Electrostatics Question 238
Question: A negatively charged plate has charge density of $ 2\times {{10}^{-6}}C/m^{2} $ . The initial distance of an electron which is moving toward plate, cannot strike the plate, if it is having energy of $ 200eV $ [RPET 1997]
Options:
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 $
Show Answer
Answer:
Correct Answer: A , C
Solution:
Here $ E=\frac{1}{4\pi {\varepsilon _{0}}}.\frac{Qz _{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{-Qqz _{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{Qqz _{0}}{R^{2}} $
therefore $ F\propto -z _{0} $ i.e., motion is S.H.M.
Here $ E=\frac{1}{4\pi {\varepsilon _{0}}}.\frac{Qz _{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{-Qqz _{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{Qqz _{0}}{R^{2}} $
therefore $ F\propto -z _{0} $ i.e., motion is S.H.M.
