How many electrons should be removed from a coin of mas 1.6 g, so that it may float in an electric field of intensity 109 N / C directed upward?

Question

How many electrons should be removed from a coin of mas 1.6 g, so that it may float in an electric field of intensity 109 N / C directed upward? (A) 9.8 × 107 (B) 9.8 × 105 (C) 9.8 × 103 (D) 9.8 × 101

Tardigrade Admin · Accepted Answer

q E=m g neE =m g Use n=(m q/e E)

Q. How many electrons should be removed from a coin of mas $1.6 g$ , so that it may float in an electric field of intensity $1 0^{9} N / C$ directed upward?

$9.8 \times 1 0^{7}$

$9.8 \times 1 0^{5}$

$9.8 \times 1 0^{3}$

$9.8 \times 1 0^{1}$

$qE = m g$
$n e E = m g$
Use $n = \frac{m q}{e E}$

Q. How many electrons should be removed from a coin of mas 1.6g, so that it may float in an electric field of intensity 109N/C directed upward?

9.8×107

9.8×105

9.8×103

9.8×101