A) \[T=2\pi r\sqrt{\frac{m}{2k\lambda q}}\]
B) \[{{T}^{2}}=\frac{4{{\pi }^{2}}m}{2k\lambda q}.{{r}^{3}}\]
C) \[T=\frac{1}{2\pi r}\sqrt{\frac{2k\lambda q}{m}}\]
D) \[T=\frac{1}{2\pi r}\sqrt{\frac{m}{2k\lambda q}}\] where, k =\[\frac{1}{4\pi {{\varepsilon }_{0}}}\]
Correct Answer: A
Solution :
Idea Here, the electric force on the charge due to infinitely long wire .acts as o. centripetal force. Electric field around an infinite line charge is calculated using Gauss 'law and it is \[E=\frac{\lambda }{2\pi {{\varepsilon }_{0}}r}\] Now, particle is moving in a circle. So, centripetal force is provided by electric force \[\Rightarrow \]\[\frac{m{{v}^{2}}}{r}=\frac{\lambda }{2\pi {{\varepsilon }_{0}}r}.q\] \[\Rightarrow \]\[v={{\left( \frac{2k\lambda q}{m} \right)}^{1/2}}\left[ \text{using}\frac{1}{4\pi {{\varepsilon }_{0}}}=k \right]\] Now time period T\[=\frac{2\pi r}{v}\] \[=2\pi r\times \sqrt{\frac{m}{2k\lambda q}}\] TEST Edge The questions based on the electric field of a charged sheet or charged conducting shell could be asked.
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