Electric Field Help: Find Magnitude & Direction

[Linco]

Can someone tell me if I got this right?

If a charge of 3 x 10–6 coulombs sets up an electric field, find:

a.The magnitude of the field 9 meters away.
F=K(q1q2/d*2) F= 9 X 10*9(3 x 10*6/9*2) =3.33N

b.The directive of the field.
The directive would be outwards.

 

[Pyrrhus]

Do you know what's the electric field?, on a) you wrote the coulombian force definition, applied the electric field definition, and used the wrong unit and used the wrong unit for electric field...

 

[Linco]

The problem didn't supply the electric field.

So I am guessing I need to use E=F/q?

im not sure how to do this without E

 

[Pyrrhus]

I suppose the problem is asking for the Electric Field supplied by the given charge at a distance of 9 meters, you found the magnitude, except your unit is wrong, it's N/C. Remember the Electric field is the amount of coulombian force per charge. You can find the direction through the electric field convention, for positive charges the electric field lines are directed outwards.