A conducting spherical shell with inner radius a and outer radius b has a positive point charge Q...

A conducting spherical shell with inner radius a and outer radius b has a positive point charge Q located at its center. The total charge on the shell is -3 Q, and it is insulated from its surroundings. Derive expressions for the electric field magnitude E in terms of the distance from the center r for the regions r<a, a<r<b, and r>b.

a b

-3Q+Q

Bold type denotes vector quantities

Bold type denotes vector quantities

• Question 1• Question 2• Question 3• Question 4• Question 5• Question 6• Question 7• Question 8• Question 9

• A : I• B : II & IV• C : III• D : III & IV

1. Which of the following physics principles should one use to solve this problem?

I. Ampere’s LawII. Faraday’s Law of InductionIII. Gauss’s LawIV.Superposition of Electric Fields

This law deals with magnetic fields produced by electric current.

Choice: A

Incorrect

Faraday’s law deals with the time rate of change of magnetic flux, so this is not applicable to our situation.

On the other hand, the principle of superposition of electric fields is very helpful here. Considering the

contributions to the electric field from each charge will make this task easier to evaluate.

Choice: B

Incorrect

We can use this law to solve for E as we exploit symmetry, but another

physics principle will also be helpful.

Choice: C

Incorrect

Choice: DCorrect

Gauss’s law and the principle of superposition of electric fields are very helpful here. Considering the

contributions to the electric field from the charge in each region will make this task easier to evaluate.

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 total charge inside the surface divided by o (permittivity of free space).

Gauss’s Law relates the electric flux through a surface to the total charge enclosed by the surface.

Choice: A

Incorrect

€

E ≠Qenc

Choice: B

Incorrect

According to Gauss’s Law, the total electric flux through a closed surface is

equal to the total charge inside the surface divided by

o , not by the area.

Choice: C

Correct

€

Φnet = E • dA∫ =Qenc

o

In mathematical form, Gauss’s Law is expressed as:

3. What is a convenient Gaussian surface for this system?

• A: circle• B: cube• C: sphere

Choice: A

Incorrect

The system is three dimensional.

The magnitude of the electric field is not the same at all points

on a cubical surface.

Choice: B

Incorrect

Choice: C

Correct

This is a good choice, because the system shows spherical

symmetry.

Please get out a piece of scratch paper and sketch the charge configuration of the situation described in the problem statement. For now, only draw the arrangement due to the charge +Q inside the cavity. Draw your Gaussian surface with a radius r<a.

Example: Your sketch should look something like this:

a

r+Q-

-

--

-

-

- --

-Q+

+

+

+

+

+

++

+

+Q

1. A negative charge of magnitude -Q is induced uniformly around the inner surface of the cavity.

1. A negative charge of magnitude -Q is induced uniformly around the inner surface of the cavity.

2. The negative charge is drawn to the inner surface and a positive charge remains on the outer surface of the conducting shell.

2. The negative charge is drawn to the inner surface and a positive charge remains on the outer surface of the conducting shell.

Gaussian surface

Gaussian surface

4. For r<a, as depicted in our sketch, the field strength is equal in magnitude everywhere on the surface and is radially outward in direction. Choose all of the following that are correct expressions for the total flux in this case.

I)

II)

III)

A: I only

C: III only

B: II only

D: I and II only

€

E4πr2

€

+Qo

€

+Q4πr2o

E: I and III only

This is a correct expression, but it is not the only one.

Choice: A

Incorrect

This is a correct expression, but it is not the only one.

Choice: B

Incorrect

Check the units. This is an expression for electric field.

Choice: C

Incorrect

Both these expressions are correct.

Choice: D

Correct

€

Φnet =+Qo

€

Φnet = E • dA∫ =E dA∫ =EA =E4πr2

Since the electric field is perpendicular to the infinitesimal area dA, the magnitude of the electric field E can be pulled out of the integral.

Mathematical expression of Gauss’s Law:

The total charge enclosed in our Gaussian surface is only +Q, therefore:

€

Φnet = E • dA∫ =Qenc

o

One of these expressions is not correct. Try Again.

Choice: E

Incorrect

5. Which one of the following expressions describes the electric field E in the region

r<a?

• A:

• B:

• C:

€

−Q4πr2o

€

+Q4πr2o

€

−3Q4πr2o

Choice: A

Incorrect

The total charge enclosed in the Gaussian sphere is +Q.

Choice: B

Correct

€

E4πr2 =+Qo

E = +Q4πr2o

From the previous question we have:

Simple algebra shows that:

Choice: C

Incorrect

Since our Gaussian surface has a radius r less than a, the only charge that is enclosed is the

point charge +Q.

6. For electrostatic equilibrium, the electric field inside the conductor (metal) is which of the following?

• A: uniform but non-zero• B: zero• C: non-uniform

Choice: A

Incorrect

This is in violation of the equilibrium condition.

Choice: B

Correct

This implies equilibrium.

Choice: C

Incorrect

The charges are uniformly distributed.

• Please make another sketch of the charge configuration of the spherical shell, this time only consider the -3Q charge that is placed on the outer surface of the conductor.

Example

-

-

-

-

-

-

-

--

-

-3Q

The charge -3Q distributes uniformly on the surface of the conductor, making the field everywhere inside of the conductor equal to zero.

The charge -3Q distributes uniformly on the surface of the conductor, making the field everywhere inside of the conductor equal to zero.Notice that

there is no charge enclosed if the gaussian surface is placed anywhere inside of the surface of the conductor.

Notice that there is no charge enclosed if the gaussian surface is placed anywhere inside of the surface of the conductor.

• Considering a superposition of the electric fields produced by the point charge +Q and the surface charge -3Q will help us find expressions for the electric field in the regions a<r<b and r>b.

• Sketch the charge configuration of the spherical shell considering all sources of electric field (combine your two previous drawings).

Superposition of electric fields and charge summation

+ =-

--

-

-

--

+

+

+

+

+

+

+Q

++Q

-Q

+Q

-Q

-3Q

+Q

-

--

--

--

-

-

- -

-

-

-

-

- -

+

+

++

The red arrows depict electric field lines. Notice that there are more field lines in the second drawing pointing inwards than there are in the first pointing outwards.

The red arrows depict electric field lines. Notice that there are more field lines in the second drawing pointing inwards than there are in the first pointing outwards.

7. Add a Gaussian surface to your drawing with a radius a<r<b.

What is the electric field in this region (inside of the conductor).

• A: E=0

• B: E=

• C: E=

€

+Q4πr2o

€

−Q4πr2o

Choice: A

Correct

€

E4πr2 =Qenc

o

Qenc =(+Q) + (−Q) = 0

E = 0

The electric field inside of a conductor is always equal to zero.

r

a

b

Choice: B

Incorrect

Remember that the electric field inside of a conductor is always zero.

Your Gaussian surface should enclose more than just the point

charge +Q.

Remember that the electric field inside of a conductor is always zero.

Your Gaussian surface should enclose more than just the induced charge -Q on the inside of the shell.

Choice: C

Correct

8. What is the net charge enclosed by a Gaussian sphere in the r>b region?

• A: +Q

• B: -3Q

• C: -Q

• D: -2Q

Choice: A

Incorrect

br

€

Qenc =(+Q) + (−Q) + (+Q) + (−3Q)Qenc = −2Q

Gaussian surface

Gaussian surface

The point charge in the center, the induced charges -Q and +Q on the inner and outer surfaces of the shell, and the -3Q spread around the outside of the shell are all enclosed by the Gaussian sphere.

The point charge in the center, the induced charges -Q and +Q on the inner and outer surfaces of the shell, and the -3Q spread around the outside of the shell are all enclosed by the Gaussian sphere.

Choice: B

Incorrect

br

€

Qenc =(+Q) + (−Q) + (+Q) + (−3Q)Qenc = −2Q

Gaussian surface

Gaussian surface

The point charge in the center, the induced charges -Q and +Q on the inner and outer surfaces of the shell, and the -3Q spread around the outside of the shell are all enclosed by the Gaussian sphere.

The point charge in the center, the induced charges -Q and +Q on the inner and outer surfaces of the shell, and the -3Q spread around the outside of the shell are all enclosed by the Gaussian sphere.

Choice: C

Incorrect

br

€

Qenc =(+Q) + (−Q) + (+Q) + (−3Q)Qenc = −2Q

Gaussian surface

Gaussian surface

The point charge in the center, the induced charges -Q and +Q on the inner and outer surfaces of the shell, and the -3Q spread around the outside of the shell are all enclosed by the Gaussian sphere.

The point charge in the center, the induced charges -Q and +Q on the inner and outer surfaces of the shell, and the -3Q spread around the outside of the shell are all enclosed by the Gaussian sphere.

Choice: D

Correct

br

€

Qenc =(+Q) + (−Q) + (+Q) + (−3Q)Qenc = −2Q

Gaussian surface

Gaussian surface

The point charge in the center, the induced charges -Q and +Q on the inner and outer surfaces of the shell, and the -3Q spread around the outside of the shell are all enclosed by the Gaussian sphere.

The point charge in the center, the induced charges -Q and +Q on the inner and outer surfaces of the shell, and the -3Q spread around the outside of the shell are all enclosed by the Gaussian sphere.

9. Which one of the following is the expression for the electric field E in the region r>b?

A: (radially outward)

C: (radially inward)

B: (radially outward)

D: (radially inward)

€

+Q(4πr2εo )

€

−Q2πr2o

€

−2Qo

€

−Q2πr2o

The total enclosed charge is not +Q. Also, field lines from a negative

source charge are directed inward.Please check your sketch and try

again.

Choice: A

Incorrect

Choice: B

Incorrect

Field lines from a negative source charge are directed inward.

Field lines from a negative source charge are directed inward.

Choice: C

Correct

€

E =Qenc

4πr2ε0=−2Q

4πr2ε0=−Q

2πr2ε0

Reasoning:

Choice: D

Incorrect

This is the total flux through the Gaussian sphere.