Question

Two tiny spheres of mass m = 8.80mg 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 50.0?in the following figure

Answer 1

E = force/unit charge

so you know the mass, and you know g, and you know the attraction between the spheres (kqq/r^2)

sum of the Forces = ma... i.e

but i'm thinking torques now, honestly... based on a pivot point at the hook... (r)X(mg) would be your torque in this case, you'll have to use trig to find the r between the 2 spheres [2(.530m(sin(25deg)))]

so actually you're gonna wanna use sum of the torques = I(alpha) = 0
both the kqq/r^2 and the force from the field are going to have components when they act as torques . (you want the tangential component)

so basically you have a torque from gravity, from the spheres, and from the field. and the sum of those = 0
so lets see if i can set this up...

you have:
(.530m)(.0086kg)(9.8m/s^2)(sin(65deg)) -
(.530m) (K[ (72.0*[10^-6]^2 C) ]^2 / [2 (.530m [sin (25deg) ] ) ]^2) -
(.530m)( [Electric Field] N/C)(72.0[10^-6]C) = 0 N*m

I really don't feel like crunching and solving for [Electric Field], but I do believe that's your Newton's second law application for this problem, draw a picture and check the angles though. I think they're right, but I'm not 100%
*EDIT*
Remember its r [cross] f
for torque, you should do alright.