Chapter 20.2, Problem 17SSC

To determine

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

Answer to Problem 17SSC

The force is inversely proportional to the square of the distance, thus if the distance is tripled keeping the charges constant, the force will reduce 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 ${\epsilon }_0=\text{ permittivity of free space}=8.85\times {10}^{-12}\text{F/m}$

As can be seen from above formula, the force between the charges is inversely related to the square of the distance between the charges.

  $F\propto \frac{1}{r^2}$

Now, if the distance between the charges is increased keeping the charges same, then the force will decrease.

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 $9$ .