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\[\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}-\frac{1}{2}\frac{{{q}^{2}}{{E}^{2}}}{k}\]
Correct Answer: A
Solution :
When electric field is applied, at equilibrium \[F=kx=qE\] \[x=\frac{qE}{k},x-extension\,\,of\,\,spring\] Total Energy of system = Kinetic energy + Potential energy \[T.E.=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}k{{x}^{2}}\] \[=\frac{1}{2}\frac{{{q}^{2}}{{E}^{2}}}{k}+\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\]
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