Solve the following problems on your own paper. Please:

• Show all work.
• Put a box around your final answer to each problem.
• Be sure proper significant figures (SF) used for all answers.
• Separate all solutions with a horizontal line on your paper.
• A random selection of five problems will be graded for 20 points each.
• Solutions for suggested odd problems can be seen by following links.
• Solutions for assigned problems will be linked from this page after the due date.

Chapter 23

#2

This may help, depending upon how you approach the problem: the equation defining the right-hand face of the cube is (1.40m)^2 j.

#10

The trick here is realizing whether either shell contributes zero to the field.

#18

Notice that the flux is equation to zero within the material of a conducting shell but not zero inside the inner surface of the shell since it isn't a solid.

#28

I'll help set this up in class.

#34

You don't need to derive anything here.

#36

Assume that the only force is the electric force. If you don't know all of your metric prefixes, check the front cover of the text.

#44

Justify your answer for part (a)!

#50

To find the total charge for the last part, you need to know the volume. That's just the total volume enclosed by the outer surface minus the volume of the inner cavity.

Suggested problems from chapter 23:

1, 3, 4, 5, 6, 7, 8, 9, 12, 13, 15, 16, 17, 19, 20, 21, 22, 23, 24, 25, 27, 29, 31, 32, 35, 37, 38, 39, 40, 42, 43, 44, 45, 46, 49, 50, 56, 57, 58, 59, 60, 61, 62

Chapter 24

#4

Use equation 24-7. Remind yourself of the definition of equipotential.

#8

You have to integrate to get this right since the electric field varies along the x-axis.

#12

The electric potential depends upon the magnitude of the radial distance, not direction.

#32

Answer with 2 SF.

#40

Again, potential is not a vector, so the angles aren't really important, just the distances.

#42

A drawing helps visualize this. For the ring, the total charge is what matters, not the distribution, because all of the charge is exactly the same distance from either point in question (this would be a MUCH more difficult problem if we were looking at points off-axis of the ring).

#46

Use the work-energy theorem to find the final velocity. This is similar to problems 44 and 45.

#54

See figure 24-18 for help with this one. The +400V can be imagined to be the potential on the surface of the sphere if you like.

Suggested problems for chapter 24:

1, 2, 3, 4, 5, 7, 10, 11, 13, 14, 15, 16, 17, 30, 31, 33, 37, 38, 39, 41, 43, 44, 45, 47, 53, 55, 56, 57, 58