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

Given data: In a consistent field of oil, a charged oil drop is suspended $3\times {10}^4 \mathrm{V}/\mathrm{m}.$

Concept used: Electric field and electric field lines.

Detailed solution:

Electric force of drop balances weight of drop.

$$\begin{gathered} q E=m g \\ q=\frac{mg}{E} \end{gathered}$$

now,

$$\begin{gathered} q=\frac{9.9\times {10}^{-15}\times 10}{3\times {10}^4} \\ =3.3\times {10}^{-18}\mathrm{C} \end{gathered}$$