Question

How many electrons should be removed from a coin of mas 1.6g, so that it may float in an electric field of intensity ${10}^9$ N/C directed upward?

• A. $9.8\times {10}^7$
• B. $9.8\times {10}^5$
• C. $9.8\times {10}^3$
• D. $9.8\times {10}^1$

Answer 1

The correct option is A $9.8\times {10}^7$

Here, $F=mg=neE$ where $q=ne$, $E$ is the electric field.

So, $n=\frac{mg}{eE}=\frac{(1.6\times {10}^{-3})\left(9.8\right)}{(1.6\times {10}^{-19})\left({10}^9\right)}=9.8\times {10}^7$