A charged oil drop is suspended in a uniform field of $3\times 10^4\text{V/m}$ so that it neither falls nor rises. The charge on the drop will be: (Take mass of the drop $=9.9\times 10^{-15}\text{kg}$ and $g=10{\text{m/s}}^2$)

1. A
   $3.3\times 10^{-18}\text{C}$
2. B
   $3.2\times 10^{-18}\text{C}$
3. C
   $1.6\times 10^{-18}\text{C}$
4. D
   $4.8\times 10^{-18}\text{C}$