How many electrons should be removed from a coin of mass 1.6 g, so that it may float in an electric field of intensity ${10}^9 {\mathrm{NC}}^{-1}$ directed upward. (Take g = 10 $\mathrm{m}/{\mathrm{s}}^2$)

Let n be the number of electrons removed from the
coin.
Then, charge on coin, q = +ne

$$\begin{gathered} \text{ Now, } \mathrm{qE}=\mathrm{mg} \\ \text{ Or } \left(\mathrm{ne}\right)\mathrm{E}=\mathrm{mg} \\ \text{ Or } \mathrm{n}=\frac{\mathrm{mg}}{\mathrm{eE}}=\frac{1.6\times {10}^{-3}\times 10}{1.6\times {10}^{-19}\times {10}^9}={10}^8 \end{gathered}$$