# What’s the magnitude of the charge on the oil doplet?

May I know why’s the answer A instead or B or C?

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- NCSLv 73 months agoFavorite Answer
The field is constant between the plates:

E = V/x

When the field points down, Newton's Second gives us

ma = mg - qE = mg - qV/x

which rearranges to

mg = ma + qV/x ← #1

And when the field points up,

m*3a = mg + qV/x

which rearranges to

mg = 3ma - qV/x ← #2

Subtrace #1 from #2 to get

0 = 2ma - 2qV/x

and so

q = max / V

Hope this helps!

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