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JEE Mains · Physics · STD 12 - 1. Electric charges and fields

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?

  1. A The total energy of the system is \(\frac{1}{2}m{\omega ^2}{A^2} + \frac{1}{2}\frac{{{q^2}{E^2}}}{k}\)
  2. B The new equilibrium position is at a distance: \(\frac{{2qE}}{k}\) from \(x = 0\)
  3. C The new equilibrium position is at a distance: \(\frac{{qE}}{{2k}}\) from \(x = 0\)
  4. D The total energy of the system is \(\frac{1}{2}m{\omega ^2}{A^2} - \frac{1}{2}\frac{{{q^2}{E^2}}}{k}\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

Equilibrium position will shifto point where resultant force \(=0\) \(k{x_{eq}} = {\text{qE}} \Rightarrow {{\text{x}}_{{\text{eq}}}} = \frac{{{\text{qE}}}}{{\text{k}}}\) Total energy…