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JEE Main201815 Apr 2018Morning ShiftPhysicsElectrostaticsActual

A body of mass M and charge q is connected to a spring of spring constant k . It is oscillating along x -direction about its equilibrium position, taken to be at x=0 , with an amplitude A . An electric field E is applied along the x -direction. Which of the following statements is correct?

Options

  1. AThe total energy of the system is 1 2 m ^2 A^2+ 1 2 q^2 E^2 k
  2. BThe new equilibrium position is at a distance: 2 q E k from x=0
  3. CThe new equilibrium position is at a distance: q E 2 k from x =0
  4. DThe total energy of the system is 1 2 m ^2 A^2- 1 2 q^2 E^2 k

Correct answer

A. The total energy of the system is 1 2 m ^2 A^2+ 1 2 q^2 E^2 k

Step-by-step solution

Equilibrium position will shift to point where resultant force =0 kx _ eq = qE x _ eq = qE k Total energy = 1 2 ~m ^2 ~A ^2+ 1 2 kx _ eq ^2 Total energy = 1 2 m ^2 A^2+ 1 2 q^2 E^2 k

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