Hanging Conducting Spheres

Mar 2020
Southampton UK
Two small conducting spheres P and Q, each of mass 1.5 X 10^-5 kg, are suspended from the same point O by insulating threads 0.10 m long. When the spheres are charged with equal charges they come to rest with both threads inclined at 30 degrees to the vertical.

(a) Show that the force of repulsion is 85uN between the spheres.

I understand that vertical component of tension will be size of gravity and, horizontal component will be size of electric force, both in opposite directions to the actual forces experienced by each sphere, in order for the spheres to be stationary.

As a result, I think the answer will be Fe = Fg *sin30, = (1.5*10^-5) *g * sin 30, however I do not get this to be 85 uN, but 74 uN to 2s.f.?

Is my method correct?
Jan 2019
horizontal equilibrium
$T\sin(30) = F_e$

vertical equilibrium
$T\cos(30) = mg$

substituting $\dfrac{mg}{\cos(30)}$ for $T$ in the horizontal equation

$F_e = mg\tan(30) = 8.5 \times 10^{-5} \text{ N}$
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