Two tiny spheres of mass = 8.70 mg carry charges of equal magnitude, 72.0 nC , but opposite sign. They are tied to the same ceiling hook by light strings of length 0.530 m . When a horizontal uniform electric field E that is directed to the left is turned on, the spheres hang at rest with the angle between the strings equal to 50 in the following figure. (note that q1 is positive and it is the left once)

http://session.masteringphysics.com/problemAsset/1331112/7/21-82.jpg

What is the magnitude E of the field?

Question

Two tiny spheres of mass  = 8.70 mg  carry charges of equal magnitude, 72.0 nC , but opposite sign. They are tied to the same ceiling hook by light strings of length 0.530 m . When a horizontal uniform electric field E that is directed to the left is turned on, the spheres hang at rest with the angle  between the strings equal to 50 in the following figure. (note that q1 is positive and it is the left once)
	
http://session.masteringphysics.com/problemAsset/1331112/7/21-82.jpg

What is the magnitude E of the field?

anonymous · Answer

I understand why their solution is correct, but I *don't* understand why mine is incorrect. 

The field must have a magnitude strong enough to overcome the attraction of the two sphers AND separate them a distance of 2Lsin([theta]/2), right?

So the Electric Field magnitude equals the Force of Q1 on Q2 + w/e force is required to keep the two that distance apart. The answer I come up with is double the force of Q1 on Q2. 

But it's not right.

... Why not?

anonymous · Answer

$$k q ^{2}/r ^{2}=q*E $$ will give the required answer.
@above you are not correct because your explanation does not take into account the fact that the electric field is uniform and not equivalent to a sphere