Electric field of circular ring wire

[Acuben]

http://yfrog.com/ea1c281p
http://img514.imageshack.us/img514/3479/1c281.png

Uploaded with ImageShack.us

Homework Statement

A circular ring of fine wire carries a uniformly distributed positive charge q.
Find the magnitutde and diretion of the electric field at the center of the ring caused by just the charge on a
portion of the ring subtending an angle a at the center, in terms of q, a, and radius r
(Total charge of the wire is Q)

Homework Equations

formula for circular wire E=2k(Lambda)*sin (a/2) / r
note: sin (a/2) assuming the wire is of angle a.
k=(4(pi)e)^-1
note: e does not stand for exponent! It's not 2.72...

The Attempt at a Solution

I simply plug the E=2k(Lambda)*sin (a/2) / r

so I get

[ 2kq sin (a/2) ] / [ar^2]
==Simplified: 2kq/arr times sin (a/2)
or
[q sin (a/2) ] / [ 2(pi)*e*a*r^2]
==Simplified: q/2earrpi times sin (a/2)


but
answer: [q sin (a/2) ] / [ 4(pi)^2*e*r^2]
==Simplified: q/4(pi)(pi)err times sin (a/2)
which is very similar to the answer except the difference of having 2pi instead of a.

 

[BigJDubs]

I figured out that it should be a instead of 2pi. Now I understand the answer to be:[q sin (a/2) ] / [ 4(pi)^2*e*r^2] ==Simplified: q/4(pi)(pi)earr times sin (a/2)Is this correct?