If r >> a, wouldn't we be able to treat the rod as a point charge? And in that case, wouldn't F = ke*((Q of sphere)/r^2)*(Q of the rod)? because the charge wouldn't have to be distributed over the rod if it's being considered as a point charge. Or am I looking at this all wrong?
Shouldn't the volume being used be 4/3pi(b-a)^3 because it's an insulating sphere? and after your force integration, what happened to the 2a? Shouldn't it be F = ke*(Q of sphere)*(Q of rod)/2a[1/r-a - 1/r+a]