Version 001 – Exam 1 – shih – (57480)
µN
5
50
16
1.
9m
5
+
5 nC
µN
9m
◦
14.0362
17
+
7 nC
9m
+
7 nC
4 6 µN
q1 q2
r2
~ net = F
~1 + F
~3
F
q
2
2
Fnet = Fnet,x
+ Fnet,y
Felectric = kC
9m
+
5 nC
Find the magnitude of the net electric force
on the 7 nC charge. The Coulomb constant is
8.98755 × 109 N · m2 /C2 .
1. 1.33978e-08
2. 2.01278e-08
3. 5.23443e-09
4. 1.74007e-09
5. 3.88861e-09
6. 2.53508e-09
7. 2.13158e-08
8. 2.26446e-09
9. 2.83544e-08
10. 1.48865e-09
Correct answer: 2.26446 × 10−9 N.
Explanation:
Let :
2. 264
94
3 nC
+
9m
1.
Holt SF 17Rev 21
001 (part 1 of 2) 10.0 points
Three positive point charges are arranged in a
triangular pattern in a plane, as shown below.
3 nC
+
9m
This print-out should have 26 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
1
q1 = +3 nC = +3 × 10−9 C ,
q2 = +7 nC = +7 × 10−9 C ,
q3 = +5 nC = +5 × 10−9 C ,
(x1 , y1 ) = (0 m, 9 m) ,
(x2 , y2 ) = (9 m, 0 m) ,
(x3 , y3 ) = (0 m, −9 m) , and
kC = 8.98755 × 109 N · m2 /C2 .
The distances between the 7 nC charge and
q1 and q2 are
r12 = (x2 − x1 )2 + (y2 − y1 )2
= (x2 − 0)2 + (0 − y1 )2
= x22 + y12 = (9 m)2 + (−9 m)2
= 162 m2 , and
r32 = (x3 − x2 )2 + (y3 − y2 )2
= (0 − x2 )2 + (y3 − 0)2
= x22 + y32 = (−9 m)2 + (−9 m)2
= 162 m2 .
Consider the magnitudes of the forces. The
repulsive force
q2 q1
F1 = kC 2
r1
= 8.98755 × 109 N · m2 /C2
7 × 10−9 C 3 × 10−9 C
×
162 m2
= 1.16505 × 10−9 N
acts downward and to the right along the line
connecting q2 and q1 , and the repulsive force
q2 q3
F3 = kC 2
r3
= 8.98755 × 109 N · m2 /C2
7 × 10−9 C 5 × 10−9 C
×
162 m2
= 1.94175 × 10−9 N
Version 001 – Exam 1 – shih – (57480)
acts upward and to the right along the line
connecting q2 and q3 .
Now, lets consider the directions of the
forces.
Since |x2 | = |y1 | = |y3 | ,
θ21 = −θ23 = 45◦ , and
Fx = F1 cos 45◦ + F3 cos 45◦
= (1.16505 × 10−9 N) cos 45◦
+ (1.94175 × 10−9 N) cos 45◦
= 2.19684 × 10−9 N
Fy = −F1 sin 45◦ + F3 sin 45◦
= −(1.16505 × 10−9 N) sin 45◦
+ (1.94175 × 10−9 N) sin 45◦
=q
5.49211 × 10−10 N
Fx2 + Fy2
Fnet =
=
h
2.19684 × 10−9 N
2
+ 5.49211 × 10−10 N
= 2.26446 × 10−9 N .
2 i 21
002 (part 2 of 2) 10.0 points
What is the direction of this force (measured
from the positive x-axis as an angle between
−180◦ and 180◦ , with counterclockwise positive)?
1. 18.4349
2. 30.9638
3. 26.5651
4. -23.1986
5. 11.3099
6. 14.0362
7. -15.9454
8. 12.9946
9. 21.8014
10. -32.4712
Correct answer: 14.0362 .
◦
Explanation:
tan θ =
Fy
Fx
θ = arctan
Fy
Fx
2
= tan
−1
5.49211 × 10−10 N
2.19684 × 10−9 N
= 14.0362◦ .
The net force on the 7 nC charge is
2.26446 × 10−9 N acting 14.0362 ◦ above the
positive x-axis.
Magnitude of Force
003 10.0 points
There are two identical small metal spheres
with charges 64.3 µC and −41.622 µC. The
distance between them is 6 cm. The spheres
are placed in contact then set at their original
distance.
Calculate the magnitude of the force
between the two spheres at the final
position.
The Coulomb constant is
8.98755 × 109 N · m2 /C2 .
1. 2952.86
2. 67.3856
3. 320.989
4. 9.64534
5. 8.28741
6. 57.464
7. 133.051
8. 45.6884
9. 3.22108
10. 39.5515
Correct answer: 320.989 N.
Explanation:
Let :
qA = 64.3 µC
= 6.43 × 10−5 C ,
qB = −41.622 µC
= −4.1622 × 10−5 C ,
d = 6 cm = 0.06 m .
and
When the spheres are in contact, the
charges will rearrange themselves until equilibrium is reached. Each sphere will then have
half of the original total charge:
q=
qA + qB
2
Version 001 – Exam 1 – shih – (57480)
6.43 × 10−5 C−4.1622 × 10−5 C
2
−5
= 1.1339 × 10 C .
3
=
The force between the two spheres is
q2
d2
= 8.98755 × 109 N · m2 /C2
F =k
(1.1339 × 10−5 C)2
×
(0.06 m)2
Let : L = 18.8 cm = 0.188 m and
q = −8.63 µC = −8.63 × 10−6 C .
Call the length of the rod L and its charge
q. Due to symmetry
Z
Ey = dEy = 0
and
Ex =
= 320.989 N .
Z
dE sin θ = ke
Z
dq sin θ
,
r2
where dq = λ dx = λ r dθ, so that
18.8 cm
Uniformly Charged Bent Wire 01
004 10.0 points
A uniformly charged insulating rod of length
18.8 cm is bent into the shape of a semicircle
as in the figure.
The value of the Coulomb constant is
8.98755 × 109 N m2 /C2 .
−8.63 µC
O
y
θ
b
x
ke λ
Ex = −
r
=−
If the rod has a total charge of −8.63 µC,
find the horizontal component of the electric
field at O, the center of the semicircle. Define
right as positive.
1. -46219500.0
2. -8843070.0
3. -13788500.0
4. -15471100.0
5. -65540400.0
6. -11306900.0
7. -50340000.0
8. -31098000.0
9. -18157300.0
10. -58994900.0
Correct answer: −1.37885 × 107 N/C.
Explanation:
=2
where λ =
Z
3π/2
cos θ dθ
π/2
ke λ
(sin θ)
r
3π/2
π/2
ke λ
,
r
q
L
and r = . Therefore,
L
π
2 ke q π
L2
2 (8.98755 × 109 N m2 /C2 )
=
(0.188 m)2
× (−8.63 × 10−6 C) π
Ex =
= −1.37885 × 107 N/C .
Since the rod has negative charge, the field is
pointing to the left (towards the charge distribution). A positive test charge at O would
feel an attractive force from the semicircle,
pointing to the left.
Version 001 – Exam 1 – shih – (57480)
4
T cos θ = m g − q Ey
T sin θ
q Ex
tan θ =
=
T cos θ
m g − q Ey
An Oblique Field
005 (part 1 of 2) 10.0 points
A charged cork ball is suspended on a light
string in the presence of a uniform electric
field as in the figure. The ball is in equilibrium.
(m g − q Ey ) tan θ = q Ex
q Ex
m g − q Ey =
tan θ
~ = (2.5 × 105 N/C)ı̂ + (3.4 × 105 N/C) ̂ .
E
q Ex
tan θ
mg
q Ex
q=
−
Ey
Ey tan θ
m g tan θ
q=
Ey tan θ + Ex
The acceleration of gravity is 9.8 m/s2 .
q Ey = m g −
̂
0. 5
ı̂
E
m
q=
34◦
1.3 g
(0.0013 kg) (9.8 m/s2 ) tan 34◦
(3.4 × 105 N/C) tan 34◦ + (2.5 × 105 N/C)
= 17.9275 nC .
Find the charge on the ball.
1. 11.1438
2. 6.80226
3. 11.5353
4. 12.0853
5. 15.6364
6. 17.9275
7. 19.3324
8. 9.39524
9. 4.79127
10. 10.6053
006 (part 2 of 2) 10.0 points
Find the tension in the string.
1. 0.00976323
2. 0.00685318
3. 0.00384903
4. 0.00303988
5. 0.00432643
6. 0.00962097
7. 0.0080149
8. 0.00511639
9. 0.0086204
10. 0.00491469
Correct answer: 17.9275 nC.
Explanation:
Correct answer: 0.0080149 N.
Let : Ex
Ey
θ
mb
5
= 2.5 × 10 N/C ,
= 3.4 × 105 N/C ,
= 34◦ , and
= 1.3 g = 0.0013 kg .
Explanation:
The first equation for force equilibrium
gives
In the ı̂ and ̂ directions, force equilibrium
tells us
q Ex − T sin θ = 0
q Ey + T cos θ − m g = 0 ,
T sin θ = q Ex
(1)
(2)
q Ex
sin θ
(17.9275 nC) (2.5 × 105 N/C)
=
sin 34◦
= 0.0080149 N .
T =
Charge Inside a Box 02
Version 001 – Exam 1 – shih – (57480)
007 10.0 points
A cubic box of side a, oriented as shown, contains an unknown charge. The vertically directed electric field has a uniform magnitude
E at the top surface and 2 E at the bottom
surface.
E
5
(inward is negative) and the bottom
Φbottom = 2 E a2 .
The total electric flux is
ΦE = −E a2 + 2 E a2 = E a2 .
Using Gauss’s Law, the charge inside the box
is
Qencl = ǫ0 ΦE = ǫ0 E a2 .
a
2E
How much charge Q is inside the box?
1. Qencl = 0
2. Qencl = ǫ0 E a2 correct
3. Qencl = 2
E
ǫ0 a2
4. Qencl = 2 ǫ0 E a2
5. insufficient information
E
ǫ0 a2
E
=3
ǫ0 a2
6. Qencl =
7. Qencl
8. Qencl = 3 ǫ0 E a2
9. Qencl = 6 ǫ0 E a2
1
ǫ0 E a2
2
Explanation:
Electric flux through a surface S is, by convention, positive for electric field lines going
out of the surface S and negative for lines
going in.
Here the surface is a cube and no flux goes
through the vertical sides. The top receives
10. Qencl =
Φtop = −E a2
Energy gained from A to B
008 10.0 points
A proton is released from rest in a uniform
electric field of magnitude 1.9 × 105 V/m directed along the positive x axis. The proton
undergoes a displacement of 0.2 m in the direction of the electric field as shown in the
figure.
1.9 × 105 V/m
+
−
+
−
+
−
+
−
+
−
+
−
+ vA = 0 v0
−
+ +
−
+
−
0.2 m
+
−
B −
+ A
+
−
+
−
Apply the principle of energy conservation to find the amount of the kinetic energy
gained after it has moved 0.2 m.
1. 8.65176e-15
2. 9.13241e-15
3. 9.61306e-15
4. 8.33132e-15
5. 1.92261e-15
6. 6.08827e-15
7. 1.53809e-14
8. 6.72914e-15
9. 4.80653e-15
10. 5.28719e-15
Correct answer: 6.08827 × 10−15 J.
Explanation:
Version 001 – Exam 1 – shih – (57480)
The change in the potential energy of the
proton is
∆U = qp ∆V
= (1.60218 × 10−19 C) (−38000 V)
= −6.08827 × 10−15 J .
Conservation of energy in this case is
∆K + ∆U = Kf − Ki + ∆U = 0 .
So
Kf − Ki = −∆U = 6.08827 × 10
−15
J .
Q r 2
9. Φ =
ǫ0 R
Q R
10. Φ =
ǫ0 r
Explanation:
Gauss’ law states that
I
~ · dA
~ = qin = Φ .
E
ǫ0
The net charge inside the sphere of radius
r is
qin = ρ
Uniformly Charged Sphere 05 p
009 (part 1 of 2) 10.0 points
Consider the uniformly charged sphere with
radius R, shown in the figure below.
Q is the total
charge inside
the sphere
Find the total flux passing through the
Gaussian surface (a spherical shell) with radius r.
2. Φ =
3. Φ =
4. Φ =
5. Φ =
6. Φ =
7. Φ =
8. Φ =
Φ=
Q r3
.
ǫ0 R3
010 (part 2 of 2) 10.0 points
Find the electric field at radius r .
r
Q
ǫ0
Q r2
ǫ0 R3
3
Q R
ǫ0 r
Qr
ǫ0 R
Q R2
ǫ0 r 3
Q r 3
correct
ǫ0 R
2
Q R
ǫ0 r
Q
ǫ0 r
4
Q 4
r3
π r3 =
π r3 = Q 3 ,
4
3
3
R
π R3
3
so the total flux through the surface is
R
p
1. Φ =
6
QR
r3
Q R2
2. E = k
r3
Q
3. E = k
r
Q
4. E = k 2
r
Q r2
5. E = k
R3
Q
6. E = k
R
Q
7. E = k 2
R
Q R2
8. E = k
r4
Qr
9. E = k 3 correct
R
Q r2
10. E = k
R4
Explanation:
1. E = k
Version 001 – Exam 1 – shih – (57480)
7
From Gauss’ law,
2. 6.77 eV
3. 101.3 eV
4. 709.1 eV
5. 4.37 eV
6. 5.14 eV correct
q = 1.6 × 10−19 C ,
k = 8.99 × 109 N · m2 /C2 ,
d = 2.8 × 10−10 m .
W = ∆K + ∆U = 0 − Ui
k (−e)e
k q2
=−
=
d
d
9
(8.99 × 10 N · m2 /C2 )
=
2.8 × 10−10 m = 5.14 eV .
− 5.8 × 10−9 C
1.2 cm
Calculate the electric potential at the midpoint of the base if the magnitude of the positive charge is 4.4 × 10−9 C and the magnitude
of the negative charges are 5.8 × 10−9 C.
1. -10456.8
2. -4738.42
3. -6892.22
4. -4802.96
5. -7145.57
6. -8588.8
7. -16657.1
8. -7645.36
9. -5124.04
10. -3688.97
Explanation:
and
Let : q1 = 4.4 × 10−9 C ,
q2 = q3 = −5.8 × 10−9 C ,
a = 5.5 cm ,
b = 1.2 cm , and
ke = 8.98755 × 109 N · m2 /C2 .
The energy is
× (1.6 × 10
5.8 × 10−9 C −
Correct answer: −16657.1 V.
Explanation:
Let :
cm
1. 8.91 eV
5. 5
Tipler PSE5 23 24
011 10.0 points
The distance between the K+ and Cl− ions in
KCl is 2.8 × 10−10 m.
Find the energy required to separate the
two ions to an infinite distance apart, assuming them to be point charges initially at rest.
The elemental charge is 1.6 × 10−19 C and the
Coulomb constant is 8.99 × 109 N · m2 /C2 .
5. 5
so the electric field is
Φ
Q r3 1
E=
=
4 π r2
ǫ0 R3 4 π r 2
Qr
=
4 π ǫ0 R3
kQr
=
.
R3
cm
Holt SF 18Rev 39
012 10.0 points
The three charges shown in the figure are
located at the vertices of an isosceles triangle.
The Coulomb constant is 8.98755 ×
109 N · m2 /C2 and the acceleration of gravity is 9.8 m/s2 .
4.4 × 10−9 C
+
E · 4 π r2 = Φ ,
−19
C)
2
1 eV
1.6 × 10−19 J
Let r1 , r2 and r3 be the distances of the
respective charges from the midpoint of the
base of the triangle.
Version 001 – Exam 1 – shih – (57480)
+
a
−
r1
r2
r3
−
b
r2 = r3 =
2
s
2
b
1p 2
2
4a − b2
r1 = a −
=
2
2
q
r
Then the total electric potential is
V = ke
Vtot = V1 + V2 + V3
q1
q2
q3
= ke
+ ke
+ ke
r
r2
r3
1
2 q2 2 q3
2 q1
+
+
= ke √
b
b
4a2 − b2
q1
q2 + q3
= 2 ke √
+
b
4a2 − b2
q1
2 q2
= 2 ke √
+
b
4a2 − b2
9
= 2 (8.98755 × 10 N · m2 /C2 )
"
4.4 × 10−9 C
× p
4 (0.055 m)2 − (0.012 m)2
2 (−5.8 × 10−9 C)
+
0.012 m
= −16657.1 V .
Connected Spheres 02
013 (part 1 of 2) 10.0 points
Two charged spherical conductors are connected by a long conducting wire, and a
charge of 15 µC is placed on the combination. One sphere has a radius of 4.05 cm and
the other a radius of 6.05 cm.
What is the electric field near the surface of
the sphere with the smaller radius?
1. 31520800.0
2. 40954200.0
8
3. 85309100.0
4. 36123200.0
5. 41620400.0
6. 44828500.0
7. 35028000.0
8. 29420600.0
9. 32957700.0
10. 19693800.0
Correct answer: 3.29577 × 107 V/m.
Explanation:
Let : r1 = 4.05 cm = 0.0405 m ,
r2 = 6.05 cm = 0.0605 m ,
Q = 15 µC = 1.5 × 10−5 C , and
ke = 8.98755 × 109 N · m2 /C2 .
Since the two spheres are connected by a long
conducting wire, they are at the same potential, and
q1
q2
=
(1)
r1
r2
We are also given the total charge of the combination
Q = q1 + q2
(2)
With two equations and two unknowns, we
can solve for the charge q1 on the smaller
sphere.
From (1),
r1 (Q − q1 )
r1 q2
=
r2
r2
q1 r2 = r1 Q − r1 q1
q1 =
q1 (r1 + r2 ) = r1 Q
r1 Q
q1 =
.
r1 + r2
Hence the electric field near the surface of the
sphere with radius r1 is
q1
E = ke 2
r1
Q
= ke
(r1 + r2 ) r1
= 8.98755 × 109 N · m2 /C2
×
1.5 × 10−5 C
(0.0405 m + 0.0605 m) (0.0405 m)
= 3.29577 × 107 V/m .
Version 001 – Exam 1 – shih – (57480)
014 (part 2 of 2) 10.0 points
What is the potential of the larger sphere?
1. 1051180.0
2. 1334780.0
3. 1158920.0
4. 2662000.0
5. 1517900.0
6. 3300280.0
7. 2015180.0
8. 1948630.0
9. 1781320.0
10. 2105990.0
Correct answer: 1.33478 × 106 V.
Explanation:
Since the two spheres are at the same potential, it suffices to consider the potential of
either sphere. Since q1 was obtained in the
first part of the question, we may simply determine the potential of the smaller sphere.
V = ke
q1
r1
Q
r1 + r2
= 8.98755 × 109 N · m2 /C2
= ke
×
1.5 × 10−5 C
0.0405 m + 0.0605 m
= 1.33478 × 106 V .
Tipler PSE5 24 19
015 10.0 points
What is the electrostatic potential energy of
an isolated spherical conductor of 10 cm radius that is charged to 1000 V? The Coulomb
constant is 8.99 × 109 N · m2 /C2 .
1. 208.565
2. 16.6852
3. 11.1235
4. 44.4939
5. 13.9043
6. 266.963
7. 75.0834
8. 278.087
9. 5.56174
10. 100.111
9
Correct answer: 5.56174 µJ.
Explanation:
Let :
r = 10 cm = 0.1 m ,
V = 1 kV = 1000 V , and
k = 8.99 × 109 N · m2 /C2 .
The potential on a spherical conductor is
kQ
r
rV
Q=
,
k
so the electrostatic potential energy is
V =
1
QV
2 1 rV
=
V
2
k
rV2
=
2k
106 µJ
(0.1 m) (1000 V)2
·
=
2 (8.99 × 109 N · m2 /C2 )
1J
U=
= 5.56174 µJ .
Sphere Within a Shell A01
016 10.0 points
Consider a solid conducting sphere with an
inner radius R1 surrounded by a concentric
thick conducting spherical shell which has an
inner radius R2 and outer radius R3 . There is
a charge Q on the sphere and no net charge
on the shell.
For this problem, we adopt the standard
convention of setting the electric potential at
infinity to zero.
q2 = 0
R1
R2
Ob
R3
b
q1 = Q
B
b
A
Version 001 – Exam 1 – shih – (57480)
Find the potential at O, where RO < R1 .
1. VO = 0
2. VO =
kQ kQ kQ
−
+
correct
R3
R2
R1
3. VO = ∞
10
for a thin shell of radius a.
Thus the potential due to the inner shell
(radius R2 ) is
V2 = k
qinner
Q
= −k
R2
R2
and that due to the outer shell (radius R3 ) is
kQ kQ
+
R3
R1
2kQ
5. VO =
R1
kQ kQ
6. VO =
+
R2
R1
√
2kQ
7. VO =
R1
kQ
8. VO =
R1
2kQ
9. VO =
R1 + R2
√
2 2kQ
10. VO =
R1
Explanation:
We are still inside the spherical shell. The
potential due to the shell requires knowing
something about where charge exists on the
shell. To this end, consider a Gaussian spherical surface inside the shell. This surface can
have no flux through it since no electric field
can be upheld inside a conductor. By Gauss’s
Law, there can therefore be no charge enclosed. But we are already enclosing Q on
the sphere, so a charge −Q must reside on the
inner surface of the shell:
4. VO =
qinner = −Q .
Since the net charge on the shell is zero, the
charge on the outer surface must be +Q for
the inner and outer charges to add up to zero:
V3 = k
Q
qouter
=k
,
R3
R3
so the contribution from the two surfaces of
the shell is
V1 = k
Q
Q
−k
.
R3
R2
However, we are now looking at a point
inside the sphere. The charge is all on the
surface of the sphere, so similarly to the situation for the shell we have
V =k
Q
R1
everywhere inside the sphere.
Thus the total potential at O is
VO = k
Q
Q
Q
−k
+k
.
R3
R2
R1
This potential is the same at O as it is everywhere inside the sphere. A conductor is an
equipotential object.
Add a Charge to Four 3
017 (part 1 of 2) 10.0 points
Four charges are placed at the corners of a
square of side a, with q1 = q2 = −q, q3 = q4 =
+q, where q is positive. Initially there is no
charge at the center of the square.
q2 = −q
q3 = +q
qouter = +Q .
Now that we know the exact distribution
of charge, we can utilize the expression for
potential inside of a spherical charge distribution:
q
V =k ,
a
q
q1 = −q
q4 = +q
Version 001 – Exam 1 – shih – (57480)
Find the work required to bring the charge
q from infinity and place it at the center of
the square.
1. W =
2. W =
3. W =
4. W =
5. W =
6. W =
−2 k q 2
a2
−2 k q 2
a
4 k q2
a
2 k q2
a2
8 k q2
a2
−4 k q 2
a2
7. W = 0 correct
−4 k q 2
a
4 k q2
9. W =
a2
2 k q2
10. W =
a
Explanation:
Based on the superposition principle, the
potential at the center due to the charges at
the corners is
8. W =
V = V1 + V2 + V3 + V4
kq
=
(−1 − 1 + 1 + 1) = 0 ,
r
where r is the common distance from the
center to the corners. The work required to
bring the charge q from infinity to the center
is then W = q V = 0.
018 (part 2 of 2) 10.0 points
What is the magnitude of the total electrostatic energy of the final 5 charge system? It
may be useful to consider the symmetry property of the charge distribution which leads to
cancellations among several terms.
k q2
.
1. U = 4
a
11
√ k q2
2
. correct
a
√ k q2
.
3. U = 4 2
a
k q2
4. U = 2
.
a
√ k q2
5. U = 2 2 .
a
√ k q2
6. U = 4 2 2 .
a
√ k q2
7. U = 2 2 2 .
a
2
kq
8. U = 4 2 .
a
k q2
9. U = 2 2 .
a
k q2
10. U = 8 2 .
a
Explanation:
No work is done in bringing in the fifth
charge and placing it at the center, so the
total electrostatic energy is contributed only
by 4 corner charges:
2. U =
U = U12 + U13 + U14 + U23 + U24 + U34
1
1
k q2
+1 − √ − 1 − 1 − √ + 1
=
a
2
2
2
√ kq
=− 2
.
a
Cylindrical Capacitor 03 v2
019 10.0 points
A 75 m length of coaxial cable has a solid
cylindrical wire inner conductor with a diameter of 1.168 mm and carries a charge of
5.21 µC. The surrounding conductor is a
cylindrical shell and has an inner diameter of
6.552 mm and a charge of −5.21 µC.
Assume the region between the conductors
is air. The Coulomb constant is 8.98755 ×
109 N · m2 /C2 .
What is the capacitance of this cable?
1. 3.5117
2. 2.62668
3. 1.73136
Version 001 – Exam 1 – shih – (57480)
4. 1.01978
5. 2.41954
6. 2.89136
7. 1.15634
8. 1.9711
9. 2.26507
10. 2.05129
12
4.43 µF
8.41 µF
a
b
8.66 µF
1.84 µF
c
d
11.8 V
Correct answer: 2.41954 nF.
Explanation:
Let : ke = 8.98755 × 109 N · m2 /C2 ,
Q = 5.21 µC ,
ℓ = 75 m ,
a = 1.168 mm , and
b = 6.552 mm .
The charge per unit length is λ ≡
Z
Q
.
ℓ
b
~ · d~s
E
a
Z b
dr
= −2 ke λ
a r b
Q
.
= −2 ke ln
ℓ
a
V =−
Find the equivalent capacitance between
points a and d.
1. 2.54995
2. 2.21028
3. 1.29119
4. 1.55152
5. 2.3927
6. 1.32972
7. 1.52131
8. 1.82451
9. 1.35359
10. 1.77146
Correct answer: 1.35359 µF.
Explanation:
Let :
The capacitance of a cylindrical capacitor
is given by
Q
V
ℓ
=
2 ke
C≡
1
b
ln
a
75 m
=
2 (8.98755 × 109 N · m2 /C2 )
1
1 × 109 nF
·
× 6.552 mm
1F
ln
1.168 mm
C1
C2
C3
C4
EB
= 8.41 µF ,
= 4.43 µF ,
= 8.66 µF ,
= 1.84 µF ,
= 11.8 V .
and
C2
a
C1
b
C3
c
C4
d
EB
For capacitors in series,
= 2.41954 nF
1
Capacitor Circuit 04 shortest
020 (part 1 of 2) 10.0 points
Consider the group of capacitors shown in the
figure.
Cseries
Vseries
X 1
C
X i
=
Vi ,
=
and the individual charges are the same.
Version 001 – Exam 1 – shih – (57480)
For parallel capacitors,
X
Cparallel =
Ci
X
Qparallel =
Qi ,
and the individual voltages are the same.
In the given circuit C2 and C3 are connected
parallel with equivalent capacitance
C23 = C2 + C3
= 4.43 µF + 8.66 µF
= 13.09 µF .
C1 , C23 , and C4 are connected in series with
equivalent capacitance
−1
1
1
1
Cba =
+
+
C1 C23 C4
−1
1
1
1
+
+
=
8.41 µF 13.09 µF 1.84 µF
= 1.35359 µF .
021 (part 2 of 2) 10.0 points
Determine the charge on the 4.43 µF capacitor
at the top center part of the circuit.
1. 34.3037
2. 3.95722
3. 16.2366
4. 16.4891
5. 10.9564
6. 11.6865
7. 5.40545
8. 14.9974
9. 5.05784
10. 10.1086
Correct answer: 5.40545 µC.
Explanation:
The voltage drops across C1 and C4 are
then
V1 =
Q1
15.9723 µC
=
= 1.8992 V
C1
8.41 µF
The remaining voltage is
Vremain = Vtotal − V1 − V4
= 11.8 V − 1.8992 V − 8.6806 V
= 1.22019 V .
This remaining voltage is the same across the
parallel capacitors, so
Q2 = C2 Vremain
= (4.43 µF) (1.22019 V)
= 5.40545 µC .
Two Dielectric Slabs v1
022 (part 1 of 3) 10.0 points
A parallel plate capacitor has capacitance C0 ,
plate separation d and plate area A. Two
dielectric slabs of dielectric constants κ1 and
d
κ2 , each of thickness , are inserted between
2
the plates. Charges Q and −Q are put on the
left and right plates, respectively.
dielectric κ2
constant
left
dielectric
constant κ1
right
V4 =
Q4
15.9723 µC
=
= 8.6806 V .
C4
1.84 µF
ℓ
A = ℓ2
d
Without the dielectrics, what is the field E0
between the plates?
1. E0 =
2. E0 =
3. E0 =
4. E0 =
5. E0 =
and
13
6. E0 =
Q
d
2 ǫ0 A
Q
4 ǫ0 A
2Q
d
ǫ0 A
Q
correct
ǫ0 A
Q
d
4 ǫ0 A
Q
2ǫ0 A
Version 001 – Exam 1 – shih – (57480)
2Q
7. E0 =
ǫ0 A
Explanation:
Solution:
Qencl
.
From Gauss’ Law, we know that Φ =
ǫ0
For a flat surface, like a plate, the flux is the
normal component of the electric field times
the area. Also, for a parallel plate capacitor,
the field is only in the gap (neglecting edge effects), so
Φ = E0 A
Q
Φ
=
.
E0 =
A
ǫ0 A
023 (part 2 of 3) 10.0 points
With the dielectrics inserted, the fields in κ1
and κ2 are respectively
What is the resultant capacitance?
1. 2.06894e-07
2. 1.44507e-08
3. 8.05613e-09
4. 1.06711e-08
5. 3.21764e-08
6. 4.75003e-08
7. 9.74561e-09
8. 1.68571e-08
9. 2.02877e-08
10. 2.86385e-08
Correct answer: 2.02877 × 10−8 F.
Explanation:
The potential difference between the two
plates is the sum of the potential differences
across the two dielectrics. The potential difference across a region with a constant electric
field is the product of the field and the distance, so
V = V1 + V2
E0 d E0 d
=
+
κ1 2
κ 2
2
Q d
1
1
=
.
+
ǫ0 A 2 κ1 κ2
1. E1 = E0 κ2 and E2 = E0 κ1
E0
E0
and E2 =
κ2
κ1
E0
E0
3. E1 =
and E2 =
correct
κ1
κ2
E0 κ2
E0 κ1
4. E1 =
and E2 =
κ1 + κ2
κ1 + κ2
2. E1 =
5. E1 = E0 κ1 and E2 = E0 κ2
E0 κ1 κ2
E0 κ1 κ2
and E2 =
κ1 + κ2
κ1 + κ2
E0 κ1
E0 κ2
7. E1 =
and E2 =
κ1 + κ2
κ1 + κ2
Explanation:
In any region with dielectric constant κ, the
permittivity ǫ0 becomes κǫ0 , so the fields in
κ1 and κ2 are
6. E1 =
E1 =
Q
E0
=
κ1 ǫ0 A
κ1
E2 =
Q
E0
=
.
κ2 ǫ0 A
κ2
024 (part 3 of 3) 10.0 points
Let κ1 = 3, κ2 = 4.41, A = 2.74 m2 , and
d = 4.27 mm.
14
The capacitance is given by
Q
V
2 ǫ0 A
=
d
C=
!
1
1
1
κ1 + κ2
κ1 κ2
ǫ0 A
=2
·
κ + κ2
d
1
(3) (4.41)
=2
3 + 4.41
8.854 × 10−12 F/m (2.74 m2 )
×
0.00427 m
= 2.02877 × 10−8 F .
Charge in a Wire
025 10.0 points
The current in a wire decreases with time
according to the relationship
I = (2.32 mA) × e−a t
Version 001 – Exam 1 – shih – (57480)
where a = 0.13328 s−1 .
Determine the total charge that passes
through the wire from t = 0 to the time the
current has diminished to zero.
1. 0.0114046
2. 0.017407
3. 0.0105792
4. 0.0216086
5. 0.0281363
6. 0.0112545
7. 0.0197329
8. 0.0277611
9. 0.0129802
10. 0.0296369
15
Correct answer: 5.83517 A.
Explanation:
Let :
V = 0.59 V ,
ℓ = 1.3 m ,
A = 0.72 mm2 = 7.2 × 10−7 m2 ,
ρ = 5.6 × 10
−8
The resistance is
R=
Correct answer: 0.017407 C.
VA
ρℓ
(0.59 V) (7.2 × 10−7 m2 )
=
(5.6 × 10−8 Ω · m) (1.3 m)
dq
I=
dt
q=
I=
t
I dt
Zt=0
∞
= 5.83517 A .
(0.00232 A)e−0.13328 s
t
e−0.13328 s t
= (0.00232 A )
−0.13328 s−1
∞
=
V
ρℓ
=
,
I
A
so the current is
Explanation:
Z
Ω · m.
−1
dt
t=0
−1
0
= 0.017407 C .
Current in Tungsten Wire
026 10.0 points
A 0.59 V potential difference is maintained
across a 1.3 m length of tungsten wire that
has a cross-sectional area of 0.72 mm2 and the
resistivity of the tungsten is 5.6 × 10−8 Ω · m.
What is the current in the wire?
1. 5.83517
2. 1.72556
3. 11.873
4. 8.71763
5. 13.7143
6. 16.7545
7. 8.0
8. 1.62175
9. 3.69643
10. 14.5357
and