Gauss's law and vectors

[600burger]

I'm encountering a problem with a practice test and previous homework problem in my electrostatics class.

Charges exist at three corners of a square as shown. Edge length = L. What is the Electric Field Strength (E) at the corner with no charge.


-3q
O------------O +2q
|
|
|
|
O-------------
+2q


so i break it into vectors using E= (k q)/(r^2) (Form of gauss) But i dont know what the final answer should look like...

[Tide]

Remember, the electric field is a vector - it has magnitude and direction!

[robphy]

Using symmetry of this problem,
you should be able to determine the direction,
and [almost by inspection] write down the algebraic answer.

Hint: if you can determine the direction (using symmetry), it is advantageous to use a coordinate system that has an axis along this direction [instead of the standard x- and y- axes]. The only components you'll need to consider are those along this axis.

[600burger]

Right a vector. So i would come out with an answer in unit vector form...like....



(x-component)i+(y-component)j. Or is there a simpiler way to write it?