Exercise 22.2 A flat sheet is in the shape of a rectangle with sides of lengths 0.400 and 0.600 . The sheet is immersed in a uniform electric field of magnitude 60.0 that is directed at 20 from the plane of the sheet . Part A Find the magnitude of the electric flux through the sheet. Express your answer to two significant figures and include the appropriate units.

Exercise 22.2
A flat sheet is in the shape of a rectangle with sides of lengths 0.400 and 0.600 . The sheet is immersed in a uniform electric field of magnitude 60.0 that is directed at 20 from the plane of the sheet .
Part A
Find the magnitude of the electric flux through the sheet.
Express your answer to two significant figures and include the appropriate units.
ANSWER:
Exercise 22.4
It was shown in Example 21.11 (Section 21.5) in the textbook that the electric field due to an infinite line of charge is perpendicular to the line and has magnitude . Consider an imaginary cylinder with a radius of = 0.130 and a length of = 0.430 that has an infinite line of positive charge running along its axis. The charge per unit length on the line is = 7.20 .
Part A
What is the electric flux through the cylinder due to this infinite line of charge?
ANSWER:
Part B
m m N/C ∘
= Φ
E = λ/2π rϵ0 r m l m λ
μC/m
= Φ /CN ⋅ m2
What is the flux through the cylinder if its radius is increased to 0.505 ?
ANSWER:
Part C
What is the flux through the cylinder if its length is increased to 0.810 ?
ANSWER:
Exercise 22.9
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 18.0 , giving it a charge of -49.0 .
Part A
Find the electric field just inside the paint layer.
Express your answer with the appropriate units. Enter positive value if the field is directed away from the center of the sphere and negative value if the field is directed toward the center of the sphere.
ANSWER:
Part B
Find the electric field just outside the paint layer.
Express your answer with the appropriate units. Enter positive value if the field is directed away from the center of the sphere and negative value if the field is directed toward the center of the sphere.
ANSWER:
Part C
Find the electric field 5.00 outside the surface of the paint layer.
Express your answer with the appropriate units. Enter positive value if the field is directed away from the center of the sphere and negative value if the field is directed toward the center of the sphere.
ANSWER:
r = m
= Φ /CN ⋅ m2
l = m
= Φ /CN ⋅ m2
cm μC
= E
= E
cm
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Gauss’s Law
Learning Goal:
To understand the meaning of the variables in Gauss’s law, and the conditions under which the law is applicable.
Gauss’s law is usually written
where is the permittivity of vacuum.
Part A
How should the integral in Gauss’s law be evaluated?
ANSWER:
Part B Complete previous part(s)
Exercise 22.12
The nuclei of large atoms, such as uranium, with protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately .
Part A
What is the electric field this nucleus produces just outside its surface?
Express your answer using two significant figures.
ANSWER:
Part B
What magnitude of electric field does it produce at the distance of the electrons, which is about 1.9×10−10 ?
Express your answer using two significant figures.
= E
= ∮ ⋅d = ,ΦE E⃗ A⃗ qencl ϵ0
= 8.85 × /(N ⋅ )ϵ0 10 −12 C2 m2
around the perimeter of a closed loop
over the surface bounded by a closed loop
over a closed surface
92 7.4 × m10−15
= E N/C
m
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ANSWER:
Part C
The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?
Express your answer using two significant figures.
ANSWER:
A Charged Sphere with a Cavity
An insulating sphere of radius , centered at the origin, has a uniform volume charge density .
Part A
Find the electric field inside the sphere (for < ) in terms of the position vector .
Express your answer in terms of , (Greek letter rho), and .
You did not open hints for this part.
ANSWER:
Part B Complete previous part(s)
± The Charge on a Thundercloud
In a thunderstorm, charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considered to be distributed uniformly throughout the cloud. For the purposes of this problem, take the cloud to be a sphere of diameter 1.00 kilometer. The point of this problem is to estimate the maximum amount of charge that this cloud can contain, assuming that the charge builds up until the electric field at the surface of the cloud reaches the value at which the surrounding air breaks down. This breakdown means that the air becomes highly ionized, enabling it to conduct the charge from the cloud to the ground or another nearby cloud. The ionized air will then emit light due to the recombination of the electrons and atoms to form excited molecules that radiate light. In addition, the large current will heat up the air, resulting in its rapid expansion. These two phenomena account for the appearance of lightning and the sound of thunder. Take the breakdown electric field of air to be
.
Part A
Estimate the total charge on the cloud when the breakdown of the surrounding air is reached.
= E N/C
= Enet N/C
a ρ
( )E ⃗ r ⃗ r a r ⃗
r ⃗ ρ ϵ0
= ( )E ⃗ r ⃗
= 3.00 × N/CEb 10 6
q
Express your answer numerically, to three significant figures, using .
You did not open hints for this part.
ANSWER:
Charge Distribution on a Conducting Shell – 2
A positive charge is kept (fixed) off-center inside a fixed spherical conducting shell that is electrically neutral, and the charges in the shell are allowed to reach electrostatic equilibrium.
Part A
The large positive charge inside the shell is roughly 16 times that of the smaller charges shown on the inner and outer surfaces of the spherical shell. Which of the following figures best represents the charge distribution on the inner and outer walls of the shell?
You did not open hints for this part.
ANSWER:
Charge Distribution on a Conducting Shell – 1
A positive charge is kept (fixed) at the center inside a fixed spherical neutral conducting shell.
= 8.85 × /(N ⋅ )ϵ0 10 −12 C2 m2
= Coulombs q
1
2
3
4
5
Part A
The positive charge is equal to roughly 16 of the smaller charges shown on the surfaces of the spherical shell. Which of the pictures best represents the charge distribution on the inner and outer walls of the shell?
You did not open hints for this part.
ANSWER:
Conceptual Question 22.01
Part A
If the electric flux through a closed surface is zero, the electric field at points on that surface must be zero.
ANSWER:
Conceptual Question 22.02
1
2
3
4
5
True
False
Part A
The figure shows four Gaussian surfaces surrounding a distribtuion of charges.
(a) Which Gaussian surfaces have an electric flux of through them?
ANSWER:
Part B
(b) Which Gaussian surfaces have no electric flux through them?
ANSWER:
Conceptual Question 22.04
Part A
Consider a spherical Gaussian surface of radius centered at the origin. A charge is placed inside the sphere. To maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located
ANSWER:
+q/ϵ0
b
c
d
a
a
c
b
d
R Q
Conceptual Question 22.09
Part A
An uncharged conductor has a hollow cavity inside of it. Within this cavity there is a charge of +10µC that does not touch the conductor. There are no other charges in the vicinity. Which statement about this conductor is true? (There may be more than one correct choice.)
ANSWER:
Conceptual Question 22.08
Part A
A charge is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length . At a point that is 1.25 cm outside the sheet, the magnitude of the electric field due to the sheet is . If the sheet is now
stretched so that its sides have length 2 , what is the magnitude of the electric field at ?
ANSWER:
at = 0, = /2, = 0.x y R z
at = /2, = 0, = 0.x R y z
at = 0, = 0, = /2.x y z R
at the origin.
The charge can be located anywhere, since flux does not depend on the position of the charge as long as it is inside the sphere.
The inner surface of the conductor carries a charge of -10µC and its outer surface carries no excess charge.
The inner and outer surfaces of the conductor each contain charges of -5µC.
The net electric field within the material of the conductor points away from the +10µC charge.
The outer surface of the conductor contains +10µC of charge and the inner surface contains -10µC.
Both surfaces of the conductor carry no excess charge because the conductor is uncharged.
Q d P E
d P
Prelecture Concept Question 22.06
Part A
Five point charges q and four Gaussian surfaces S are shown in the figure. What is the total electric flux through surface S2?
ANSWER:
4E
/2E
E
/4E
2E
Prelecture Concept Question 22.05
Part A
Five point charges q and four Gaussian surfaces S are represented in the figure shown. Through which of the Gaussian surfaces are the total electric flux zero?
Check all that apply.
ANSWER:
zero
3q/εo
5q/εo
q/εo
2q/εo
4q/εo
Problem 22.16
Part A
Electric charge is uniformly distributed inside a nonconducting sphere of radius 0.30 m. The electric field at a point , which is 0.50 m from the center of the sphere, is 15,000 N/C and is directed radially outward. What is the maximum magnitude of the electric field due to this sphere?
ANSWER:
Problem 22.23
Part A
A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The upper face carries +95.0 nC/m2 while the lower face carries -25.0 nC/ m2. What is the magnitude of the electric field at a point within the sheet 2.00 cm below the upper face? ( = 8.85 × 10-12 C2/N · m2)
ANSWER:
S2
S3
S4
S1
The total electric flux is not zero through any of the Gaussian surfaces.
P
36,000 N/C
42,000 N/C
48,000 N/C
30,000 N/C
25,000 N/C
ε0
Problem 22.22
Part A
A very large sheet of a conductor carries a uniform charge density of 4.00 pC/mm2 on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor? ( = 8.85 × 10-12 C2/N · m2)
ANSWER:
Problem 22.38
A long line carrying a uniform linear charge density runs parallel to and from the surface of a large, flat plastic sheet that has a uniform surface charge density of on one side.
Part A
Find the location of all points where an particle would feel no force due to this arrangement of charged objects.
ANSWER:
Part B
Choose an appropriate location of these points at a distance, calculated in part A.
0.00 N/C
1.36 × 104 N/C
7.91 × 103 N/C
3.95 × 103 N/C
6.78 × 103 N/C
ε0
9.04 × 105 N/C
4.52 × 105 N/C
2.26 × 105 N/C
0.226 N/C
0.452 N/C
+50.0μC/m 10.0 cm −100μC/m2
α
= from the line. L m
ANSWER:
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above the line
between the line and the sheet