Thursday February 4, 1999
Announcements:  Quizzes on Friday and Monday will cover Chapters 23 and 24

Review Sessions for the first Midterm:
Sunday February 7
    2:30 to 4:00        103 Osmond        Mr. Takemae
    4:00 to 5:30        101 Osmond        Mr. Wertz
    6:00 to 7:30        101 Osmond        Mr. Minnick
Monday February 8
    3:30 to 5:00        117 Osmond        Mr. Nelson
    6:00 to 7:30        101 Osmond        Mr. Zadorozhny

Extended Office Hours next week.
Professor Samarth        February 8        1:00 to 3:30        338 Davey Lab
Professor Chan            February 9        12:45 to 3:30       327 Davey Lab

Lecture notes:

• dA and E are parallel
• The charge per unit area is l .
• E is radially outward from the center; there is no component of E in the y direction ( no electric field comes out of the end caps )
• F = I E (dot product) dA = E I dA = E (2pr) h
• Gauss' Law says : F = (Amount of charge enclosed)/ e_o =  lh / e_o
• Therefore, (E)(2pr h) = lh / e_o
• E = l / 2pre_o
• Compare this with problem # 33 in Chapter 23

• The charge per unit area is s.
• Again, I E (dot product) dA = (Amount of charge enclosed)/ e_o.
• With a gaussian cylinder, A = the area of the end caps
• There is no contribution to I E (dot product) dA except at the two end caps, where the electric field and the area are parallel.
• Therefore, (EA +EA) = sA/ e_o.
• So, E = s /2 e_o.
• See also section 23.7 from the text.

• The electric field is zero everywhere inside a conductor
• Before the external field is applied, electrons are uniformly distributed throughout the conductor.
• Under the external field E, electrons will accumulate at the left; there will be a deficit of electrons on the right.
• These charges set up there won field, which opposes the external field.

• The surface charge density increases until the net field equal zero and the charges stop moving
• E + E^l = 0

• If there are excess charges on a conductor, these charges will reside entirely on its surface.

• Assume there are excess charges deposited onto the conductor, and that the gaussian surface inside the conductor is as close to the surface as possible
• Then, by Gauss' Law, I E (dot product) dA = Amount of charge enclosed)/ e_o. But E inside a conductor equals zero and q enclosed equals zero.
• Therefore, the charges reside on the surface.

• A conducting sphere with +Q is surrounded by a spherical conducting shell
• What is the net charge on the inner surface of the shell?

• If the I E (dot product) dA = (Amount of charge enclosed)/ e_o, then the total charge enclosed must equal zero.
• +Q + Q(of the inner surface) = 0
• Therefore the charge on the inner surface is -Q
• Another positive charge is placed outside the shell. What is the net charge on the inner surface of the shell?
• The net charge on the inner surface doesn't change. The inner surface is still zero due to Gauss' Law.
• If +q is moved to a position between the shell and the sphere, what is the net charge on the inner surface of the shell?
• I E (dot product) dA = (Amount of charge enclosed)/ e_o= 0
• +Q + q + Q( inner surface) = 0
• Q ( inner surface) = - Q -q
• Are the previous answers valid if the sphere and shell are not concentric?
• Yes, because even if they weren't concentric the middle can still be enclosed by a Gaussian Surface

• What is the electric field just outside a charged conductor? What is the direction of this field?

• s is the charge per unit area
• The tiny cylindrical gaussian surface is imbedded into the conductor.
• There is no flux through the end cap inside the conductor, since the electric field is zero inside the conductor.
• The electric field outside the conductor is perpendicular to the surface; if it is not, the charges will move. Therefore, there is no flux through the cylindrical wall.
• Gauss' Law says that I E (dot product) dA = (Amount of charge enclosed)/ e_o
• EA = sA / e_o ; E = s / e_o
• The electric field is perpendicular to the surface.