Question

Question: Two tiny, spherical water drops, with identical charges of$\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{16}}\mathbf{C}$, have a center-to-center separation of 1.00 cm. (a) What is the magnitude of the electrostatic force acting between them? (b) How many excess electrons are on each drop, giving it its charge imbalance?

Short answer

• a)The magnitude of the electrostatic force acting between them is$9\times {10}^{-19}N$
• b) There are excess electrons on each drop.

Step-by-step solution

The given data 

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

Understanding the concept of Coulomb’s law 

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)

a) Calculation of the magnitude of the force

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

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

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

b) Calculation of the number of excess electrons

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:

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

Hence, the value of the excess electrons is 625