Chapter 20, Problem 5STP

To determine

To Choose: The correct option.

Answer to Problem 5STP

C. $2.7\times {10}^2\text{}\text{N}$

Explanation of Solution

Given:

The initial force between the two equally charged objects, $F=90\text{}\text{N}$

Formula used:

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

Calculation:

Let the initial charges be $q$ , then the force will be,

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

Now, if one object is changed of same size but triple charge, then the new force will be,

  $\begin{array}{c}F\text{'}=\frac{1}{4\pi {\epsilon }_0}\frac{q\left(3q\right)}{r^2}\\ =3\left(\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}\right)\\ =3F\\ =270\text{}\text{N}\end{array}$

Conclusion:

Hence, the new force the objects will exert on each other is $2.7\times {10}^2\text{}\text{N}$ .