Question

A charge Q is transferred from an initially uncharged plastic ball to an identical ball 17 cm away. The force of attraction is then 68nM .
How many electrons were transferred from one ball to the other?

Answer 1

Answer:

2.92 x 10¹² electrons

Explanation:

given,

distance between two plastic ball, r = 17 cm

                                      r = 0.17 m

Force of attraction = F = 68 mN

                F = 0.068 N

number of electron transferred from one ball to another.

using Coulomb Force equation

$F = \dfrac{kq^2}{r^2}$

$0.068 = \dfrac{9\times 10^9\times q^2}{0.17^2}$

q² = 2.1835 x 10⁻¹³

q = 4.67 x 10⁻⁷ C

now, number of electron

 $N = \dfrac{q}{e}$

e is the charge of electron which is equal to 1.6 x 10⁻¹⁹ C

  $N = \dfrac{4.67\times 10^{-7}}{1.6\times 10^{-19}}$

       N = 2.92 x 10¹² electrons

electrons were transferred from one ball to the other is 2.92 x 10¹²