Question

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?

Answer 1

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 \(n_{missing} = \frac{q}{e^-}\)

Answer 2

and we have\[Eq = mg\] since we can apply the first law of Newton