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COMEDK · Physics · 17. Electrostatics

Two point charges \(\mathrm{M}\) and \(\mathrm{N}\) having charges \(+q\) and \(-q\) respectively are placed at a distance apart. Force acting between them is \(\mathrm{F}\). If \(30 \%\) of charge of \(\mathrm{N}\) is transferred to \(\mathrm{M}\), then the force between the charges becomes:

  1. A \(\dfrac{100}{49} F\)
  2. B \(\dfrac{49}{100} F\)
  3. C F
  4. D \(\dfrac{9}{16} F\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\dfrac{49}{100} F\)

Step-by-step Solution

Detailed explanation

Initial force: \(F = \dfrac{kq^2}{r^2}\) 30% of charge of N transferred to M: Charge transferred \(= 0.3q\) New charge on M \(= +q + (-0.3q) = 0.7q\) New charge on N \(= -q - (-0.3q) = -0.7q\) \(F' = \dfrac{k(0.7q)(0.7q)}{r^2} = 0.49 \cdot \dfrac{kq^2}{r^2} = \dfrac{49}{100}F\)