Question

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 $10^{9} N C ^{-1}$ directed upward. (Take $g=10 \,m / s ^{2}$ )

• A. $10^{6}$
• B. $10^{7}$
• C. $10^{9}$
• D. $10^{8}$

Answer 1

Answer: (D)

Let $n$ be the number of electrons removed from the coin.
Then, charge on coin, $q=+n e$
Now, $q E=m g$
Or $(n e) E=m g$
Or $n=\frac{m g}{e E}=\frac{1.6 \times 10^{-3} \times 10}{1.6 \times 10^{-19} \times 10^{9}}=10^{8}$