Answer
The drop has $~~120~~$ excess electrons.
Work Step by Step
Since the drop is suspended, the upward force on the drop due to the electric field must be equal in magnitude to the drop's weight. In part (a), we found that the drop's weight is $8.87\times 10^{-15}~N$
We can find the magnitude of the charge of the water drop:
$F = 8.87\times 10^{-15}~N$
$q~E = 8.87\times 10^{-15}~N$
$q = \frac{8.87\times 10^{-15}~N}{E}$
$q = \frac{8.87\times 10^{-15}~N}{462~N/C}$
$q = 1.92\times 10^{-17}~C$
We can find the number $N$of excess electrons:
$N\times (1.6\times 10^{-19}~C) = 1.92\times 10^{-17}~C$
$N = \frac{1.92\times 10^{-17}~C}{1.6\times 10^{-19}~C}$
$N = 120$
The drop has $~~120~~$ excess electrons.
