an insulating sphere of radius b has a spherical cavity of radius a located within its volume and...
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An insulating sphere of radius b has a spherical cavity of radius a located within its volume and
centered a distance R from the center of the sphere. A cross section of the sphere is shown below.
The solid part of the sphere has a uniform volume charge density . Find the electric field inside the
cavity.
€
ρ
R
b a
€
ρ
Vectors are denoted by arrows.
€
r V
€
r E
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• Question 1• Question 2• Question 3• Question 4• Question 5
• Question 6• Question 7• Question 8• Question 9• Question 10
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1. Which of the following principles should be used to
solve this problem?i) Ampere’s Lawii) Superposition of electric fieldiii) Gauss’ Lawiv) Coulomb’s Law
A: i onlyB: ii and iii only C: ii and iv onlyD: iii only
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This law deals with magnetic fields produced by current. We are
interested in the charge enclosed by a surface, not enclosed current.
Choice: A
Incorrect
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Gauss’s Law allows us to determine the electric field at a point when there is sufficient symmetry. However, due to the cavity, the object under consideration does not possess spherical symmetry.
Considering a superposition will make this problem easier to solve. The electric field is a superposition of the field due to a uniformly
charged (let it be positive) sphere of radius b and that of a uniformly charged (negative) sphere of radius a.
Equivalently, the superposition can be treated as a subtraction involving two positively charged spheres.
Choice: B
Correct
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Since we are dealing with a uniform distribution of charge throughout a
volume, using Coulomb’s Law will prove to be very difficult. There is a much better
way to solve this problem.
Choice: C
Incorrect
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Choice: DIncorrect
We should use Gauss’s Law to solve this problem, but we must use another principle
along with it to take advantage of symmetries.
The use of superposition is required here because the overall problem lacks spherical symmetry,
but can be broken up into two problems, each of which is spherically symmetric.
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2. Which statement correctly describes Gauss’s Law?
A: The total electric field through a closed surface is equal to the net charge inside the surface.
B: The total electric flux through a closed surface is equal to the total charge inside the surface divided by the area
C: The total electric flux through a closed surface is equal to the net charge enclosed by the surface divided by o (permittivity of free space).
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Choice: A
Incorrect
Mathematically this would be expressed as Enet=Qenc which is not true. They don’t even have the
same units.
This concept does not make sense.
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Choice: B
Incorrect
The enclosed charge should be divided by o, not by the
area.
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Choice: C
Correct
€
Φnet = r E • d
r A∫ =
Qenc
o
This is correct, and is expressed mathematically as:
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3. What is a geometrically convenient Gaussian surface for
this system?
A: Circle
B: Cube
C: Sphere
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The system is 3-dimensional, so our Gaussian surface should be
also.
Choice: A
Incorrect
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The electric field is not equal everywhere on this surface.
Choice: B
Incorrect
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This is the correct choice, because the system can be broken into two subsystems which both show spherical
symmetry.
Choice: C
Correct
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Considering a superposition of electric field, we will first determine the field at a point P inside the
insulator as if there was no cavity. Then, we will determine the electric field that would be in a
sphere the size of the cavity only.
On a piece of scratch paper, please draw a diagram of the insulator without a cavity. Include
your Gaussian surface.
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Example:
Gaussian surface
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4. Assuming that the cavity is filled with the same uniform volume charge density as the solid sphere, what is the charge enclosed by
the Gaussian sphere.
A:
B:
C:
D:
€
ρ4πr2
€
ρ43πb3
€
ρ43πr3
€
ρ
€
ρ
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Choice: A
Incorrect
We are dealing with a volume charge density, not a surface charge.
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Choice: B
Incorrect
We should not use b as the radius of our sphere, because we are only concerned with the portion of the insulator inside the Gaussian surface.
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Choice: C
Correct
€
ρ=Qenc
V
volume of a sphere V =4
3πR3
Qenc = ρV = ρ4
3πr3
Notice the we use r as the radius because we are not concerned with the charge enclosed by the outer surface of the insulator, but only the charge within the Gaussian surface.
Notice the we use r as the radius because we are not concerned with the charge enclosed by the outer surface of the insulator, but only the charge within the Gaussian surface.
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This is the charge density. Use the relation below to find the
enclosed charge.
Choice: D
Incorrect
€
ρ=Qenc
V
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5. What is the total flux in the case described in question 4?
i.
ii.
iii.
A: i and iii
B: i and ii
C: i only
D: iii only
€
Eb4πr2
€
4ρπr3
3o
€
ρr30
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Option iii. is the electric field strength.
Choice: A
Incorrect
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This follows from Gauss’s Law:Choice: B
Correct
€
Φnet = r E • d
r A= E∫∫ dAcos θ( )
θ = 0o so ,
Φnet = E dA = EA = E∫ 4πr2 = Er 4πr2
€
Φnet = r E • d
r A∫ =
Qenc
o
=4ρπr3
3o
Notice that here, E is constant and can be pulled out of the integral
Notice that here, E is constant and can be pulled out of the integral
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This is not the only correct expression for the total electric
flux.
Choice: C
Incorrect
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Check the units.
Choice: D
Incorrect
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6. The magnitude of the electric field Er due to the charge enclosed in question
4 is equal to which of the following?
A:
B:
C:
€
ρ4πr3
30
€
ρ4πr30
€
ρr30
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This is the total flux. We are looking for the magnitude of the
electric field at any point P inside the insulating sphere of radius b.
Choice: A
Incorrect
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Choice: B
Correct
€
Er4πr2 =4ρπr3
3o
Er =4ρπr3
4πr23o
= ρr3o
From the previous question:
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Choice: C
Incorrect
€
Er4πr2 =4ρπr3
3o
Try again. Start with the information obtained from the previous question:
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7. Recall that when we calculated the electric flux using Gauss’s Law in question 5, the angle between E and dA
was 0. The electric field is directed radially outward.
Which of the following equations is the correct representation of the electric field vector?
A:
B:
C:
€
r Er =
ρ r r
3o
€
r Er =
ρr3o
€
r Er =
ρr r r
3o
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Choice: ACorrect
The electric field and the distance from the insulating center of the sphere to our point of interest are both directed radially outward.
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Choice: BIncorrect
We must use the vector , which is directed radially outward from the center of
the insulating sphere, not just the magnitude r.
€
r r
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Choice: CIncorrect
We must not add a factor of r, we just replace the magnitude r with the vector , since it has a direction (radially outward).
€
r r
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Now consider the electric field Er´ that would come from the cavity alone, if it were full of the insulating
material with uniform charge density . Notice the similarity to the previous questions.
€
ρ
a
r
€
ρ
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8. What is the charge enclosed by a small Gaussian sphere of radius , with the same center as the small sphere of radius a?
€
ρ4πa3
3
€
ρ4π r 3
3
€
ρ4πr3
3
€
r <a
A: C:B:
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This is the total charge in the cavity (if it were full of the
insulting material), not within the Gaussian surface.
Choice: A
Incorrect
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is the radius of the Gaussian surface that we used earlier to find the electric flux
through part of the solid insulator. Here we are only concerned with the cavity (if it were full of the insulting material), so we
should use .
Choice: B
Incorrect
€
r
€
r
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This is the enclosed charge.
Choice: C
Correct
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9. What is the magnitude of the electric field Er´ due to the enclosed charge in
question 7?A:
B:
C:
€
ρ4π r 3
30
€
ρa30
€
ρ r30
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This is the total flux. We need an expression for electric field.
Choice: A
Incorrect
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We use the same reasoning as in question 6:
Choice: B
Correct
€
E r 4π r 2 =4ρπ r 3
3o
E r = 4ρπ r 3
4π r 23o
=ρ r3o
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This is the field at the surface of the uniformly charged sphere that
replaced the empty cavity. We want an expression that can give us the strength of the electric field at any
point inside the filled cavity.
Choice: C
Incorrect
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Notice that similarly to question 7, the electric field vector is expressed as:
€
r E r
€
r E r =
ρ r r
3o
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10. Using the principle of superposition and the results from the previous questions, which expression will give us the electric field at any point in the cavity, as asked for in the original
problem statement?
€
r E =
ρ(r r − ′
r r )
3ε0
€
r E =
ρ(r r −
r b )
3ε0
€
r E =
ρ(r r + ′
r r )
3ε0
A: B: C:
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This is not the actual field.
We should subtract the electric field that would be from the cavity
if it we full of the charged insulating material.
Choice: A
Incorrect
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Choice: B
CorrectThe superposition of electric fields can be
shown mathematically as:
€
r E =
r Er −
r E r
r E =ρ
r r
3o
−ρ r r
3o
= ρ3o
r r −
r r( )
Notice thatNotice thatr r´
€
r R +
r r =
r r
r r −
r r =
r R
The electric field is the same everywhere inside the cavity!!!
The electric field is the same everywhere inside the cavity!!!
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Choice: C
Incorrect
We must subtract the electric field that would have been in the cavity if it were full of the same insulating material. The radius of the insulating sphere b should
not appear in our expression.