A charged oil drop is suspended in a uniform field of $3 \times 10 ^ { 4 } \mathrm {~V} / \mathrm { m }$ so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge $= 9.9 \times 10 ^ { - 15 } \mathrm {~kg}$ and $\mathrm { g } = 10 \mathrm {~m} / \mathrm { s } ^ { 2 }$ )\\
(1) $3.3 \times 10 ^ { - 18 } \mathrm { C }$\\
(2) $3.2 \times 10 ^ { - 18 } \mathrm { C }$\\
(3) $1.6 \times 10 ^ { - 18 } \mathrm { C }$\\
(4) $4.8 \times 10 ^ { - 18 } \mathrm { C }$