A) 0
B) \[\frac{e}{\sqrt{{{\varepsilon }_{0}}a{{ }_{0}}m}}\]
C) \[\frac{e}{\sqrt{4\pi {{\varepsilon }_{0}}a{{ }_{0}}m}}\]
D) \[\frac{\sqrt{4\pi {{\varepsilon }_{0}}a{{ }_{0}}m}}{e}\]
Correct Answer: C
Solution :
[c] Centripetal force = Colombian force \[\frac{m{{v}^{2}}}{{{a}_{0}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{e\times e}{a_{0}^{2}}\] \[\Rightarrow {{v}^{2}}=\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}.{{a}_{0}}.m}\Rightarrow v=\frac{e}{\sqrt{4\pi {{\varepsilon }_{0}}.{{a}_{0}}.m}}\]You need to login to perform this action.
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