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Brad Fitzpatrick

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my day in review [Oct. 3rd, 2000|09:02 pm]
Brad Fitzpatrick
after my unproductive morning, i decided to get my life in order. i went to office depot and bought a bunch of organizational furniture... desk stand, file cabinets, file folders, pencils, etc... then I cleaned my room, my desk, made my bed, and did a bunch of homework.

sarah (livejournal user #8 and friend from first year in college) came by and visited me about 6:00 pm... that was cool. hadn't seen her in awhile. made me happy.

afterwards blythe came over and I cooked us macaroni and cheese and dr. pepper. gourmet.

now i'm working on physics. the homework is all web-based now, so we know right away if we got it right or not... no showing our work, kick ass. i'm stuck on these two:

"Two fixed particles, of charges q1 = +2.0 µC and q2 = -7.0 µC, are 14 cm apart. How far from each should a third charge be located so that no net electrostatic force acts on it? " Seemed easy (4.88 cm), but I keep getting it wrong.

"Two engineering students, John with a weight of 180 lb and Mary with a weight of 90 lb, are 100 ft apart. Suppose each has a 0.04% imbalance in the amount of positive and negative charge, one student being positive and the other negative. Estimate roughly the electrostatic force of attraction between them by replacing each student with a sphere of water having the same mass as the student." No clue. Need to read the book I guess. But what's with the stupid units?

[User Picture]From: visions
2000-10-03 11:45 pm (UTC)
to the first question, it would depend on the placement of the particle. if they are linear or if they are planar. i couldnt help without knowing that. the basic idea though, find the net electrostatic force acting in the direction of the third charge. then, just figure out the distance at which that charge is disapated, assuming that q3 is uncharged or the point at which q3's charge counteracts the net force exhibited by the other charges.

to the second question, simply convert the weights to a mass... then the percentage of imbalance is the amount of charge exerted by each.. which will be attractive since they are opposite charges.

im not sure if that helps at all... but maybe. :)
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[User Picture]From: d4b
2000-10-05 01:18 pm (UTC)


John and Mary are attracted to each other because they are madly in love with each other. But I guess they forgot to mention that in the textbook. ;-)
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[User Picture]From: shortypoke
2000-10-03 11:51 pm (UTC)


The famous macncheese and dr.pepper,
i name it the Brad's special
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[User Picture]From: whitaker
2000-10-04 08:23 am (UTC)
God... I must be a REAL dumbass if I failed MY Physics test. I'd shit myself if I had problems like those.
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[User Picture]From: mattrope
2000-10-04 09:30 am (UTC)

Problem #1

You said you got 4.88, right? If you are doing it the same way I am, I think you should get two answers: 4.88 and -16.08. I think the -16.08 might be the one you want (so the third charge would be 16.08 cm from the positive charge and 30.08 cm from the negative charge). That configuration seems logical to me...

If you choose the 4.88 number, the third charge will be between the positive and negative charges, so it will almost certainly want to move one way or the other...
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[User Picture]From: bradfitz
2000-10-04 09:33 am (UTC)

Re: Problem #1

Yeah, it was the 16.08 one.

As for the other problem, turns out I did have it right, but the website was wrong.
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