Wednesday, July 14, 2010

HW3 is out

Homework 3 is now out. Since I'm late by about a day getting it out, I'll push the deadline back one day ... so this is due by 11:59pm on Friday, 16 July.

We'll go over most of these problems Thursday and Friday in class, at least enough to get you started.

(This means you have no HW or quiz tomorrow, but you have both on Friday ...)

2 comments:

Anonymous said...

On number four, I am guessing that the charges of the points (positive and negative) make a difference in the potential energies of the two. I am just confused at how we go about putting the positive and negative in our formula?

pleclair said...

We'll go over this one in class. Basically, in the formula you just put in (-q) in and (+q) for positive and negative charges. For a positive q and negative charge (-q) separated by a distance a, the potential energy is

PE = k(q)(-q)/a