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 109 N C -1 directed upward. (Take g=10 m / s 2 )

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 109 N C -1 directed upward. (Take g=10 m / s 2 ) (A) 106 (B) 107 (C) 109 (D) 108

Tardigrade Admin · Accepted Answer

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=(m g/e E)=(1.6 × 10-3 × 10/1.6 × 10-19 × 109)=108

Q. 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 $1 0^{9} N C^{- 1}$ directed upward. (Take $g = 10 m / s^{2}$ )

$1 0^{6}$

$1 0^{7}$

$1 0^{9}$

$1 0^{8}$

Let $n$ be the number of electrons removed from the coin.
Then, charge on coin, $q = + n e$
Now, $qE = m g$
Or $(n e) E = m g$
Or $n = \frac{m g}{e E} = \frac{1.6 \times 1 0 ^{- 3} \times 10}{1.6 \times 1 0 ^{- 19} \times 1 0 ^{9}} = 1 0^{8}$

Q. How many electrons should be removed from a coin of mass 1.6g, so that it may float in an electric field of intensity 109NC−1 directed upward. (Take g=10m/s2 )

106

107

109

108