Recent content by bjornert

ok, yeah got it now...thanks

ok, i think i got it now... \int^{d}_{-d}\frac{k\lambda_{o}zdz}{d(3d-z)^{2}} and then removing constants... \frac{k\lambda_{o}}{d}\int^{d}_{-d}\frac{zdz}{(3d-z)^{2}} but still the integral doesn't turn out nicely...should i be using the length of the vector r |\overline{r}| =...

not exactly, the point and the insulator are both on the same axis...the insulator spans (-d) to (d) and the point is at 3d all along the z axis (z-hat) also yes, the charge density is non uniform so it changes at the origin as you stated...lambda z/d changing sign as a function of it's...

d is the length of the insulator (from -d to d), as stated in the question.

Ok, so here's the question. I've pretty much got it except I'm having trouble with one part of constructing the integral. The problem is... An insulator which lies between the positions -d\hat{z} and d\hat{z} has a nonuniform linear charge density \lambda = \lambda_{o}\frac{z}{d} . Find the...