Foo - My turn for a physics problem:

I got A and B right, but cannot get C.

Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1.80 millimeters thick, with a dielectric constant of K=3.60. The resultant electric field in the dielectric is 1.20*10^6 volts per meter.

PART A) Compute the magnitude of the charge per unit area (sigma) on the conducting plate. Express your answer in coulombs per square meter to three significant figures.
ANSWER: sigma = 3.82×10^-5 C/m^2

PART B) Compute the magnitude of the charge per unit area (sigma_1) on the surfaces of the dielectric. Express your answer using three significant figures.
sigma_1 =2.76×10^-5 C/m^2

PART C) Find the total electric-field energy U stored in the capacitor. Express your answer in joules to three significant figures.
U = J

I have tried: 1.91×10^-5, 26.8, .149, 1.97×10^-3 with no success.

#2 (and more importantly): A parallel-plate vacuum capacitor, with its plates separated by a distance of x_1 has an amount of energy equal to U stored in it. The separation is then decreased to x_2.

Part A) What is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed?

I've only tried "2U," and the message, "The correct answer involves the variable x_1 which was not part of your answer."

Part B) What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed?

Physics E&M blows. This is what computers and electrical engineers are for, not mechanical engineers.

Thanks for the help,
James

Argh... I don't remember much E&M. And it doesn't help that the class was in cgs units, not MKS units. A few years ago, that would have looked like a really easy problem. Now, I'd have to dig up my physics textbook. Sorry... that commentary is of little help to you, I know.

Argh is right. My college physics book is at work. Are there still homework help sites and forums? I remember when aol had chat rooms set up where you could get help from teachers.