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When the distance between two charged particle is halved, the force between them becomes
  • a)
    One fourth
  • b)
    One half
  • c)
    Double
  • d)
    Four times
Correct answer is option 'D'. Can you explain this answer?
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Explanation:

Coulomb’s law states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically,

F = k(q1q2) / r^2

where F is the force between the charges, q1 and q2 are the charges of the particles, r is the distance between the particles, and k is the constant of proportionality.

Now, let’s assume that the distance between the two charges is halved, i.e., r/2. Substituting this value in the above equation, we get:

F’ = k(q1q2) / (r/2)^2

F’ = k(q1q2) / (r^2/4)

F’ = 4k(q1q2) / r^2

Comparing this equation with the original equation, we can see that the force between the charges has become four times the original force. Hence, the correct answer is option D, i.e., the force between the charges becomes four times when the distance between them is halved.
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