A charged oil drop is suspended in a uniform field of $3 \times 10^4 \mathrm{~V} / \mathrm{m}$ so that it…
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 $\mathrm{g}=$ $10 \mathrm{~m} / \mathrm{s}^2$ )
$3.3 \times 10^{-18} \mathrm{C}$
$3.2 \times 10^{-18} \mathrm{C}$
$1.6 \times 10^{-18} \mathrm{C}$
$4.8 \times 10^{-18} \mathrm{C}$
Solution
Since ball is hanging in equilibrium, force by gravity is balanced by electric force.
$
\begin{aligned}
& \mathrm{qE}=\mathrm{mg} \\
& \Rightarrow \mathrm{q}=\frac{\mathrm{m} \times \mathrm{g}}{\mathrm{E}} \\
& \Rightarrow \frac{9.9 \times 10^{-15} \times 10}{3 \times 10^4} \\
& \therefore \mathrm{q}=3.3 \times 10^{-18} \mathrm{C}
\end{aligned}
$