Recent content by nikkou

but my integration is done only on the vertical components! E=k\lambda d*\int((\frac{1}{(\frac{l}{2})^{2}+(\frac{w}{2})^{ 2}+d^{2}})* cos\alpha) dl the cos\alpha takes care of that. I don't have the exact official answer to the the numbers given on my original...

Ok, after reading your post and starting over with the intention to recreate what you wrote there, here's what i ended up with: as i said the distance i used in the integration is: \sqrt{(\frac{l}{2})^{2}+(\frac{w}{2})^{2}+d^{2}} (this is the distance from dl/dw to p) and my integral...

WOW! That's amazing nick, thanks so much! you definitely got what it takes to explain this stuff, it worked perfectly! the only thing i did different is that i expressed the distance as \sqrt{ (\frac{l}{2})^{2}+(\frac{w}{2})^{2}+d^{2} } which is the "real" distance from the wire to the...

There's something i don't get - the distance you're both using isn't the right one. it's true just for the middle of one of the sides - the distance from [dl] to the point of measurement is changing for any [dl]. or am i totally wrong?

How would i use the angle? i integrated using the distance which i got by the pythagorean theorem.

Hi guys, here's a question i thought i knew how to solve but somehow it doesn't work. been trying and trying for almost a week :-( here it is: "A thin wire is bent in the form of a rectangle with length l = 12 cm and width w = 10 cm. The wire is charged up to a uniform charge density of...

man, that's embarrassing... It's such a simple question, I just messed up the units. please excuse my inattention, I promise not to be so hasty next time and make sure i really need help with something substantial before posting :-) Thank you very very much!

Well, I did use 1/4 of the given mass for each rod. sorry for not mentioning it. and the formula i wrote is the same one you did: (ML^2)/3 . still it doesn't work... The real answer, which i forgot to give, is: 3.97×10-3 kg*m^2 I never get it...

[SOLVED] a tricky Moment of Inertia question, help needed... The question is in the JPG attachment. This is what i tried: Obviously the side which is right on the axis has no moment of inertia so we're left with the 2 horizontal ones and the other vertical. for the two horizontal sides we've...