Recent content by Slightly Odd Guy

Well, I'm not going to give you a hard time about it. Thank you very much for your help!

Okay, I've checked the integral for p, and it works, and I can appreciate how this condenses to a dipole field, since the rod and dipole would look the same given enough distance. So that means I'm trying to condense Erod to the form E=c*k*p/d3, c being a numerical multiplier, correct?

Homework Statement A thin rod of length 2L has a linear charge density that isλ0 at the left end but decreases linearly with distance going from left to right in such a way that the charge on the entire rod is zero. Given E = −kλ0/L(d/(L−d)−ln(d−L)+d/(L+d)+ln(L+d)) for a point P that is...

Ok, the correct answer is d=0.36m. I don't know how they come up with that answer.

I don't, unfortunately. My online homework only tells me that it's wrong.

I switched some numbers around in my second post, so it screw some more things up. I think it's supposed to be: 0.05=|d2-(d2+0.013)|/(d2+0.013) in which case it gives d=0.501m, which is wrong.

0.013m2 is ~ (230mm/2)2.

Well... I could probably compare the radii for both. In which case, I would have: 0.05 = |d2-(d2+0.013)|/d2 0.05=0.013/d2 d=0.51 m Am I on the right track there?

Homework Statement You wish to determine the electric field magnitude along the perpendicular bisector of a 230-mm line along which35 nC of charge is distributed uniformly. You want to get by with a minimal amount of work, so you need to know when it is sufficient to approximate the line of...

Gravity! Of course! So now 67m/s2 = aE -g => aE = 76 m/s2 Which gives us E=(m/q)(76m/s2)=(10kg/C)(76m/s2) = 760 N/C Thank you very much, rude man!

Homework Statement A positively charged particle initially at rest on the ground accelerates upward to 180 m/s in 2.70 s .The particle has a charge-to-mass ratio of 0.100 C/kg and the electric field in this region is constant and uniform. What are the magnitude and direction of the electric...