there is a square on the XoY plane, centered at the origin (just outlines of the square) it has a charge Q (Q>0) and side 2L, i must evaluate the electric field along the z axis. see attached image
The Attempt at a Solution
So first i divided the square into 4 lines, and noticed that due to the simmetry there is only a field in the z direction, and all the lines of the square contribute the same to that field.
dE= k(Q/8L)1* dx/(x2+y2+z2)2*(z/(x2+y2+z2))½)3
where subscrip1 is Q/8L because the full Q is for the whole square, so we divide it by 4 to get the charge on the line and divide again by 2L to get the charge density.
subscript 2 is the distance of the charge to the line.
subscript 3 is cosine of E with Ez so we get only the z component
Now the integral, i get an not-so-straightforward integral, so if there is a simplification i could do here please advise.
so i get
and i integrate either over -L to L or from 0 to L and multiply it by 2.
and i get
This result makes sense to me physically, as field is 0 when z is 0 (makes sense since field would no longer have a z component) and also goes to 0 when z goes to infinity.
So the total field is 4*E. Am i correct? any simplification suggestions for the integral
EDIT: oh and y should be L since its fixed.