question_answer

A small charged particle of mass m and charge q is suspended by an insulated thread in front of a very large sheet of charge density a. The angle made by the treat with the vertical is equilibrium is:

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             \[\tan \theta =\left( \frac{q\sigma }{2{{\varepsilon }_{0}}mg} \right)\]                 \[\theta ={{\tan }^{-1}}\left( \frac{q\sigma }{2{{\varepsilon }_{0}}mg} \right)\]