# Gauss's law and vectors

1. Sep 12, 2004

### 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...

2. Sep 12, 2004

### Tide

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

3. Sep 12, 2004

### 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.

4. Sep 12, 2004

### 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?