A) \[\frac{\lambda }{16\,{{\varepsilon }_{0}}}\]
B) \[\frac{\lambda }{8\,{{\varepsilon }_{0}}}\]
C) \[\frac{\lambda }{12\,{{\varepsilon }_{0}}}\]
D) \[\frac{\lambda }{4\,{{\varepsilon }_{0}}}\]
Correct Answer: C
Solution :
The given situation can be shown as Length of the arc, \[l=r\theta =r\frac{\pi }{3}\] \[\therefore \] Charge of the arc \[=\frac{r\pi }{3}\times \lambda \] So, the potential at the centre \[=\frac{kq}{r}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{r\pi }{3}\times \frac{\lambda }{r}\] \[=\frac{\lambda }{12{{\varepsilon }_{0}}}\]You need to login to perform this action.
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