Chapter 4, Problem 1SAQ

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

• is one-fourth as high as it was before the separation
• is twice as high as it was before the distance separation
• does not change
• is one-half as high as it was before the separation

Interpretation Introduction

Interpretation:

“Potential energy after doubling the separation” should be determined.

Concept introduction:

The potential energy of the two charges separated by distance r is inversely propotional to distance and can be shown as follows:

E = K q₁q₂ /r

E is the potential energy

K is the force constant

q₁ q₂ are the charges

r is the separation between charges

Answer to Problem 1SAQ

Correct answer:

Option d is correct.

Explanation of Solution

Reason for correct answer:

On doubling the separation between the charges, the potential energy can be expressed as:

E₁ = K q₁q₂/ 2r = E /2

Hence, the potential energy after doubling the separation between the charges becomes half times higher than the initial energy.

Conclusion

Reason for incorrect answer:

The other options a, b, and c are not the exact solution.