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, \[\tan \theta =\frac{{{F}_{e}}}{mg}=\frac{q\sigma }{\frac{{{\varepsilon }_{0}}}{mg}}=\frac{q\sigma }{{{\varepsilon }_{0}}mg}\]You need to login to perform this action.
You will be redirected in
3 sec