A charged oil drop is suspended in uniform field of $3×10^4 V/m$, so that it neither falls nor rises. The charge on the drop will be

(take the mass of the charge = $9.9 × 10^{-15} kg$ and $g = 10 m/s^2$)

$3.3 × 10^{-18}C$

$\text{Net Force on the oil drop is zero}$

$\Rightarrow qE = mg$

$\Rightarrow q = \frac{mg}{E} = \frac{9.9\times 10^{-15}\times 10}{3\times 10^4}= 3.3\times 10^{-18} C$