Question

An arc of radius r carries change. The linear density of charge is $\lambda$ and the arc subtends a angle $\frac{\pi }{3}$ at the center. What is electric potential at the center

• A. $\frac{\lambda }{4{\epsilon }_0}$
• B. $\frac{\lambda }{8{\epsilon }_0}$
• C. $\frac{\lambda }{12{\epsilon }_0}$
• D. $\frac{\lambda }{16{\epsilon }_0}$

Answer 1

The correct option is C $\frac{\lambda }{12{\epsilon }_0}$

An arc of radius $=r$

linear density of charge $=\lambda$

subtends a angle $=\frac{\pi }{3}$ the centre.

$V=\frac{1}{4}\frac{\lambda }{12{ϵ}_0}$

$V=\frac{1}{4}\times \frac{\lambda }{3{ϵ}_0}=\frac{\lambda }{12{ϵ}_0}$