## College Physics (4th Edition)

$q = 2 e$
We can find the magnitude of the charge on the drop: $F = q~E$ $F = q~(\frac{V}{d})$ $q = \frac{F~d}{V}$ $q = \frac{(9.6\times 10^{-16}~N)(0.16~m)}{480~J/C}$ $q = 3.2\times 10^{-19}~C$ We can express the magnitude of the charge in terms of $e$: $q = \frac{3.2\times 10^{-19}~C}{1.6\times 10^{-19}~C} = 2 e$