Chapter 20.2, Problem 12PP

To determine

The effect on the electrostatic force between the charges if the distance between the two charges is tripled.

Answer to Problem 12PP

The electrostatic force will be reduced by nine times.

Explanation of Solution

Introduction:

The electrostatic force between two charges, $q_1$ and $q_2$ , $r$ distance apart is given by,

  $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$

Where ${\epsilon }_0=\text{ permittivity of free space}=8.85\times {10}^{-12}\text{F/m}$

Let the initial force between the charges be,

  $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$

Now, if the distance between them is tripled keeping the charges same, then the new force will be,

  $\begin{array}{l}F\text{'}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{{\left(r\text{'}\right)}^2}\\ F\text{'}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{{\left(3r\right)}^2}\\ F\text{'}=\frac{F}{9}\end{array}$

Conclusion:

Hence, the new force between the charges will be equal to the initial force divided by $9$ .