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