Chapter 20, Problem 31A

To determine

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

Answer to Problem 31A

The force is inversely proportional to the square of the distance, thus if the distance is decreased keeping the charges constant, the force will increase.

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 decreased keeping the charges same, then the force will increase.

For example:

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 halved keeping the charges same, then the new force will be,

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

Conclusion:

Hence, the force between the charges will be increase if the distance between the charges is decreased.