Finding the electric field at point - Can someone check my work?

[emhelp100]

Homework Statement

Four equivalent charges are placed at (0,0), (a,0), (0,a), and (a,a). What is the electric field at point B (a, a/2)?

Homework Equations

The Attempt at a Solution

My attempt:
Charges at (a,a) and (a,0) cancel each other out.
[itex]E_{(0,a)}= \frac{Q(\hat{x}a +\hat{y}\frac{a}{2})}{4\pi e_0 (a^2+\frac{a^2}{4})^{3/2}}[/itex]

[itex]E_{(0,0)}= \frac{Q(\hat{x}a -\hat{y}\frac{a}{2})}{4\pi e_0 (a^2+\frac{a^2}{4})^{3/2}}[/itex]

[itex]\sum E = E_{(0,a)}+E_{(0,0)}= \frac{Q(\hat{x}2a)}{4\pi e_0 (a^2+\frac{a^2}{4})^{3/2}}[/itex]
[itex]E_B = \frac{Q(\hat{x}2a)}{4\pi e_0 (a^2+\frac{a^2}{4})^{3/2}}[/itex]
Is my answer correct?

 

[haruspex]

Is my answer correct?

Yes, but you could simplify it some.

 

[Dr Dr news]

Where did the a in the numerator come from?

 

[haruspex]

Where did the a in the numerator come from?

Same place the extra power of 1/2 came from in the denominator: taking the component in the x direction.