## anonymous one year ago A charged oil drop, weighing 9.6 x 10^-12 N, drops at a constant speed in a 2 x 10^7 N/C electric field. a) What is the charge on the drop? b) How many electrons are missing?

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1. IrishBoy123

a) clearly the E field acts upwards to offset gravitational acceleration and you can say that $$Eq = mg$$ to calculate q. b) divide that q amongst as many electrons as add up to that number of Coulumbs worth of charge. ie $$n_{missing} = \frac{q}{e^-}$$

2. Michele_Laino

and we have$Eq = mg$ since we can apply the first law of Newton