Answer:
a
[tex]F = 2.32*10^{-17} \ N[/tex]
b
[tex]n =3869 \ electrons[/tex]
Explanation:
From the question we are told that
The charge on each water drop is [tex]q_1=q_2=q = - 6.19*10^{-16} \ C[/tex]
The distance of separation is [tex]d = 1.22\ cm = 0.0122 \ m[/tex]
Generally the electrostatic force between the water drops is mathematically represented as
[tex]F = \frac{k * q_1 * q_2 }{ d^ 2}[/tex]
Here k is the coulombs constant with value [tex]k = 9*10^9 \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
So
[tex]F = \frac{9*10^9 * -6.19 *10^{-16} * (-6.19*10^{-16}) }{ 0.0122^ 2}[/tex]
[tex]F = 2.32*10^{-17} \ N[/tex]
Generally the quantity of charge is mathematically represented as
[tex]q = n * e[/tex]
Here n is the number of electron present
and e is the charge on one electron with value [tex]e = 1.60*10^{-19} \ C[/tex]
So
[tex]n = \frac{6.19 *10^{-16}}{1.60*10^{-19}}[/tex]
[tex]n =3869 \ electrons[/tex]
