Chapter 20.2, Problem 15SSC

To determine

To describe:The relationship between the magnitude of the electrostatic force, charge on two objects and the distance between the objects.

To write: The equation for this relationship.

Explanation of Solution

Introduction:

The force of attraction or repulsion between the electrically charged particles is known as electrostatic force. Electric charges of same polarity opposes to each other and the charges with different polarity attracts with each other. The range of electrostatic force is very small which is in the order of one tenth of the nucleus diameter, that is ${10}^{-16}\text{ m}$ .

If $q_A$ and $q_B$ are the two electric charges separated by a distance $d$ then the magnitude of electrostatic force $\left(F\right)$ between these two charges varies as,

  $F\alpha q_A{q}_B$ $\left(1\right)$

And also the magnitude of electrostatic force $\left(F\right)$ between the two charges varies inversely with the square of the separation between them. This is given by,

  $F\alpha \frac{1}{d^2}$ $\left(2\right)$

Combining the expressions $\left(1\right)$ and $\left(2\right)$ , the expression for the electrostatic force between the two charges $q_A$ and $q_B$ separated by a distance $d$ can be written as,

  $F\alpha \frac{q_A{q}_B}{d^2}$

OR

  $F\text{ = K}\frac{q_A{q}_B}{d^2}$

Where $\text{ K}$ is a constant, $\text{ K}=9\times {10}^9\text{ N}{\text{.m}}^2/C^2$

This is called Coulomb’s law.