A charged oil drop is suspended in unifor m 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 $g=10 \mathrm{~m} / \mathrm{s}^2$ )

• Correct answer
  A

  $3.3 \times 10^{-18} \mathrm{C}$
• B

  $3.2 \times 10^{-18} \mathrm{C}$
• C

  $1.6 \times 10^{-18} \mathrm{C}$
• D

  $4.8 \times 10^{-18} \mathrm{C}$

Solution

$\begin{aligned} & \text { In steady state, } \\ & \text { electric force on drop } \\ & \quad \begin{aligned} \quad & =\text { weight of drop } \\ \begin{aligned} \therefore\end{aligned} \quad q E= & m g \\ & q=\frac{m g}{E} \\ & =\frac{9.9 \times 10^{-15} \times 10}{3 \times 10^4} \\ & =3.3 \times 10^{-18} \mathrm{C}\end{aligned}\end{aligned}$

Hence, the correct answer is option: 'A'.