What is the Electric Field Magnitude at a Charged Ball's Position?

[Zypheron]

I'm having a problem with one of my physics problems.

A plastic ball of radius 5 cm is hanging from a string 2m away from a charged glass rod. The ball has been charged to -5.13 microCoulombs. If the electric force on the ball is 0.6672 milliNewtons in the horizontal direction, what is the magnitude of the electric field at the balls position?

I've worked at it by using E = F/Qo, but it just doesn't seem to be working out ><. I'm not looking for an answer, I'm looking to see if i am going about this problem in the right way.

 

[HallsofIvy]

I presume this is part of a more extensive problem because, the way the problem is phrased, neither the radius of the ball nor the distance to the glass rod is relevant.
Yes, E= F/Q₀= 0.6672 milliNewtons/(-5.13 microCoulombs). If that's not the correct answer, check to see if you need to convert units.

 

[Zypheron]

Yeah, those tricksy tricksy physics professors like to give the useless info. I ended up getting it right in the end and such (stupid lab quiz required that i use a direction, although it was implied by the positive and negative x-axis ><. Thanks for the help though. I was getting really really stressed out about it (needed to get that one so i could go to my lab today)