A) Periodic for all values of \[{{z}_{0}}\] satisfying \[0<{{z}_{0}}<\infty \]
B) Simple harmonic for all values of satisfying \[0<{{z}_{0}}<R\]
C) Approximately simple harmonic provided \[{{z}_{0}}<<R\]
D) Such that \[P\] crosses \[O\] and continues to move along the negative \[z-\]axis towards \[z=-\infty \]
Correct Answer: A , C
Solution :
Here \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q{{z}_{0}}}{{{({{R}^{2}}+z_{0}^{2})}^{3/2}}}\] where Q is the charge on ring and \[{{z}_{0}}\]is the distance of the point from origin. Then \[F=qE=\frac{-\,Qq{{z}_{0}}}{4\pi {{\varepsilon }_{0}}{{({{R}^{2}}+z_{0}^{2})}^{3/2}}}\] When charge ? q crosses origin, force is again towards centre i.e., motion is periodic. Now if \[{{z}_{0}}<<R\] \ \[F=-\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Qq{{z}_{0}}}{{{R}^{2}}}\] Þ\[F\propto -{{z}_{0}}\] i.e., motion is S.H.M.Solution :
Here \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q{{z}_{0}}}{{{({{R}^{2}}+z_{0}^{2})}^{3/2}}}\] where Q is the charge on ring and \[{{z}_{0}}\]is the distance of the point from origin. Then \[F=qE=\frac{-\,Qq{{z}_{0}}}{4\pi {{\varepsilon }_{0}}{{({{R}^{2}}+z_{0}^{2})}^{3/2}}}\] When charge ? q crosses origin, force is again towards centre i.e., motion is periodic. Now if \[{{z}_{0}}<<R\] \ \[F=-\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Qq{{z}_{0}}}{{{R}^{2}}}\] Þ\[F\propto -{{z}_{0}}\] i.e., motion is S.H.M.You need to login to perform this action.
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