A) it will oscillate about its original position
B) it will move further towards the positive charge
C) its electric potential energy will decrease and kinetic energy will increase
D) its total energy remains constant but is non-zero
Correct Answer: D
Solution :
Initially the force on the sphere is equal due to both \[=\frac{d\phi }{R}\] and \[\frac{{{\mu }_{0}}ni}{2r}=H\tan \theta \] charge. \[n\propto H\tan \theta \] Net force \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{\tan {{\theta }_{1}}}{\tan {{\theta }_{2}}}\] On displacing the sphere towards the \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{\tan {{60}^{o}}}{\tan {{45}^{o}}}=\frac{\sqrt{3}}{1}\] charge, force on sphere due to \[={{10}^{4}}J/T\] charge will be more than due to the \[B=4\times {{10}^{-5}}T,\theta ={{60}^{o}}\] positive charge, because it is nearer. So, sphere will move further to the charge.
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