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Physics · WORK, ENERGY AND POWER

हिन्दी

A particle moves in a conservative force field where the potential energy function is given by \( V(x) = \frac{a}{x^2} - \frac{b}{x} \), where \( a \) and \( b \) are positive constants. The magnitude of the force acting on the particle at its stable equilibrium position is:

  1. A
    \( \frac{b^2}{2a} \)
  2. B
    \( \frac{b^2}{4a} \)
  3. C
    \( \frac{b^3}{8a^2} \)
  4. D
    \( 0 \)
View Answer & Solution

Correct answer · Option D

\( 0 \)

Explanation

By definition, the force is \( F = -\frac{dV}{dx} \). At any equilibrium position, the net force is identically zero, \( F = 0 \).
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