Chapter 20, Problem 46A

To determine

New electrostatic force between the charged bodies.

Answer to Problem 46A

  $F\text{'}=2.32\text{N}$

Explanation of Solution

Given:

Force exerted by charged bodies, $F=0.145\text{N}$

Charged bodies are moved to one-fourth of original distance.

Formula Used:

From Coulomb’s law,

  $F=\frac{K{q}_1{q}_2}{d^2}$

Where,

  $F$ is electrostatic force

  $K$ is coulomb’s constant

  $q_1$ and $q_2$ are charges and

  $d$ is distance between the charges.

Calculation:

Since, $F=\frac{K{q}_1{q}_2}{d^2}$

Now, for $d\text{'}=\frac{d}{4}$

  $\begin{array}{c}F\text{'}=\frac{K{q}_1{q}_2}{{\left(\frac{d}{4}\right)}^2}\\ =\frac{K{q}_1{q}_2}{d^2}\times 16\\ F\text{'}=16F\\ =16\times 0.145\text{N}\\ =\text{2}\text{.32}\text{N}\end{array}$

Conclusion:

Therefore, the new electrostatic force between the charged bodies is $F\text{'}=2.32\text{N}$ .