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

• Question 1• Question 2• Question 3• Question 4• Question 5

• Question 6• Question 7• Question 8• Question 9• Question 10

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

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

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

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

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.

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).

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.

Choice: B

Incorrect

The enclosed charge should be divided by o, not by the

area.

Choice: C

Correct

€

Φnet = r E • d

r A∫ =

Qenc

o

This is correct, and is expressed mathematically as:

3. What is a geometrically convenient Gaussian surface for

this system?

A: Circle

B: Cube

C: Sphere

The system is 3-dimensional, so our Gaussian surface should be

also.

Choice: A

Incorrect

The electric field is not equal everywhere on this surface.

Choice: B

Incorrect

This is the correct choice, because the system can be broken into two subsystems which both show spherical

symmetry.

Choice: C

Correct

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.

Example:

Gaussian surface

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

Choice: A

Incorrect

We are dealing with a volume charge density, not a surface charge.

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.

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.

This is the charge density. Use the relation below to find the

enclosed charge.

Choice: D

Incorrect

€

ρ=Qenc

V

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

Option iii. is the electric field strength.

Choice: A

Incorrect

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

This is not the only correct expression for the total electric

flux.

Choice: C

Incorrect

Check the units.

Choice: D

Incorrect

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

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

Choice: B

Correct

€

Er4πr2 =4ρπr3

3o

Er =4ρπr3

4πr23o

= ρr3o

From the previous question:

Choice: C

Incorrect

€

Er4πr2 =4ρπr3

3o

Try again. Start with the information obtained from the previous question:

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

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.

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

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

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

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:

This is the total charge in the cavity (if it were full of the

insulting material), not within the Gaussian surface.

Choice: A

Incorrect

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

This is the enclosed charge.

Choice: C

Correct

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

This is the total flux. We need an expression for electric field.

Choice: A

Incorrect

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

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

Notice that similarly to question 7, the electric field vector is expressed as:

€

r E r

€

r E r =

ρ r r

3o

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:

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

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!!!

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.