question_answer

The line AA is on charged infinite A conducting   plane   which   is perpendicular to the plane of the paper. The plane has a surface density of charge \[\sigma \] and B is hall of mass m with a like charge of magnitude q. B is connected by string from a point on the line AA. The tangent of angle \[(\theta )\] formed between the line AA and tile string is

A)  \[\frac{q\sigma }{2{{\varepsilon }_{0}}mg}\]

B)  \[\frac{q\sigma }{4\pi {{\varepsilon }_{0}}mg}\]

C)  \[\frac{q\sigma }{2\pi {{\varepsilon }_{0}}mg}\]

D)  \[\frac{q\sigma }{{{\varepsilon }_{0}}mg}\]

Correct Answer: D

Solution :

 The diagram is as follows The electric field due to charged infinite conducting sheet is \[E=\frac{\sigma }{{{\varepsilon }_{0}}}.\] Now, force (electric force) on the charged ball is\[F=qE=\frac{q\sigma }{{{\varepsilon }_{0}}}\] The resultant of electric force and mg balance the tension produced in the string. So, \[\tan \theta =\frac{{{F}_{e}}}{mg}=\frac{q\sigma }{\frac{{{\varepsilon }_{0}}}{mg}}=\frac{q\sigma }{{{\varepsilon }_{0}}mg}\]