Bead Entering Capacitor

1. The problem statement, all variables and given/known data
A small bead of mass m= 34 g and charge of Q=140 mC has a velocity (all in the horizontal direction) of 7 m/s as it enters the gap of a parallel plate capacitor, initially traveling parallel to the plates. The plates are separated by 4 mm. If the bead enters the capacitor 2 mm above the bottom plate, and leaves the capacitor 2 mm above the bottom plate, calculate the potential difference between the two plates.

m = 0.034kg
Q = 0.140 C
vx=7m/s
y = 0
d = 0.004m

2. Relevant equations

ma = Fg + qE
V = Ed

3. The attempt at a solution

I was more or less wondering how to conceptually think about this problem. Since the bead leaves the plates at the exact same height it was initially, does that mean that the force of gravity is equal to the electric force? Therefore I can equate

mg = qE

After, I can just sub in V = Ed into the force equation and solve for potential difference.
But with this solution, does that mean that the initial speed is irrelevant?

Thanks

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