physics problem dealing with capacitors and a balance...?

Question

A capacitor balance has on one side a mass M, while on the other side is a capacitor whose two plates are separated by a be a gap of variable width. When the capacitor is charged to a voltage V, the attractive force between the plates balances the weight of the hanging mass. Calculate the voltage neccesary to balance a mass of 20mg if the plate area is 40 cm^2 & the plate separation d is 1 mm. can someone help me out..

Answer 1

Work is force times distance
W=F d
also for a capacitor
W=1/2q(V2-V1) or where V is voltage
W=1/2q V
since q=V/C c- capacitance we have
W=1/2C V ^2 while

C= e A/d
e- epselon zero
A - area of the plates
d - distance between the plates
we have

W=1/2 (e A/d) V ^2 and since F=W/d
F= e A V ^2/(2d^2)

Fg=Fe
mg=e A V ^2 /(2d^2)

Ffinally
V= sqrt( 2 mg d^2 / (e A )) where
e - Vacuum permittivity =8.854 E-12 F /m

Answer 2

From the ref, F = EQ = VQ/d = V^2C/d (where C = e*A/d) = V^2*e*A/d^2
Then V = sqrt(Fd^2/(e*A)) = sqrt(gmd^2/(e*A)) =
sqrt(9.8*.00002*1E-6/(8.854E-12*.004)) = 74.39 V.
I think the factor of 2 discrepancy in the argument to the square root between this and the previous answer lies in the fact that in a parallel-plate setup W = VQ or V^2C, not V^2C/2.