# Electric Field Done Correctly?

1. May 20, 2006

### nyclio

Two charges located along the x-axis, (-3*10^-6)C at the origin, and (6*10^-6)C at the point (2,0)

What is the electric field at point -1,0?

i did:

Etotal = Eq1+Eq2

=(k(q1)/r^2) + (k(q2)/(2-r)^2)
=(-2.09*10^4) N/C

Can someone please double check my calculations (i did it a few different ways and i always end up with a negative answer)? I plugged this into http://hyperphysics.phy-astr.gsu.edu/hbase/electric/e2p.html#c1 and for some reason their answer comes out positive. Which one is correct?

2. May 20, 2006

### Andrew Mason

The direction of the field is given by the direction of the force on a unit positive charge (ie. of +1C. ). $\vec E = \vec F/q$. So, in this case the field is positive.

AM