Equilibrium position will shift to point where resultant force =0 kxeq =qE⇒xeq =kqE Total energy =21 mω2 A2+21kxeq 2 Total energy =21mω2A2+21kq2E2
JEE Main 2018 — Physics Electromagnetism
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?
Held on 15 Apr 2018 · Verified 6 Jul 2026.
The total energy of the system is 21mω2A2+21kq2E2
The new equilibrium position is at a distance: k2qE from x=0
The new equilibrium position is at a distance: 2kqE from x=0
The total energy of the system is 21mω2A2−21kq2E2
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