Question

Two insulating small sphere are rubbed against each other and place 96cm apart if they attract each other with a firce of 0.1N hiw many electron were transfered frim one sphere ti the other during rubbing

Answer 1

F = kq^2/r^2
0.1 = 9 x 10^9 q^2/0.96x0.96
q^2 = 0.1 x 0.96 x 0.96/ 9 x 10^9
q^2 = 1.024 x 10^-11
q = 3.2 x 10^-6
q = ne ; n = q/e
n = 3.2 x 10^-6/ 1.6 x 10^-19
n = 2 x 10^13
Electron transferred = 2 x 10^13

Answer 2

Answer:

n = 2 × 10¹³ electron

Explanation:

Let the charge on the both sphere is q. So the electrostatic force is given by :

$F=k\dfrac{q^2}{r^2}$

$0.1=k\dfrac{q^2}{(0.96)^2^2}$

$q^2=\dfrac{0.1\times 0.96^2}{9\times 10^9}$

$q^2=1.024\times 10^{-11}$

$q=3.2\times 10^{-6}\ C$

So, the charge of both insulating small sphere is $3.2\times 10^{-6}\ C$

According to quantization of charge, q = n e

So, $n=\dfrac{q}{e}$

$n=\dfrac{3.2\times 10^{-6}}{1.6\times 10^{-19}}$

$n=2\times 10^{13}$

2 × 10¹³ electron were transferred from one sphere to other during rubbing.