I Got Another One By DesiredUsername in DesiredUsername's Diary Thu Mar 07, 2002 at 08:52:20 AM EST Tags: (all tags)
	Heh, that title is doubly applicable and I didn't even plan it that way. Warning: Body contains some math.

Got my birthday present from the inlaws. I was kind of thinking it would be the computer link cable for my shiny new TI-89 but instead I got the first season DVD of the Simpsons. That's cool, though the first season isn't best. They wanted to get me Futurama but I guess it isn't available in the US yet. Weird. Guess I'll have to buy the cable myself. A plastic rod having a uniformly distributed charge -Q has been bent into a circular arc of radius R and central angle 120°. With V = 0 at infinity, what is the electric potential at the center of curvature of the rod. [by which they mean the center of the circle if you extended the arc] I know I'm doing something wrong here but I can't see exactly what: V = 1/(4πε0)∫dq/r That much is easy. Then I thought I'd integrate over the angle, from 0 to θ. That means I'd need to figure out the differential charge dq which seems to clearly be the length of the subtended segment divided by the total length of the arc multipled by -Q (I pulled the sign out to the front of the integral right away to keep this simple, though) dq = (Rdθ/Rθ)Q = Q/θ dθ Lookin' good! Move the Q outside the integral and evaluate ln(θ)|02π/3. Infinity. Oops. The more I think about it the more I think that I can't just have a bare θ in there, I need some kind of trig func. The reason is that if I evaluation the integral over different endpoints I'll get different values which shouldn't be the case in this situation. I need some kind of cyclical or periodic function--like sin(). But where do I stick it? While showering I thought of trying to integrate over the length of the arc instead to eliminate any trig craziness. But that gives me a similar problem because it requires evaling ln(S)|0θR and ln(0) = -infinity. Another idea I've had is that I don't need to integrate at all. Voltage is a scalar and all the points are equidistant. Maybe V = -Q/(4πε0r)? I shaved this morning. I usually let it go a few days anyway but when I get lazy or distracted it can sometimes go for over a week. This is a dumb method because it takes a LOT longer to shave the underbrush than to just scrape off a little stubble, but there you have it. I think I mentioned this before but what the heck. My wife (re)did our taxes the other night. We're getting over $5000 back. Gotta adjust those withholdings...

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