Two similarly charged bodies are kept 5 cm apart in air. If the second body is shifted away from the first by another 5 cm, their force of repulsion will be

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Q: 30 (NDA-I/2012)
Two similarly charged bodies are kept 5 cm apart in air. If the second body is shifted away from the first by another 5 cm, their force of repulsion will be

question_subject: 

Science

question_exam: 

NDA-I

stats: 

0,1,9,4,3,2,1

keywords: 

{'force': [0, 0, 0, 2], 'second body': [0, 0, 0, 1], 'repulsion': [1, 0, 0, 1], 'bodies': [0, 0, 3, 10], 'fourth': [1, 0, 6, 4], 'air': [1, 0, 0, 0]}

The force of repulsion between two similarly charged bodies is given by Coulomb`s law, which states that the force is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Let`s analyze the given scenario step by step:

Option 1: Doubled

If the second body is shifted away from the first by another 5 cm, the distance between them would become 10 cm. According to Coulomb`s law, if the distance is doubled, the force of repulsion will become 1/4th of its original value, not doubled. Therefore, this option is incorrect.

Option 2: Halved

Similarly, if the distance is halved to 2.5 cm, the force of repulsion will become four times its original value according to Coulomb`s law. Therefore, this option is incorrect as well.

Option 3: Quadrupled

This option is incorrect because if the distance is quadrupled (20 cm), the force of repulsion will be reduced to 1/16th of its original value.

Option 4: Reduced to one-fourth

If the second body is shifted away from the first by another 5 cm, the distance between them would become 10 cm