Chapter 8, Problem 1SAQ

Q1. According to Coulomb's law, if the separation between two particles of the same charge is doubled, what happens to the potential energy of the two particles?

a) becomes twice as high as it was before the distance separation.

b) becomesone-half as high as it was before the separation.

c) does not change.

d) becomes one-fourth as high as it was before the separation.

Interpretation Introduction

Introduction: Coulomb’s law describes about the attractions and repulsions that take place between charged particles.

Answer to Problem 1SAQ

Correct answer: If the separation between two particles of the same charge is doubled, the potential energy of the two particles becomes one-half as high as it was before the separation. Therefore, option (b) is correct.

Explanation of Solution

Justify reasons for the correct statement:

Coulomb’s law states that the potential energy $\left(E\right)$ of two charged particles is dependent on their charges $\left(q_1\right)$,$\left(q_2\right)$ and on the separation between them $\left(r\right)$. The law is expressed as follows:

$E=\frac{1}{4\prod {\epsilon }_{\text{o}}}\frac{q_1{q}_2}{r}$ (1)

Here, ${\epsilon }_{\text{o}}$ is a constant.

If the separation between two particles of the same charge is doubled, calculate the new potential energy $\left(E\text{'}\right)$ by changing the value of $r$ as $2r$ in the equation (1) as follows:$\begin{array}{c}E\text{'}=\frac{1}{4\prod {\epsilon }_{\text{o}}}\frac{q_1{q}_2}{2r}\\ =\frac{1}{2}\left(\frac{1}{4\prod {\epsilon }_{\text{o}}}\frac{q_1{q}_2}{r}\right)\\ =\frac{E}{2}\end{array}$

The potential energy of the two particles becomes one-half as high as it was before the separation.

Hence, option (b) is correct.

Justify reasons for the incorrect statements:

Option (a) is that the potential energy of the two particles becomes twice as high as it was before the separation. According to Coulomb’s law, the potential energy $\left(E\right)$ is inversely proportional to the distance, so $E$ can never become twice when $r$ is doubled. So, it is a wrong answer.

Option (c) is that the potential energy of the two particles does not change. According to Coulomb’s law, the potential energy $\left(E\right)$ is inversely proportional to the distance, so $E$ will change when $r$ is doubled. So, it is a wrong answer.

Option (d) is that the potential energy of the two particles becomes one-fourth as high as it was before the separation. According to Coulomb’s law, the potential energy $\left(E\right)$ is inversely proportional to the distance, so $E$ can never become one-fourth when $r$ is doubled. So, it is a wrong answer.

Hence, options (a), (c), and (d) are incorrect.

Conclusion

If the separation between two particles of the same charge is doubled, the potential energy of the two particles becomes one-half as high as it was before the separation.