I was wondering if anyone could help me out in this exercise I've been struggling to solve.
A straight, nonconducting plastic wire $ 8.50 cm $ long carries a charge density of 175 $ nC/m$ distributed uniformly along its length. It is lying on a horizontal tabletop.
Find the magnitude and direction of the electric field this wire produces at a point $ 6.00 cm $ directly above its midpoint.
I can't use Gauss's law because we simply haven't been taught that.
$$ E_y={KdQ \over r^2}sin \phi $$
I've tried using the equation above because there's not an electric field in the x-component, but I still get nowhere.
I did use integration, but I still don't get the right answer I've got this:
$$sin \phi={D \over r} $$ Where D is the distance from the midpoint of the wire to the point. $$ r^2=(x-{L \over 2})^2+D^2$$
Therefore:
$$ E_y= \int {K\lambda D \over [(x-{L \over 2})^2+D^2]^{(3/2)} }$$
But when I solve that and evaluate from $0$ to ${L } $ I still don't get it right.