1
r1
k
1
r2
1
r1
=
6R
6
= .
7R
7
019 (part 1 of 2) 10.0 points
A solid conducting sphere of radius R1 and
total charge q1 is enclosed by a conducting
shell with an inner radius R2 and outer radius
R3 and total charge q2 .
O
2. σ =
A
R3
Find the electric field at A. Let the distance
OA = a.
1. EP = 0
2. EP =
3. EP =
4. EP =
5. EP =
6. EP =
7. EP =
q2
k a2
k q2
a2
k q1
R12
k q1
R1
k (q1 + q2 )
a
k q1
correct
a2
qencl
q1
=
ǫ0
ǫ0
q1
1 q1
=k 2.
EA =
2
4 π ǫ0 a
a
EA (4 π a2 ) =
1. σ =
q1
R2
k (q1 + q2 )
a2
q1
9. EP =
k a2
k q1
10. EP =
a
Explanation:
Consider a spherical Gaussian surface
through the point A. A is outside of the
sphere, so the enclosed charge qencl = q1 can
be treated as a point charge and
8. EP =
020 (part 2 of 2) 10.0 points
Find the surface charge density on the outer
surface of the shell.
q2
R1
10
3. σ =
4. σ =
5. σ =
6. σ =
7. σ =
8. σ =
9. σ =
10. σ =
q1 + q2
correct
4 π R32
q2
4 π R32 ǫ0
q2 − q1
4 π R32
q2
4 π R22
q1
4 π R12
q2 − q1
4 π R32 ǫ0
q2
4 π R22 ǫ0
q1
4 π R12 ǫ0
q2
4 π R32
q1 + q2
4 π R32 ǫ0
Explanation:
Inside the conductor the electric field is
zero, so the charge on the inner surface of the
shell is q2′ = −q1 , the charge on the outer
surface is q2′′ = q2 − q2′ = q2 + q1 , and the
charge density is
Version 098 – QZ1 – ditmire – (58565)
σ=
q2′′
q1 + q2
=
.
2
4 π R3
4 π R32
021 10.0 points
A rod of length 12.4 m and total charge
Q lies along the x-axis, with its left end at
the origin. The rod has a non-uniform linear
charge density λ = α x , where α = 0.02 C/m2
and x is the position. Point A lies on the xaxis a distance 4.6 m to the left of the rod, as
shown.
∆x
x
4.6 m
x
A
O
12.4 m
Find the potential at A. The Coulomb
constant is 8.98755 × 109 N m2 /C2 .
1. 675192000.0
2. 4459900000.0
3. 3291960000.0
4. 491279000.0
5. 1863470000.0
6. 1450970000.0
7. 479742000.0
8. 2413820000.0
9. 1148080000.0
10. 284467000.0
Correct answer: 1.14808 × 109 V.
Explanation:
Let :
k = 8.98755 × 109 N m2 /C2 ,
d = 4.6 m ,
α = 0.02 C/m2 , and
L = 12.4 m .
The potential is the integral over the rod
of the contribution from small elements of the
rod:
Z
Z
Z L
k dq
k α x dx
V = dV =
=
d+x
d+x
0
Z L
x dx
.
= kα
0 x+d
x
x+d−d
d
Since
=
=1−
,
x+d
x+d
x+d
11
!
1
V = kα
dx − d
dx
0
0 x+d
L+d
= k α L − d ln
d
9
= 8.98755 × 10 N m2 /C2
× 0.02 C/m2
× 12.4 m − (4.6 m)
12.4 m + 4.6 m
× ln
4.6 m
Z
L
Z
L
= 1.14808 × 109 V .
022 10.0 points
A 5.4 µC charge located at the origin of
a cartesian coordinate system is surrounded
by a nonconducting hollow sphere of radius
5.3 cm. A drill with a radius of 1.22 mm is
aligned along the z axis, and a hole is drilled
in the sphere.
Calculate the electric flux through the
hole.
The permittivity of a vacuum is
8.8542 × 10−12 C2 /N · m2 .
1. 35.2456
2. 17.8466
3. 80.789
4. 166.789
5. 7.28397
6. 50.4312
7. 22.5307
8. 12.011
9. 25.8408
10. 40.2475
Correct answer: 80.789 N · m2 /C.
Explanation:
Let : q = 5.4 µC ,
R = 5.3 cm = 0.053 m ,
r = 1.22 mm = 0.00122 m , and
ǫ0 = 8.8542 × 10−12 C2 /N · m2 .
By Gauss’ Law,
Φ=
I
~ · dA
~ = q ,
E
ǫ0
Version 098 – QZ1 – ditmire – (58565)
so the total flux emanating from the point
charge is
5.4 µC
Φtot =
8.8542 × 10−12 C2 /N · m2
= 6.0988 × 105 N · m2 /C .
The flux through the hole can be expressed in
terms of the ratio of the area of the hole to
the area of the sphere:
Ahole
π r2
r2
Φ
=
=
=
,
Φtot
4 π R2
4π R2
4 R2
where R is the area of the sphere. Since the
hole is small, the surface area of the sphere
“missing” due to the hole can be approximated by the planar area of the hole, π r 2
(i.e., we neglect the “curved part” of the
sphere since the hole is small), where r is
the radius of the drill. Our final expression is
then
r2
Φ = Φtot
4 R2
= 6.0988 × 105 N · m2 /C
(0.00122 m2 )
×
4 (0.053 m)2
= 80.789 N · m2 /C .
Note: If we do not make the approximation
that the area missing from the sphere is just
the planar area π r 2 , we find the expression
!
r
1
r2
Φ = Φtot ·
1− 1− 2 ,
2
R
which, for r ≪ R, reduces to the previous
expression. (This also gives the same answer
to within the numerical precision required in
this problem.)
11 µF
11 µF
8 µF
156 V
023 10.0 points
Consider the capacitor network
11 µF
11 µF
y
z
11 µF
11 µF
12
What is the charge on the 8 µF capacitor
centered on the left directly between points y
and z?
1. 0.001
2. 0.000676
3. 0.000576
4. 0.001314
5. 0.001248
6. 0.001332
7. 0.00152
8. 0.002064
9. 0.000549
10. 0.000696
Correct answer: 0.001248 C.
Explanation:
Let :
C≡
Cx = 8 × 10−6 F
V = 156 V .
and
q
, so
V
q = Cx V = (8 × 10−6 F) (156 V)
= 0.001248 C .