# Electric field at the center of the equilateral triangle

## Homework Statement

Consider an equilateral triangle of side 15.6 cm. A
charge of +2.0 µC is placed at one vertex and
charges of –4.0 µC each are placed at the other
two, as shown in the diagram to the right.
Determine the electric field at the centre of the
triangle.

I've found the distance from each point to the centre. Now do I just find the electric field of each compared to the centre and add all three together?

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That is the way.

Ok, I tried it and here's what I got:

Distance from the centre to the top vertice of the triangle (with a charge of +2.0 uC) = 8.4 cm
Distance from each of the two bottom vertices to the centre (each with a charge of -4.0 uC) = 3.2 cm

E = kQ1/r^2 + kQ2/r^2 + kQ3/r^2
= k (Q1/r^2 + Q2/r^2 + Q3/r^2)
= 9 x 10^9 ( 0.002/0.084 + 0.004/0.032 + 0.004/0.032)
= 2.46 x 10^9 N/C

Does that look like I did everything right? And, does the direction matter, or is that implied based on the negative and positive charges...

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Electric field is a vector.