Electrostatics

JEE Main level, single correct, medium

123 of 311

Question 123 of 311

A body of mass MM and charge qq is connected to a spring of spring constant kk. It is oscillating along x\mathrm{x}-direction about its equilibrium position, taken to be at x=0x=0, with an amplitude AA. An electric field EE is applied along the x\mathrm{x}-direction. Which of the following statements is correct?

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Answer: (A) The total energy of the system is
12mω2A2+12q2E2k\frac{1}{2} m \omega^2 A^2+\frac{1}{2} \frac{q^2 E^2}{k}

Equilibrium position will shift to point where resultant force =0=0
kxeq =qE⇒xeq =qEk\mathrm{kx}_{\text {eq }}=\mathrm{qE} \Rightarrow \mathrm{x}_{\text {eq }}=\frac{\mathrm{qE}}{\mathrm{k}}
Total energy =12 mω2 A2+12kxeq 2=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2+\frac{1}{2} \mathrm{kx}_{\text {eq }}^2
Total energy =12mω2A2+12q2E2k=\frac{1}{2} m \omega^2 A^2+\frac{1}{2} \frac{q^2 E^2}{k}

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