Question

Problem 2 A charge Q is transferred from an initially uncharged plastic ball to an identical ball 32.50 cm away. The force of attraction is then 29.60 mN. How many electrons were transferred from one ball to the other?

Answer 1

here,

sepration between the balls , r = 32.5 cm = 0.325 m

let the magnitude of charge on each ball be q

the force of attraction , F = 29.6 mN

0.0296 = K * q^2 /r^2

0.0296 = 9 * 10^9 * q^2 /0.325^2

solving for q

q = 5.89 * 10^-7 C

magnitude of charge on electron , e = 1.6 * 10^-19 C

the number of electrons transferred , N = q /e

N = 5.89 * 10^-7 /( 1.6 * 10^-19)

N = 3.68 * 10^12 electrons