Chapter 22, Problem 114A

To determine

The force between two charges for a given distance.

Answer to Problem 114A

The force between given charges is $0.41\text{ N}$

Explanation of Solution

Given:

Two charges of $+3.0\times {10}^{-6}\text{ C}$ and $+6.0\times {10}^{-5}\text{ C}$ separated by a distance 2.0 m

Formula used:

According to Coulomb’s law, two charges $q_1$ and $q_2$ separated by a distance r will experience a force which is given as,

  $F=K.\frac{q_1{q}_2}{r^2}$ …… (1)

Where,

K is the proportionality constant also known as Coulomb’s constant.

Calculation:

The value of Coulomb’s constant is $K=9.0\times {10}^9\text{ N}\cdot {\text{m}}^2/{\text{C}}^2$ . Now, by substituting the charges and the distance between them in (1) , the distance can be calculated as,

  $\begin{array}{c}F=9.0\times {10}^9\times \frac{3.0\times {10}^{-6}\times 6.0\times {10}^{-5}}{2^2}\\ =0.41\text{N}\end{array}$

Conclusion:

Therefore, the force between the charges is 0.41 N.