1. 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.
Vectors are denoted by arrows. b a R

2. Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Question 8
Question 9
Question 10

3. Which of the following principles should be used to solve this problem?
i) Ampere’s Law
ii) Superposition of electric field
iii) Gauss’ Law
iv) Coulomb’s Law
A: i only
B: ii and iii only
C: ii and iv only
D: iii only

4. Choice: A
Incorrect
This law deals with magnetic fields produced by current. We are interested in the charge enclosed by a surface, not enclosed current.

5. Choice: B
Correct
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.

6. Choice: C
Incorrect
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.

7. Choice: D Incorrect
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.

8. 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).

9. This concept does not make sense.
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.

10. The enclosed charge should be divided by o, not by the area.
Choice: B
Incorrect
The enclosed charge should be divided by o, not by the area.

11. Choice: C
Correct
This is correct, and is expressed mathematically as:

12. 3. What is a geometrically convenient Gaussian surface for this system?
A: Circle
B: Cube
C: Sphere

13. The system is 3-dimensional, so our Gaussian surface should be also.
Choice: A
Incorrect
The system is 3-dimensional, so our Gaussian surface should be also.

14. The electric field is not equal everywhere on this surface.
Choice: B
Incorrect
The electric field is not equal everywhere on this surface.

15. Choice: C
Correct
This is the correct choice, because the system can be broken into two subsystems which both show spherical symmetry.

16. 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.

17. Example:
Gaussian surface

18. 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:

19. We are dealing with a volume charge density, not a surface charge.
Choice: A
Incorrect
We are dealing with a volume charge density, not a surface charge.

20. 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.

21. Choice: C
Correct
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.

22. Choice: D
Incorrect
This is the charge density. Use the relation below to find the enclosed charge.

23. 5. What is the total flux in the case described in question 4?
A: i and iii
B: i and ii
C: i only
D: iii only

24. Option iii. is the electric field strength.
Choice: A
Incorrect
Option iii. is the electric field strength.

25. This follows from Gauss’s Law:
Choice: B
Correct
This follows from Gauss’s Law:
Notice that here, E is constant and can be pulled out of the integral

26. This is not the only correct expression for the total electric flux.
Choice: C
Incorrect
This is not the only correct expression for the total electric flux.

27. Choice: D
Incorrect
Check the units.

28. 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:

29. Choice: A
Incorrect
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.

30. Choice: B
Correct
From the previous question:

31. Choice: C
Incorrect
Try again. Start with the information obtained from the previous question:

32. 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:

33. Choice: A Correct
The electric field and the distance from the insulating center of the sphere to our point of interest are both directed radially outward.

34. Choice: B Incorrect
We must use the vector , which is directed radially outward from the center of the insulating sphere, not just the magnitude r.

35. Choice: C Incorrect
We must not add a factor of r, we just replace the magnitude r with the vector , since it has a direction (radially outward).

36. 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. a r

37. 8. What is the charge enclosed by a small Gaussian sphere of radius , with the same center as the small sphere of radius a? A: B: C:

38. Choice: A
Incorrect
This is the total charge in the cavity (if it were full of the insulting material), not within the Gaussian surface.

39. Choice: B
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 .

40. This is the enclosed charge.
Choice: C
Correct
This is the enclosed charge.

41. 9. What is the magnitude of the electric field Er´ due to the enclosed charge in question 7?B: C:

42. This is the total flux. We need an expression for electric field.
Choice: A
Incorrect
This is the total flux. We need an expression for electric field.

43. We use the same reasoning as in question 6:
Choice: B
Correct
We use the same reasoning as in question 6:

44. Choice: C
Incorrect
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.

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

46. 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? A: B: C:

47. Choice: A
Incorrect
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.

48. The superposition of electric fields can be shown mathematically as:
Choice: B
Correct
The superposition of electric fields can be shown mathematically as:
Notice that r´ r
The electric field is the same everywhere inside the cavity!!!

49. 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.