A) \[{{\tan }^{-1}}\left( \frac{\sigma q}{2{{\varepsilon }_{0}}mg} \right)\]
B) \[{{\tan }^{-1}}\left( \frac{\sigma q}{{{\varepsilon }_{0}}mg} \right)\]
C) \[\tan \left( \frac{2\sigma q}{{{\varepsilon }_{0}}mg} \right)\]
D) zero
Correct Answer: A
Solution :
In equilibrium, along X axis \[T\,\,\sin \theta =q\varepsilon \] ?..(i) Where, T is the tension is the string. Along Y-axis in equilibrium \[T\,\,\cos \theta =mg\] ??(ii) From (i) and (ii), we obtain \[\tan \theta =\left( \frac{q\sigma }{2{{\varepsilon }_{0}}mg} \right)\] \[\theta ={{\tan }^{-1}}\left( \frac{q\sigma }{2{{\varepsilon }_{0}}mg} \right)\]You need to login to perform this action.
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