Question

A charge Q is transferred from an initially uncharged plastic ball to an identical ball 24 cm away.The force of attraction is then 17 mN. How many electrons were transferred from one ball to the other?

Answer 1

Answer:

The number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons

Explanation:

Given;

magnitude of the attractive force, F = 17 mN = 0.017 N

distance between the two objects, r = 24 cm = 0.24 m

The attractive force is given by Coulomb's law;

$F = \frac{1}{4\pi \epsilon _0} \times \frac{Q^2}{r^2} = \frac{kQ^2}{r^2} \\\\Q^2 = \frac{Fr^2}{k} \\\\Q = \sqrt{ \frac{Fr^2}{k}} \\\\Q = \sqrt{ \frac{(0.017)(0.24)^2}{9\times 10^9}} \\\\Q = 3.298 \times 10^{-7} \ C$

The charge of 1 electron = 1.602 x 10⁻¹⁹ C

n(1.602 x 10⁻¹⁹ C) = 3.298 x 10⁻⁷

$n = \frac{3.298 \times 10^{-7}}{1.602 \times 10^{-19}} = 2.06 \times 10^{12} \ electrons$

Therefore, the number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons