Question

A wire is bent into semicrcular arc of radius R.The rod has a uniform linear charge distribution $\lambda$. The potential at the center is

• A. $\frac{\lambda }{2\pi {\epsilon }_{o}R}$
• B. $\frac{\lambda }{{\epsilon }_o}$
• C. $\frac{\lambda }{2{\epsilon }_o}$
• D. $\frac{\lambda }{4{\epsilon }_o}$

Answer 1

The correct option is D $\frac{\lambda }{4{\epsilon }_o}$

$\begin{array}{c}Q=\lambda .\pi R\\ v=\frac{kQ}{R}=\frac{k}{R}.\lambda \pi \\ =\frac{\lambda \pi }{4\pi {\epsilon }_0}=\frac{\pi }{4{\epsilon }_0}\\ Hence,\\ optionDiscorrectanswer.\end{array}$