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Physics · WORK, ENERGY AND POWER
हिन्दीA 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:
Correct answer · Option 4
\( 0 \)
Explanation
By definition, the force is \( F = -\frac{dV}{dx} \). At any equilibrium position, the net force is identically zero, \( F = 0 \).
