Calculating magnitude of electric field at center of square

[Physics2341313]

Homework Statement

Find the magnitude and direction of net electric field at the center of the square array of charges. Find $E_x and E_y$

The square array of charges http://postimg.org/image/4gf94ymmf/

The Attempt at a Solution

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My attempt at drawing in the force vectors http://postimg.org/image/mae0fm1d9/ . Now the +3q and +q's should contribute a net of zero to the electric field and they can be ignored. So, we have $E = k [ -2q^2 / d^2 + q^2/d^2 + 5q^2/d^2] = 4k q^2/d^2$

Taking the x-component of the field we have
$E_x = (4k q^2/d^2)cos(45)$

the y-component:
$E_x = (4k q^2/d^2)sin(45)$

This isn't right or I would not be posting here... so what am I doing wrong? I'm really not understanding how to do these types of problems for net fields in squares. Have I even drawn the vectors correctly?

 

[Bystander]

k[−2q 2 /d 2 +q 2 /d 2 +5q 2 /d 2 ]=

You're missing a term.

 

[Physics2341313]

The missing term is the -5q charge, yes? So it should be
$E=k[−2q/(\sqrt2 * d)+q/(\sqrt 2 * d) + 2(5q/(\sqrt 2 * d))= k[9q / (\sqrt 2 * d)]$

Also, changed q^2 to q since this is the electric field not force... silly mistake, and the bottom term should be $\sqrt 2 * d$ instead of d^2?

This is still incorrect is not? I'm really not seeing what I'm missing here.

 

[Bystander]

Methinks that should do almost do it (dumped the picture, so I'm going by memory); the "two" and "one" are both on axes? In which case "d."

 

[Physics2341313]

Methinks that should do almost do it (dumped the picture, so I'm going by memory); the "two" and "one" are both on axes? In which case "d."

Ah fair point forgot about those being strictly on the x-axis, so need to fix that then resolve into E_x and E_y for the respective terms and then take the square root of those squared for the magnitude.