91Ó°ÊÓ

• a.Charges of the identical water drops,$q=-1.00×{10}^{-16}C$
• b. The separation between the water drops,$r=1cm\left(\frac{1m}{100cm}\right)=0.01m$

Using the concept of Coulomb's law, we can find the magnitude of the required force between the particles. Again, the number of electrons can be found by dividing the net charge by the value of the electronic charge.

Formulae:

The magnitude of the electrostatic force between any two particles,

$F=K\frac{\left|q_1\right|\left|q_2\right|}{r^2}$ (1)

The number of electrons present,

$n=q/e$ (2)

Using the given data in equation (1), we can get the magnitude of the force given as:

$$\begin{gathered} F=\frac{\left(9×{10}^9{\displaystyle \frac{N.m^2}{C^2}}\right){\left(1.00×{10}^{-16}C\right)}^2}{{\left(0.01m\right)}^2} \\ =9×{10}^{-19}N \end{gathered}$$

Hence, the value of the force is$9×{10}^{-19}N$

If n is the number of excess electrons (of charge –e each) on each drop then the number of excess electrons can be given using equation (2) as follows:

$$\begin{gathered} n=\frac{1.00×{10}^{-16}C}{1.60×{10}^{-19}C} \\ =625 \end{gathered}$$

Hence, the value of the excess electrons is 625