Electric field. Linear charge density 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.
 A: I will just give you a hint. After then try out yourself. 
Integration is the key technique to be used in this question. Take a general point on the perpendicular bisector at a distance R from the rod. Find the general formula for net electric field and integrate it.  
A: The main idea of problems like these (determining the electric field due to some charged shape) is to integrate. You know the density of the electric charge along the wire, and so given some point in space $\vec y$ you could compute (magnitude and direction of) the electric field at $\vec y$ due to a tiny piece of the wire $dx$ using Coulomb's law. All that remains is to integrate along the wire w.r.t $x$ in order to compute the total field. You might have to think a little carefully, however, as you will be integrating vectors (use components)!
As a note for when you do learn Gauss' law -- it wouldn't have helped in this case anyway, as Gauss' law is most useful when there is a large amount of symmetry in the system. Here the wire is finite, which would be a dealbreaker.
