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devil might be in detail.
what are you using for the electrical field of the sheet?

the situation described in your problem is:
|dw:1436512391791:dw|

|dw:1436513140003:dw|

Then use v^2 = 2as to find the velocity where a = Ft/m and s = 0.6-0.25

Yes you're right! E = σ/2ε for a very large flat sheet. my mistake, thanks

of course those equation are compatible with your data, if:
\[\Large \frac{{qE}}{m} - g < 0\]

forgive any typo's